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Integrate a polynomial beside a logarithm of a quotient of linears by its logarithmic derivative - #2005
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Rafael-SOWNet merged 1 commit intoOct 21, 2026
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… its logarithmic derivative By parts, the integral of P F is S F minus the integral of S F', and F' is now read with the power of the quotient written apart, B n (b/(a + b x) - d/(c + d x)), since (Q^n)'/Q^n is n Q'/Q. Taken whole, the derivative was two terms over a cubic, and each remainder was divided by it with simplified coefficients. Over a power of a linear, the question in u = f + g x is now asked without the `0 provided c != 0` the pole's own fraction carries, which the substitution search spent seconds declining. Pocket of 1,202 rows in Rubi's 3.x: 1,171 -> 1,179 solved, 23 -> 15 timeouts, 0 wrong on either arm. Samples of every family unchanged, except 3.x 220 -> 221. The ten rows that moved, re-run alone: 4/10 -> 10/10 solved; 3.2.1:258 20 s -> 58 ms (46.9 GB -> 0.03 GB), 3.2.1:48 17 s -> 2.8 s. No row's allocation grew past 1.3x. 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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…ower of u The rule that reads an exponential's exponent P as the variable, beside its derivative, wrote P as u only where it stood as written. Taking the leading coefficient's power out of a power below the bar hands on e^P P'/(P/c)^k, where P/c is another polynomial, so e^(a + b x + c x^2) (b + 2 c x)/(a + b x + c x^2)^3 went on to the ansatz and ran out of time, where it is e^u/u^3. A polynomial equal to k P coefficient by coefficient is now read as k u, and a whole power of one as the power of k times the power of u. Rubi's 2.3: the cube, past 60 s on master f7f2a7a, is answered in 27 ms (197 characters); the square, 14.4 s and 9.9 GB there, takes 0.3 s and 7 MB (114 characters, carrying a factor that is 1). 2.5.0 declined both. Of the other 21 rows the branch's chain moved against 4877ed2, 18 answer identically on master f7f2a7a and here, and 3 are declined or past 20 s on both: #2005 and #2012 have since written the other mechanisms this branch had, and those are not part of it any more. 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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…ower of u The rule that reads an exponential's exponent P as the variable, beside its derivative, wrote P as u only where it stood as written. Taking the leading coefficient's power out of a power below the bar hands on e^P P'/(P/c)^k, where P/c is another polynomial, so e^(a + b x + c x^2) (b + 2 c x)/(a + b x + c x^2)^3 went on to the ansatz and ran out of time, where it is e^u/u^3. A polynomial equal to k P coefficient by coefficient is now read as k u, and a whole power of one as the power of k times the power of u. Rubi's 2.3: the cube, past 60 s on master f7f2a7a, is answered in 27 ms (197 characters); the square, 14.4 s and 9.9 GB there, takes 0.3 s and 7 MB (114 characters, carrying a factor that is 1). 2.5.0 declined both. Of the other 21 rows the branch's chain moved against 4877ed2, 17 answer identically on master f7f2a7a and here, and 4 are declined or past 20 s on both: #2005 and #2012 have since written the other mechanisms this branch had, and those are not part of it any more. 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
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…ower of u (#2020) The rule that reads an exponential's exponent P as the variable, beside its derivative, wrote P as u only where it stood as written. Taking the leading coefficient's power out of a power below the bar hands on e^P P'/(P/c)^k, where P/c is another polynomial, so e^(a + b x + c x^2) (b + 2 c x)/(a + b x + c x^2)^3 went on to the ansatz and ran out of time, where it is e^u/u^3. A polynomial equal to k P coefficient by coefficient is now read as k u, and a whole power of one as the power of k times the power of u. Rubi's 2.3: the cube, past 60 s on master f7f2a7a, is answered in 27 ms (197 characters); the square, 14.4 s and 9.9 GB there, takes 0.3 s and 7 MB (114 characters, carrying a factor that is 1). 2.5.0 declined both. Of the other 21 rows the branch's chain moved against 4877ed2, 17 answer identically on master f7f2a7a and here, and 4 are declined or past 20 s on both: #2005 and #2012 have since written the other mechanisms this branch had, and those are not part of it any more. Part of #233. Claude-Session: https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura Co-authored-by: Claude Opus 5.5 (1M context) <noreply@anthropic.com>
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Part of #233.
By parts,
∫ P FisS F - ∫ S F'withS = ∫ P.F'is now read with the power of the quotient written apart,B n (b/(a + b x) - d/(c + d x)), which is exact since(Q^n)'/Q^n = n Q'/Q. Taken whole, the derivative was two terms over(c + d x)^2 (a + b x), and each remainder was divided by that cubic with its coefficients simplified.Over a power of a linear, the partial fraction at the pole the antiderivative was matched to is a zero over a constant, which the inner simplification writes
0 provided c != 0. The question inu = f + g xis now asked without that condition (the generic case). The substitution search used to spend seconds declining it.Measured against master 20a4b39, then rebased onto a567c8b:
The ten rows that moved, re-run alone (time, allocation):
No row solved on both arms grew its allocation past 1.3×.
ExponentialSubstitutionIntegralTest.ADilogarithmIsDeclinedAtOnce("x^2*sech(x)^2")failed once locally, at load 17, against its 20 s bound. It passed alone, and it allocates 0.999 GB on both master and this branch, so the change does not reach it.The entry in BREAKING-CHANGES.md has 2.5.0's column measured.
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https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura