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Integrate a power of a quotient of linears beside whole powers of them - #1953
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Rafael-SOWNet merged 1 commit intoOct 19, 2026
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Under the Moebius substitution t = (p + q x)/(r + s x), p + q x = t (r + s x), r + s x = D/(q - s t) and dx = D/(q - s t)^2 dt with D = q r - p s, so t^alpha (p + q x)^j (r + s x)^k dx is D^(j + k + 1) t^(alpha + j) (q - s t)^(-(j + k + 2)) dt: a polynomial in t times a power of it wherever j + k <= -2. A whole power of a quadratic that is a multiple of the linears' product counts as a power of each. e^(n acoth(a x)) arrives folded as ((a x + 1)/(a x - 1))^(n/2), and over (c - a^2 c x^2)^m it is that. Only for a power of the quotient that is not whole: a whole one is a rational function, the partial fractions'. Rubi's 7.4.2: e^(n acoth(a x))/(c - a^2 c x^2)^m for m = 2, 3, 4, and three x^m rows of 7.3.6 and 7.4.2 whose radical substitution leaves such a power in the new variable, go from unevaluated or timed out to solved, 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.
e^(n acoth(a x))/(c - a^2 c x^2)^mwas declined form= 2, 3 and 4. The fold rule writese^(n acoth(a x))as((a x + 1)/(a x - 1))^(n/2). Under the Moebius substitutiont = (p + q x)/(r + s x)we havep + q x = t (r + s x),r + s x = D/(q - s t)anddx = D/(q - s t)^2 dt. Sot^alpha (p + q x)^j (r + s x)^k dxbecomesD^(j + k + 1) t^(alpha + j) (q - s t)^(-(j + k + 2)) dt, a polynomial inttimes a power of it whereverj + k <= -2. The new rule,SolveByTheMoebiusSubstitutionOfAQuotientOfLinears, does this. A whole power of a quadratic that is a multiple of the two linears' product counts as a power of each. Two scope limits:Measured against master
8a1ef454:x^mrows (7.3.6:882 and 1448, 7.4.2:1037) whose radical substitution leaves such a power in the new variable. Their answers carryprovided (1 + a x)/(1 - a x) >= 0The atanh mirrors, answered since earlier in this cycle, now also come out under this substitution, written
((1 + a x)/(1 - a x))^(n/2 - 1)where they were(((1 + a x)/(1 - a x))^(1/2))^(n - 2). The two are equal for principal powers and about the same length. 2.5.0 declined them, so the entry inBREAKING-CHANGES.mdcounts them as new answers.The rebase onto #1952 conflicted at the shared anchor. I resolved it by keeping both rules; the diff is the branch's own, and the tests of #1950–#1952 together with this one pass on the combined tree (27/27).
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https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura