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SolveByDividingByTheRoot(AngouriMath.Entity,AngouriMath.Entity.Variable)

Method (no overloads)

Summary

A rational function of x^n times (c + d x^n)^(k - 1/n) for a whole
k, rationalised by u = x/(c + d x^n)^(1/n).

Remarks

du/dx is c (c + d x^n)^(-(n + 1)/n), so dx/(c + d x^n)^(1/n) is
(c + d x^n) du/c; and u^n = x^n/(c + d x^n) gives x^n = c u^n/(1 - d u^n) and c + d x^n = c/(1 - d u^n), so everything in x^n is rational in
u^n and the integrand is a rational function of u. Timofeev's
1/((1 + x^4)(2 + x^4)^(1/4)) is 1/(1 + u^4) under it, and his
(1 + x^4)^(3/4)/(2 + x^4)^2, whose root is (1 + x^4)^(1 - 1/4), is
rational the same way; Welz's 1/((1 - x^3)(a + b x^3)^(1/3)) is
1/(1 - (a + b) u^3). Rubi's rule for (c + d x^n)^p/(a + b x^n) at
p = -1/n is this substitution.
The identity holds wherever the root is real, which is where the answer is
asked; u goes back in as x (c + d x^n)^(-1/n).
And beside a power of x that is not one of x^n: x^j times a
rational function of x^n times (c + d x^n)^(k - (j + 1)/n) is
u^j times the same in u, since x^j (c + d x^n)^(-j/n) is
u^j. Rubi's 1.1.3.4: x/((a + b x^3)^(2/3) (c + d x^3)) is
u/(c - (b c - a d) u^3), and x^4 (a + b x^3)^(1/3)/(c + d x^3) is the
same with x^3 beside it; each was declined. A power of x below the bar is read
the same way, with the multiple of n it leaves going into the rational function:
(a + b x^3)^(1/3)/(x^2 (c + d x^3)) is x times 1/(x^3 (c + d x^3)) times (a + b x^3)^(1 - 2/3), Rubi's 1.1.3.4:763.
https://github.com/asc-community/AngouriMath/issues/718

AngouriMath

Table of contents

AngouriMath

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

SolveByDividingByTheRoot(AngouriMath.Entity,AngouriMath.Entity.Variable)