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Write a trinomial with conjugate roots over its roots in the square - #1948
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Rafael-SOWNet merged 1 commit intoOct 19, 2026
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The pieces over x^2 - r beside the root of a quadratic are case-free now, so one writing over the two roots holds for complex r as well as real, and the conjugate pieces sum to a real, continuous antiderivative. Answers over a + b x^2 + c x^4 with b^2 < 4 a c drop from 9-31K characters to 1-4K. Fixes #1885. 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 00:44
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Fixes #1885.
When
a + b x^2 + c x^4has complex roots in x^2, the integrator wrote it as two real quadratics,c (x^2 + p x + m)(x^2 - p x + m), and that writing produced the long answers. Writing it over its two roots in x^2,c (x^2 - r)(x^2 - s), was ruled out because the piece answers were piecewise on signs and so became NaN for complex r. They no longer have sign cases: each piece overx^2 - risarctan(w x / sqrt(d + e x^2))withw^2 = (d + e r)/(-r), which holds on either branch. So one writing now serves real and complex roots, and the conjugate pieces add up to a real antiderivative.Characters in the answer, on master 4e5ae5f:
x^2 sqrt(d + k x^2)/(a + b x^2 + c x^4)sqrt(d + k x^2)/(a + b x^2 + c x^4)1/((a + b x^2 + c x^4) sqrt(d + k x^2))x^4/((a + b x^2 + c x^4) sqrt(d + k x^2))(d + k x^2)^(3/2)/(a + b x^2 + c x^4)Checks:
Rubi checks, against master, with no wrong answers anywhere:
(f x)^m (d + e x^2)^q (a + b x^2 + c x^4)^p: 299 → 300 solved. The row that moved was re-run on its own: 0/1 on master, 1/1 here.2.5.0 answered none of these integrands, so no BREAKING-CHANGES entry.
🤖 Generated with Claude Code
https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura