Integrate a power of a quotient of linears over a half-odd power of their product

[Rafael-SOWNet]

The parser folds e^(n acoth(a x)) into ((a x + 1)/(a x - 1))^(n/2), a power T^alpha of a quotient of linears. Over a half-odd power of a multiple Q of their product, with a power of x beside it, the antiderivative has Liouville's shape T^alpha P(x) Q^(1/2 - K). Since T'/T = D/((p + q x)(r + s x)), asking F' to equal the integrand gives a three-term recurrence for the coefficients of P. Where the equation past the last one fails, the antiderivative is hypergeometric and the rule declines.

Rubi's 7.4.2 e^(n acoth(a x))/(c - a^2 c x^2)^(3/2) to ^(9/2), and the same beside x, x^2 and x^3, were declined. They are answered now.

Measured on 8a1ef454, branch against master:

Run	Problems	Master	Branch
Pocket: Rubi's 7.3 and 7.4	2,010	1,965 solved	1,975 solved
Family 1, 40 a file	1,381	1,362	1,362
Family 0, whole	1,814	1,789	1,789
Families 2 to 8, sampled	2,410	2,362	2,359

Re-run alone on both builds, 10 of the 13 problems that moved go from declined to answered (7.4.2:858–861, 864, 865, 868–871). The other three are load:

• 3.2.1:292 and 7.4.1:369 time out on both builds;
• 7.3.5:50 took master 24 s against a 5 s budget, and other branches' runs have had it going either way.

No run had a wrong answer.

On this master the dispatch line sits after SolveAFunctionOfTheHyperbolicTangent, the last rule of its group. The Moebius rule (#1953) and its tests pass beside it.

APowerOfAQuotientOfLinearsOverAHalfOddPowerOfTheirProductIntegralTest differentiates the answers back. BREAKING-CHANGES.md has the entry, with the 2.5.0 column measured on v2.5.0.

Part of #233.

🤖 Generated with Claude Code

https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura