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

 Method (no overloads)

Summary

x^m arsech(sqrt(x)) or x^m arcsch(sqrt(x)), m >= 0, by parts. The
logarithm ln(p + s) each is written as, p = 1/sqrt(x) and s = sqrt(1/x -+ 1),
has the derivative -p^3/(2 s), since p' = -p^3/2 and s^2 = p^2 -+ 1; so the
remainder is integral x^m p/s/(2 (m + 1)). And K = (p/s) sqrt(r), r = 1 -+ x,
has K^2 = 1: it is 1 or -1, constant where it is continuous, and the remainder is
K integral x^m/sqrt(r), a polynomial times the root, with K written as it is
defined. Rubi's 7.5.2 and 7.6.2.
https://github.com/asc-community/AngouriMath/issues/233

Remarks

Under u = sqrt(x) the substitution asked 2 u^7 arsech(u) one level down, where
the rules that answer it at the top decline, and x^3 arsech(sqrt(x)) ran out of time;
x^2 arsech(sqrt(x)) took thirteen seconds and x^3 arcsch(sqrt(x)) twenty-four. For the
secant's K is -1 where x < 0, and for the cosecant's where -1 < x < 0: left
out, the answer is wrong there, as the one the substitution gave x arcsch(sqrt(x)) was.

























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