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Integrate secant powers of a logarithm two apart as the derivative of a product - #1977
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Rafael-SOWNet merged 1 commit intoOct 20, 2026
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… a product With u = a + b ln(c x^n), B = b n and k = m + 1, the derivative of x^k sec(u)^(p - 2) (B (p - 2) tan(u) - k) is x^m (D sec(u)^p - E sec(u)^(p - 2)), where D = B^2 (p - 1)(p - 2) and E = B^2 (p - 2)^2 + k^2. So x^m (alpha sec(u)^p + beta sec(u)^(p - 2)) is a multiple of that product wherever beta D + alpha E = 0, and a lone power is one where E is zero. The cosecant's twin is the same, and k = 0 is left to the substitution. Rubi pocket (1,525 rows of family 4): 1480 -> 1487 solved, none lost. The moves re-run alone answer nine rows of 4.7.5: seven verified, and 4.7.5:337 and 395, whose integrands are complex along the whole real line, agree with Simpson's rule to 1e-13. 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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With
u = a + b ln(c x^n),B = b nandk = m + 1, the derivative ofx^k sec(u)^(p - 2) (B (p - 2) tan(u) - k)isx^m (D sec(u)^p - E sec(u)^(p - 2)), whereD = B^2 (p - 1)(p - 2)andE = B^2 (p - 2)^2 + k^2. Sox^m (alpha sec(u)^p + beta sec(u)^(p - 2))is a multiple of that product whereverbeta D + alpha E = 0. A lone power is the case whereEis zero, which is Rubi'ssec(a + i ln(c x^n)/(n (p - 2)))^p. The cosecant's product is-x^k csc(u)^(p - 2) (B (p - 2) cot(u) + k). The rule does not apply atk = 0, where the logarithm's derivative is beside the power and the substitution answers.Measured on
8a1ef454, branch against master:Re-run alone on both builds, nine problems of Rubi's 4.7.5 that were unevaluated or timed out are answered, and none goes the other way. Seven are verified at sampled points, in 18–362 ms: 4.7.5:334, 341, 342, 392, 398, 399 and 400. The integrands of 4.7.5:337 and 395 are complex at every real point, through
x^(sqrt(-(1+m)^2)/2), so the harness leaves them unverified. Under the probe's quadrature, both answers agree with Simpson's rule over [0.5, 1.5] to 1e-13, in the real part and in the imaginary part. None of the problems answered on both builds allocates more on the branch. No run had a wrong answer.SecantPowersOfALogarithmTwoApartAsTheDerivativeOfAProductIntegralTestdifferentiates 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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https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura