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SolveAPowerOfXBesidePowersOfAQuotientOfLinearsAndTheirProductAlgebraically(AngouriMath.Entity,AngouriMath.Entity.Variable)

Method (no overloads)

Summary

x^j T^alpha Q^beta, for a power of a quotient of linears T = (p + q x)/(r + s x) and one of a multiple of their product, Q = kappa (p + q x)(r + s x), where the
antiderivative is algebraic: T^alpha Q^(beta + 1) P(x) with P of degree
j - 1. With D = q r - p s, T'/T = kappa D/Q, so the derivative is
T^alpha Q^beta (alpha kappa D P + (beta + 1) Q' P + Q P'), and asking the bracket
to be x^j is, with Q = A x^2 + B x + C, the three-term system
A (m + 2 beta + 1) p_(m-1) + (alpha kappa D + B (m + beta + 1)) p_m + C (m + 1) p_(m+1) = [m = j] for m from 0 to j: solved from the top, p_(j-1) = 1/(A (j + 2 beta + 1)),
down to p_0, and the equation at m = 0 checked. Rubi's 7.3.6
e^(n artanh(a x)) x^2 (c - a^2 c x^2)^(-1 - n^2/2) is
e^(n artanh(a x)) (1 - a n x)/(a^3 c n (1 - n^2) (c - a^2 c x^2)^(n^2/2)).
https://github.com/asc-community/AngouriMath/issues/233

Remarks

One equation more than unknowns: where the check fails the antiderivative is not this
one term -- it has a logarithm, or is hypergeometric -- and the rule declines. Where it
holds, the rules after it ran past the budget: e^(6 artanh(a x)) x^2/(c - a^2 c x^2)^19 was written as a rational function over a denominator of degree 38 and handed to the
partial fractions, and is (6 a x - 1)/(210 a^3 c^19 (1 - a x)^21 (1 + a x)^15);
e^(5 artanh(a x)) x^2/(c - a^2 c x^2)^(27/2) the same, and
e^(artanh(a x)/2) x^2/(c - a^2 c x^2)^(9/8) was declined. Where the denominator
is small the partial fractions answer it too, as a sum over its two roots:
e^(2 artanh(a x)) x^2/(c - a^2 c x^2)^3 is here (2 a x - 1) T Q^(-2)/(6 a^3 c).
The Moebius substitution after it takes a power of Q that fixes the degree of
P; this takes the degree from x^j.
The powers are differentiated as principal powers wherever T and Q are not 0,
on a cut as off it, so the answer holds where the integrand is defined, and is checked
at sampled points as well. Only for x above the bar: without it P has no
degree to take.

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AngouriMath

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

SolveAPowerOfXBesidePowersOfAQuotientOfLinearsAndTheirProductAlgebraically(AngouriMath.Entity,AngouriMath.Entity.Variable)