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Read hyperbolic integrands through cosh^2 = (1 + i sinh)(1 - i sinh), a symbolic power's own variable, and a phase - #2018
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… a symbolic power's own variable, and a phase Five shapes of Rubi's 6.1.1 and 6.7.1, each through an identity the substitutions did not use: - (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 is cosh(x)^4/(1 ± i sinh(x))^4, which the same identity cancels to (1 ∓ i sinh)^2/(1 ± i sinh)^2; cosh^(2m) over (1 ± i sinh)^(2m) and the other way up only. - A symbolic power of sinh, cosh, sech or csch beside an odd power of the other, under u = sinh, cosh, sech or csch; a power of csch stays a power of u, since it is not sinh^(-n) where the sine is negative. - (cosh(x) coth(x))^(3/2) and (csch(x) + sinh(x))^(3/2) are (cosh(x)^2/sinh(x))^(3/2), cosh^3 sinh^(-3/2) times a constant on each side of zero. - Half-odd powers of S = b cosh(x) + c sinh(x) ± sqrt(b^2 - c^2): with T = b sinh + c cosh, d/dx (T S^(n-1)) = n S^n - a (2n - 1) S^(n-1) steps the power to ±1/2. The whole powers are left to the exponential substitution, which writes 1/S^2, 1/S^3 and 1/S^4 in 182, 272 and 416 characters, where the recurrence gave 350, 539 and 728. Chain against master 4877ed2: pocket of 1,756 rows, 1,686 -> 1,723 solved, 0 wrong; samples unchanged or better; the 43 rows that moved, alone, 4 -> 42 solved. Re-measured on master f7f2a7a after the rebase: 14 of the 43 are new there (12 of (e + f x)^k cosh^m/(a + i a sinh) past 20 s -> 0.1-0.9 s, (csch(x) + sinh(x))^(3/2) and ^(5/2) declined -> answered), and the other 29 answer on both, most much cheaper here: sech(a + b x)^n tanh^3 9.12 -> 0.02 GB, 1/(sech(x) + i tanh(x))^4 10.5 -> 0.07 GB and 551 -> 241 characters. Part of #233. Co-Authored-By: Claude Opus 5.5 (1M context) <noreply@anthropic.com> Claude-Session: https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura
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Part of #233.
Five shapes of Rubi's 6.1.1 and 6.7.1, each answered through an identity of the hyperbolic functions that the substitutions did not use. One test class per shape.
(g + f x)^m cosh(c + d x)^k/(a + i a sinh(c + d x))and itssinh^krows. Sincecosh^2 = (1 + i sinh)(1 - i sinh), the numerator divides by the denominator, leaving a polynomial and a constant over it.1/(sech(x) ± i tanh(x))^4iscosh(x)^4/(1 ± i sinh(x))^4, which the same identity cancels to(1 ∓ i sinh)^2/(1 ± i sinh)^2. It applies only tocosh^(2m)over(1 ± i sinh)^(2m)and the other way up; cancelling where something else is left made other answers longer.sinh,cosh,sechorcschbeside an odd power of the other, underu= that function. A power ofcschstays a power ofu, since it is notsinh^(-n)where the sine is negative.(cosh(x) coth(x))^(3/2)and(csch(x) + sinh(x))^(3/2)are both(cosh(x)^2/sinh(x))^(3/2), which iscosh^3 sinh^(-3/2)times a constant on each side of zero.S = b cosh(x) + c sinh(x) ± sqrt(b^2 - c^2). WithT = b sinh + c cosh,d/dx (T S^(n-1)) = n S^n - a (2n - 1) S^(n-1)steps the power to±1/2. The branch first also stepped whole powers up toT/(a S). On today's master the exponential substitution writes1/S^2,1/S^3and1/S^4in 182, 272 and 416 characters, where the recurrence gave 350, 539 and 728, so whole powers are now left to it.Measured.
(e + f x)^k cosh^m/(a + i a sinh)go from past 20 s to 0.1–0.9 s;(csch(x) + sinh(x))^(3/2)and^(5/2)go from declined to answered.sech(a+b*x)^n*tanh(a+b*x)^3cosh(a+b*x)^3*sinh(a+b*x)^n1/(sech(x)+i*tanh(x))^4(sech(x)+i*tanh(x))^4(g+f*x)*cosh(c+d*x)^2/(a+i*a*sinh(c+d*x))1/sqrt(b*cosh(x)+c*sinh(x)+sqrt(b^2-c^2))Every answer was checked by differentiating back at the probe's points; the test classes pin each shape the same way. 426 tests pass locally across the five classes, the hyperbolic tests and the corpus gate.
The BREAKING-CHANGES.md entry has 2.5.0's column measured. Its "Now" column and its comparison with the unreleased master are re-measured on this build and on f7f2a7a.
🤖 Generated with Claude Code
https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura