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Write an exponential of a whole multiple of an inverse hyperbolic tangent algebraically - #1931
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Rafael-SOWNet merged 5 commits intoOct 17, 2026
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…f the two linears Co-Authored-By: Claude Opus 5.5 (1M context) <noreply@anthropic.com> Claude-Session: https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura
An odd multiple is written with sqrt(1 - u^2), which beyond u = 1 has the other sign from e^(artanh(u)); beside a power that is not whole the integrand is real there, and 191 of 7.3.6's answers came out wrong. A power of a product splits into its factors' powers only where it is whole. Beside a sum with a reciprocal of x in it the integrand as it was is answered in under half a second, and written so took a minute or more. Co-Authored-By: Claude Opus 5.5 (1M context) <noreply@anthropic.com> Claude-Session: https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura
sqrt(1 - u^2) is e^(artanh(u)) times (1 + u)^(-1) for u < 1 and its negative beyond, where the integrand is not real: e^(artanh(x)) x (1+x)^2 differentiated back to the negative of the integrand at x = 1.9. The sign is constant on each side of the pole at u = 1 and goes in front. The test's x^2/(e^(5 artanh(a x)) (q - a^2 q x^2)^(27/2)) is left to the other rules now, with its odd multiple beside a half-odd power. Co-Authored-By: Claude Opus 5.5 (1M context) <noreply@anthropic.com> Claude-Session: https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura
…n-inverse-hyperbolic-tangent-written-algebraically # Conflicts: # Sources/AngouriMath/Functions/Continuous/Integration/IndefiniteIntegralSolver.cs # Sources/AngouriMath/Functions/Continuous/Integration/Integration.Definition.cs
…ANGES 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 00:43
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Part of #718.
e^(n artanh(u))with u linear arrives ase^((n/2) ln((1 + u)/(1 - u))), and the rules after it timed out on Rubi's 7.3.6, for examplee^(artanh(a x)) x^2/(c - a^2 c x^2)^2. This writes it algebraically and asks the result as a question of its own:((1 + u)/(1 - u))^(n/2).(1 + u)^n/(1 - u^2)^(n/2), timessgn(1 - u). Beyond u = 1 that form is the negative of the principale^(artanh(u)); the sign is constant on each side of the pole and goes in front.1 - u^2,1 + uand1 - ubeside it are gathered into the two linears, but only where they are whole.It declines in two cases:
(c - c/(a x))^3. Master answers those in under half a second, and written algebraically they took a minute.Rubi corpus, against master 67d8e8d (budget 5 s): the 2,771-row pocket (7.3.x) goes from 2,673 to 2,685 solved, with the same 3 wrong on both arms. Of the 275 rows that moved in the first chain, re-run alone, 12 go from timeout or unsolved to solved and none from solved to anything else. Family 0 whole and families 1–8 sampled are unchanged, the allocation gate passes, and the suite passes (15,549 passed, 0 failed) after merging master.
2.5.0 leaves all twelve unsolved. They are recorded in BREAKING-CHANGES with their sizes, 79 to 9,684 characters.
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