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Read x times a rational function of an exponential of a logarithm off the integrand - #2047
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… the integrand Rubi's 6.5.3 (1 - b^2 n^2) sech(u) + 2 b^2 n^2 sech(u)^3 for u = a + b ln(c x^n) is (x (-sech(u) - b n tanh(u) sech(u)))', and its 6.6.3 twin in the cosecant the same way; both were declined. With w = e^u the Euler derivative x w' is b n w, so (x G(w))' = G + b n w G'(w): the integrand P/Q in w is that for G = N/Q exactly where N Q + c w (N' Q - N Q') = P Q, linear in N's coefficients, read at sampled values of w and solved with the symbols in; the homogeneous solution w^(-1/c) is no rational function for a symbolic c, so the solution is unique. Only where x stands in the exponentials alone and the exponent's Euler derivative is a constant, at the question asked; the quotient is simplified in w to lowest terms and the answer checked. A symbolic power K B^p of one function of w goes through x K B^(p - 2) H(w), since (x B^(p - 2) H)' = B^(p - 2) (H + c w H' + (p - 2) c (w B'/B) H): Rubi's 6.5.3 sech(a - ln(c x^n)/(n (p - 2)))^p, whose c is -1/(p - 2), is x (p - 2)/(p - 1) sech(u)^(p - 2) (1 - tanh(u)), and its cosecant twin in 6.6.3 the same way. There the system is square, (p - 2) c is simplified once, every row is scaled by c's denominator and the entries taken bare: with 1/(p - 2) and a provided not p - 2 = 0 in them, the elimination over p did not settle. 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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Part of #233.
(1 - b^2 n^2) sech(u) + 2 b^2 n^2 sech(u)^3, foru = a + b ln(c x^n), is(x (-sech(u) - b n tanh(u) sech(u)))'. It was declined, and so was its 6.6.3 twin in the cosecant.w = e^u,x w' = b n w, so(x G(w))' = G + b n w G'(w). Where x stands only in the exponentials, the integrand isP/Qin w. ForG = N/Q,N Q + c w (N' Q - N Q') = P Qis linear in N's coefficients, and it is read at sampled values of w. A symboliccmakes the solution unique: the homogeneous one isw^(-1/c).K B^pof one function B of w goes throughx K B^(p - 2) H(w):H + c w H' + (p - 2) c (w B'/B) H = B^2, as a square system with the rows scaled by the rate's denominator. That answers 6.5.3'ssech(a - ln(c x^n)/(n (p - 2)))^pand 6.6.3's cosecant twin, where Rubi's closed form holds atb^2 n^2 (p - 2)^2 = 1.(1 - b^2 n^2) sech(u) + 2 b^2 n^2 sech(u)^3sech(a - ln(c x^n)/(n (p - 2)))^psech(u)^(p - 2)csch(a - ln(c x^n)/(n (p - 2)))^pChain against master 78cd31b:
Rebased onto ef4150e with one conflict, against #2040's rule at the same place in the dispatch. The resolution keeps both, #2040's first, and both rules' tests pass (9 of 9). The suite is left to CI. The BREAKING-CHANGES lengths were measured on the branch.
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https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura