AngouriMath
SolveANegativePowerOfALinearBesideAPowerOfAnotherByTheRecurrence(AngouriMath.Entity,AngouriMath.Entity.Variable,System.Boolean)
Method (no overloads)
Summary
A whole power m <= -2 of a linear beside a power that is not whole of another,
times a polynomial:1/((a + b x)^4 sqrt(c + d x)) ,
(A + B x + C x^2)/((a + b x)^3 (c + d x)^(3/2)) . The polynomial is written in powers
of the first linear, and each negative power is raised to-1 by the recurrence
int L1^m L2^n = L1^(m + 1) L2^(n + 1)/((m + 1) D) - (m + n + 2) d/((m + 1) D) int L1^(m + 1) L2^n ,
L1 = a + b x , L2 = c + d x , D = b c - a d ; what is left are the
questionsint L2^n/L1 and int L1^k L2^n for k >= 0 , which the
integrator answers in a step.
times a polynomial:
of the first linear, and each negative power is raised to
questions
integrator answers in a step.
Remarks
it, and the reduction of that took four seconds at the square and twenty at the fourth
power; and the partial fractions of a rational function beside a root are terms of
exactly this shape, so it is their cost too.
https://github.com/asc-community/AngouriMath/issues/718
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