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Integrate x^m (d + e x^2)^p times an inverse trigonometric function by parts - #1973
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Rafael-SOWNet merged 1 commit intoOct 20, 2026
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…y parts x^m (d + e x^2)^p (a + b F(c x)), with F an inverse trigonometric function and m + 2p + 3 = 0, is integrated by parts against the antiderivative of x^m (d + e x^2)^p that is one term, v = x^(m + 1) (d + e x^2)^(p + 1)/(d (m + 1)): the integral is v (a + b F(c x)) - b ∫ v F(c x)', and F(c x)' is algebraic. Rubi's 5.1.4:789, 5.3.4:1399, 5.5.1:144 and 5.6.1:149 were past the budget. Part of #233. Co-Authored-By: Claude Opus 5.5 (1M context) <noreply@anthropic.com> Claude-Session: https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura
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October 29, 2026 00:43
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x^m (d + e x^2)^p (a + b F(c x)), withFan inverse trigonometric function andm + 2p + 3 = 0, is integrated by parts. It is taken against the antiderivative ofx^m (d + e x^2)^pthat is a single term,v = x^(m + 1) (d + e x^2)^(p + 1)/(d (m + 1)). The integral is thenv (a + b F(c x)) - b ∫ v F(c x)', andF(c x)'is algebraic.Measured on
8a1ef454, branch against master:Re-run alone on both builds, 4 problems go from timed out to answered, in 33–800 ms: Rubi's 5.1.4:789, 5.3.4:1399, 5.5.1:144 and 5.6.1:149. None goes the other way. One problem that master answered at 24 s timed out on the branch in the pocket (5.3.6:227); it sits at the edge of the budget and is not among the moves re-run alone. No run had a wrong answer.
APowerOfXOverAPowerOfAQuadraticTimesAnInverseTrigonometricFunctionByPartsIntegralTestdifferentiates the answers back.BREAKING-CHANGES.mdhas the entry, with the 2.5.0 column measured on v2.5.0.Part of #233.
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