AngouriMath
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)) ,
undert = 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 isK/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,
negativex 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 ofe^t , sqrt(sinh(2 t)) say, is real there as well; a radical one
has its integral int found for a real t , under which e^t is
positive, so that answer is givenprovided 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
under
is 0 wherever it is defined. So the answer is
negative
powers of
has its integral in
positive, so that answer is given
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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