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

x/(sqrt(1 + x^2) sqrt(1 - x^2)) had no antiderivative and x/sqrt(1 - x^4) is one substitution. Nothing reads two radicals: the substitution wants one subtree
to replace, Euler's rule wants one root of one quadratic, and each rule looked at
this integrand and saw two of what it takes one of. It is what by parts leaves from
arcsin(x)/(1 + x^2)^(3/2) and from ln(x + sqrt(1 + x^2))/(1 - x^2)^(3/2),
and from every other pairing of an inverse function's radical with the one in
the power it stands over.
When the identity holds. On the principal branch, sqrt(P) sqrt(Q) = sqrt(PQ) whenever at most one of P and Q is negative: with both negative the
left is i sqrt|P| * i sqrt|Q| = -sqrt|PQ| and the right is +sqrt|PQ|.
For k roots the same count applies -- i^k against i^(k mod 2) --
so the rule asks whether there is a real x at which two of the bases are
negative, and only rewrites when there is not. The bases are polynomials with
numeric coefficients; their real roots cut the line into intervals on each of which
every base keeps its sign, so the count at one point of each interval is the count
on the interval. Where the solver cannot give the roots, the rule declines.
A rewriting rule, unscoped: what it hands on has strictly fewer radicals than what
it was given, and it is asked one level down more often than at the top, since the
two-radical shape is what a by-parts step produces.
https://github.com/asc-community/AngouriMath/issues/718

Square roots only: an odd root of a power is the power of the root on the reals
with no sign to keep. The sign is a symbol to the integration, so a symbol's
square is one and its odd powers the symbol, and the answer holds wherever the
factor is not zero, where the integrand is singular anyway. At the top only: a
substitution's own variable is one it knows the sign of.
https://github.com/asc-community/AngouriMath/issues/718