Integrand size = 66, antiderivative size = 256 \[ \int \frac {\left (2-k^2\right ) x-2 x^3+k^2 x^5}{\left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3} \left (1-d+\left (-2+d k^2\right ) x^2+x^4\right )} \, dx=-\frac {\sqrt {3} \arctan \left (\frac {\sqrt {3} \left (1+\left (-1-k^2\right ) x^2+k^2 x^4\right )^{2/3}}{2 \sqrt [3]{d}-2 \sqrt [3]{d} k^2 x^2+\left (1+\left (-1-k^2\right ) x^2+k^2 x^4\right )^{2/3}}\right )}{2 \sqrt [3]{d}}-\frac {\log \left (-\sqrt [3]{d}+\sqrt [3]{d} k^2 x^2+\left (1+\left (-1-k^2\right ) x^2+k^2 x^4\right )^{2/3}\right )}{2 \sqrt [3]{d}}+\frac {\log \left (d^{2/3}-2 d^{2/3} k^2 x^2+d^{2/3} k^4 x^4+\left (\sqrt [3]{d}-\sqrt [3]{d} k^2 x^2\right ) \left (1+\left (-1-k^2\right ) x^2+k^2 x^4\right )^{2/3}+\left (1+\left (-1-k^2\right ) x^2+k^2 x^4\right )^{4/3}\right )}{4 \sqrt [3]{d}} \]
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\[ \int \frac {\left (2-k^2\right ) x-2 x^3+k^2 x^5}{\left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3} \left (1-d+\left (-2+d k^2\right ) x^2+x^4\right )} \, dx=\int \frac {\left (2-k^2\right ) x-2 x^3+k^2 x^5}{\left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3} \left (1-d+\left (-2+d k^2\right ) x^2+x^4\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {x \left (2-k^2-2 x^2+k^2 x^4\right )}{\left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3} \left (1-d+\left (-2+d k^2\right ) x^2+x^4\right )} \, dx \\ & = \int \frac {x \left (2-k^2-2 x^2+k^2 x^4\right )}{\left (1-d+\left (-2+d k^2\right ) x^2+x^4\right ) \left (1+\left (-1-k^2\right ) x^2+k^2 x^4\right )^{2/3}} \, dx \\ & = \frac {1}{2} \text {Subst}\left (\int \frac {2-k^2-2 x+k^2 x^2}{\left (1-d+\left (-2+d k^2\right ) x+x^2\right ) \left (1+\left (-1-k^2\right ) x+k^2 x^2\right )^{2/3}} \, dx,x,x^2\right ) \\ & = \frac {1}{2} \text {Subst}\left (\int \left (\frac {k^2}{\left (1+\left (-1-k^2\right ) x+k^2 x^2\right )^{2/3}}+\frac {2-(2-d) k^2-\left (2-2 k^2+d k^4\right ) x}{\left (1-d+\left (-2+d k^2\right ) x+x^2\right ) \left (1+\left (-1-k^2\right ) x+k^2 x^2\right )^{2/3}}\right ) \, dx,x,x^2\right ) \\ & = \frac {1}{2} \text {Subst}\left (\int \frac {2-(2-d) k^2+\left (-2+2 k^2-d k^4\right ) x}{\left (1-d+\left (-2+d k^2\right ) x+x^2\right ) \left (1+\left (-1-k^2\right ) x+k^2 x^2\right )^{2/3}} \, dx,x,x^2\right )+\frac {1}{2} k^2 \text {Subst}\left (\int \frac {1}{\left (1+\left (-1-k^2\right ) x+k^2 x^2\right )^{2/3}} \, dx,x,x^2\right ) \\ & = \frac {1}{2} \text {Subst}\left (\int \frac {2-(2-d) k^2+\left (-2+2 k^2-d k^4\right ) x}{\left (1-d+\left (-2+d k^2\right ) x+x^2\right ) \left (1+\left (-1-k^2\right ) x+k^2 x^2\right )^{2/3}} \, dx,x,x^2\right )+\frac {\left (3 k^2 \sqrt {\left (-1-k^2+2 k^2 x^2\right )^2}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {-4 k^2+\left (-1-k^2\right )^2+4 k^2 x^3}} \, dx,x,\sqrt [3]{\left (-1+x^2\right ) \left (-1+k^2 x^2\right )}\right )}{2 \left (-1-k^2+2 k^2 x^2\right )} \\ & = -\frac {3^{3/4} \sqrt {2+\sqrt {3}} k^{4/3} \sqrt {\left (-1-k^2+2 k^2 x^2\right )^2} \left (\left (-1+k^2\right )^{2/3}+2^{2/3} k^{2/3} \sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}\right ) \sqrt {\frac {\left (-1+k^2\right )^{4/3}-2^{2/3} k^{2/3} \left (-1+k^2\right )^{2/3} \sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}+2 \sqrt [3]{2} k^{4/3} \left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3}}{\left (\left (1+\sqrt {3}\right ) \left (-1+k^2\right )^{2/3}+2^{2/3} k^{2/3} \sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1-\sqrt {3}\right ) \left (-1+k^2\right )^{2/3}+2^{2/3} k^{2/3} \sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}}{\left (1+\sqrt {3}\right ) \left (-1+k^2\right )^{2/3}+2^{2/3} k^{2/3} \sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}}\right ),-7-4 \sqrt {3}\right )}{2^{2/3} \left (1+k^2-2 k^2 x^2\right ) \sqrt {\left (-1-k^2 \left (1-2 x^2\right )\right )^2} \sqrt {\frac {\left (-1+k^2\right )^{2/3} \left (\left (-1+k^2\right )^{2/3}+2^{2/3} k^{2/3} \sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}\right )}{\left (\left (1+\sqrt {3}\right ) \left (-1+k^2\right )^{2/3}+2^{2/3} k^{2/3} \sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}\right )^2}}}+\frac {1}{2} \text {Subst}\left (\int \frac {2-(2-d) k^2+\left (-2+2 k^2-d k^4\right ) x}{\left (1-d+\left (-2+d k^2\right ) x+x^2\right ) \left (1+\left (-1-k^2\right ) x+k^2 x^2\right )^{2/3}} \, dx,x,x^2\right ) \\ \end{align*}
Time = 13.84 (sec) , antiderivative size = 197, normalized size of antiderivative = 0.77 \[ \int \frac {\left (2-k^2\right ) x-2 x^3+k^2 x^5}{\left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3} \left (1-d+\left (-2+d k^2\right ) x^2+x^4\right )} \, dx=\frac {\left (-1+x^2\right )^{2/3} \left (-1+k^2 x^2\right )^{2/3} \left (-2 \sqrt {3} \arctan \left (\frac {\sqrt {3} \left (-1+x^2\right )^{2/3}}{\left (-1+x^2\right )^{2/3}-2 \sqrt [3]{d} \sqrt [3]{-1+k^2 x^2}}\right )-2 \log \left (\left (-1+x^2\right )^{2/3}+\sqrt [3]{d} \sqrt [3]{-1+k^2 x^2}\right )+\log \left (\left (-1+x^2\right )^{4/3}-\sqrt [3]{d} \left (-1+x^2\right )^{2/3} \sqrt [3]{-1+k^2 x^2}+d^{2/3} \left (-1+k^2 x^2\right )^{2/3}\right )\right )}{4 \sqrt [3]{d} \left (\left (-1+x^2\right ) \left (-1+k^2 x^2\right )\right )^{2/3}} \]
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\[\int \frac {\left (-k^{2}+2\right ) x -2 x^{3}+k^{2} x^{5}}{{\left (\left (-x^{2}+1\right ) \left (-k^{2} x^{2}+1\right )\right )}^{\frac {2}{3}} \left (1-d +\left (d \,k^{2}-2\right ) x^{2}+x^{4}\right )}d x\]
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Timed out. \[ \int \frac {\left (2-k^2\right ) x-2 x^3+k^2 x^5}{\left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3} \left (1-d+\left (-2+d k^2\right ) x^2+x^4\right )} \, dx=\text {Timed out} \]
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Timed out. \[ \int \frac {\left (2-k^2\right ) x-2 x^3+k^2 x^5}{\left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3} \left (1-d+\left (-2+d k^2\right ) x^2+x^4\right )} \, dx=\text {Timed out} \]
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\[ \int \frac {\left (2-k^2\right ) x-2 x^3+k^2 x^5}{\left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3} \left (1-d+\left (-2+d k^2\right ) x^2+x^4\right )} \, dx=\int { \frac {k^{2} x^{5} - 2 \, x^{3} - {\left (k^{2} - 2\right )} x}{{\left (x^{4} + {\left (d k^{2} - 2\right )} x^{2} - d + 1\right )} \left ({\left (k^{2} x^{2} - 1\right )} {\left (x^{2} - 1\right )}\right )^{\frac {2}{3}}} \,d x } \]
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\[ \int \frac {\left (2-k^2\right ) x-2 x^3+k^2 x^5}{\left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3} \left (1-d+\left (-2+d k^2\right ) x^2+x^4\right )} \, dx=\int { \frac {k^{2} x^{5} - 2 \, x^{3} - {\left (k^{2} - 2\right )} x}{{\left (x^{4} + {\left (d k^{2} - 2\right )} x^{2} - d + 1\right )} \left ({\left (k^{2} x^{2} - 1\right )} {\left (x^{2} - 1\right )}\right )^{\frac {2}{3}}} \,d x } \]
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Timed out. \[ \int \frac {\left (2-k^2\right ) x-2 x^3+k^2 x^5}{\left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3} \left (1-d+\left (-2+d k^2\right ) x^2+x^4\right )} \, dx=-\int \frac {x\,\left (k^2-2\right )-k^2\,x^5+2\,x^3}{{\left (\left (x^2-1\right )\,\left (k^2\,x^2-1\right )\right )}^{2/3}\,\left (x^4+\left (d\,k^2-2\right )\,x^2-d+1\right )} \,d x \]
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