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SolveByPartsAgainstAnInverseHyperbolicOverAHalfOddPowerOfAQuadratic(AngouriMath.Entity,AngouriMath.Entity.Variable,System.Boolean)

Method (no overloads)

Summary

A + B F(a x) for F the inverse hyperbolic sine, cosine or tangent, over a
half-odd power Q^(n + 1/2) of Q = d + e x^2, n >= 1, by parts against
F. Q^(-n - 1/2) is ((x Q^(1/2 - n))' + (2n - 2) Q^(1/2 - n))/((2n - 1) d), so
its antiderivative is S = x sum alpha_k Q^(1/2 - k), k = 1..n; and the logarithm
F is written as has the derivative 2a/(1 - a^2 x^2), a/sqrt(a^2 x^2 + 1) or
a/sqrt(a^2 x^2 - 1), even in x. So the remainder F' S is x times a
function of x^2, and under w = x^2 each of its terms is a power of the linear
d + e w over 1 - a^2 w or the root of another linear, closed by rationalising.
Rubi's 7.1.4, 7.2.4, 7.2.5 and 7.3.4, atanh(a x)/(c + d x^2)^(9/2) and the like.
https://github.com/asc-community/AngouriMath/issues/233

Remarks

By parts in the search differentiates the logarithm as written, and the remainder's
numerator, a (1 - a x) + a (1 + a x) uncollected, was split into terms even in
x, each a harder question than the whole: from (c + d x^2)^(7/2) on they were
declined or ran out of time. Asked whole the remainder in w ran out of time too, and
each term of it is answered in half a second at most. Not where Q is the function's own quadratic, which the
substitution under the hyperbolic function answers; and only an argument a x, which
keeps the remainder odd. The identities hold for real x on every branch the
logarithms take, the cuts included, where each is a constant i pi off its modulus.

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AngouriMath

SolveByPartsAgainstAnInverseHyperbolicOverAHalfOddPowerOfAQuadratic​(AngouriMath.​Entity,​AngouriMath.​Entity.​Variable,​System.​Boolean)

SolveByPartsAgainstAnInverseHyperbolicOverAHalfOddPowerOfAQuadratic(AngouriMath.Entity,AngouriMath.Entity.Variable,System.Boolean)