A rational function of e^y with a symbolic denominator is integrated by the hyperbolic half angle - #1466
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…by the hyperbolic half angle `csch(x)^2/(a + b sinh(x))^2` was a search past eighty seconds: under u = e^x its denominator is a symbolic quartic nothing factors. Under u = tanh(x/2) it is a rational function whose written factors the symbolic partial fractions answer, and the route took that half only where every exponential was written with an even factor. It is taken now for a root-free integrand of a plain e^y as well, where the denominator's shape depends on a symbol: read as a polynomial in v = e^y, a coefficient that is not a number times the first, and two distinct factors of v below the bar. Both conditions are measured, not supposed. `tanh(x)^6/(a + a sech(x))` has `a` for its content and the rest factors over the rationals: it answers a page shorter under u = e^x, and the suite's bound on the length of that answer is what caught the first version. `e^asinh(a + b x)/x^4` leaves `e^u (e^u + e^-u)/(e^u - e^-u - 2a)^4` -- one symbolic quadratic to the fourth, whose roots the quadratic formula gives -- in a second under u = e^x against thirty under the half; bare v is no factor, being where an odd power of it was gathered. And Bioche's hyperbolic rule goes before this route, not after: with the half taken for a symbolic denominator, `cosh(x)^7/(a + b cosh(x)^2)` and `sech(x)^3/(a + b sinh(x))^2` are half a second by u = cosh(x) and past the budget by the half, which makes a polynomial of twice the degree. Family 6 of the Rubi suite: 363 -> 366 of 417, 0 wrong; family 7 215/270 and the 1774-problem suite 1707 unchanged. Suite 12619 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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csch(x)^2/(a + b sinh(x))^2was a search past eighty seconds: underu = e^xits denominator is a symbolic quartic nothing factors. Underu = tanh(x/2)it is a rational function whose written factors the symbolic partial fractions answer — andSolveARationalFunctionOfTheHyperbolicTangenttook that half only where every exponential was written with an even factor (coth(c + d x)'se^(2(c + d x))). It is taken now for a root-free integrand of a plaine^yas well, where the denominator's shape depends on a symbol: read as a polynomial inv = e^y, a coefficient that is not a number times the first, and two distinct factors ofvbelow the bar.Both conditions come from measurements, and each cost a round:
tanh(x)^6/(a + a sech(x))hasafor its content and the rest factors over the rationals, sou = e^xanswers it a page shorter. The first version took the half there and produced 12,502 characters against the 8,000 the suite pins (SymbolicFactorsPartialFractionsTest.TheCoefficientsAreInLowestTerms) — the corpus was green and the unit suite caught it.e^asinh(a + b x)/x^4leavese^u (e^u + e^-u)/(e^u - e^-u - 2a)^4after the inverse hyperbolic substitution: root-free, symbolic, but a single quadratic invto the fourth, whose roots the quadratic formula gives — one second underu = e^xagainst thirty under the half. Barevis not counted, being where an odd power of it was gathered.SolveByBiochesOddHyperbolicSubstitution(A rational function of the hyperbolic functions odd in the sine or the cosine is integrated by the other #1464) must come before this route:cosh(x)^7/(a + b cosh(x)^2)andsech(x)^3/(a + b sinh(x))^2are half a second byu = cosh(x)and past the budget by the half, which makes a polynomial of twice the degree. Measured: without the reorder family 6 fell to 359, below master.Measured (final build):
familycap.sh 6 20): 363 → 366/417, 0 wrong, timeouts 22 → 21; 6.1.5:116 and 6.6.3:112/157 gained, nothing lost.bench.sh: PASSED on all 19 gated benchmarks.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