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SolveAPseudoEllipticQuotientOverTheRootOfACubicWithARootMidway​(AngouriMath.​Entity,​AngouriMath.​Entity.​Variable)

 Method (no overloads)

Summary

A linear over a linear beside the reciprocal of the square root of a cubic C,
(x + A - 2 r)/((x - A) sqrt(C)), where C vanishes at r, midway between
the pole and the zero, and C + m (x - r)^2 is lambda (x - r) (x - A)^2 for a
constant m, lambda the leading coefficient: its integral is
-(2/sqrt(m)) atan(sqrt(m) (x - r)/sqrt(C)). Welz's
(a + x)/((x - a) sqrt(a^2 x - (1 + a^2) x^2 + x^3)).

Remarks

With z = (x - r)/sqrt(C), 1 + m z^2 is (C + m (x - r)^2)/C, which is
lambda (x - r) (x - A)^2/C, and 2 C - (x - r) C' is
-lambda (x - r) (x - A) (x + A - 2 r): so
d/dx atan(sqrt(m) z) = -sqrt(m) (x + A - 2 r)/(2 (x - A) sqrt(C)). That uses only
sqrt(C)^2 = C, so it holds on either branch of the root and of sqrt(m), and
the arctangent's argument is i or -i only where C or x - A vanishes. Where m is negative the same is atanh(sqrt(-m) z)/sqrt(-m). A
linear P in atan(sqrt(m) P/sqrt(C)) gives a linear over a linear in one
other way, where C + m P^2 is lambda (x - A)^3, which is not read here.
 Rubi's answer to Welz's (x - a - sqrt(1 + a^2))/((x - a + sqrt(1 + a^2)) sqrt((x - a) (1 + x^2))) is this, and to the one in the summary this with the root split as sqrt(x) sqrt(C/x) and the factor sqrt(x) sqrt(C/x)/sqrt(C) beside it. With numbers for the
coefficients the root is split so, x taken out of it, or the ansatz over Padé
approximants answers; with symbols neither did, and the search past them ran out of time
on Welz's at r = a^2/(2 a - 1). So with a symbol in the integrand only, the
conditions decided by value, and the answer differentiated back at sampled points.
https://github.com/asc-community/AngouriMath/issues/233

























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