Repository navigation
Integrate cube roots of two linears beside a power of a third, and a root over a power of x (b + c x) - #1976
Merged
Rafael-SOWNet merged 1 commit intoOct 20, 2026
Conversation
…root over a power of x (b + c x) (a + b x)^(1/3) (c + d x)^(2/3)/(e + f x)^2 and its kin in Rubi's 1.1.1.3 are rational in t = (a + b x)^(1/3)/(c + d x)^(1/3). The substitution that reached it left a binomial in t^3 spelled as k (b - d t^3) + f (c t^3 - a), and the rational rules ran out of time on it. It is written as that binomial now, with the compound coefficients named, where the denominator in t is not squarefree. (A + B x) sqrt(d + e x)/(b x + c x^2)^3 and its kin in 1.2.1.3 are rational in t = sqrt(d + e x) with the quadratic kept as x (b + c x). The substitution that reached it late in the chain multiplied it out. Both rules decline where the linears share a root, decided as a value. Rubi pocket (2,824 rows of family 1): 2777 -> 2794 solved. The moves re-run alone gain 19 rows (1.1.1.3 x12, 1.2.1.2:447, 1.2.1.3:1394-1398). Two sampled rows master answers in 23-24 s timed out under load; at a 60 s budget both builds answer them with the same allocation to 0.0001%. Part of #233. Co-Authored-By: Claude Opus 5.5 (1M context) <noreply@anthropic.com> Claude-Session: https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura
Rafael-SOWNet
added a commit
that referenced
this pull request
Oct 21, 2026
… x) over its two linears (#1994) (d + e x)^(7/2)/(b x + c x^2)^2 was answered by a rule late in the chain that writes the whole integrand as one rational function in t = sqrt(d + e x) with b e - c d written out, and whose reduction never put its coefficients in lowest terms: denominators like b^35 c^5 d^31 k^35, one coefficient of 155,000 characters copied into four arms, 937,000 characters in all. The rule that keeps x (b + c x) for a root above the bar now takes the second power as well as the third and up: in s = t^2 the integrand is a polynomial part and partial fractions over s - d and s + rho, rho = (b e - c d)/c, whose coefficients come from the Taylor series at each root, each linear's fractions asked as one question, with the atoms d, b e/c, e and P's coefficients named -- naming whole coefficients hid a cancellation. Rubi's 1.2.1.2:435 is 2,568 characters from 936,939, and 1.2.1.3:1386 3,933 from 939,897, in about 10 MB from 10-11 GB. Of 83 rows of 1.2.1.2 and 1.2.1.3 with a half-odd power of a linear over a power of b x + c x^2 or a + b x + c x^2, 20 are shorter and 63 unchanged, character for character, and no verdict is lost. The #1976 entry's two third-power figures follow. Part of #233. Claude-Session: https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura Co-authored-by: Claude Opus 5.5 (1M context) <noreply@anthropic.com>
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
1 participant
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.
Two seams of Rubi's family 1:
(a + b x)^(1/3) (c + d x)^(2/3)/(e + f x)^2and its kin in 1.1.1.3 are rational int = (a + b x)^(1/3)/(c + d x)^(1/3)wherever the two powers add to a whole number, for every complexx. The substitution that reached it left a binomial int^3spelled ask (b - d t^3) + f (c t^3 - a), and the rational rules ran out of time on it. It is now written as that binomial from the coefficients, each compound coefficient named and written back into the answer, where the denominator intis not squarefree.(A + B x) sqrt(d + e x)/(b x + c x^2)^3and its kin in 1.2.1.3 are rational int = sqrt(d + e x)with the quadratic kept asx (b + c x). The substitution that reached it, late in the chain, multiplied the quadratic out. It is asked that way now, with the root above the bar and a third power or higher.Each declines where its linears share a root, decided as a value.
Measured on
8a1ef454, branch against master:Re-run alone on both builds, 19 problems go from timed out to answered, in 0.2–2.0 s: Rubi's 1.1.1.3:3415–3418, 3425–3427, 3439, 3440 and 3472–3474; 1.2.1.2:447; and 1.2.1.3:1394–1398. 1.2.1.2:444 goes from 21 s to 0.4 s.
The three sampled problems that master answered and the branch did not, 3.2.1:292, 6.4.2:50 and 6.6.1:25, are the ones master answers in 17–24 s, against a 5 s budget. Re-run alone, both builds answer 6.6.1:25. At a 60 s budget, both answer the other two and allocate the same to 0.0001% (34.2 GB and 28.4 GB), so the branch does not reach them. None of the problems answered on both builds allocates more on the branch. No run had a wrong answer.
The two new test classes differentiate the answers back.
BREAKING-CHANGES.mdhas the entry, with the 2.5.0 column measured on v2.5.0.Part of #233.
🤖 Generated with Claude Code
https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura