A whole power of a product below the bar is read distributed after a radical substitution - #1438
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…radical substitution The quotient-of-linears substitution u = sqrt((a x + 1)/(a x - 1)) leaves (a (1 - u^2))^5 below the bar, which no rational reader reads: the same rational function was declined after five seconds in that spelling and answered in 27 ms as .../(a^5 (1 - u^2)^5). The integrand was then split into its terms and each decomposed on its own, through the scaling route, which is where the nine seconds of Rubi's x^3 sqrt((a x + 1)/(a x - 1)) went (dotnet-stack samples: SolveByLinearRadicalSubstitution, SolveBySplittingSum, SolveByScalingTheVariable). The exponential substitution already distributes such powers for the same reason; the loop is a helper now and the linear-radical substitution calls it too, where the integrand in u is rational -- with a root of u still in it the distributed spelling sent Timofeev's 314 down a path that answered with a sign function. x^3 sqrt((a x + 1)/(a x - 1)): 8.7 s to 0.5 s; e^acoth(a x) x^3: 17.7 s to 0.1 s; two more 7.4.2 rows answered within the probe's budget. #718 Co-Authored-By: Claude Opus 5 (1M context) <noreply@anthropic.com> Claude-Session: https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura
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Where the nine seconds of Rubi's
x^3 sqrt((a x + 1)/(a x - 1))went, bydotnet-stacksamples: the quotient-of-linears substitutionu = sqrt((a x + 1)/(a x - 1))leaves(12 a (u^4 + u^6) + 4 a (u^8 + u^2)) / (a (1 - u^2))^5inu, and(a (1 - u^2))^5below the bar is something no rational reader reads — the same rational function is declined after 5 s in that spelling and answered in 27 ms as.../(a^5 (1 - u^2)^5). So the integrand was split into its terms (SolveBySplittingSum) and each decomposed on its own through the scaling route, four times over the same denominator.The exponential substitution already distributes whole powers of products for the same reason (Timofeev's
1/(a^2 + b^2 cosh(x)^2)); that loop is a helper now,WithWholePowersOfProductsDistributed, and the linear-radical substitution calls it — only where the integrand inuis rational: with a root ofustill in it (a second base, Timofeev's 314) the distributed spelling sent the radical rules down a path that answered with a sign function where the undistributed one is answered exactly, which the suite caught.x^3 sqrt((a x + 1)/(a x - 1))e^acoth(a x) x^3(7.4.2)1/(e^(3 acoth(a x)) x^4)(7.4.2)Family 7 on the corpus: 162/270, 0 wrong, 46 timeouts (159 and 49 before). 1774 suite unchanged: 1706, 0 wrong, 0 timeout. Four rows added to
ExponentialOfALogarithmIntegralTest, pinned as solved, not as timed. Full suite 12171 passed, 0 failed.Part of #718.
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https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura