x^2 over a three-quarter power of a quadratic binomial beside another is integrated where that is elementary - #1728
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… is integrated where that is elementary x^2/((A + B x^2)^(3/4) (C + D x^2)) at B C - 2 A D = 0, Rubi's 1.1.2.4, is the difference of the two functions whose sum answers the quarter power at that ratio, with another constant, and was declined. It is answered by the signs of A and B now, the constants solved for from the two functions' derivatives and each form checked at points in its sign case. Part of #718. Co-Authored-By: Claude Opus 5.5 (1M context) <noreply@anthropic.com> Claude-Session: https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura
…ree-quarter-power-beside-another-quadratic
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Part of #718.
x^2/((A + B x^2)^(3/4) (C + D x^2))atB C - 2 A D = 0, Rubi's 1.1.2.4, is the difference of the two functions whose sum answers1/((A + B x^2)^(1/4) (C + D x^2))at that ratio, with another constant, and was declined; 2.5.0 declined it too:959c2d2ex^2/((a + b x^2)^(3/4) (2 a + b x^2)),aandbpositivebnegativeanegativex^2/((-2 + 3 x^2) (-1 + 3 x^2)^(3/4))x^2/((2 - 3 x^2)^(3/4) (4 - 3 x^2))What changes. The rule that answers
1/((A + B x^2)^(1/4) (C + D x^2))atB C = 2 A Dreadsx^2above the bar and(A + B x^2)^(3/4)below it as well, and answers it by the signs ofAandB: an arctangent less an inverse hyperbolic tangent, overA^(1/4)and the cube of the root ofBor-B-- of-AwhereAis negative, where the two share one argument -- timesB/D, as the quarter power's answer is. The constants are solved for from the two functions' derivatives, and each form is checked at points in its sign case.Tests:
EllipticLookingQuotientOfBinomialsTest.XSquaredOverTheThreeQuarterPower, six rows across the sign cases ofAandB, each differentiated back at points in its case, the negative side among them; all six fail on master.Measured first on the 446 problems of Rubi's 1.1.2.3 and 1.1.2.4 with a quarter power in them or in their answers, 196 run, at the corpus's 5-second budget, against master
8f3757cd, the branch's base:Measured then on the Rubi corpus against master
8f3757cd:No answer is wrong in the sample on either build. Of the 25 problems the two builds disagreed on, run again one build at a time, 20 are answered here and not on master, 18 of them 1.1.2.4's of this shape, and four are answered on both. One went the other way in that run,
1/((a g + b g x)^3 (A + B ln(e (a + b x)/(c + d x)))), which has no quarter power in it and takes 35 to 43 s on either build without a budget, measured alternately, at the edge of the corpus's.The suite passes on the commit measured,
c1fe3336, 14,597 tests, and so does the allocation gate. Master959c2d2eis merged in since, without conflicts; the calculus tests pass on the merge, 4,010 of them.🤖 Generated with Claude Code
https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura