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

Exact on the domain the integrand has. With a fractional n, x^n is real
only for a positive x -- at a negative one it is a complex number or nothing --
and on x > 0 the identity holds for a positive c. That condition
travels with the answer as provided c > 0, unless c is a positive
number already; the (e x)^(n - 1) of Rubi's 6.5.2 carries the same one.
A symbolic n confines nothing: it can be an even number, and then
(x^n)^b is real for a negative x too, where x^(n b) is not it --
(x^2)^(1/2) is |x| and x^(2 (1/2)) is x, so Rubi's 3.5
(ln(a x^n)^m)^p/x, answered ln(a x^n)^(m p + 1)/(n (m p + 1)), had the
wrong sign wherever the logarithm is negative for m = 2, p = 1/2. There
the power is written K x^(n b) with K = (c x^n)^b/x^(n b), whose logarithmic
derivative is b n/x - n b/x = 0, so it is constant wherever the power is
continuous, for every c, n and b: the answer is K times the
answer for x^(n b), x (x^n)^b/(n b + 1) for the power alone, which is
Rubi's, and no condition on c is owed. Only for a factor of the integrand:
inside a sum K does not come out in front, and with it as a symbol there
1/(1 + 4 (x^(2 n))^(1/n)) was a 568-character piecewise on the sign of
K where it is arctan(2 x)/2; there the power is split as for a
fractional n, which holds for x > 0. A whole exponent is
SolveByDistributingWholePowersOfProducts(AngouriMath.Entity,AngouriMath.Entity.Variable,System.Boolean)'s, and needs no condition.
The two are asked from two places: the monomial where it always was, and the constant
alone after the rules that answer its shapes for any real constant, since this one
owes provided c > 0 and, asked first, answered them weaker.
A constant alone that is written with a minus in front, as Rubi's
(a + b tan(x))/sqrt(-e tan(x)) writes it, has the other sign's arm beside it:
(c g)^b is (-c)^b (-g)^b for a negative c, and the answer is the
two, each where it holds. With the first alone it said nothing for a positive e,
where the integrand is real wherever the tangent is negative: the arm it gave is the one
where the symbol is negative. A constant written without a minus keeps the one arm,
which is the case of its symbols positive. With negativeArm the
constant is taken out the other way, `-c` assumed positive.
https://github.com/asc-community/AngouriMath/issues/718