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

 Method (no overloads)

Summary

A power of x times a function of one logarithm of a monomial, x^m G(ln(c x^n)),
under t = ln(c x^n): dx = x dt/n, and x^(m + 1) is
K e^((m + 1) t/n) with K = x^(m + 1) (c x^n)^(-(m + 1)/n), whose derivative
is 0 wherever it is defined. So the answer is K/n times the integral of
e^((m + 1) t/n) G(t) at t = ln(c x^n), an antiderivative wherever
c x^n is positive -- for an even n and a positive c the whole line,
negative x included, where ln(c x^n) is not ln(c) + n ln(x). Where
c x^n is negative, t is ln(-c x^n) + i pi, and a function of whole
powers of e^t, sqrt(sinh(2 t)) say, is real there as well; a radical one
has its integral in t found for a real t, under which e^t is
positive, so that answer is given provided c x^n > 0. A power of a monomial,
(e x)^p, is x^p times a factor of the same kind. What by parts leaves of
(e x)^m Si(d (a + b ln(c x^n))) is this, with G(t) a sine over a linear in
t. Rubi's answers to 8.3 to 8.5 are written in exactly this K.
https://github.com/asc-community/AngouriMath/issues/1501

Remarks

After SolveByLogarithmSubstitution(AngouriMath.Entity,AngouriMath.Entity.Variable,System.Boolean), which answers ln(x) alone.

























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