AngouriMath
SolveALinearBesideCubeRootsOfTwoLinearsAsOneOfAQuadratic(AngouriMath.Entity,AngouriMath.Entity.Variable)
Method (no overloads)
Summary
A linear below the bar beside the cube roots of two other linears whose product is a
quadratic,K/(L (A^(1/3) B^(1/3))) , as the linear below the cube root of
A B that SolveALinearBelowACubeRootOfAQuadraticInClosedForm(AngouriMath.Entity,AngouriMath.Entity.Variable) answers,
withA^(1/3) B^(1/3) written back for that root in the answer: Rubi's
1/((a + b x) (c + d x)^(1/3) (b c + a d + 2 b d x)^(1/3)) and
1/((d - 3 e x)^(1/3) (d + e x) (d + 3 e x)^(1/3)) .
quadratic,
with
Remarks
The closed form is F(x, R) with logarithms and an arctangent of rational functions
of x andR , and its derivative is the integrand wherever R^3 = A B and
R' = (A B)' R/(3 A B) ; A^(1/3) B^(1/3) is such an R on every branch,
where(A B)^(1/3) is it only where A and B are not both negative. Both were declined:
the closed form reads one cube root, and SolveByCombiningRadicals(AngouriMath.Entity,AngouriMath.Entity.Variable,System.Boolean), which
joins two, does it for roots it can sign. The answer is differentiated back.
https://github.com/asc-community/AngouriMath/issues/233
of x and
where
the closed form reads one cube root, and SolveByCombiningRadicals(AngouriMath.Entity,AngouriMath.Entity.Variable,System.Boolean), which
joins two, does it for roots it can sign. The answer is differentiated back.
https://github.com/asc-community/AngouriMath/issues/233
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