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

 Method (no overloads)

Summary

A polynomial of degree two or more beside two powers of linears whose exponents add up
to -1, P(x) (a + b x)^m (c + d x)^n with m + n = -1, by the reduction that
leaves a constant times the two powers alone:
int P L1^m L2^n dx = R L1^(m + 1) L2^(n + 1) + C int L1^m L2^n dx, R a polynomial
of degree one less than P. Rubi's (e + f x)^3/((a + b x)^(1/3) (c + d x)^(2/3)).

Remarks

The derivative of R L1^(m + 1) L2^(n + 1) is
L1^m L2^n (R' L1 L2 + R ((m + 1) b L2 + (n + 1) d L1)), which takes x^j to
j a c x^(j - 1) + (j (a d + b c) + (m + 1) b c + (n + 1) a d) x^j + (j + 1) b d x^(j + 1);
the last coefficient is never zero, so R is read off P's coefficients from the top down,
and what is left of P is the constant C.
Written out term by term, each power of x beside the two roots was answered by the
substitution for a quotient of two linear radicals in two to six seconds, and the cube
past the budget, where the two powers alone take a tenth of one. Only for a polynomial of
degree two or more, which is where that ran out of time, and with a symbol in it; the
answer is differentiated back at sampled points.
https://github.com/asc-community/AngouriMath/issues/233

























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