AngouriMath
SolveTwoEqualHalfOddPowersOfLinearsAsOneBesideAPowerOfAQuadratic(AngouriMath.Entity,AngouriMath.Entity.Variable,System.Boolean)
Method (no overloads)
Summary
The same half-odd power of two linears beside a power of a quadratic below the bar,
A^r B^r R(x) , as one power of their product, the integral of (A B)^r R(x) asked,
andsqrt(A) sqrt(B) written back in it for the root of A B : Rubi's
1/((a + b x + c x^2)^2 sqrt(1 - d x) sqrt(1 + d x)) .
and
Remarks
The answer is a function of x and the root whose derivative holds for any root whose
square isA B , and sqrt(A) sqrt(B) is one on every branch; so it holds as
written back, where(A B)^r and A^r B^r differ in sign. Asked of the two roots,
the polynomial term and the substitutions after it ran past the budget, where of the one
it is answered in two seconds. Only beside a quadratic to the second power below the bar
or past it, where it was declined, and at the question asked; the answer is differentiated
back at sampled points.
https://github.com/asc-community/AngouriMath/issues/233
square is
written back, where
the polynomial term and the substitutions after it ran past the budget, where of the one
it is answered in two seconds. Only beside a quadratic to the second power below the bar
or past it, where it was declined, 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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