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Integrate a power of x times an artanh whose derivative is a root, by parts - #1952
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Rafael-SOWNet merged 1 commit intoOct 19, 2026
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… parts atanh(A x/sqrt(d + B x^2)) with A^2 = B is written ln((1 + u)/(1 - u))/2, whose derivative is A/sqrt(d + B x^2): 1 - u^2 = d/(d + B x^2) and u' = A d/(d + B x^2)^(3/2). So by parts what is left is x^(m + 1)/sqrt(d + B x^2), which the radical rules answer. Differentiated as written, the logarithm was a quotient of two sums with a root in each, and the search on what was left ran past the budget. Rubi's 7.3.7: 13 rows go from unevaluated or timed out to solved (7.3 1752 -> 1765 of 1791), 0 wrong. Families 0 and 1 and samples of 2-8 otherwise identical; gate and suite pass. 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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Part of #233.
x^m atanh(sqrt(e) x/sqrt(d + e x^2))was declined or ran out of time for every power fromx^6to1/x^9. The parser writesatanh(u)asln((1 + u)/(1 - u))/2. Foru = A x/sqrt(d + B x^2)withA^2 = B, that logarithm's derivative is2 A/sqrt(d + B x^2), because1 - u^2 = d/(d + B x^2)andu' = A d/(d + B x^2)^(3/2). By parts, what is left isx^(m + 1)/sqrt(d + B x^2). The new rule,SolveByPartsAgainstAnArtanhWhoseDerivativeIsARoot, takes it that way. Differentiated as written, the logarithm was a quotient of two sums with a root in each, and the search ran past its budget. The identity is algebraic, with no branch in it.Measured against master
8a1ef454:Rebased onto #1950 and #1951 without conflict. All 13 rows were declined by 2.5.0 too, and
BREAKING-CHANGES.mdhas the entry.🤖 Generated with Claude Code
https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura