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Reduce a power of a linear over a power of a quadratic a power at a time - #1892
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Rafael-SOWNet merged 4 commits intoOct 15, 2026
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Rafael-SOWNet merged 4 commits into
Rafael-SOWNet merged 4 commits into
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L^m N/Q^n, for m not whole and N linear, is (alpha + beta x) L^(m + 1) Q^(1 - n) plus L^m W/Q^(n - 1) with W linear again, alpha and beta from two equations that make N - (alpha + beta x)(1 - n) L Q' a multiple of Q; over the first power the integrator is asked with the two coefficients named. Co-Authored-By: Claude Opus 5.5 (1M context) <noreply@anthropic.com> Claude-Session: https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura
…top to the rules before Co-Authored-By: Claude Opus 5.5 (1M context) <noreply@anthropic.com> Claude-Session: https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura
…duced gives Co-Authored-By: Claude Opus 5.5 (1M context) <noreply@anthropic.com> Claude-Session: https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura
…-over-a-power-of-a-quadratic-is-reduced # Conflicts: # Sources/AngouriMath/Functions/Continuous/Integration/IndefiniteIntegralSolver.cs # Sources/AngouriMath/Functions/Continuous/Integration/Integration.Definition.cs
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Part of #718.
A power of a linear that is not whole, times a linear, over a whole power of a quadratic —
L^m N / Q^n— is reduced one power ofQat a time (SolveAPowerOfALinearOverAPowerOfAQuadratic). Each step writes the numerator againstQ'andQby a 2×2 Cramer solve; what is left is the same shape withnone lower, and overQ^1the integrator already answers it. The last integral is asked with its two coefficients named, so the solve that ends the chain sees atoms rather than compound coefficients.Guards: the rule answers only the question asked; it declines a bare
xto a positive power, imaginary coefficients, and a quadratic whose discriminant is a polynomial square (that one factors, and the factored route is the better answer).Measured
Pocket of 7,020 Rubi rows, 5 s per row, with both arms on master 37b4798. The branch has merged #1891 since then, and the unit test and build pass on the merge:
All nine new answers held when each row was rerun alone. Two other rows read as moves in the pocket and were timeouts on both sides when rerun alone (1.2.1.4:1120, 1.3.1:568).
(d+e*x)^(7/2)/(a+b*x+c*x^2)^21/((d+e*x)^(3/2)*(a+b*x+c*x^2)^2)(d+e*x)^(k/2)/(a+b*x+c*x^2)^3, k = 7, 5, 3, 1, −1(A+B*x)/(x^(3/2)*(a+b*x+c*x^2)^2)On the 6,939 rows that both solve, total time is 1,170 s against 1,530 s on master. Answers run from 1.2K to 7.1K characters. Exact checks at 50–2000 digits, with every symbol pinned, passed 168 of 168 over the targets, and 8 of 8 on Welz 102. The families 0–8 corpora have no losses. The gate passed, and the suite run in two halves passed 15,401 tests.
BREAKING-CHANGES.mdrecords the new answers.🤖 Generated with Claude Code
https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura