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SolveAPolynomialTimesPowersOfPolynomialsByRaisingThePowers(AngouriMath.Entity,AngouriMath.Entity.Variable)

Method (no overloads)

Summary

A polynomial times powers of polynomials, P R_1^(s_1) ... R_m^(s_m) with an
exponent that is not a whole number -- a fraction or a symbol -- answered as
Q R_1^(s_1 + 1) ... R_m^(s_m + 1) for a polynomial Q where there is one:
with R = R_1 ... R_m, the derivative of that is the integrand's powers times
Q' R + Q sum_i (s_i + 1) R_i' R/R_i, and that equal to P is a linear
system in the coefficients of Q.

Remarks

Rubi's (a g + e x - b g x^4)/(a + b x^4)^(3/2) is
(2 a g x + e x^2)/(2 a sqrt(a + b x^4)); its
(d + f x^2)^p (2 c d f + 2 b f^2 (3 + 2 p) x + 2 c f^2 (3 + 2 p) x^2) is
(b f (3 + 2 p)/(1 + p) + 2 c f x) (d + f x^2)^(p + 1); and
x^2 (1 - a x)^(-1 - n (1 + n)/2) (1 + a x)^(-1 - (n - 1) n/2) is a linear times both
powers raised. Each was declined: SolveByAPolynomialTimesAPowerOfTheBase(AngouriMath.Entity,AngouriMath.Entity.Variable) reads one base, holding a function of x, and solves for rational coefficients;
SolveAsTheDerivativeOfAProductOfPowers(AngouriMath.Entity,AngouriMath.Entity.Variable,System.Boolean) wants a constant for Q; and the
rules for the root of a quadratic read no quartic and no symbolic exponent.
The degree of Q. The leading coefficient of the bracket is that of Q times lc(R) (deg Q + sum_i (s_i + 1) deg R_i), so deg Q is
deg P - deg R + 1, or sigma = -sum_i (s_i + 1) deg R_i where that is a whole
number above it, at which the top equation leaves the top coefficient free: the quartic
above has sigma = 2 against 1.
Where it holds. Wherever no base is zero: (z^w)' = w z^(w - 1) z' and
z^(w + 1) = z z^w hold for the principal power at every z but zero, on the
negative real axis too. The system is solved with the symbols pinned first, which
declines one with no solution at once, and the answer is differentiated back at sampled
points. A whole exponent of -1 is a logarithm's, and a rational function, with no
exponent that is not whole, is the rational integrator's; both are declined. Asked late,
beside the base's rule, so that the substitutions and the rules for the roots of
quadratics answer what they read in their own forms first.
https://github.com/asc-community/AngouriMath/issues/233

AngouriMath

Table of contents

AngouriMath

SolveAPolynomialTimesPowersOfPolynomialsByRaisingThePowers​(AngouriMath.​Entity,​AngouriMath.​Entity.​Variable)

SolveAPolynomialTimesPowersOfPolynomialsByRaisingThePowers(AngouriMath.Entity,AngouriMath.Entity.Variable)