Integrand size = 80, antiderivative size = 212 \[ \int \frac {-3+\left (1-2 k^2\right ) x+3 k^2 x^2+k^2 x^3}{\sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (-1+d-(2+d) x-\left (1+d k^2\right ) x^2+d k^2 x^3\right )} \, dx=\frac {\sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{d} \sqrt [3]{1+\left (-1-k^2\right ) x^2+k^2 x^4}}{2+2 x+\sqrt [3]{d} \sqrt [3]{1+\left (-1-k^2\right ) x^2+k^2 x^4}}\right )}{d^{2/3}}+\frac {\log \left (1+x-\sqrt [3]{d} \sqrt [3]{1+\left (-1-k^2\right ) x^2+k^2 x^4}\right )}{d^{2/3}}-\frac {\log \left (1+2 x+x^2+\left (\sqrt [3]{d}+\sqrt [3]{d} x\right ) \sqrt [3]{1+\left (-1-k^2\right ) x^2+k^2 x^4}+d^{2/3} \left (1+\left (-1-k^2\right ) x^2+k^2 x^4\right )^{2/3}\right )}{2 d^{2/3}} \]
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\[ \int \frac {-3+\left (1-2 k^2\right ) x+3 k^2 x^2+k^2 x^3}{\sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (-1+d-(2+d) x-\left (1+d k^2\right ) x^2+d k^2 x^3\right )} \, dx=\int \frac {-3+\left (1-2 k^2\right ) x+3 k^2 x^2+k^2 x^3}{\sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (-1+d-(2+d) x-\left (1+d k^2\right ) x^2+d k^2 x^3\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {-3+\left (1-2 k^2\right ) x+3 k^2 x^2+k^2 x^3}{\left (-1+d-(2+d) x-\left (1+d k^2\right ) x^2+d k^2 x^3\right ) \sqrt [3]{1+\left (-1-k^2\right ) x^2+k^2 x^4}} \, dx \\ & = \int \left (\frac {1}{d \sqrt [3]{1+\left (-1-k^2\right ) x^2+k^2 x^4}}+\frac {1-4 d+2 \left (1+d \left (1-k^2\right )\right ) x+\left (1+4 d k^2\right ) x^2}{d \left (-1+d-(2+d) x-\left (1+d k^2\right ) x^2+d k^2 x^3\right ) \sqrt [3]{1+\left (-1-k^2\right ) x^2+k^2 x^4}}\right ) \, dx \\ & = \frac {\int \frac {1}{\sqrt [3]{1+\left (-1-k^2\right ) x^2+k^2 x^4}} \, dx}{d}+\frac {\int \frac {1-4 d+2 \left (1+d \left (1-k^2\right )\right ) x+\left (1+4 d k^2\right ) x^2}{\left (-1+d-(2+d) x-\left (1+d k^2\right ) x^2+d k^2 x^3\right ) \sqrt [3]{1+\left (-1-k^2\right ) x^2+k^2 x^4}} \, dx}{d} \\ & = \frac {\int \frac {-1+4 d-2 \left (1+d-d k^2\right ) x-\left (1+4 d k^2\right ) x^2}{\left (1-d+(2+d) x+\left (1+d k^2\right ) x^2-d k^2 x^3\right ) \sqrt [3]{1+\left (-1-k^2\right ) x^2+k^2 x^4}} \, dx}{d}+\frac {\left (\sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2}\right ) \int \frac {1}{\sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2}} \, dx}{d \sqrt [3]{1+\left (-1-k^2\right ) x^2+k^2 x^4}} \\ & = \frac {x \sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{3},\frac {1}{3},\frac {3}{2},x^2,k^2 x^2\right )}{d \sqrt [3]{1-\left (1+k^2\right ) x^2+k^2 x^4}}+\frac {\int \left (\frac {4 \left (1-\frac {1}{4 d}\right ) d}{\left (1-d+(2+d) x+\left (1+d k^2\right ) x^2-d k^2 x^3\right ) \sqrt [3]{1+\left (-1-k^2\right ) x^2+k^2 x^4}}+\frac {2 \left (-1-d \left (1-k^2\right )\right ) x}{\left (1-d+(2+d) x+\left (1+d k^2\right ) x^2-d k^2 x^3\right ) \sqrt [3]{1+\left (-1-k^2\right ) x^2+k^2 x^4}}+\frac {\left (-1-4 d k^2\right ) x^2}{\left (1-d+(2+d) x+\left (1+d k^2\right ) x^2-d k^2 x^3\right ) \sqrt [3]{1+\left (-1-k^2\right ) x^2+k^2 x^4}}\right ) \, dx}{d} \\ & = \frac {x \sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{3},\frac {1}{3},\frac {3}{2},x^2,k^2 x^2\right )}{d \sqrt [3]{1-\left (1+k^2\right ) x^2+k^2 x^4}}+\frac {(-1+4 d) \int \frac {1}{\left (1-d+(2+d) x+\left (1+d k^2\right ) x^2-d k^2 x^3\right ) \sqrt [3]{1+\left (-1-k^2\right ) x^2+k^2 x^4}} \, dx}{d}+\frac {\left (-1-4 d k^2\right ) \int \frac {x^2}{\left (1-d+(2+d) x+\left (1+d k^2\right ) x^2-d k^2 x^3\right ) \sqrt [3]{1+\left (-1-k^2\right ) x^2+k^2 x^4}} \, dx}{d}+\frac {\left (2 \left (-1-d \left (1-k^2\right )\right )\right ) \int \frac {x}{\left (1-d+(2+d) x+\left (1+d k^2\right ) x^2-d k^2 x^3\right ) \sqrt [3]{1+\left (-1-k^2\right ) x^2+k^2 x^4}} \, dx}{d} \\ \end{align*}
\[ \int \frac {-3+\left (1-2 k^2\right ) x+3 k^2 x^2+k^2 x^3}{\sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (-1+d-(2+d) x-\left (1+d k^2\right ) x^2+d k^2 x^3\right )} \, dx=\int \frac {-3+\left (1-2 k^2\right ) x+3 k^2 x^2+k^2 x^3}{\sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (-1+d-(2+d) x-\left (1+d k^2\right ) x^2+d k^2 x^3\right )} \, dx \]
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\[\int \frac {-3+\left (-2 k^{2}+1\right ) x +3 k^{2} x^{2}+k^{2} x^{3}}{{\left (\left (-x^{2}+1\right ) \left (-k^{2} x^{2}+1\right )\right )}^{\frac {1}{3}} \left (-1+d -\left (2+d \right ) x -\left (d \,k^{2}+1\right ) x^{2}+d \,k^{2} x^{3}\right )}d x\]
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Timed out. \[ \int \frac {-3+\left (1-2 k^2\right ) x+3 k^2 x^2+k^2 x^3}{\sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (-1+d-(2+d) x-\left (1+d k^2\right ) x^2+d k^2 x^3\right )} \, dx=\text {Timed out} \]
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Timed out. \[ \int \frac {-3+\left (1-2 k^2\right ) x+3 k^2 x^2+k^2 x^3}{\sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (-1+d-(2+d) x-\left (1+d k^2\right ) x^2+d k^2 x^3\right )} \, dx=\text {Timed out} \]
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\[ \int \frac {-3+\left (1-2 k^2\right ) x+3 k^2 x^2+k^2 x^3}{\sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (-1+d-(2+d) x-\left (1+d k^2\right ) x^2+d k^2 x^3\right )} \, dx=\int { \frac {k^{2} x^{3} + 3 \, k^{2} x^{2} - {\left (2 \, k^{2} - 1\right )} x - 3}{{\left (d k^{2} x^{3} - {\left (d k^{2} + 1\right )} x^{2} - {\left (d + 2\right )} x + d - 1\right )} \left ({\left (k^{2} x^{2} - 1\right )} {\left (x^{2} - 1\right )}\right )^{\frac {1}{3}}} \,d x } \]
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\[ \int \frac {-3+\left (1-2 k^2\right ) x+3 k^2 x^2+k^2 x^3}{\sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (-1+d-(2+d) x-\left (1+d k^2\right ) x^2+d k^2 x^3\right )} \, dx=\int { \frac {k^{2} x^{3} + 3 \, k^{2} x^{2} - {\left (2 \, k^{2} - 1\right )} x - 3}{{\left (d k^{2} x^{3} - {\left (d k^{2} + 1\right )} x^{2} - {\left (d + 2\right )} x + d - 1\right )} \left ({\left (k^{2} x^{2} - 1\right )} {\left (x^{2} - 1\right )}\right )^{\frac {1}{3}}} \,d x } \]
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Timed out. \[ \int \frac {-3+\left (1-2 k^2\right ) x+3 k^2 x^2+k^2 x^3}{\sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (-1+d-(2+d) x-\left (1+d k^2\right ) x^2+d k^2 x^3\right )} \, dx=\int -\frac {3\,k^2\,x^2+k^2\,x^3-x\,\left (2\,k^2-1\right )-3}{{\left (\left (x^2-1\right )\,\left (k^2\,x^2-1\right )\right )}^{1/3}\,\left (x^2\,\left (d\,k^2+1\right )-d+x\,\left (d+2\right )-d\,k^2\,x^3+1\right )} \,d x \]
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