AngouriMath
SolveARationalFunctionTimesAPowerOfALogarithm(AngouriMath.Entity,AngouriMath.Entity.Variable,System.Boolean)
Method (no overloads)
Summary
A rational function of x times a whole power of a factor whose
derivative is rational --A + B ln(R) or A + B arctan(R) with R rational in x -- by parts with the power differentiated:
F^p Q = (F^p int Q)' - p F^(p-1) F' int Q , and the remainder is a rational
function times the next lower power, rational outright for the first. Rubi's
(f + g x)^m (A + B ln(e ((a + b x)/(c + d x))^n))^p for whole m and
p , of which (f + g x)(A + B ln(e (a + b x)^2/(c + d x)^2)) and
(A + B ln(e (a + b x)/(c + d x)))/(f + g x)^5 were a substitution search past
its budget on the symbols, simplifying the integrand over each candidate's
derivative, before the same step of parts was reached below it.
derivative is rational --
function times the next lower power, rational outright for the first. Rubi's
its budget on the symbols, simplifying the integrand over each candidate's
derivative, before the same step of parts was reached below it.
Remarks
The step is exact and closed: what is integrated is rational, what is left is
asked as a question of its own with the power one lower, and the power is what
the recursion decreases on. At any depth where the rational function holds
x , since the step is closed and its own remainder is two levels down once
a constant has been taken out in front; against a constant alone it is by parts
against one, asked at the top or one below it like the rest of by parts.
https://github.com/asc-community/AngouriMath/issues/718
asked as a question of its own with the power one lower, and the power is what
the recursion decreases on. At any depth where the rational function holds
a constant has been taken out in front; against a constant alone it is by parts
against one, asked at the top or one below it like the rest of by parts.
https://github.com/asc-community/AngouriMath/issues/718
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