Read hyperbolic integrands through cosh^2 = (1 + i sinh)(1 - i sinh), a symbolic power's own variable, and a phase

BREAKING-CHANGES.md

@@ -741,6 +741,41 @@ power, and the integrand is answered in a fraction of a second
 | `"csch(c+d*x)*(a+b*sinh(c+d*x)^4)^2".ToEntity().Integrate("x")` | past fifteen minutes | 426 characters |
 | `"csch(c+d*x)*(a+b*sinh(c+d*x)^4)^3".ToEntity().Integrate("x")` | past fifteen minutes | 646 characters |
 
+### Hyperbolic integrands are read through `cosh^2 = (1 + i sinh)(1 - i sinh)`, a symbolic power's own variable, and a phase
+
+**Answers where there were none, and shorter ones.** Rows of these shapes in Rubi's 6.1.1 and 6.7.1 were declined or ran
+out of time. Each is answered now through an identity of the hyperbolic functions that the substitutions did not use
+([#233](https://github.com/asc-community/AngouriMath/issues/233)):
+- `(g + f x)^m cosh(c + d x)^k/(a + i a sinh(c + d x))` and its `sinh^k` rows. `cosh^2 = (1 + i sinh)(1 - i sinh)`, so
+  the numerator divides by the denominator, leaving a polynomial and a constant over it.
+- `1/(sech(x) ± i tanh(x))^4`. It is `cosh(x)^4/(1 ± i sinh(x))^4`, which the same identity cancels to
+  `(1 ∓ i sinh(x))^2/(1 ± i sinh(x))^2`. Only `cosh^(2m)` over `(1 ± i sinh)^(2m)` and the other way up: cancelling
+  where a polynomial or another power of the binomial is left made other answers longer.
+- A symbolic power of `sinh`, `cosh`, `sech` or `csch` beside an odd power of the other function, with `tanh` and
+  `coth` read as quotients. `cosh(a + b x)^3 sinh(a + b x)^n` is integrated under `u = sinh`, and
+  `sech(a + b x)^n tanh(a + b x)^3` under `u = sech`. The power of `csch` stays a power of `u`, because it is not
+  `sinh^(-n)` where the sine is negative.
+- `(cosh(x) coth(x))^(3/2)` and `(csch(x) + sinh(x))^(3/2)`. Both are `(cosh(x)^2/sinh(x))^(3/2)`, which is
+  `cosh(x)^3 sinh(x)^(-3/2)` times a constant on each side of zero, odd in the cosine.
+- Half-odd powers of `S = b cosh(x) + c sinh(x) ± sqrt(b^2 - c^2)`. With `T = b sinh(x) + c cosh(x)` and `a` the
+  constant term, `d/dx (T S^(n - 1)) = n S^n - a (2n - 1) S^(n - 1)` steps the power to `±1/2`. The whole powers stay
+  the exponential substitution's, which writes them shorter.
+
+2.5.0 declined or ran out of time on all of the inputs below. The unreleased master answered some of them, and
+those answers are now two to twelve times shorter: `(sech(x) + i tanh(x))^2` and `^4` were 437 and 2,916
+characters, and `1/(sech(x) + i tanh(x))^2` and `^4` were 372 and 551.
+
+| Input | Was (2.5.0) | Now |
+|---|---|---|
+| `"(g+f*x)*cosh(c+d*x)^2/(a+i*a*sinh(c+d*x))".ToEntity().Integrate("x")` | `integral(...)` | 366 characters |
+| `"cosh(a+b*x)^3*sinh(a+b*x)^n".ToEntity().Integrate("x")` | `integral(...)` | 149 characters |
+| `"sech(a+b*x)^n*tanh(a+b*x)^3".ToEntity().Integrate("x")` | past two minutes | 183 characters |
+| `"coth(a+b*x)^3*csch(a+b*x)^n".ToEntity().Integrate("x")` | past a minute | 242 characters |
+| `"(cosh(x)*coth(x))^(3/2)".ToEntity().Integrate("x")` | `integral(...)` | 247 characters |
+| `"1/sqrt(b*cosh(x)+c*sinh(x)+sqrt(b^2-c^2))".ToEntity().Integrate("x")` | `integral(...)` | 409 characters |
+| `"1/(sech(x)+i*tanh(x))^4".ToEntity().Integrate("x")` | `integral(...)` | 241 characters |
+| `"(sech(x)+i*tanh(x))^4".ToEntity().Integrate("x")` | `integral(...)`, after 54 seconds | 251 characters, where the unreleased master gave 2,916 |
+
 ### A power of a polynomial whose leading coefficient is zero spelled otherwise is not divided by it
 
 **Answers where there were none.** `1/((a (1 + x^2))^2 - (a (1 - x^2))^2)^3` was left unevaluated. That is

Sources/AngouriMath/Functions/Continuous/Integration/Integration.Definition.cs

@@ -1134,6 +1134,8 @@ private static Entity WithItsPiecewiseSumsCombined(Entity expr)
             // And a power of `a + b cos(y) + c sin(y)` with `a^2 = b^2 + c^2`, the same square
             // turned by a phase, in closed form: the substitution search spent the budget on it.
             if ((answer = IndefiniteIntegralSolver.SolveAPowerOfACosineAndASinePlusTheirAmplitude(expr, x)) is { }) return answer;
+            // And `a + b cosh(y) + c sinh(y)` with `a^2 = b^2 - c^2`, the hyperbolic one.
+            if ((answer = IndefiniteIntegralSolver.SolveAPowerOfAHyperbolicCosineAndSinePlusTheirAmplitude(expr, x)) is { }) return answer;
             // And symbolic powers of `a ± a sin(y)` beside a power of `g cos(y)`, by u = sin(y), where
             // `(1 + u)(1 - u)` is the cosine's square: the half angle wants numeric powers, and the
             // substitution search spent the budget on these. Numeric ones too beside a fractional
@@ -1158,6 +1160,20 @@ private static Entity WithItsPiecewiseSumsCombined(Entity expr)
             if ((answer = IndefiniteIntegralSolver.SolveAHalfOddPowerOfOnePlusAnImaginaryHyperbolicSine(expr, x, integrateByParts)) is { }) return answer;
             // And the cosecant's: a ± i a csch(y) is a (w ± i)^2/(w^2 - 1) in w = e^y.
             if ((answer = IndefiniteIntegralSolver.SolveAPowerOfAnImaginaryHyperbolicCosecantSumInTheExponential(expr, x, integrateByParts)) is { }) return answer;
+            // And a polynomial in sinh(y) and cosh(y) over a ± i a sinh(y) beside a power of x,
+            // divided out, cosh(y)^2 being (1 + i sinh(y))(1 - i sinh(y)).
+            if ((answer = IndefiniteIntegralSolver.SolveByDividingOutOnePlusAnImaginaryHyperbolicSine(expr, x, integrateByParts)) is { }) return answer;
+            // A symbolic power of one hyperbolic function beside an odd power of the other, under
+            // u = the first: Bioche's substitution reads whole and rational powers. The first may be
+            // sech(y) or csch(y), and tanh(y) and coth(y) beside it are read as quotients.
+            if ((answer = IndefiniteIntegralSolver.SolveASymbolicPowerOfAHyperbolicFunctionBesideAnOddPowerOfTheOther(expr, x, integrateByParts)) is { }) return answer;
+            // And cosh(y)^2 taken out of a fractional power of a quotient, `cosh(x) coth(x)` being
+            // `cosh(x)^2/sinh(x)`: Bioche's substitution reads the radicands that are polynomials.
+            if ((answer = IndefiniteIntegralSolver.SolveByTakingTheHyperbolicCosineSquaredOutOfAPower(expr, x, integrateByParts)) is { }) return answer;
+            // And cosh(y)^(2m) over (1 ± i sinh(y))^(2m), or the other way up, cancelled through
+            // cosh(y)^2 = (1 + i sinh(y))(1 - i sinh(y)): `1/(sech(x) + i tanh(x))^4` is
+            // `(1 - i sinh(x))^2/(1 + i sinh(x))^2`, where the substitutions spent the budget.
+            if ((answer = IndefiniteIntegralSolver.SolveByCancellingOnePlusAnImaginaryHyperbolicSineAgainstTheCosineSquared(expr, x, integrateByParts)) is { }) return answer;
             // A quotient of linears as a function's argument, written over its denominator: a
             // constant plus a multiple of the reciprocal of a linear, which the rules read.
             if ((answer = IndefiniteIntegralSolver.SolveByWritingAQuotientOfLinearsOverItsDenominator(expr, x, integrateByParts)) is { }) return answer;

Sources/Tests/UnitTests/Calculus/APowerOfAHyperbolicCosineAndSinePlusTheirAmplitudeIntegralTest.cs

@@ -0,0 +1,94 @@
+//
+// Copyright (c) 2019-2026 Angouri.
+// AngouriMath is licensed under MIT.
+// Details: https://github.com/asc-community/AngouriMath/blob/master/LICENSE.md.
+// Website: https://am.angouri.org.
+//
+
+using System;
+using AngouriMath.Extensions;
+using Xunit;
+
+namespace AngouriMath.Tests.Calculus
+{
+    /// <summary>
+    /// A half-odd power of <c>a + b cosh(y) + c sinh(y)</c> with <c>a^2 = b^2 - c^2</c>, a square of
+    /// the half angle turned by a phase, in closed form through <c>T = b sinh(y) + c cosh(y)</c>.
+    /// Rubi's 6.7.1.
+    /// <a href="https://github.com/asc-community/AngouriMath/issues/233">#233</a>
+    /// </summary>
+    /// <remarks>
+    /// Compared as complex numbers: with <c>b &lt; |c|</c> the constant is imaginary and so is the
+    /// base everywhere. Away from the zero of the base where <c>a</c> is negative.
+    /// </remarks>
+    [Trait("Area", "Calculus")]
+    public sealed class APowerOfAHyperbolicCosineAndSinePlusTheirAmplitudeIntegralTest
+    {
+        [Theory]
+        [InlineData("(b*cosh(x) + c*sinh(x) + sqrt(b^2 - c^2))^(5/2)", 2.0, 1.0)]
+        [InlineData("sqrt(b*cosh(x) + c*sinh(x) - sqrt(b^2 - c^2))", 2.0, 1.0)]
+        [InlineData("1/sqrt(b*cosh(x) + c*sinh(x) + sqrt(b^2 - c^2))", 2.0, 1.0)]
+        [InlineData("1/(b*cosh(x) + c*sinh(x) - sqrt(b^2 - c^2))^(5/2)", 2.0, -1.0)]
+        [InlineData("1/(b*cosh(g + f*x) + c*sinh(g + f*x) + sqrt(b^2 - c^2))^(3/2)", 0.7, 1.1)]
+        [InlineData("(b*cosh(g + f*x) + c*sinh(g + f*x) - sqrt(b^2 - c^2))^(3/2)", 0.7, 1.1)]
+        [InlineData("1/sqrt(5*cosh(2*x + 1) + 3*sinh(2*x + 1) + 4)", 0.0, 0.0)]
+        [InlineData("3*(2*cosh(x) - sinh(x) - sqrt(3))^(-3/2)", 0.0, 0.0)]
+        public void IsWrittenThroughTheDerivativeOfItsBase(string integrand, double b, double c)
+            => DifferentiatesBack(integrand, b, c);
+
+        // A whole power is left to the exponential substitution, which writes `1/S^2`, `1/S^3` and
+        // `1/S^4` shorter than the same recurrence stepped up to `T/(a S)`: answered all the same.
+        [Theory]
+        [InlineData("1/(b*cosh(x) + c*sinh(x) + sqrt(b^2 - c^2))^2", 2.0, 1.0)]
+        [InlineData("1/(b*cosh(x) + c*sinh(x) + sqrt(b^2 - c^2))^3", 2.0, -1.0)]
+        [InlineData("1/(b*cosh(x) + c*sinh(x) + sqrt(b^2 - c^2))^4", 0.7, 1.1)]
+        [InlineData("1/(b*cosh(g + f*x) + c*sinh(g + f*x) - sqrt(b^2 - c^2))^3", 2.0, 1.0)]
+        [InlineData("3*(2*cosh(x) - sinh(x) - sqrt(3))^(-2)", 0.0, 0.0)]
+        [InlineData("1/(5*cosh(2*x + 1) + 3*sinh(2*x + 1) + 4)^3", 0.0, 0.0)]
+        public void AWholePowerIsAnsweredToo(string integrand, double b, double c)
+            => DifferentiatesBack(integrand, b, c);
+
+        private static void DifferentiatesBack(string integrand, double b, double c)
+        {
+            var integral = integrand.ToEntity().Integrate("x");
+            Assert.DoesNotContain("integral(", integral.Stringize());
+            Entity Pinned(Entity e) => e.Substitute("b", b).Substitute("c", c).Substitute("g", 0.4).Substitute("f", 1.3);
+            var derivative = Pinned(integral.Substitute("C", 0)).Differentiate("x");
+            var original = Pinned(integrand.ToEntity());
+            foreach (var at in new[] { -1.9, -1.1, -0.2, 0.4, 1.1, 1.9 })
+            {
+                var want = original.Substitute("x", at).EvalNumerical();
+                var got = derivative.Substitute("x", at).EvalNumerical();
+                Assert.True(Math.Abs((double)(got - want).RealPart) + Math.Abs((double)(got - want).ImaginaryPart)
+                            < 1e-9 * Math.Max(1, Math.Abs((double)want.RealPart) + Math.Abs((double)want.ImaginaryPart)),
+                    $"d/dx of the antiderivative of {integrand} is {got} at x = {at}, where the integrand is {want}");
+            }
+        }
+
+        /// <summary>
+        /// Real and continuous where the integrand is: across the point where <c>T</c> vanishes and
+        /// the base does not, and on either side of the base's zero where <c>a</c> is negative.
+        /// The difference of the answer against Simpson's rule.
+        /// </summary>
+        [Theory]
+        [InlineData(1.0, -2.0, 1.0)]
+        [InlineData(-1.0, 0.1, 2.0)]
+        [InlineData(-1.0, -3.0, -1.0)]
+        public void IsAContinuousAntiderivativeOnTheRealLine(double sign, double from, double to)
+        {
+            var root = sign > 0 ? "sqrt(3)" : "-sqrt(3)";
+            var integral = $"1/sqrt(2*cosh(x) + sinh(x) + {root})".ToEntity().Integrate("x").Substitute("C", 0);
+            Assert.DoesNotContain("integral(", integral.Stringize());
+            double Integrand(double t) => 1 / Math.Sqrt(2 * Math.Cosh(t) + Math.Sinh(t) + sign * Math.Sqrt(3));
+            const int steps = 400;
+            var h = (to - from) / steps;
+            var simpson = Integrand(from) + Integrand(to);
+            for (var k = 1; k < steps; k++)
+                simpson += (k % 2 == 1 ? 4 : 2) * Integrand(from + k * h);
+            simpson *= h / 3;
+            var difference = integral.Substitute("x", to).EvalNumerical() - integral.Substitute("x", from).EvalNumerical();
+            Assert.True(Math.Abs((double)difference.ImaginaryPart) < 1e-9, $"F({to}) - F({from}) is {difference}");
+            Assert.True(Math.Abs((double)difference.RealPart - simpson) < 1e-7, $"F({to}) - F({from}) is {difference}, Simpson's rule {simpson}");
+        }
+    }
+}

Sources/Tests/UnitTests/Calculus/ASymbolicPowerOfAHyperbolicFunctionBesideAnOddPowerOfTheOtherIntegralTest.cs

@@ -0,0 +1,80 @@
+//
+// Copyright (c) 2019-2026 Angouri.
+// AngouriMath is licensed under MIT.
+// Details: https://github.com/asc-community/AngouriMath/blob/master/LICENSE.md.
+// Website: https://am.angouri.org.
+//
+
+using System;
+using AngouriMath.Extensions;
+using Xunit;
+
+namespace AngouriMath.Tests.Calculus
+{
+    /// <summary>
+    /// A symbolic power of <c>sinh(y)</c> beside an odd power of <c>cosh(y)</c>, under
+    /// <c>u = sinh(y)</c>, and the other way about under <c>u = cosh(y)</c>. Rubi's 6.7.1.
+    /// <a href="https://github.com/asc-community/AngouriMath/issues/233">#233</a>
+    /// </summary>
+    /// <remarks>Compared as complex numbers: a symbolic power of a negative sine is complex.</remarks>
+    [Trait("Area", "Calculus")]
+    public sealed class ASymbolicPowerOfAHyperbolicFunctionBesideAnOddPowerOfTheOtherIntegralTest
+    {
+        [Theory]
+        [InlineData("cosh(a + b*x)^3*sinh(a + b*x)^n")]
+        [InlineData("cosh(a + b*x)^5*sinh(a + b*x)^n")]
+        [InlineData("cosh(a + b*x)^m*sinh(a + b*x)^3")]
+        [InlineData("cosh(a + b*x)^m*sinh(a + b*x)^5")]
+        [InlineData("2*cosh(x)^3/sinh(x)^n")]
+        public void IsAPolynomialTimesAPowerInTheOther(string integrand)
+        {
+            var integral = integrand.ToEntity().Integrate("x");
+            Assert.DoesNotContain("integral(", integral.Stringize());
+            Entity Pinned(Entity e) => e.Substitute("a", 0.4).Substitute("b", 1.3).Substitute("n", 0.37).Substitute("m", -1.6);
+            var derivative = Pinned(integral.Substitute("C", 0)).Differentiate("x");
+            var original = Pinned(integrand.ToEntity());
+            foreach (var at in new[] { -1.9, -1.1, -0.7, 0.4, 1.1, 1.9 })
+            {
+                var want = original.Substitute("x", at).EvalNumerical();
+                var got = derivative.Substitute("x", at).EvalNumerical();
+                Assert.True(Math.Abs((double)(got - want).RealPart) + Math.Abs((double)(got - want).ImaginaryPart)
+                            < 1e-9 * Math.Max(1, Math.Abs((double)want.RealPart) + Math.Abs((double)want.ImaginaryPart)),
+                    $"d/dx of the antiderivative of {integrand} is {got} at x = {at}, where the integrand is {want}");
+            }
+        }
+
+        /// <summary>
+        /// The reciprocals to the symbolic power and the tangent and cotangent beside them, read as
+        /// powers of the two: <c>sech(y)^n tanh(y)^3</c> under <c>u = sech(y)</c>, and
+        /// <c>coth(y)^3 csch(y)^n</c> under <c>u = csch(y)</c>, whose power is complex where the sine
+        /// is negative and not that of <c>sinh(y)^(-n)</c>. Rubi's 6.7.1. Both signs of the
+        /// argument at each slope.
+        /// </summary>
+        [Theory]
+        [InlineData("sech(a + b*x)^n*tanh(a + b*x)^3", 0.4, 1.3)]
+        [InlineData("coth(a + b*x)^3*csch(a + b*x)^n", 0.4, 1.3)]
+        [InlineData("coth(a + b*x)^3*csch(a + b*x)^n", 0.9, -0.8)]
+        [InlineData("sech(a + b*x)^m*tanh(a + b*x)^5", -0.4, 1.3)]
+        [InlineData("coth(a + b*x)^5*csch(a + b*x)^m", 0.9, -0.8)]
+        [InlineData("cosh(a + b*x)^n*tanh(a + b*x)^3", 0.4, 1.3)]
+        [InlineData("sinh(a + b*x)^m*coth(a + b*x)^3", -0.4, 1.3)]
+        [InlineData("csch(a + b*x)^n*cosh(a + b*x)^3", 0.4, 1.3)]
+        [InlineData("2*sech(x)^n*sinh(x)^3", 0.0, 1.0)]
+        public void ReadsTheReciprocalsAndTheQuotients(string integrand, double a, double b)
+        {
+            var integral = integrand.ToEntity().Integrate("x");
+            Assert.DoesNotContain("integral(", integral.Stringize());
+            Entity Pinned(Entity e) => e.Substitute("a", a).Substitute("b", b).Substitute("n", 0.37).Substitute("m", -1.6);
+            var derivative = Pinned(integral.Substitute("C", 0)).Differentiate("x");
+            var original = Pinned(integrand.ToEntity());
+            foreach (var at in new[] { -1.9, -1.1, -0.7, 0.4, 1.1, 1.9 })
+            {
+                var want = original.Substitute("x", at).EvalNumerical();
+                var got = derivative.Substitute("x", at).EvalNumerical();
+                Assert.True(Math.Abs((double)(got - want).RealPart) + Math.Abs((double)(got - want).ImaginaryPart)
+                            < 1e-9 * Math.Max(1, Math.Abs((double)want.RealPart) + Math.Abs((double)want.ImaginaryPart)),
+                    $"d/dx of the antiderivative of {integrand} is {got} at x = {at}, where the integrand is {want}");
+            }
+        }
+    }
+}

Sources/Tests/UnitTests/Calculus/CancellingOnePlusAnImaginaryHyperbolicSineAgainstTheCosineSquaredIntegralTest.cs

@@ -0,0 +1,50 @@
+//
+// Copyright (c) 2019-2026 Angouri.
+// AngouriMath is licensed under MIT.
+// Details: https://github.com/asc-community/AngouriMath/blob/master/LICENSE.md.
+// Website: https://am.angouri.org.
+//
+
+using System;
+using AngouriMath.Extensions;
+using Xunit;
+
+namespace AngouriMath.Tests.Calculus
+{
+    /// <summary>
+    /// <c>cosh(y)^(2m)</c> over <c>(1 ± i sinh(y))^(2m)</c>, or the other way up, cancelled through
+    /// <c>cosh(y)^2 = (1 + i sinh(y))(1 - i sinh(y))</c>: <c>1/(sech(x) + i tanh(x))^4</c> is
+    /// <c>(1 - i sinh(x))^2/(1 + i sinh(x))^2</c>. Rubi's 6.7.1.
+    /// <a href="https://github.com/asc-community/AngouriMath/issues/233">#233</a>
+    /// </summary>
+    /// <remarks>Compared as complex numbers: the integrand is complex wherever it is defined.</remarks>
+    [Trait("Area", "Calculus")]
+    public sealed class CancellingOnePlusAnImaginaryHyperbolicSineAgainstTheCosineSquaredIntegralTest
+    {
+        [Theory]
+        [InlineData("1/(sech(x) + i*tanh(x))^4")]
+        [InlineData("1/(sech(x) - i*tanh(x))^4")]
+        [InlineData("1/(sech(x) + i*tanh(x))^6")]
+        [InlineData("1/(sech(a + b*x) - i*tanh(a + b*x))^4")]
+        [InlineData("cosh(x)^4/(1 + i*sinh(x))^4")]
+        [InlineData("3*cosh(a + b*x)^6/(a - i*a*sinh(a + b*x))^6")]
+        [InlineData("(sech(x) + i*tanh(x))^6")]
+        [InlineData("cosh(x)^2/(i + sinh(x))^2")]
+        public void IsTheQuotientTheCancellationLeaves(string integrand)
+        {
+            var integral = integrand.ToEntity().Integrate("x");
+            Assert.DoesNotContain("integral(", integral.Stringize());
+            Entity Pinned(Entity e) => e.Substitute("a", -0.4).Substitute("b", 1.3);
+            var derivative = Pinned(integral.Substitute("C", 0)).Differentiate("x");
+            var original = Pinned(integrand.ToEntity());
+            foreach (var at in new[] { -1.9, -1.1, -0.2, 0.4, 1.1, 1.9 })
+            {
+                var want = original.Substitute("x", at).EvalNumerical();
+                var got = derivative.Substitute("x", at).EvalNumerical();
+                Assert.True(Math.Abs((double)(got - want).RealPart) + Math.Abs((double)(got - want).ImaginaryPart)
+                            < 1e-9 * Math.Max(1, Math.Abs((double)want.RealPart) + Math.Abs((double)want.ImaginaryPart)),
+                    $"d/dx of the antiderivative of {integrand} is {got} at x = {at}, where the integrand is {want}");
+            }
+        }
+    }
+}