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Integrate three pseudo-elliptic shapes from Welz's problems in closed form - #1979
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Rafael-SOWNet merged 1 commit intoOct 20, 2026
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… form - (x + A - 2 r)/((x - A) sqrt(C)), for a cubic C vanishing at r midway between the pole and that zero, with C + m (x - r)^2 = lambda (x - r)(x - A)^2: the integral is -(2/sqrt(m)) atan(sqrt(m) (x - r)/sqrt(C)), which uses only sqrt(C)^2 = C. Symbolic integrands only; with numbers the other rules answer. - sqrt(a + b x^2 + c x^4)/(d + e x^4) with c d + a e = 0 and a c < 0: under u = x/R it is (a/d) du/(1 - 2 b u^2 + w^2 u^4), an artanh and an arctangent. - (x + s)/((x^2 + s x + s^2) (a + b x^3)^(1/3)): multiplied above and below by s - x, a rational function of t = x/y beside dt and a function of x^3 beside its derivative. Only over x + s; any other linear leaves a non-elementary part. Each answer is differentiated back at sampled points before it is returned. Rubi pocket (1,138 rows of the independent suites): 1113 -> 1120 solved; the moves re-run alone gain Welz 109, 133, 138, 162, 163, 164 and 192. 3.5:37, which master answers in 23 s, timed out alone; at a 60 s budget both builds answer it, with the same allocation to 0.04%. 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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Seven of Welz's problems were declined. Each is now answered by a closed form for its shape, and the answer is differentiated back at sampled points before it is returned:
(x + A - 2 r)/((x - A) sqrt(C)), for a cubicCthat vanishes atr, whererlies midway between the pole and that zero, andC + m (x - r)^2islambda (x - r)(x - A)^2withma constant. The integral is-(2/sqrt(m)) atan(sqrt(m) (x - r)/sqrt(C)). It uses onlysqrt(C)^2 = C, so it holds on either branch. The rule takes symbolic integrands only; with numbers, the other rules answer.sqrt(a + b x^2 + c x^4)/(d + e x^4)wherec d + a e = 0anda c < 0. Underu = x/R, withRthe root, it is(a/d) du/(1 - 2 b u^2 + w^2 u^4)withw = sqrt(b^2 - 4 a c). That gives an inverse hyperbolic tangent and an arctangent, whose arguments reach±1and±ionly where the binomial or the trinomial vanishes.(x + s)/((x^2 + s x + s^2) (a + b x^3)^(1/3)). Multiplied above and below bys - xit becomes(s^2 - x^2)/((s^3 - x^3) y), withythe cube root. That splits into a rational function oft = x/ybesidedt, and a function ofx^3beside its derivative. The rule applies only overx + s: any other linear leaves a part that is not elementary.Measured on
8a1ef454, branch against master:Re-run alone on both builds, seven of Welz's problems go from unevaluated to answered, in 32–650 ms: 109, 133, 138, 162, 163, 164 and 192. Welz goes from 97 to 104 of 112.
3.5:37, which master answers in 23 s against a 5 s budget, timed out on the branch when re-run alone. At a 60 s budget both builds answer it, and the allocations agree to 0.04% (37.0 GB), so the branch does not reach it. None of the problems answered on both builds allocates more on the branch. No run had a wrong answer.
The three new test classes differentiate the answers back.
BREAKING-CHANGES.mdhas the entry, with the 2.5.0 column measured on v2.5.0.Part of #233.
🤖 Generated with Claude Code
https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura