Repository navigation
The integrand the root's substitution leaves is cancelled as written, not simplified - #1714
Merged
Rafael-SOWNet merged 4 commits intoOct 3, 2026
Merged
Conversation
…ncelled as written, not simplified A rational function of x^n beside (c + d x^n)^(k - 1/n) is a rational function of u = x/(c + d x^n)^(1/n), and the integrand in u was simplified before it was integrated. What that found is the factor the substitution leaves on both sides of the bar, and with symbols in it the search grew with the power of the binomial: 1/((a + b x^2)^(5/2)(1 + x^2)) took 5 s, the 7/2 power past 30 s. Each factor written alike above and below is cancelled power for power now, and the rest is left as it is written, which the rules after it read better than an expansion: (a + b x^4)^(7/4)/(c + d x^4) is answered in a second. Part of #718 Co-Authored-By: Claude Opus 5.5 (1M context) <noreply@anthropic.com> Claude-Session: https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura
The substitution writes a base twice over where the radicand is a binomial in a half-angle tangent: under a secant's, 1 - (-u^2) - u^2, which is 1 and which nothing after it collects, so the integrand went on with a degree-eight numerator that is not there. sqrt(a + a sec(e + f x))/(c - c sec(e + f x))^4 ran past a minute; it is 1.2 s with each base written as its polynomial in u before the factors written alike on both sides are cancelled. Part of #718. Co-Authored-By: Claude Opus 5.5 (1M context) <noreply@anthropic.com> Claude-Session: https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura
…tegrand-is-inner-simplified
…tegrand-is-inner-simplified
Rafael-SOWNet
deleted the
the-root-divided-integrand-is-inner-simplified
branch
October 29, 2026 00:44
This file contains hidden or bidirectional Unicode text that may be interpreted or compiled differently than what appears below. To review, open the file in an editor that reveals hidden Unicode characters.
Learn more about bidirectional Unicode characters
Sign up for free
to join this conversation on GitHub.
Already have an account?
Sign in to comment
Add this suggestion to a batch that can be applied as a single commit.This suggestion is invalid because no changes were made to the code.Suggestions cannot be applied while the pull request is closed.Suggestions cannot be applied while viewing a subset of changes.Only one suggestion per line can be applied in a batch.Add this suggestion to a batch that can be applied as a single commit.Applying suggestions on deleted lines is not supported.You must change the existing code in this line in order to create a valid suggestion.Outdated suggestions cannot be applied.This suggestion has been applied or marked resolved.Suggestions cannot be applied from pending reviews.Suggestions cannot be applied on multi-line comments.Suggestions cannot be applied while the pull request is queued to merge.Suggestion cannot be applied right now. Please check back later.
Part of #718.
A rational function of
x^nbeside(c + d x^n)^(k - 1/n)is a rational function ofu = x/(c + d x^n)^(1/n), and the integrand inuwas simplified before it was integrated. With symbols in it the simplifier's search grew with the power of the binomial, where integrating what it returned took milliseconds:97c4b6921/((a + b x^2)^(7/2) (1 + x^2))x^2/((a + b x^2)^(7/2) (c + d x^2))(a + b x^4)^(7/4)/(c + d x^4)1/((a + b x^2)^(3/2) (c + d x^2)^3)1/((a + b x^2)^(5/2) (c + d x^2)^3)What changes. The integrand in
uis put over one bar and not simplified. Two things the simplifier did are done instead, since what follows reads them. Each base that is a sum is written as its polynomial inu, which collects what the substitution writes twice:1 - (-u^2) - u^2is 1 under a secant's half angle. And each factor written alike above and below the bar is cancelled, power for power, the factor the substitution leaves on both sides among them; the rest stays as written, which the rules after it read better than an expansion.The last two rows also lean on #1712: their integrands in
uare a polynomial over a power of a quadratic with sums of symbols among its coefficients.Tests:
DividedByTheRootIntegralTest, Timofeev's quartic beside a fourth root and three half-odd powers of a symbolic quadratic, each differentiated back.Measured first on the 397 corpus problems of 1.1.2.3
(a + b x^2)^p (c + d x^2)^qand 1.1.3.3(a + b x^n)^p (c + d x^n)^q, at the corpus's 5-second budget, against master599dcb7f, the branch's base:Measured then on the Rubi corpus against master
599dcb7f:No answer is wrong on either build. The 14 problems the two builds disagreed on, run again one build at a time, are all answered here and 5 of them on master: 7 in 1.1.3.3, 3 in 1.1.2.3, and one each in 1.1.2.5, 1.2.1.3, 3.2.1 and 4.3.7.
The suite passes on the commit measured,
6bc973cb, 14,579 tests, and so does the allocation gate. Master97c4b692is merged in since, without conflicts, and the calculus tests pass on the merge, 3,948 of them.🤖 Generated with Claude Code
https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura