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Rafael-SOWNet merged 1 commit into from
Oct 21, 2026
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Integrate four inverse-hyperbolic shapes #2028

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59 changes: 59 additions & 0 deletions BREAKING-CHANGES.md
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Expand Up @@ -129,6 +129,23 @@ since semantic versioning admits it.
Measured on the suite: **7 of 8,525 tests** referenced the old reading — five boolean-solver inputs
written `A | B`, and two that existed to record this spelling as at risk.

### An inverse hyperbolic function over a half-odd power of an even quadratic is integrated by parts against it

**Answers where there were none.** `atanh(a x)/(c + d x^2)^(3/2)` was declined, and so were its powers up to `^(9/2)`,
`acosh(a x)` over the same powers, and `(a + b asinh(c x))` and `(a + b acosh(c x))` over `(d + k x^2)^(3/2)` to
`^(7/2)`: Rubi's 7.1.4, 7.2.4, 7.2.5 and 7.3.4. By parts, the antiderivative of the power is `x` times powers of
`(c + d x^2)^(-1/2)`, and the logarithm's derivative is even in `x`, so the remainder is `x` times a function of
`x^2`; under `w = x^2` each of its terms is a power of `c + d w` over `1 - a^2 w` or over the root of a linear,
answered at once. Not where the quadratic is the function's own, `1 - a^2 x^2` for the artanh, which the
substitution under the hyperbolic function answers ([#233](https://github.com/asc-community/AngouriMath/issues/233)).

| Input | Was (2.5.0) | Now |
|---|---|---|
| `"atanh(a*x)/(c+d*x^2)^(3/2)".ToEntity().Integrate("x")` | `integral(...)` | 1,034 characters |
| `"atanh(a*x)/(c+d*x^2)^(9/2)".ToEntity().Integrate("x")` | `integral(...)` | 7,136 characters |
| `"acosh(a*x)/(c+d*x^2)^(9/2)".ToEntity().Integrate("x")` | `integral(...)` | 2,409 characters |
| `"(a+b*asinh(c*x))/(d+k*x^2)^(7/2)".ToEntity().Integrate("x")` | `integral(...)` | 1,843 characters |

### A power of x times `atanh(sqrt(e) x/sqrt(d + e x^2))` is integrated by parts against the root

**Answers where there were none.** `x atanh(sqrt(e) x/sqrt(d + e x^2))` was declined, and so were twelve more of Rubi's
Expand All @@ -142,6 +159,21 @@ quotient of two sums with a root in each, which the search never brought down to
| `"atanh(x*sqrt(k)/sqrt(d+k*x^2))".ToEntity().Integrate("x")` | `integral(...)` | `x atanh(...) - sqrt(d + k x^2)/sqrt(k)`, written in the logarithm |
| `"atanh(x*sqrt(k)/sqrt(d+k*x^2))/x^5".ToEntity().Integrate("x")` | `integral(...)` | by parts onto `1/(x^4 sqrt(d + k x^2))` |

### A power of x times the arsech or arcsch of `sqrt(x)` is integrated by parts

**Answers where there were none.** Every power from `x^0` to `x^3` times `asech(sqrt(x))` or `acsch(sqrt(x))`, Rubi's
7.5.2 and 7.6.2, was declined or ran past thirty seconds. By parts the remainder is `K/(2 (m + 1))` times the
integral of `x^m/sqrt(1 - x)`, or of `x^m/sqrt(1 + x)` for the arcsch, where `K = sqrt(1 - x)/(sqrt(x) sqrt(1/x - 1))` is `1` for `x > 0` and `-1` for `x < 0`, and the arcsch's
`K = sqrt(1 + x)/(sqrt(x) sqrt(1/x + 1))` is `-1` between `-1` and `0` and `1` elsewhere. `K` is written in the answer
as it is defined, so the answer holds on the negative half-line too
([#233](https://github.com/asc-community/AngouriMath/issues/233)).

| Input | Was (2.5.0) | Now |
|---|---|---|
| `"x^3*asech(sqrt(x))".ToEntity().Integrate("x")` | no answer in thirty seconds | 210 characters |
| `"x*acsch(sqrt(x))".ToEntity().Integrate("x")` | `integral(...)` | 146 characters |
| `"x^3*acsch(sqrt(x))".ToEntity().Integrate("x")` | no answer in thirty seconds | 205 characters |

### A power of x is taken out of a bracket to find the derivative of a product of powers

**Answers where there were none.** `x^(2n) (c + d x)^n (2 c x + 3 d x^2)`, Rubi's 1.3.1, was declined. A bracket that
Expand Down Expand Up @@ -283,6 +315,19 @@ the working digits, or `MathS.Settings.PrecisionErrorZeroRange` where a caller h
still that integer, so `e^(-123.456)` is 0 as it was.
[#1338](https://github.com/asc-community/AngouriMath/issues/1338).

### A polynomial over a power of the linear of an artanh is written in powers of the linear

**Answers where there were none.** `x (a + b atanh(c x))^2/(d + c d x)^3`, Rubi's 7.3.4, ran out of time. In powers of
`d + c d x`, `x` is `((d + c d x) - d)/(c d)`, and the integrand is two terms, each a power of one of the logarithm's
linears beside a function of it, which are answered under `t = (1 + c x)/(1 - c x)`. Where the polynomial's degree is
at most the power less two, so that no term is over the linear alone, and the function is not the logarithm's first
power, which by parts answers as before ([#233](https://github.com/asc-community/AngouriMath/issues/233)).

| Input | Was (2.5.0) | Now |
|---|---|---|
| `"x*(a+b*atanh(c*x))^2/(d+c*d*x)^3".ToEntity().Integrate("x")` | `integral(...)` | 731 characters |
| `"x*(a+b*acoth(c*x))^2/(d+c*d*x)^3".ToEntity().Integrate("x")` | `integral(...)` | 756 characters |

### An arctangent over a power of `d (1 + i c x)` is integrated under the tangent

**Answers where there were none.** `(a + b atan(c x))^2/(d + i c d x)^2` ran out of time, and so did seven more of
Expand All @@ -297,6 +342,20 @@ becomes a polynomial in `u` times exponentials and trigonometric functions, and
| `"(a+b*atan(c*x))^2/(1+i*c*x)^4".ToEntity().Integrate("x")` | `integral(...)` | 731 characters |
| `"(a+b*atan(c*x))^3/(d+i*c*d*x)^4".ToEntity().Integrate("x")` | `integral(...)` | 1,202 characters |

### An arcoth of a linear is integrated as its artanh plus a constant

**Answers where there were none.** `acoth(x)^2/(1 - x^2)^2`, Rubi's 7.4.1, was declined, where
`(atanh(x) + k)^2/(1 - x^2)^2` is answered. `acoth(L) - atanh(L)` is `i pi/2` where `|L| < 1` and `-i pi/2` where `|L| > 1`:
constant on each interval, and not the same constant on each. Such an integrand is now asked with a symbol `K` for it,
as one in `atanh(L) + K`, and the answer is written back with `K = acoth(L) - atanh(L)`, so that it holds on every
interval without a case for each. Only where nothing else answers the integrand as written
([#233](https://github.com/asc-community/AngouriMath/issues/233)).

| Input | Was (2.5.0) | Now |
|---|---|---|
| `"acoth(x)^2/(1-x^2)^2".ToEntity().Integrate("x")` | `integral(...)` | 1,304 characters, with `(ln((x + 1)/(x - 1)) - ln((1 + x)/(1 - x)))/2` for `K` |
| `"acoth(c*x)^2/(1-c^2*x^2)^2".ToEntity().Integrate("x")` | `integral(...)` | 1,681 characters |

### A function of the hyperbolic tangent alone is integrated under `u = tanh`

**Answers where there were none.** `tanh(x)/sqrt(a + b tanh(x)^2 + c tanh(x)^4)` was declined, and so were 35 more of
Expand Down
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