A fractional power of a ± a cosh(y) is integrated by the half angle at which it is a square - #1465
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…t which it is a square `x^2 sqrt(a + a cosh(c + d x))` was left as written: no substitution rationalises a root of a hyperbolic function beside a power of x, and beside the sine and cosine of y the root is of a quotient under every one. 1 + cosh(y) is 2 cosh(y/2)^2, so the root is sqrt(2a) cosh(y/2) with no sign to write, the hyperbolic cosine being positive; 1 - cosh(y) is -2 sinh(y/2)^2, so (a - a cosh(y))^(3/2) is (-2a)^(3/2) sgn(sinh(y/2)) sinh(y/2)^3, the sign a constant between the zeros that comes out in front -- SolveByTheHalfAngleWhereOnePlusASineIsASquare's identity for the hyperbolic cosine. cosh(y) and sinh(y) beside the power are written in the half angle and the question is asked again in x. Exact for any a, the constants 2a and -2a kept whole under the power; at the top only, as every rule that writes a sign for a function. The reading of the six hyperbolic functions from their exponential spellings is lifted out of SolveByBiochesOddHyperbolicSubstitution into ReadTheHyperbolicFunctions and shared. Family 6 of the Rubi suite: 357 -> 363 of 417, 0 wrong; family 7 215/270 and the 1774-problem suite 1707 unchanged. Suite 12616 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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x^2 sqrt(a + a cosh(c + d x))was left as written: no substitution rationalises a root of a hyperbolic function beside a power ofx, and beside the sine and cosine ofythe root is of a quotient under every one.1 + cosh(y)is2 cosh(y/2)^2, so the root issqrt(2a) cosh(y/2)with no sign to write (the hyperbolic cosine is positive);1 - cosh(y)is-2 sinh(y/2)^2, so(a - a cosh(y))^(3/2)is(-2a)^(3/2) sgn(sinh(y/2)) sinh(y/2)^3, the sign a constant between the zeros that comes out in front —SolveByTheHalfAngleWhereOnePlusASineIsASquare's identity for the hyperbolic cosine.cosh(y)andsinh(y)beside the power are written in the half angle (2 cosh(y/2)^2 - 1,2 sinh(y/2) cosh(y/2)) and the question is asked again inx(ComputeAsTheSameQuestion), where by parts, the hyperbolic Bioche rule (#1464) and the exponential substitution answer it.Exact for any
a:cosh(y/2)^2andsinh(y/2)^2are not negative, so(a q)^p = a^p q^pfor the principal powers whateverais, with2aand-2akept as the one constant they are ((-2a)^pis not(-2)^p a^pfor a symbolica). Whole products2ponly; at the top only, as every rule that writes a sign for a function. The reading of the six hyperbolic functions from their exponential spellings is lifted out ofSolveByBiochesOddHyperbolicSubstitutionintoReadTheHyperbolicFunctionsand shared.Measured (final build):
familycap.sh 6 20): 357 → 363/417, 0 wrong — 6.2.1:165/170/176/178 and 6.2.5:75/141 answered; 6.2.5:64cosh(x)/sqrt(a - a cosh(x))answered too, graded unverifiable on the reals by the harness (the integrand is complex for a positivea; the test pinsa = -3as well as2and checks both sides of zero).bench.sh: PASSED on all 19 gated benchmarks.IP_QUAD=1) as well as by differentiating back witha = 2anda = -3.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