3.12.59 \(\int \frac {-2-(-1+k) (1+k) x+2 k^2 x^2}{\sqrt [4]{(1-x^2) (1-k^2 x^2)} (-1+d-(3+d) x-(3+d k^2) x^2+(-1+d k^2) x^3)} \, dx\) [1159]

3.12.59.1 Optimal result

Integrand size = 76, antiderivative size = 86 \[ \int \frac {-2-(-1+k) (1+k) x+2 k^2 x^2}{\sqrt [4]{\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (-1+d-(3+d) x-\left (3+d k^2\right ) x^2+\left (-1+d k^2\right ) x^3\right )} \, dx=\frac {\arctan \left (\frac {\sqrt [4]{d} \sqrt [4]{1+\left (-1-k^2\right ) x^2+k^2 x^4}}{1+x}\right )}{d^{3/4}}-\frac {\text {arctanh}\left (\frac {\sqrt [4]{d} \sqrt [4]{1+\left (-1-k^2\right ) x^2+k^2 x^4}}{1+x}\right )}{d^{3/4}} \]

output

arctan(d^(1/4)*(1+(-k^2-1)*x^2+k^2*x^4)^(1/4)/(1+x))/d^(3/4)-arctanh(d^(1/ 
4)*(1+(-k^2-1)*x^2+k^2*x^4)^(1/4)/(1+x))/d^(3/4)
 

3.12.59.2 Mathematica [F]

\[ \int \frac {-2-(-1+k) (1+k) x+2 k^2 x^2}{\sqrt [4]{\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (-1+d-(3+d) x-\left (3+d k^2\right ) x^2+\left (-1+d k^2\right ) x^3\right )} \, dx=\int \frac {-2-(-1+k) (1+k) x+2 k^2 x^2}{\sqrt [4]{\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (-1+d-(3+d) x-\left (3+d k^2\right ) x^2+\left (-1+d k^2\right ) x^3\right )} \, dx \]

input

Integrate[(-2 - (-1 + k)*(1 + k)*x + 2*k^2*x^2)/(((1 - x^2)*(1 - k^2*x^2)) 
^(1/4)*(-1 + d - (3 + d)*x - (3 + d*k^2)*x^2 + (-1 + d*k^2)*x^3)),x]
 

output

Integrate[(-2 - (-1 + k)*(1 + k)*x + 2*k^2*x^2)/(((1 - x^2)*(1 - k^2*x^2)) 
^(1/4)*(-1 + d - (3 + d)*x - (3 + d*k^2)*x^2 + (-1 + d*k^2)*x^3)), x]
 

3.12.59.3 Rubi [F]

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {2 k^2 x^2-(k-1) (k+1) x-2}{\sqrt [4]{\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (x^3 \left (d k^2-1\right )-x^2 \left (d k^2+3\right )-(d+3) x+d-1\right )} \, dx\)

\(\Big \downarrow \) 2048

\(\displaystyle \int \frac {2 k^2 x^2-(k-1) (k+1) x-2}{\sqrt [4]{k^2 x^4+\left (-k^2-1\right ) x^2+1} \left (x^3 \left (d k^2-1\right )-x^2 \left (d k^2+3\right )-(d+3) x+d-1\right )}dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {-2 k^2 x^2-(1-k) (k+1) x+2}{\sqrt [4]{k^2 x^4+\left (-k^2-1\right ) x^2+1} \left (x^3 \left (1-d k^2\right )+x^2 \left (d k^2+3\right )+(d+3) x-d+1\right )}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (\frac {2 k^2 x^2}{\sqrt [4]{k^2 x^4+\left (-k^2-1\right ) x^2+1} \left (-\left (x^3 \left (1-d k^2\right )\right )-x^2 \left (d k^2+3\right )-(d+3) x+d-1\right )}+\frac {\left (k^2-1\right ) x}{\sqrt [4]{k^2 x^4+\left (-k^2-1\right ) x^2+1} \left (x^3 \left (1-d k^2\right )+x^2 \left (d k^2+3\right )+(d+3) x-d+1\right )}+\frac {2}{\sqrt [4]{k^2 x^4+\left (-k^2-1\right ) x^2+1} \left (x^3 \left (1-d k^2\right )+x^2 \left (d k^2+3\right )+(d+3) x-d+1\right )}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle 2 k^2 \int \frac {x^2}{\left (-\left (\left (1-d k^2\right ) x^3\right )-\left (d k^2+3\right ) x^2-(d+3) x+d-1\right ) \sqrt [4]{k^2 x^4+\left (-k^2-1\right ) x^2+1}}dx+2 \int \frac {1}{\left (\left (1-d k^2\right ) x^3+\left (d k^2+3\right ) x^2+(d+3) x-d+1\right ) \sqrt [4]{k^2 x^4+\left (-k^2-1\right ) x^2+1}}dx-\left (1-k^2\right ) \int \frac {x}{\left (\left (1-d k^2\right ) x^3+\left (d k^2+3\right ) x^2+(d+3) x-d+1\right ) \sqrt [4]{k^2 x^4+\left (-k^2-1\right ) x^2+1}}dx\)

input

Int[(-2 - (-1 + k)*(1 + k)*x + 2*k^2*x^2)/(((1 - x^2)*(1 - k^2*x^2))^(1/4) 
*(-1 + d - (3 + d)*x - (3 + d*k^2)*x^2 + (-1 + d*k^2)*x^3)),x]
 

output

$Aborted
 

3.12.59.3.1 Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2048

Int[(u_.)*((e_.)*((a_.) + (b_.)*(x_)^(n_.))*((c_) + (d_.)*(x_)^(n_.)))^(p_) 
, x_Symbol] :> Int[u*(a*c*e + (b*c + a*d)*e*x^n + b*d*e*x^(2*n))^p, x] /; F 
reeQ[{a, b, c, d, e, n, p}, x]
 

rule 7292

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =! 
= u]
 

rule 7293

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v] 
]
 

3.12.59.4 Maple [F]

\[\int \frac {-2-\left (-1+k \right ) \left (1+k \right ) x +2 k^{2} x^{2}}{{\left (\left (-x^{2}+1\right ) \left (-k^{2} x^{2}+1\right )\right )}^{\frac {1}{4}} \left (-1+d -\left (3+d \right ) x -\left (d \,k^{2}+3\right ) x^{2}+\left (d \,k^{2}-1\right ) x^{3}\right )}d x\]

input

int((-2-(-1+k)*(1+k)*x+2*k^2*x^2)/((-x^2+1)*(-k^2*x^2+1))^(1/4)/(-1+d-(3+d 
)*x-(d*k^2+3)*x^2+(d*k^2-1)*x^3),x)
 

output

int((-2-(-1+k)*(1+k)*x+2*k^2*x^2)/((-x^2+1)*(-k^2*x^2+1))^(1/4)/(-1+d-(3+d 
)*x-(d*k^2+3)*x^2+(d*k^2-1)*x^3),x)
 

3.12.59.5 Fricas [F(-1)]

Timed out. \[ \int \frac {-2-(-1+k) (1+k) x+2 k^2 x^2}{\sqrt [4]{\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (-1+d-(3+d) x-\left (3+d k^2\right ) x^2+\left (-1+d k^2\right ) x^3\right )} \, dx=\text {Timed out} \]

input

integrate((-2-(-1+k)*(1+k)*x+2*k^2*x^2)/((-x^2+1)*(-k^2*x^2+1))^(1/4)/(-1+ 
d-(3+d)*x-(d*k^2+3)*x^2+(d*k^2-1)*x^3),x, algorithm="fricas")
 

output

Timed out
 

3.12.59.6 Sympy [F(-1)]

Timed out. \[ \int \frac {-2-(-1+k) (1+k) x+2 k^2 x^2}{\sqrt [4]{\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (-1+d-(3+d) x-\left (3+d k^2\right ) x^2+\left (-1+d k^2\right ) x^3\right )} \, dx=\text {Timed out} \]

input

integrate((-2-(-1+k)*(1+k)*x+2*k**2*x**2)/((-x**2+1)*(-k**2*x**2+1))**(1/4 
)/(-1+d-(3+d)*x-(d*k**2+3)*x**2+(d*k**2-1)*x**3),x)
 

output

Timed out
 

3.12.59.7 Maxima [F]

\[ \int \frac {-2-(-1+k) (1+k) x+2 k^2 x^2}{\sqrt [4]{\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (-1+d-(3+d) x-\left (3+d k^2\right ) x^2+\left (-1+d k^2\right ) x^3\right )} \, dx=\int { \frac {2 \, k^{2} x^{2} - {\left (k + 1\right )} {\left (k - 1\right )} x - 2}{{\left ({\left (d k^{2} - 1\right )} x^{3} - {\left (d k^{2} + 3\right )} x^{2} - {\left (d + 3\right )} x + d - 1\right )} \left ({\left (k^{2} x^{2} - 1\right )} {\left (x^{2} - 1\right )}\right )^{\frac {1}{4}}} \,d x } \]

input

integrate((-2-(-1+k)*(1+k)*x+2*k^2*x^2)/((-x^2+1)*(-k^2*x^2+1))^(1/4)/(-1+ 
d-(3+d)*x-(d*k^2+3)*x^2+(d*k^2-1)*x^3),x, algorithm="maxima")
 

output

integrate((2*k^2*x^2 - (k + 1)*(k - 1)*x - 2)/(((d*k^2 - 1)*x^3 - (d*k^2 + 
 3)*x^2 - (d + 3)*x + d - 1)*((k^2*x^2 - 1)*(x^2 - 1))^(1/4)), x)
 

3.12.59.8 Giac [F]

\[ \int \frac {-2-(-1+k) (1+k) x+2 k^2 x^2}{\sqrt [4]{\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (-1+d-(3+d) x-\left (3+d k^2\right ) x^2+\left (-1+d k^2\right ) x^3\right )} \, dx=\int { \frac {2 \, k^{2} x^{2} - {\left (k + 1\right )} {\left (k - 1\right )} x - 2}{{\left ({\left (d k^{2} - 1\right )} x^{3} - {\left (d k^{2} + 3\right )} x^{2} - {\left (d + 3\right )} x + d - 1\right )} \left ({\left (k^{2} x^{2} - 1\right )} {\left (x^{2} - 1\right )}\right )^{\frac {1}{4}}} \,d x } \]

input

integrate((-2-(-1+k)*(1+k)*x+2*k^2*x^2)/((-x^2+1)*(-k^2*x^2+1))^(1/4)/(-1+ 
d-(3+d)*x-(d*k^2+3)*x^2+(d*k^2-1)*x^3),x, algorithm="giac")
 

output

integrate((2*k^2*x^2 - (k + 1)*(k - 1)*x - 2)/(((d*k^2 - 1)*x^3 - (d*k^2 + 
 3)*x^2 - (d + 3)*x + d - 1)*((k^2*x^2 - 1)*(x^2 - 1))^(1/4)), x)
 

3.12.59.9 Mupad [F(-1)]

Timed out. \[ \int \frac {-2-(-1+k) (1+k) x+2 k^2 x^2}{\sqrt [4]{\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (-1+d-(3+d) x-\left (3+d k^2\right ) x^2+\left (-1+d k^2\right ) x^3\right )} \, dx=\int \frac {x\,\left (k-1\right )\,\left (k+1\right )-2\,k^2\,x^2+2}{{\left (\left (x^2-1\right )\,\left (k^2\,x^2-1\right )\right )}^{1/4}\,\left (\left (1-d\,k^2\right )\,x^3+\left (d\,k^2+3\right )\,x^2+\left (d+3\right )\,x-d+1\right )} \,d x \]

input

int((x*(k - 1)*(k + 1) - 2*k^2*x^2 + 2)/(((x^2 - 1)*(k^2*x^2 - 1))^(1/4)*( 
x^2*(d*k^2 + 3) - x^3*(d*k^2 - 1) - d + x*(d + 3) + 1)),x)
 

output

int((x*(k - 1)*(k + 1) - 2*k^2*x^2 + 2)/(((x^2 - 1)*(k^2*x^2 - 1))^(1/4)*( 
x^2*(d*k^2 + 3) - x^3*(d*k^2 - 1) - d + x*(d + 3) + 1)), x)