Hyperbolic powers two apart whose coefficients kill the reduction's residual are integrated - #1467
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…esidual are integrated `cosh(x)^(5/2) - 3 sqrt(cosh(x))/5` was left as written, and neither of its terms has an elementary antiderivative -- both are elliptic. Their combination has one. From the reduction `int cosh^p = sinh cosh^(p - 1)/p + (p - 1)/p int cosh^(p - 2)`, a sum of powers two apart is elementary exactly when the walk from the highest exponent down carries nothing past the lowest, and the answer is what the walk accumulated; the hyperbolic sine's reduction subtracts where the cosine's adds, and a gap in the exponents is a term with coefficient zero. The chain may be several steps long -- Rubi's `x/sech(x)^(7/2) - 5 x sqrt(sech(x))/21` closes over two, its coefficient being (5/7)(1/3). One linear factor in front is taken by parts inside the rule, `x F - int F` with `int F` the next power over `p^2 a^2`, since a second power of x would need `int f^p` itself. The rule runs before the sum is split, which would hand each elliptic term to the whole chain and spend a minute finding that out, and it reads each term over its own bar, so that `1/sech(y)^(5/2)`, `(c^(-5/2))^(-1)` and `c^(5/2)` are one exponent. The gathering of a common linear factor of a sum is shared with the route that hands such a sum on as a product. Family 6 of the Rubi suite: 366 -> 375 of 417, 0 wrong; family 7 215/270 and the 1774-problem suite 1707 unchanged. Suite 12627 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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cosh(x)^(5/2) - 3 sqrt(cosh(x))/5was left as written, and neither of its terms has an elementary antiderivative — both are elliptic. Their combination has one, and that is the whole rule: froma sum of powers two apart is elementary exactly when the walk from the highest exponent down carries nothing past the lowest, and the answer is what the walk accumulated. The hyperbolic sine's reduction subtracts where the cosine's adds (
cosh^2 = sinh^2 + 1againstsinh^2 = cosh^2 - 1), and a gap in the exponents is simply a term with coefficient zero — which is what lets the chain run several steps: Rubi'sx/sech(x)^(7/2) - 5 x sqrt(sech(x))/21closes over two, its coefficient being(5/7)(1/3).Three things the measurements decided:
x F - int F, whereFis the accumulated antiderivative andint Fis the next power overp^2 a^2. A second power ofxwould needint f^pitself, which is elliptic, so it is declined.1/sech(y)^(5/2)arrives as(c^(-5/2))^(-1)or1/(1/c)^(5/2)depending on what rewrote it, and both arec^(5/2); reading spellings instead of exponents is what made the first three versions decline the Rubi rows while answering the same integrand written ascosh^(5/2).The gathering of a common linear factor of a sum is shared with a small new route that hands such a sum on as a product (
x cosh(x)^(3/2) - x sqrt(cosh(x))/3→x (…)), for the shapes where parts, not this rule, does the work.Measured (final build):
familycap.sh 6 20): 366 → 375/417, 0 wrong, timeouts 21 → 17 — 6.1.1:124, 6.2.1:98/127, 6.5.1:24/25/26, 6.6.1:24/25/26; nothing lost.bench.sh: PASSED on all 19 gated benchmarks.cosh^(5/2) - cosh^(1/2)/2) is still declined, and the test pins that.BREAKING-CHANGES.mdcarries the entry with the 2.5.0 column measured on a 2.5.0 build.Part of #718.
🤖 Generated with Claude Code
https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura