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

 Method (no overloads)

Summary

An algebraic function whose symbols a scaling of the variable takes out, x = s y with s = c/d for a linear c + d x in it, or s = (a/b)^(1/n) for an
odd n and a binomial a + b x^n: where every polynomial in it is then a constant
times one with rational coefficients, the integral is a constant times one with numbers,
which the logarithmic ansatz for a cube root answers. Rubi's
1/((c + d x) (2 c^3 + d^3 x^3)^(1/3)) is 1/(c d) int dy/((1 + y) (2 + y^3)^(1/3)),
and (a + b x^3)^(1/3)/(a - b x^3) is a^(-2/3) (a/b)^(1/3) int (1 + y^3)^(1/3)/(1 - y^3) dy.

Remarks

With P(s y) = mu P~(y) for each polynomial P, the integrand is
s prod mu^e g(y), where a root P^(k/m) stands as R^k for
R = P^(1/m)/mu^(1/m), whose m-th power is P~(y) on every branch. The answer
in y is a function of y and the roots of the P~ whose derivative holds wherever their
m-th powers are the P~, so it is written back with each root of a P~ as that
R and y as x/s, rather than as a root of P(x)/mu, which is it only
on the principal branch where mu is positive. A polynomial above the bar that does
not scale to numbers is written out in y, each term asked on its own.
 The ansatz reads one cube root with numbers for coefficients, and these were declined
with symbols in them; with numbers, 1/((1 + 2^(1/3) x) (1 + x^3)^(2/3)) ran past the
budget in it, where in y = 2^(1/3) x it is 1/((1 + y) (2 + y^3)^(2/3)) and
is answered in a third of a second. Late, before the ansatz, and at the question asked;
the answer is differentiated back at sampled points.
https://github.com/asc-community/AngouriMath/issues/233

























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