Integrand size = 64, antiderivative size = 90 \[ \int \frac {\left (-3+k^2\right ) x+2 k^2 x^3}{\sqrt [4]{\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (-1+d+\left (3-d k^2\right ) x^2-3 x^4+x^6\right )} \, dx=\frac {\arctan \left (\frac {\sqrt [4]{d} \sqrt [4]{1+\left (-1-k^2\right ) x^2+k^2 x^4}}{-1+x^2}\right )}{d^{3/4}}-\frac {\text {arctanh}\left (\frac {\sqrt [4]{d} \sqrt [4]{1+\left (-1-k^2\right ) x^2+k^2 x^4}}{-1+x^2}\right )}{d^{3/4}} \]
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Timed out. \[ \int \frac {\left (-3+k^2\right ) x+2 k^2 x^3}{\sqrt [4]{\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (-1+d+\left (3-d k^2\right ) x^2-3 x^4+x^6\right )} \, dx=\text {\$Aborted} \]
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Rubi steps Aborted
Time = 20.01 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.20 \[ \int \frac {\left (-3+k^2\right ) x+2 k^2 x^3}{\sqrt [4]{\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (-1+d+\left (3-d k^2\right ) x^2-3 x^4+x^6\right )} \, dx=\frac {\sqrt [4]{-1+x^2} \sqrt [4]{-1+k^2 x^2} \left (\arctan \left (\frac {\sqrt [4]{d} \sqrt [4]{-1+k^2 x^2}}{\left (-1+x^2\right )^{3/4}}\right )-\text {arctanh}\left (\frac {\sqrt [4]{d} \sqrt [4]{-1+k^2 x^2}}{\left (-1+x^2\right )^{3/4}}\right )\right )}{d^{3/4} \sqrt [4]{\left (-1+x^2\right ) \left (-1+k^2 x^2\right )}} \]
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\[\int \frac {\left (k^{2}-3\right ) x +2 k^{2} x^{3}}{{\left (\left (-x^{2}+1\right ) \left (-k^{2} x^{2}+1\right )\right )}^{\frac {1}{4}} \left (-1+d +\left (-d \,k^{2}+3\right ) x^{2}-3 x^{4}+x^{6}\right )}d x\]
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Timed out. \[ \int \frac {\left (-3+k^2\right ) x+2 k^2 x^3}{\sqrt [4]{\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (-1+d+\left (3-d k^2\right ) x^2-3 x^4+x^6\right )} \, dx=\text {Timed out} \]
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Timed out. \[ \int \frac {\left (-3+k^2\right ) x+2 k^2 x^3}{\sqrt [4]{\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (-1+d+\left (3-d k^2\right ) x^2-3 x^4+x^6\right )} \, dx=\text {Timed out} \]
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\[ \int \frac {\left (-3+k^2\right ) x+2 k^2 x^3}{\sqrt [4]{\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (-1+d+\left (3-d k^2\right ) x^2-3 x^4+x^6\right )} \, dx=\int { \frac {2 \, k^{2} x^{3} + {\left (k^{2} - 3\right )} x}{{\left (x^{6} - 3 \, x^{4} - {\left (d k^{2} - 3\right )} x^{2} + d - 1\right )} \left ({\left (k^{2} x^{2} - 1\right )} {\left (x^{2} - 1\right )}\right )^{\frac {1}{4}}} \,d x } \]
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\[ \int \frac {\left (-3+k^2\right ) x+2 k^2 x^3}{\sqrt [4]{\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (-1+d+\left (3-d k^2\right ) x^2-3 x^4+x^6\right )} \, dx=\int { \frac {2 \, k^{2} x^{3} + {\left (k^{2} - 3\right )} x}{{\left (x^{6} - 3 \, x^{4} - {\left (d k^{2} - 3\right )} x^{2} + d - 1\right )} \left ({\left (k^{2} x^{2} - 1\right )} {\left (x^{2} - 1\right )}\right )^{\frac {1}{4}}} \,d x } \]
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Timed out. \[ \int \frac {\left (-3+k^2\right ) x+2 k^2 x^3}{\sqrt [4]{\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (-1+d+\left (3-d k^2\right ) x^2-3 x^4+x^6\right )} \, dx=-\int \frac {x\,\left (k^2-3\right )+2\,k^2\,x^3}{{\left (\left (x^2-1\right )\,\left (k^2\,x^2-1\right )\right )}^{1/4}\,\left (-x^6+3\,x^4+\left (d\,k^2-3\right )\,x^2-d+1\right )} \,d x \]
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