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Hold an answer through a root taken real where a second root makes the integrand real (#2017)
Three places take a root as a nonnegative real u and attach `radicand >= 0`
to the answer: the factoring in SolveByLinearRadicalSubstitution and the "as
the root" and factored branches of SolveBySubstitution. Where a second root
is imaginary on the same side, the integrand is real there as a product of
two imaginary roots, and the answer said nothing. Timofeev 926 and 927,
asec(x) beside sqrt(x^2 - 1), are real on 0 < x < 1; Rubi's
sqrt(cot) (a + b tan)^(k/2) and its kin in 4.3.2.1 and 4.3.3.1 are real
where tan < -a/b.
ConditionedOnTheRadicand, where the answer holds an even root besides,
asks PartialFractions.WhereTheDerivativeHoldsBeyondTheRadicand. That sorts
15 points per sign pattern by interval arithmetic on the integrand and the
radicand, and only where one is real beyond the zero differentiates the
answer, lazily and real points first. An answer that holds at every real
point there, beside exactly one other half-odd root, carries the integrand's
real domain -- `tan(c + d x) >= 0 or a + b tan(c + d x) <= 0`; one that also
holds at every complex point carries no condition; anything else keeps the
old one, and so does an integrand real nowhere beyond the zero.
40 answers on part of the line -> whole (26 of 4.3.3.1, 12 of 4.3.2.1,
Timofeev 926, 927), each checked with mpmath on both sides; the condition
grows 28-60 characters. Chain against master a567c8b (base a4e7656): pocket
of 2,104 rows, exactly those 40 Partial -> Solved, 0 wrong; samples unchanged
in verdicts. The check costs 24-111 MB on the rows it changes; Bondarenko 22
and 23, which the first version re-answered at 2.4x the allocation, are
master's again, byte for byte, at 0.582 and 0.505 GB.
Part of #233.
Claude-Session: https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura
Co-authored-by: Claude Opus 5.5 (1M context) <noreply@anthropic.com>
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