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A symbolic cubic's linear factors are found before the partial fractions - #1841
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Part of #718. Co-Authored-By: Claude Opus 5.5 (1M context) <noreply@anthropic.com> Claude-Session: https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura
… rows 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.
Partial fractions over a symbolic polynomial need its linear factors, and nothing looked for them when the coefficients had symbols in them.
WithSymbolicLinearFactorsFoundnow tries a polynomial of degree three or more with symbols in it, when its leading and constant coefficients are single terms. The candidates are linearsu ± v x, withua divisor of the constant term andvone of the leading coefficient, as for rational roots over the integers. The ones that divide exactly are written apart. If the numerator holds one of them too, it is cancelled from both sides, once per power, before the partial fractions run. Without that cancellation,1/(a sec x + b tan x)^4went from a 3-second decline to a timeout.2.5.0 and master
3c86dda7declined all six of these rows:1/((a + b x)(c + d x)(h + f x)), multiplied outsec(c + d x)^6/(a cos(c + d x) + b sin(c + d x))sec(c + d x)^3/(a cos(c + d x) + b sin(c + d x))^41/(a sec x + b tan x)^4Lengths are characters of
Stringize(). All six ran in one process, in this order.Rubi corpus, both arms built from
88354428, fix against master:net10.0).BREAKING-CHANGES.mdhas an entry, A symbolic cubic's linear factors are found before the partial fractions.SymbolicLinearFactorsOfACubicIntegralTestintegrates the cubic and its square, and checks each answer differentiates back at sampled points.🤖 Generated with Claude Code
https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura