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Read the root's quadratic at a symbolic pole without the slope's condition - #1888
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Rafael-SOWNet merged 4 commits intoOct 14, 2026
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…ition Q(p) for p = -n/m comes back with `provided not m = 0` where Q has no linear term, and the table read no quadratic in t through the condition: `1/((d + k x) sqrt(a + c x^2))` was declined where `1/((d + x) sqrt(a + c x^2))` was not. The slope is not zero for a linear, so the value is read bare. Co-Authored-By: Claude Opus 5.5 (1M context) <noreply@anthropic.com> Claude-Session: https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura
… p is written With the slope's condition gone, the rule for a linear beside a root read `1 + i a x` as a pole at i/a, and wrote sgn(x - i/a) with nothing checking it after; `1/(e^(3 i atan(a x)) x^3)` lost its answer. Co-Authored-By: Claude Opus 5.5 (1M context) <noreply@anthropic.com> Claude-Session: https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura
…lic-slope-beside-a-root-is-read-bare
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 #718
sqrt(a + c x^2)/(x (d + e x))and its kin in Rubi's 1.2.1.4 were declined, thoughsqrt(a + b x + c x^2)/(x (d + e x))was answered. The difference was inSolveALinearBesideTheRootOfAQuadratic, which works at the linear's rootp = -d/e. With nob xterm,(a p^2 + c).InnerSimplifiedcomes back as3 (-d/k)^2 + 2 provided not k = 0.TryStandardIntegralsreads no quadratic intthrough thatProvidedf, so1/((d + k x) sqrt(2 + 3 x^2))was declined where1/((d + x) sqrt(2 + 3 x^2))was not. The condition only says the linear's slope is not zero, so the value is now read withBare, here and inSolveAPolynomialOverAPowerOfALinearBesideTheRoot.The condition had also been stopping the rule on linears with the imaginary unit in them. With it gone,
1 + i a x(whate^(3 i atan(a x))is written as) reached the rule as a pole ati/a. The rule then wrotesgn(x - i/a)and nothing checked it afterwards, and 5.3.6:69,1/(e^(3 i atan(a x)) x^3), lost its answer. The rule now declines a linear holding the imaginary unit, which is what it did by accident before. The other rule checks its own derivative, and master's route for 5.3.6:69 goes through it, so it keeps complex poles.Measured
intbench, this branch against master 1763843, 5 s budget. The pocket was 1.2.1.2–1.2.1.4, 1.2.1.6, 1.2.1.9, 1.2.2.4, 1.3.2, 5.3.6, 7.3.6 and 4.3.2.1 in full (9,239 rows), plus the usual family sample:
All 15 moves held when re-run alone (master solves 1 of them), and none went from solved to anything else:
sqrt(a + c/x^2)/(d + e x)I checked the 14 algebraic rows exactly with
exactcheck(exact pins, 50 and 2,000 digits, four points, three pin sets with negative coefficients): 336 of 336 diff 0. The gate and the suite pass (15,389).🤖 Generated with Claude Code
https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura