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SolveAPseudoEllipticQuotientBesideACubeRootWithALinearFactor(AngouriMath.Entity,AngouriMath.Entity.Variable)

Method (no overloads)

Summary

A polynomial over a polynomial beside the reciprocal of the cube root of a quadratic or
a cubic P with a linear factor L, P = L Q, or of its square: R/y with R a constant times W/(Q - mu L^2), or R/y^2 with R a
constant times L W/(L^2 - mu Q), for y = P^(1/3), W = 2 Q L' - L Q' and
a constant mu. With m^3 = mu and T = m L/y, or T = m y/L, it
is a constant times T'/(1 - T^3), whose integral is
-ln(1 - T)/3 + ln(1 + T + T^2)/6 + atan((1 + 2 T)/sqrt(3))/sqrt(3). Welz's
1/(x ((x - 1) (q - 2 q x + x^2))^(1/3)) and
(1 - k x)/((1 - (2 - k) x) ((1 - x) x (1 - k x))^(2/3)).

Remarks

For P = L Q, 3 P L' - L P' is L W, so T = m L/y has
T' = m L W/(3 P y), and 1 - T^3 = L (Q - mu L^2)/P: the quotient is
m W/(3 (Q - mu L^2) y). And T = m y/L has T' = -m W/(3 L y^2) and
1 - T^3 = (L^2 - mu Q)/L^2: the quotient is -m L W/(3 (L^2 - mu Q) y^2). Each
uses no more about the root than y^3 = P, and about m than m^3 = mu,
so it holds on every branch of either. 1 - T is the y - m L of Rubi's
answers over y, and the norm P - mu L^3 = L (Q - mu L^2) carries the poles of
the integrand. Matching R is a linear system in the constant's reciprocal and
mu times it, solved for each linear factor of P as written.
With numbers in P the ansatz over logarithms of L - y answers these, and is
asked first; with symbols in it, as in Welz's, it and every other rule declined them. The
answer is differentiated back at sampled points.
https://github.com/asc-community/AngouriMath/issues/233

AngouriMath

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AngouriMath

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

SolveAPseudoEllipticQuotientBesideACubeRootWithALinearFactor(AngouriMath.Entity,AngouriMath.Entity.Variable)