3.27.6 \(\int \frac {(-2+k^2) x+k^2 x^3}{\sqrt [3]{(1-x^2) (1-k^2 x^2)} (1-d+(-2+d k^2) x^2+x^4)} \, dx\) [2606]

3.27.6.1 Optimal result

Integrand size = 59, antiderivative size = 226 \[ \int \frac {\left (-2+k^2\right ) x+k^2 x^3}{\sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (1-d+\left (-2+d k^2\right ) x^2+x^4\right )} \, dx=\frac {\sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{d} \sqrt [3]{1+\left (-1-k^2\right ) x^2+k^2 x^4}}{2-2 x^2+\sqrt [3]{d} \sqrt [3]{1+\left (-1-k^2\right ) x^2+k^2 x^4}}\right )}{2 d^{2/3}}+\frac {\log \left (-1+x^2+\sqrt [3]{d} \sqrt [3]{1+\left (-1-k^2\right ) x^2+k^2 x^4}\right )}{2 d^{2/3}}-\frac {\log \left (1-2 x^2+x^4+\left (\sqrt [3]{d}-\sqrt [3]{d} x^2\right ) \sqrt [3]{1+\left (-1-k^2\right ) x^2+k^2 x^4}+d^{2/3} \left (1+\left (-1-k^2\right ) x^2+k^2 x^4\right )^{2/3}\right )}{4 d^{2/3}} \]

output

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

3.27.6.2 Mathematica [A] (verified)

Time = 13.30 (sec) , antiderivative size = 209, normalized size of antiderivative = 0.92 \[ \int \frac {\left (-2+k^2\right ) x+k^2 x^3}{\sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (1-d+\left (-2+d k^2\right ) x^2+x^4\right )} \, dx=\frac {\sqrt [3]{-1+x^2} \sqrt [3]{-1+k^2 x^2} \left (2 \sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{d} \sqrt [3]{-1+k^2 x^2}}{-2 \left (-1+x^2\right )^{2/3}+\sqrt [3]{d} \sqrt [3]{-1+k^2 x^2}}\right )+2 \log \left (\left (-1+x^2\right )^{2/3}+\sqrt [3]{d} \sqrt [3]{-1+k^2 x^2}\right )-\log \left (\left (-1+x^2\right )^{4/3}-\sqrt [3]{d} \left (-1+x^2\right )^{2/3} \sqrt [3]{-1+k^2 x^2}+d^{2/3} \left (-1+k^2 x^2\right )^{2/3}\right )\right )}{4 d^{2/3} \sqrt [3]{\left (-1+x^2\right ) \left (-1+k^2 x^2\right )}} \]

input

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

output

((-1 + x^2)^(1/3)*(-1 + k^2*x^2)^(1/3)*(2*Sqrt[3]*ArcTan[(Sqrt[3]*d^(1/3)* 
(-1 + k^2*x^2)^(1/3))/(-2*(-1 + x^2)^(2/3) + d^(1/3)*(-1 + k^2*x^2)^(1/3)) 
] + 2*Log[(-1 + x^2)^(2/3) + d^(1/3)*(-1 + k^2*x^2)^(1/3)] - Log[(-1 + x^2 
)^(4/3) - d^(1/3)*(-1 + x^2)^(2/3)*(-1 + k^2*x^2)^(1/3) + d^(2/3)*(-1 + k^ 
2*x^2)^(2/3)]))/(4*d^(2/3)*((-1 + x^2)*(-1 + k^2*x^2))^(1/3))
 

3.27.6.3 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.

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

\(\Big \downarrow \) 2027

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

\(\Big \downarrow \) 2048

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

\(\Big \downarrow \) 7266

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 1375

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

input

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

output

$Aborted
 

3.27.6.3.1 Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 1375

Int[((g_.) + (h_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_)*((d_.) + 
(e_.)*(x_) + (f_.)*(x_)^2)^(q_), x_Symbol] :> Unintegrable[(g + h*x)*(a + b 
*x + c*x^2)^p*(d + e*x + f*x^2)^q, x] /; FreeQ[{a, b, c, d, e, f, g, h, p, 
q}, x]
 

rule 2027

Int[(Fx_.)*((a_.)*(x_)^(r_.) + (b_.)*(x_)^(s_.))^(p_.), x_Symbol] :> Int[x^ 
(p*r)*(a + b*x^(s - r))^p*Fx, x] /; FreeQ[{a, b, r, s}, x] && IntegerQ[p] & 
& PosQ[s - r] &&  !(EqQ[p, 1] && EqQ[u, 1])
 

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 7266

Int[(u_)*(x_)^(m_.), x_Symbol] :> Simp[1/(m + 1)   Subst[Int[SubstFor[x^(m 
+ 1), u, x], x], x, x^(m + 1)], x] /; FreeQ[m, x] && NeQ[m, -1] && Function 
OfQ[x^(m + 1), u, x]
 

3.27.6.4 Maple [F]

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

input

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

output

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

3.27.6.5 Fricas [F(-1)]

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

input

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

output

Timed out
 

3.27.6.6 Sympy [F(-1)]

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

input

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

output

Timed out
 

3.27.6.7 Maxima [F]

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

input

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

output

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

3.27.6.8 Giac [F]

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

input

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

output

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

3.27.6.9 Mupad [F(-1)]

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

input

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

output

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