AngouriMath
SolveByWritingTwoRootsAsOneUpToAConstantFactor(AngouriMath.Entity,AngouriMath.Entity.Variable,System.Boolean)
Method (no overloads)
Summary
Two square roots written as one up to a factor whose square is a constant: where the
quotient of the radicands is a constant times an even power of x,sqrt(A)/sqrt(B) is K x^k , and where their product is free of a root of x that each holds,
sqrt(A) sqrt(B) is K sqrt(A B) . Rubi's
(g + h x) sqrt(a + b x + c x^2)/(a d + b d x + c d x^2)^(3/2) ,
sqrt(b - a/x^2)/sqrt(a - b x^2) and
1/(sqrt(1 + x) sqrt(-1 - sqrt(x)) sqrt(-1 + sqrt(x))) .
quotient of the radicands is a constant times an even power of x,
Remarks
interval where it is continuous: it goes in front of the integral of the rest, as Rubi
writes it. For powers
declined: SolveByCombiningRadicals(AngouriMath.Entity,AngouriMath.Entity.Variable,System.Boolean) combines two roots of polynomials it
can sign, and reads no root inside a radicand, and
SolveByWritingAConstantMultipleOfARadicandOverIt(AngouriMath.Entity,AngouriMath.Entity.Variable,System.Boolean) wants a logarithm to name
the radicand. Late and at the question asked, so that every rule reading the roots as
written goes first. The rewriting is checked at sampled points rather than the answer
differentiated back: the decimal evaluation of the derivative of K, zero, disagreed where
a root of x is imaginary, and the answer there was right.
https://github.com/asc-community/AngouriMath/issues/233
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