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Take a root of a perfect square with a leading coefficient of no known sign - #1884
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Rafael-SOWNet merged 3 commits intoOct 14, 2026
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…n sign (a (w + h)^2)^r is a^r |w + h|^(2r) for every nonzero a where w + h is real, since its square is positive: a negative number was read so already, and a symbol of no known sign is now. And a linear of the integrand that is a multiple of the square's root is taken as the root, so that its power and the square's gather into one. Co-Authored-By: Claude Opus 5.5 (1M context) <noreply@anthropic.com> Claude-Session: https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura
…t-square-is-integrated-as-one
Co-Authored-By: Claude Opus 5.5 (1M context) <noreply@anthropic.com> Claude-Session: https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura
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Part of #718
SolveByTakingARootOfAPerfectSquaretook(a (w + h)^2)^rasa^r |w + h|^(2r)only for a leading coefficient whose sign it could establish, sosqrt(c d^2 + 2 c d e x + c e^2 x^2)was declined. The identity holds for every nonzeroaas long asw + his real:(w + h)^2is positive, soLog(a (w + h)^2)isLog(a) + 2 ln|w + h|. A negative number was already read that way, and a symbol of unknown sign now is too.SolveByWritingAPowerOfASquareAsAPowerOfItsRootalso takes a linear factor of the integrand that is a multiple of the square's root as the root itself. For(d + e x)^(-1 - 2p) (c d^2 + 2 c d e x + c e^2 x^2)^p, that gathers the two powers into(d + e x)^(-1), where before it had two powers of two linears.Measured
intbench, this branch against master 790feec, 5 s budget. The pocket was 1.2.1.2, 1.2.1.3, 1.2.1.4, 1.2.1.6, 1.2.1.9, 1.2.2.2, 1.3.2, 5.3.5 and 7.1.5 in full (7,422 rows), plus the usual family sample:
The moves, re-run alone: 13 rows went from unsolved to solved, and none went the other way.
sqrt(a + a sech(c + d x))asinh(sqrt(b x^2 - 1))beside its root, where the inverse's derivative holdssqrt(b x^2)Every new answer was checked exactly with
exactcheck(exact pins, 50 and 2,000 digits, four points, three pin sets with the leading coefficient negative in two): 276 of 276 diff 0. The gate passes. The suite passes in two halves, 1,856 and 13,522, which is master's 15,373 plus the 5 new tests. Run whole, the test host hits the 8 GB memory guard here as on master (#1883).1/sqrt(c d^2 + 2 c d e x + c e^2 x^2)was answered by 2.5.0 in 411 characters and declined on master. It is 61 characters now.Still declined in the same file, and left for another PR: 1.2.1.2:2389–2393,
(d + e x)^(-3 - 2p) ((d + e x)(a e + c d x))^p, which wantst = (a e + c d x)/(d + e x).🤖 Generated with Claude Code
https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura