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

 Method (no overloads)

Summary

x^j T^alpha/Q^(K + 1/2), a power of a quotient of linears T = (p + q x)/(r + s x) over a half-odd power of a multiple of their product, Q = kappa (p + q x)(r + s x),
closed by Liouville's ansatz F = T^alpha P(x) Q^(1/2 - K). With D = q r - p s,
T'/T = D/((p + q x)(r + s x)), so F' = T^alpha Q^(-K - 1/2) (alpha kappa D P + Q P' - (K - 1/2) 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 - 2K) p_(m-1) + (alpha kappa D + B (m - K + 1/2)) p_m + C (m + 1) p_(m+1) = [m = j],
solved from p_0 up, the last equation fixing p_0 and the one past it checked.
e^(n acoth(a x)) arrives folded as ((a x + 1)/(a x - 1))^(n/2): Rubi's 7.4.2
e^(n acoth(a x))/(c - a^2 c x^2)^(3/2) is e^(n acoth(a x)) (a x - n)/(a c (1 - n^2) sqrt(c - a^2 c x^2)).
https://github.com/asc-community/AngouriMath/issues/233

Remarks

The principal powers are differentiated as such wherever T and Q are not 0,
on a cut as off it, so the answer holds where the integrand is defined. Where the check
past the last equation fails, the antiderivative is not of this shape -- a hypergeometric
one -- and the rule declines. Only for a power of T that is not whole, and
Q without a root at 0, which the system divides by.

























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