AngouriMath
SolveALinearBelowACubeRootOfAQuadraticInClosedForm(AngouriMath.Entity,AngouriMath.Entity.Variable)
Method (no overloads)
Summary
A linear below the cube root of a quadratic, 1/((d + e x) (a + b x + c x^2)^(1/3)) ,
in closed form where the shifted quadratica' + c y^2 , y = x + b/(2 c) , and
the lineard' + e y satisfy c d'^2 = 3 a' e^2 or c d'^2 = -9 a' e^2 .
in closed form where the shifted quadratic
the linear
Remarks
With q^3 = 6 c^2 e^2/d'^2 and R = (a + b x + c x^2)^(1/3) the integral is
-sqrt(3) c e/(d'^2 q^2) arctan(1/sqrt(3) + 2 c (d' - e y)/(sqrt(3) d' q R))
- 3 c e/(2 d'^2 q^2) ln(d + e x) + 3 c e/(2 d'^2 q^2) ln(c d' - c e y - d' q R) , Rubi's
1.2.1.2 rule. Wherec d'^2 = -9 a' e^2 , with t = e y/d' , u = 1 - 3 t and k^3 = a' , it is (-3/8 ln(2 R + k u) + 1/4 ln(d + e x) + 1/8 ln(u)
- sqrt(3)/4 arctan(sqrt(3) k u/(4 R - k u)))/(e k) : on the curve
w^3 = 1 - 9 t^2 the tangent at each point over t = -1 meets it again at
(1/3, 0) , so w + r (1 - 3 t)/2 for the cube roots of unity r are the
functions whose logarithms carry the residues. Otherwise the integral is elliptic. The
answer is differentiated back.
https://github.com/asc-community/AngouriMath/issues/718
- 3 c e/(2 d'^2 q^2) ln(d + e x) + 3 c e/(2 d'^2 q^2) ln(c d' - c e y - d' q R)
1.2.1.2 rule. Where
- sqrt(3)/4 arctan(sqrt(3) k u/(4 R - k u)))/(e k)
functions whose logarithms carry the residues. Otherwise the integral is elliptic. The
answer is differentiated back.
https://github.com/asc-community/AngouriMath/issues/718
Angouri © 2019-2023 · Project's repo · Site's repo · Octicons · Transparency · 4862 pages online