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Split an odd half power of a linear over the second power of x (b + c x) over its two linears - #1994
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Rafael-SOWNet merged 1 commit intoOct 21, 2026
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… x) over its two linears (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. Co-Authored-By: Claude Opus 5.5 (1M context) <noreply@anthropic.com> Claude-Session: https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura
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(d + e x)^(7/2)/(b x + c x^2)^2was answered correctly but at 937,000 characters, where Rubi's answer is about 400. The answering rule, late in the chain, writes the whole integrand as one rational function int = sqrt(d + e x)withb e - c dwritten out. Its reduction never puts its coefficients in lowest terms, which gives denominators likeb^35 c^5 d^31 k^35, and one coefficient of 155,000 characters is copied into all four arms of the answer. The size does not grow steadily with the power: k = 1, 3 and 5 give about 3K, 7/2 gives 937K, and 9/2 gives 12.8K.#1976's rule keeps
x (b + c x)for a root above the bar, but took only the third power and higher, because master already answered the second. It now takes the second as well. Ins = t^2the integrand splits into a polynomial part and partial fractions overs - dands + rho, withrho = (b e - c d)/c. Their coefficients come from the Taylor series at each root, and each linear's fractions are asked as one question. Only atoms are named:d,b e/c,eand the polynomial's coefficients. Naming whole coefficients hid a cancellation, and one neighbour came out longer that way.Before and after, in characters:
(d+e x)^(7/2)/(b x+c x^2)^2(A+B x)(d+e x)^(7/2)/(b x+c x^2)^2^(9/2)/()^2^(k/2)/()^3(A+B x)…^(9/2)/()^3Measured on
8270086f, branch against master:435 and 1386 time out on master at the harness's 5 s budget and are answered on the branch in 37 and 104 ms. That is the pocket's +2. Of the 83 rows of 1.2.1.2 and 1.2.1.3 with this shape, 20 are shorter, 63 are the same character for character, and no verdict is lost. Every changed answer was checked under four pin sets, with points in each interval between 0 and
-b/cwhere the root is real. None of the problems answered on both builds allocates more on the branch. No run had a wrong answer.The rule's test file gains seven cases. Each is answered under a length bound and differentiated back at two points in each interval under two pin sets; 1,262 related tests pass.
BREAKING-CHANGES.mdhas the entry, with the 2.5.0 column measured on v2.5.0. It also updates the two third-power figures in #1976's entry, which this change makes stale.Part of #233.
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https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura