SolveARationalFunctionBesideAnIrrationalOddRootOfALinearSplitFirst​(AngouriMath.​Entity,​AngouriMath.​Entity.​Variable,​System.​Boolean)

The same with numbers for coefficients, beside a cube root or a higher one, where a linear
below the bar has the radicand at its root no rational power of that order: Rubi's
1/(u^2 (1 - u)^(1/3) (3 + u)), which x^3 (1 - x^2)^(1/3) (3 + x^2) is under
u = x^2. In v = (1 - u)^(1/3) the linear 3 + u is 4 - v^3, and
the partial fractions over the cube root of 4 beside the rest ran past the budget, where
split over the rationals first each term beside the root is answered in a tenth of a
second. Before the cancellation with functions as indeterminates, which spent five seconds
declining it, and the substitution, which this pre-empts only there.
https://github.com/asc-community/AngouriMath/issues/233