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Integrate a polynomial over a symbolic quartic that splits into two quadratics - #1915
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Rafael-SOWNet merged 3 commits intoOct 17, 2026
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…uadratics in a hidden variable A palindromic quartic c x^4 + b x^3 + e x^2 + b x + c is c (x^2 - p x + 1)(x^2 - q x + 1), p and q the roots of c w^2 + b w + e - 2 c, being x^2 times a quadratic in w = x + 1/x; one whose odd terms vanish under y = x + c3/(4 c4) is c4 (y^2 - p)(y^2 - q). The integral over the factored quartic is asked with p and q named and they are written back. Only a quartic written as one sum is read: the factored one is palindromic again, and read again the answer ran to 803,488 characters. Rubi's 1.3.1 (A + B x + C x^2 + D x^3)/(a + b x + c x^2 + b x^3 + a x^4) and 1/(4 a c + 4 c^2 x^2 + 4 c d x^3 + d^2 x^4) were declined. Part of #718 Co-Authored-By: Claude Opus 5.5 (1M context) <noreply@anthropic.com> Claude-Session: https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura
…that-splits-into-two-quadratics # Conflicts: # Sources/AngouriMath/Functions/Continuous/Integration/IndefiniteIntegralSolver.cs # Sources/AngouriMath/Functions/Continuous/Integration/Integration.Definition.cs
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.
A quartic with a symbol in its coefficients that nothing factors over the rationals can still be two quadratics in a hidden variable:
c x^4 + b x^3 + e x^2 + b x + c: it isx^2times a quadratic inw = x + 1/x, so it isc (x^2 - p x + 1)(x^2 - q x + 1)withp,qthe roots ofc w^2 + b w + e - 2c;y = x + c3/(4 c4)takes the odd terms away:c4 (y^2 - p)(y^2 - q).SolveAPolynomialOverASymbolicQuarticThatSplitsIntoTwoQuadraticsasks the integral over the factored quartic withpandqnamed, and writes them back. The shifted coefficients come from the binomial theorem in lowest terms. Only a quartic written as one sum is read. The factored one is palindromic again, and when it was read again the answer ran to 803,488 characters.Checks
ASymbolicQuarticThatSplitsIntoTwoQuadraticsIntegralTest: 4 cases, differentiated back at four points under two pin sets.Corpus, measured against master 87be872 (this branch then merged 67d8e8d):
Both solve on this branch re-run alone, and neither on master. 3.2.3:49 read as Unsolved → Timeout, and 3.2.3:51 as slow. Re-timed alone twice per arm, 3.2.3:49 declines in 26.8–27.3 s on master and 27.5–27.9 s here, and 51 solves in 1,375–1,376 vs 1,335–1,463 ms. No losses. Gate passed; suite 15,481 passed, 0 failed.
Sizes against 2.5.0 are in the BREAKING-CHANGES entry: every case was
integral(...)there, and is 1,620–3,920 characters here.🤖 Generated with Claude Code
https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura