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A polynomial over a power of a linear beside a radical is written in powers of the linear at its root - #1391
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…or coefficients and the quadratic's numerator by the residue modulo it, and a decomposition is integrated term by term Every rational function of the tangent with a power of a linear in it -- Rubi's tan^4 (A + B tan)/(a + b tan)^4 is u^4 (A + B u)/((a + b u)^4 (1 + u^2)) under the substitution -- went to the Hermite reduction, because the partial-fraction gate for symbolic linear factors asked for linears only, and the reduction answered it in a^63 b^10, right and unusable: the answer did not evaluate within the corpus's budget. The gate now admits quadratic factors to the first power beside the linears, and the decomposition is computed where each piece lives. The coefficient of 1/F^j over a linear F of multiplicity k is beta^(j-k) times the (k-j)th Taylor coefficient at the root of the numerator over the other factors, by the quotient of the two polynomials' own series at the root -- differentiating the quotient symbolically gives the same numbers as quotients of derivatives the gcd does not cancel, in b^249 at the third order. The numerator over a quadratic is the residue of the numerator times the inverses of the other factors in the ring modulo the quadratic: two coefficients, each a small quotient, with nothing expanded but the quadratic's own reductions. Two things around it. The gcd that puts a coefficient in lowest terms is bounded at 256 nodes for the simplifier's sake, and a fourth-order block's coefficient is eleven hundred; the partial-fraction caller may raise the bound, and where the gcd still declines past its degree and step limits, the written factors of the denominator -- a series leaves b^9 (a^2 + b^2)^7 -- are divided out of the numerator exactly for as long as that goes. And a decomposition is now integrated one term at a time as written: handed to the chain whole, a sum with a term past the size taken as written went through the expanding gather, which combined D^(-1) (P + Q x)/(1 + x^2) into (P + Q x)/(D + D x^2), a quadratic with symbols in every coefficient, integrated as a piecewise on the sign of its discriminant. x^4 (A + B x)/((a + b x)^4 (1 + x^2)) is a line of arctangents and logarithms in a quarter of a second, and the tangent form with it. Part of #718. Co-Authored-By: Claude Opus 5 (1M context) <noreply@anthropic.com> Claude-Session: https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura
…powers of the linear at its root, and what the power divides goes over the radical alone P(x)/((a + b x)^k S(x)) with S not a polynomial in x: P is p_0 + p_1 (a + b x) + ... + (a + b x)^k Q(x), a polynomial identity whose coefficients are the Taylor coefficients of P at -a/b over the powers of b, so the integrand is Q/S + sum p_j/((a + b x)^(k-j) S) exactly and everywhere, and each piece is integrated on its own. (A + B x + C x^2 + D x^3)/((a + b x) sqrt(c + d x)) was answered through the substitution u = sqrt(c + d x) term by term, a cubic in u^2 - c over a symbolic quadratic in u each time, in a hundred kilobytes of piecewise that did not evaluate within the corpus's budget: three of six in Rubi's P(x) (a + b x)^m (c + d x)^n file and four of six in the next were timeouts of that kind. Reduced, it is a quadratic over the root, three powers by the substitution, and one p_0/((a + b x) sqrt(c + d x)); under four kilobytes in half a second. Before the sum is split, so that the whole polynomial is reduced once rather than each of its terms, and only beside a factor that is not a polynomial, a quotient of polynomials being the partial fractions' to take apart. Carries #1389's commit for the Taylor coefficients at a root, which are its. Part of #718. Co-Authored-By: Claude Opus 5 (1M context) <noreply@anthropic.com> Claude-Session: https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura
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P(x)/((a + b x)^k S(x))withSnot a polynomial inx— a polynomial over a linear beside a radical — is written with the polynomial in powers of the linear at its root,P = p_0 + p_1 (a + b x) + … + (a + b x)^k Q(x), a polynomial identity whose coefficients are the Taylor coefficients ofPat−a/bover the powers ofb. The integrand is thenQ/S + Σ p_j/((a + b x)^(k−j) S)exactly and everywhere, and each piece is integrated on its own.Why.
(A + B x + C x^2 + D x^3)/((a + b x) sqrt(c + d x))was answered through the substitutionu = sqrt(c + d x)term by term, a cubic inu^2 − cover a symbolic quadratic inueach time, in a hundred kilobytes of piecewise that did not evaluate within the corpus's budget — three of six in Rubi'sP(x) (a + b x)^m (c + d x)^nfile and four of six in the next were timeouts of that kind. Reduced, it is a quadratic over the root (three powers by the substitution) and onep_0/((a + b x) sqrt(c + d x)): under four kilobytes, half a second.Placed before the sum is split, so the whole polynomial is reduced once rather than each of its terms; only beside a factor that is not a polynomial, a quotient of polynomials being the partial fractions' to take apart.
Carries #1389's commit (base
master): the Taylor coefficients at a root are that PR's helper, madeinternalhere. Rebased onto master once #1389 merges.Measured. Family 1 (6 per file = 228, same harness on both sides): 136 → 137 on top of #1389, one timeout answered; the remaining timeouts of the same two files are
(a + b x)^2 (c + d x)^(3/2)and(a + b x)^3 (c + d x)^(5/2)denominators, where the reduced piecesp_j/((a + b x)^(k−j) (c + d x)^(n/2))go through the radical substitution into a repeated symbolic quadratic, which the partial fractions do not take apart yet — the next step. Independent suites 1707 → 1707, 0 wrong, 0 timeouts. Four suites green; the test differentiates three shapes back at pinned symbols and bounds each answer at four kilobytes.Part of #718.
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https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura