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SolveAPseudoEllipticQuotientOverTheRootOfACubicBinomial(AngouriMath.Entity,AngouriMath.Entity.Variable)

Method (no overloads)

Summary

A linear over a linear, or a quadratic over a quadratic, beside the reciprocal of the
square root of a cubic binomial a + b x^3, at the coefficients where its
integral is elementary: Rubi's (1 + x + sqrt(3))/((1 + x - sqrt(3)) sqrt(1 + x^3)),
(c - 2 d x)/((c + d x) sqrt(c^3 + 4 d^3 x^3)).

Remarks

The curve y^2 = a + b x^3 is elliptic, and the integral of (e + f x)/((c + d x) y) is elementary only where the coefficients make it so. Rubi's 1.3.3 names those places,
and at each the integral is a constant times int du/(1 + K u^2) at
u = (1 + p x)/y, or (1 + p x)^2/y:

b c^3 = 4 a d^3,  2 d e + c f = 0:   (2 e/d)          int 1/(1 + 3 a u^2),        u = (1 + 2 d x/c)/y
b c^3 = -8 a d^3, 2 d e + c f = 0:   (-2 e/(9 d))     int 1/(1 - (a/9) u^2),      u = (1 + f x/e)^2/y
b^2 c^6 - 20 a b c^3 d^3 - 8 a^2 d^6 = 0, 6 a d^4 e = c f (b c^3 - 22 a d^3), k = (d e + 2 c f)/(c f):
                                     ((1 + k) e/d)    int 1/(1 + (3 + 2k) a u^2), u = (1 + (1 + k) d x/c)/y

and over a quadratic, (f + g x + h x^2)/((c + d x + e x^2) y) where
b g^3 = 8 a h^3, g^2 + 2 f h = 0 and b d f + b c g = 4 a e h:
-(g/e) int 1/(1 - M u^2), M = (b d f - 2 a e h)/(2 e h), at u = (1 + 2 h x/g)/y.
Each condition is decided at sampled values of the symbols, the inner integral is an
arctangent or an inverse hyperbolic tangent by the sign of its constant, and the answer
is differentiated back at sampled points before it is returned.
https://github.com/asc-community/AngouriMath/issues/718

AngouriMath

Table of contents

AngouriMath

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

SolveAPseudoEllipticQuotientOverTheRootOfACubicBinomial(AngouriMath.Entity,AngouriMath.Entity.Variable)