A rational integral over the roots of its denominator

[Rafael-SOWNet]

The rational integrator declined a rational function whose residues lie in a field of degree three or more. 1/(x^3 + x + 1) was left unevaluated, and so was #1285's sqrt(x)/(1 + x + x^4), which is 2u^2/(1 + u^2 + u^8) under u = sqrt(x). Their logarithms are written now as a sum over the poles they belong to, the form #1285 asked for, with the node from #1624:

sum(A(r)/D'(r) ln(x - r), r in { r : E(r) = 0 })

• Which poles. The poles whose residues are roots of the Rothstein–Trager resultant R are the common roots of D and D'^n R(A/D'), so E is their gcd over the rationals. No arithmetic in the residues' field is needed. Residues that are rational or roots of a quadratic keep the logarithms and arctangents they had.
• Only as a last resort. Asked beside the other rules, the sum answered each of Jeffrey's three terms of (-1 + 4 cos(x) + 5 cos(x)^2)/(-1 - 4 cos(x) - 3 cos(x)^2 + 4 cos(x)^3) with a sum over the roots of a sextic, where the whole is 2 arctan(tan(x/2)^3 - 2 tan(x/2)). So the rule declines as before and notes that a sum would answer. Only when the whole question comes back unanswered, with that noted, is it asked again with sums allowed, under a memo of its own. The answer found then is remembered for the question.
• Checked like any other answer. The integrator checks an answer in double intervals and then in decimals, and neither read the sum node, so every such answer would have been declined by its own check. PreciseEvaluation encloses each root now. It takes Durand–Kerner's approximations and widens each by its Weierstrass correction, n |w_k| about z_k. Where those rectangles are apart, each holds exactly one root (Braess and Hadeler), and the summand is worked out on each rectangle.
• Still declined: the same kind of denominator with a symbol among its coefficients, which the sum is not written for.

Measured

	Was (2.5.0)	Is
"1/(x^3 + x + 1)".Integrate("x")	integral(1 / (x ^ 3 + x + 1), x)	sum(1 / (3 * r ^ 2 + 1) * ln(x - r), r in { r : r ^ 3 + r + 1 = 0 }) + C
"sqrt(x)/(1 + x + x^4)".Integrate("x")	integral(sqrt(x) / (1 + x + x ^ 4), x)	sum(2 * r ^ 2 / (8 * r ^ 7 + 2 * r) * ln(x ^ (1/2) - r), r in { r : r ^ 8 + r ^ 2 + 1 = 0 }) + C
"1/(1 - x^4 + x^8)".Integrate("x")	integral(1 / (1 - x ^ 4 + x ^ 8), x)	sum(1 / (8 * r ^ 7 + (-4) * r ^ 3) * ln(x - r), r in { r : r ^ 8 + -r ^ 4 + 1 = 0 }) + C
"1/(x^4 + a*x + 1)".Integrate("x")	integral(1 / (x ^ 4 + a * x + 1), x)	the same

Jeffrey's integrand is 2 arctan(tan(x/2)^3 - 2 tan(x/2)) on master and here. Every answer above differentiates back to its integrand at six points.

• Rubi, master at ad05a79b and this change on it, run side by side, 0 wrong everywhere:
  • The independent suites: 1753 → 1756 of 1814. The gains are Bondarenko 22 and 23, and Hearn 66, 1/(1 - x^4 + x^8).
  • Family 1 at ten problems a file: 295 → 301 of 377. The gains are 1/(x^2 (1 - x^3 + x^6)), four quotients over 1 - x^4 + x^8 and 1 - 5x^4 + x^8, and 1.3.1 Update actions #361.
  • Families 2–7: 663 of 722 on both.
• Run alone, the nine gains are 0 of 9 on master and 9 of 9 here. The tenth problem that moved, 1.1.4.3 we need more limit solvers #231, is 17.9 s unevaluated on both. Its master timeout came from the load of the side-by-side run.
• A question still declined costs what it did. The problems unevaluated in both arms took:
  • 14.2 s against 13.4 s in the independent suites;
  • 65.6 s against 64.7 s in families 2–7;
  • 89.9 s against 92.3 s in the family 1 sample.
• The unit tests pass: 14,221, none failed (net10.0). The native AOT publish of the C++ wrapper builds with no trim or AOT warnings.
• The performance gate passes on 34c548a8, whose library code is this head's. Allocation matches the baseline on all 19 gated benchmarks.

Tests whose pins moved

BinomialDenominatorIntegralTest, PartialFractionsTest and RationalIntegralsTest pinned four denominators as declined: 1/(x^3 + x + 1), 1/(x^3 + x^2 + x + 2), 1/(x^4 + x + 1) and 1/(x^4 + x^3 + 1). Each is now the sum, checked by differentiating it back. The boundary pinned in their place is 1/(x^4 + a x + 1) and 1/(x^3 + x + a).

DecliningStaysCheap keeps its ten-second bound, renamed FindingThatNothingSplitsStaysCheap. The second pass starts only once the first has declined, so the decline it guards is still inside the bound.

Closes #1285.

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