AngouriMath
SolveByWritingOutAProductUnderANestedRoot(AngouriMath.Entity,AngouriMath.Entity.Variable,System.Boolean)
Method (no overloads)
Summary
A root of a product of a polynomial and a sum with a root in it, the product written out
under the root: Rubi'ssqrt(x (a x + b sqrt(-a/b^2 + a^2 x^2/b^2)))/(x sqrt(-a/b^2 + a^2 x^2/b^2)) is sqrt(a x^2 + b x sqrt(...)) over the same, which is answered.
under the root: Rubi's
Remarks
Written as a product, the rule that takes a power of x out of a root took x out and left
sqrt(x) sqrt(a x + b sqrt(...)) , which nothing read. The two radicands are one
polynomial in x and the inner root, so the identity is exact. At the question asked and
late, after every rule reading the product as written; the rewritten integrand is asked on
its own.
https://github.com/asc-community/AngouriMath/issues/233
polynomial in x and the inner root, so the identity is exact. At the question asked and
late, after every rule reading the product as written; the rewritten integrand is asked on
its own.
https://github.com/asc-community/AngouriMath/issues/233
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