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A power of a multiple of a quadratic's derivative beside a power of the quadratic is a binomial - #1691
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Rafael-SOWNet merged 2 commits intoOct 2, 2026
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…he quadratic is a binomial Rubi's 1.2.1.2 (b d + 2 c d x)^m (a + b x + c x^2)^p, with a power that is not whole, was declined or ran out of time: no substitution the search tried saw that the linear is a multiple of the quadratic's derivative. Under t = b d + 2 c d x the quadratic is (t^2/d^2 - (b^2 - 4 a c))/(4 c), exactly, since the substitution is affine, and the integrand is a binomial in t. The quadratic is written in t from its coefficients rather than by substituting, which would leave a linear term zero as a value and not as written; and each constant of the binomial that is a compound of the coefficients is named by a symbol of its own and written back into the answer. The rules answer t^(-7) sqrt(c t^2 + k) in 63 ms, and with a - b^2/(4 c) for k they ran past twenty seconds. A quadratic with no linear term is a binomial already and is left alone, else the rule would ask its own question with t scaled. Part of #718. Co-Authored-By: Claude Opus 5.5 (1M context) <noreply@anthropic.com> Claude-Session: https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura
…-quadratic-as-the-variable
This was referenced Oct 2, 2026
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Part of #718.
Rubi's 1.2.1.2 has 344 problems of the form
(b d + 2 c d x)^m (a + b x + c x^2)^p, a power of a multiple of the quadratic's derivative beside a power of the quadratic. With a power that is not whole these were declined or ran out of time, since nothing saw that the linear is a multiple of the derivative.Under
t = b d + 2 c d xthe quadratic is(t^2/d^2 - (b^2 - 4 a c))/(4 c). The substitution is affine, so this holds exactly, and the integrand is a binomial int. The new rule,SolveByTheDerivativeOfAQuadraticAsTheVariable, does that:tfrom its coefficients. Substituting instead would leave a linear term that is zero as a value and not as written.t^(-7) sqrt(c t^2 + k)in 63 ms, and witha - b^2/(4 c)written forkthey ran past twenty seconds.tit would be the same question withtscaled.(a + b x + c x^2)^(3/2)/(b d + 2 c d x)^3(a + b x + c x^2)^(1/2)/(b d + 2 c d x)^7(a + b x + c x^2)^(5/2)/(b d + 2 c d x)^5(b d + 2 c d x)^(5/2)/(a + b x + c x^2)^3Master is
e33fbab2, and the times include checking the answer at six points.On the whole 1.2.1.2 pocket against master
be009d56, the 234 of those 344 problems whose answers are in functions the library has, each run alone at the corpus's 5-second budget: 160 solved on master, 211 here, none wrong on either, and 13 timeouts where master has 17.Tests:
DerivativeOfAQuadraticAsTheVariableTest, the four rows above, differentiated back with the symbols pinned, on both sides of the derivative's root wherever the integrand is real.Measured on the Rubi corpus against master
be009d56, with both builds side by side:Both builds give the same three wrong answers: two in family 1, 1.2.1.1's roots of negative squares, which #1672 fixed on master after this branch's base, and one in family 7,
e^(2 acoth(a x)) sqrt(c - a c x)/x, which #1682 fixed.Six problems moved between the two runs, so I re-ran each alone on both builds. This branch answers three that master declines or runs out of time on, and none that master answers is lost: from 1.2.1.2,
sqrt(c d^2 + 2 c d e x + c e^2 x^2)/(d + e x)^4,sqrt(a + b x + c x^2)/(b d + 2 c d x)and1/((b d + 2 c d x)^3 (a + b x + c x^2)^(3/2)).The suite passes, 14,316 tests on a fresh build, and so does the allocation gate. Master
e33fbab2is merged in since, without conflicts; the four rows above were measured on that merge, and its tests pass.🤖 Generated with Claude Code
https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura