AngouriMath
SolveAPowerOfXBesideATranscendentalPowerOfABinomial(AngouriMath.Entity,AngouriMath.Entity.Variable,System.Boolean)
Method (no overloads)
Summary
A power x^k beside a power of a logarithm of one binomial a + b x^q , written in
t = a + b x^q , where N = (k + 1)/q is a whole number above 0. Then x^k dx is
(t - a)^(N - 1) dt/(q b^N) on every branch, since x^(k + 1) = (x^q)^N and
x^(k + 1 - q) = (x^q)^(N - 1) are whole powers of x^q , and what is left is a polynomial in t beside
the function of t:(a + b ln(c (d + e/sqrt(x))))^p/x^2 is -2/e^2 times the integral of
(t - d) (a + b ln(c t))^p . A linear a + b x is the rule above's. Rubi's 3.4.
https://github.com/asc-community/AngouriMath/issues/2075
the function of t:
https://github.com/asc-community/AngouriMath/issues/2075
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