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Integrate a trigonometric function of a + b/x beside a rational function in 1/x - #1878
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Rafael-SOWNet merged 3 commits intoOct 14, 2026
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…ion in 1/x Under x = 1/u, sin(c + d/x)/(e + f x) is -sin(c + d u)/(u (e u + f)) du, which the partial fractions and the sine and cosine integrals answer. Rubi 4.1.12:267, 268, 293, 294, 321, 330, 390-392, 395-397. Part of #718 Co-Authored-By: Claude Opus 5.5 (1M context) <noreply@anthropic.com> Claude-Session: https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura
…iprocal-is-integrated-in-the-reciprocal # Conflicts: # Sources/AngouriMath/Functions/Continuous/Integration/IndefiniteIntegralSolver.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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Under
x = 1/u, a trigonometric function ofa + b/xbeside a rational function ofxbecomes a trigonometric function of a linear beside a rational function ofu. For example,sin(c + d/x)/(e + f x)is-sin(c + d u)/(u (e u + f)) du, which the partial fractions and the sine and cosine integrals answer.x = 1/uis one-to-one on either side of 0. Every argument has to be linear in1/x, so the question inuis not read again. The substitution search couldn't do this, because it reads a candidate's derivative but never solves the candidate forx. Below a root of a linear,sin(a + b/sqrt(c + d x)), the linear radical substitution reaches this shape first.Corpus, measured against master 7df6694:
The
d/xrows are 0.2–2.5K chars. The rows below a root are 4.6–33K chars: the sine over linears that comes out of the substitution carries expanded seventh-degree coefficients, which is master's own defect, #1877.Part of #718
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