A root of a rational function of coth(x) is integrated by the hyperbolic half-angle - #1463
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…lic half-angle `coth(x)^3/sqrt(a + b coth(x)^2)` was a search past the budget: the rational-in-tanh route refused the half-angle -- u = tanh(x) for an integrand written in e^(2x) -- whenever a root was in the integrand, because for sqrt(sinh(x)) the half makes a root of 2u/(1 - u^2) where u = e^x answers with a root of a polynomial. With a root it now takes the half where every radicand comes out over a power of u: a + b coth(x)^2 is (b + a u^2)/u^2, whose root is sqrt(b + a u^2)/|u|, and |u|^k for the bare square is sgn(u) u^k at every depth -- the one half-odd power the perfect-square rule takes beyond the square root. The halved argument is simplified (InnerSimplified left `2x/2` in the answer). Family 6 of the Rubi suite: 335 -> 344 of 417, 0 wrong, timeouts 42 -> 34; family 7 215/270 and the 1774-problem suite 1707 unchanged. Suite 12600 passed; allocation gate passed on all 19 gated benchmarks. Part of #718. Co-Authored-By: Claude Opus 5 (1M context) <noreply@anthropic.com> Claude-Session: https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura
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coth(x)^3/sqrt(a + b coth(x)^2)was a search past the budget (17–50 s, unevaluated).SolveARationalFunctionOfTheHyperbolicTangentrefused the half-angle —u = tanh(x)for an integrand written ine^(2x)— whenever a root was in the integrand, because forsqrt(sinh(x))the half makes a root of2u/(1 - u^2)whereu = e^xanswers with a root of a polynomial. With a root it now takes the half where every radicand comes out over a power ofu:a + b coth(x)^2is(b + a u^2)/u^2, whose root issqrt(b + a u^2)/|u|, and|u|^kfor the bare square issgn(u) u^kat every depth — the one half-odd powerSolveByTakingARootOfAPerfectSquare(#1462) takes beyond the square root, with the radicand restricted tox^2so the(1 + 2u + u^2)^(5/2)case that PR measured stays untouched. The halved argument goes throughSimplify(InnerSimplifiedleft2x/2in the answer).Measured (final build; the suite and gate ran on a build that also carried the inert dispatcher trace,
AM_TRACE_CHAINunset, stripped before the commit; the three test classes were re-run on the stripped build, 17 passed):familycap.sh 6 20): 335 → 344/417, 0 wrong, timeouts 42 → 34 — the ninecoth^m (a + b coth²)^(p/2)rows of 6.4.7, nothing lost.bench.sh: PASSED on all 19 gated benchmarks.IP_QUAD=1) as well as by the pinned differentiate-back test.BREAKING-CHANGES.mdcarries the entry with the 2.5.0 column measured on a 2.5.0 build.Part of #718.
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https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura