Integrand size = 91, antiderivative size = 94 \[ \int \frac {\left (-2 k+(-1+k) (1+k) x+2 k x^2\right ) \left (1+2 k x+k^2 x^2\right )}{\left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{3/4} \left (-1+d+(1+3 d) k x+\left (1+3 d k^2\right ) x^2+k \left (-1+d k^2\right ) x^3\right )} \, dx=-\frac {\arctan \left (\frac {\sqrt [4]{d}+\sqrt [4]{d} k x}{\sqrt [4]{1+\left (-1-k^2\right ) x^2+k^2 x^4}}\right )}{d^{3/4}}+\frac {\text {arctanh}\left (\frac {\sqrt [4]{d}+\sqrt [4]{d} k x}{\sqrt [4]{1+\left (-1-k^2\right ) x^2+k^2 x^4}}\right )}{d^{3/4}} \]
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\[ \int \frac {\left (-2 k+(-1+k) (1+k) x+2 k x^2\right ) \left (1+2 k x+k^2 x^2\right )}{\left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{3/4} \left (-1+d+(1+3 d) k x+\left (1+3 d k^2\right ) x^2+k \left (-1+d k^2\right ) x^3\right )} \, dx=\int \frac {\left (-2 k+(-1+k) (1+k) x+2 k x^2\right ) \left (1+2 k x+k^2 x^2\right )}{\left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{3/4} \left (-1+d+(1+3 d) k x+\left (1+3 d k^2\right ) x^2+k \left (-1+d k^2\right ) x^3\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {(1+k x)^2 \left (-2 k+(-1+k) (1+k) x+2 k x^2\right )}{\left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{3/4} \left (-1+d+(1+3 d) k x+\left (1+3 d k^2\right ) x^2+k \left (-1+d k^2\right ) x^3\right )} \, dx \\ & = \int \frac {(1+k x)^2 \left (-2 k+(-1+k) (1+k) x+2 k x^2\right )}{\left (-1+d+(1+3 d) k x+\left (1+3 d k^2\right ) x^2+k \left (-1+d k^2\right ) x^3\right ) \left (1+\left (-1-k^2\right ) x^2+k^2 x^4\right )^{3/4}} \, dx \\ & = \int \frac {(1+k x)^2 \left (2 k+(1-k) (1+k) x-2 k x^2\right )}{\left (1-d-(1+3 d) k x-\left (1+3 d k^2\right ) x^2+k \left (1-d k^2\right ) x^3\right ) \left (1+\left (-1-k^2\right ) x^2+k^2 x^4\right )^{3/4}} \, dx \\ & = \int \left (-\frac {k \left (5+(1+3 d) k^2-d k^4\right )}{\left (1-d k^2\right )^2 \left (1+\left (-1-k^2\right ) x^2+k^2 x^4\right )^{3/4}}-\frac {2 k^2 x}{\left (1-d k^2\right ) \left (1+\left (-1-k^2\right ) x^2+k^2 x^4\right )^{3/4}}+\frac {k \left (7+k^2-3 d^2 k^2 \left (1-k^2\right )-d \left (5+2 k^2+k^4\right )\right )+\left (1-k^4-6 d^2 k^4 \left (1-k^2\right )-d k^2 \left (19+14 k^2-k^4\right )\right ) x-k \left (5+3 (1+8 d) k^2+3 d^2 k^4-3 d^2 k^6\right ) x^2}{\left (-1+d k^2\right )^2 \left (1-d-(1+3 d) k x-\left (1+3 d k^2\right ) x^2+k \left (1-d k^2\right ) x^3\right ) \left (1+\left (-1-k^2\right ) x^2+k^2 x^4\right )^{3/4}}\right ) \, dx \\ & = -\frac {\left (2 k^2\right ) \int \frac {x}{\left (1+\left (-1-k^2\right ) x^2+k^2 x^4\right )^{3/4}} \, dx}{1-d k^2}+\frac {\int \frac {k \left (7+k^2-3 d^2 k^2 \left (1-k^2\right )-d \left (5+2 k^2+k^4\right )\right )+\left (1-k^4-6 d^2 k^4 \left (1-k^2\right )-d k^2 \left (19+14 k^2-k^4\right )\right ) x-k \left (5+3 (1+8 d) k^2+3 d^2 k^4-3 d^2 k^6\right ) x^2}{\left (1-d-(1+3 d) k x-\left (1+3 d k^2\right ) x^2+k \left (1-d k^2\right ) x^3\right ) \left (1+\left (-1-k^2\right ) x^2+k^2 x^4\right )^{3/4}} \, dx}{\left (-1+d k^2\right )^2}-\frac {\left (k \left (5+(1+3 d) k^2-d k^4\right )\right ) \int \frac {1}{\left (1+\left (-1-k^2\right ) x^2+k^2 x^4\right )^{3/4}} \, dx}{\left (1-d k^2\right )^2} \\ & = -\frac {k^2 \text {Subst}\left (\int \frac {1}{\left (1+\left (-1-k^2\right ) x+k^2 x^2\right )^{3/4}} \, dx,x,x^2\right )}{1-d k^2}+\frac {\int \left (\frac {k \left (7+k^2-3 d^2 k^2 \left (1-k^2\right )-d \left (5+2 k^2+k^4\right )\right )}{\left (1-d-(1+3 d) k x-\left (1+3 d k^2\right ) x^2+k \left (1-d k^2\right ) x^3\right ) \left (1+\left (-1-k^2\right ) x^2+k^2 x^4\right )^{3/4}}+\frac {\left (1-k^4-6 d^2 k^4 \left (1-k^2\right )-d k^2 \left (19+14 k^2-k^4\right )\right ) x}{\left (1-d-(1+3 d) k x-\left (1+3 d k^2\right ) x^2+k \left (1-d k^2\right ) x^3\right ) \left (1+\left (-1-k^2\right ) x^2+k^2 x^4\right )^{3/4}}+\frac {k \left (-5-3 (1+8 d) k^2-3 d^2 k^4+3 d^2 k^6\right ) x^2}{\left (1-d-(1+3 d) k x-\left (1+3 d k^2\right ) x^2+k \left (1-d k^2\right ) x^3\right ) \left (1+\left (-1-k^2\right ) x^2+k^2 x^4\right )^{3/4}}\right ) \, dx}{\left (-1+d k^2\right )^2}-\frac {\left (k \left (5+(1+3 d) k^2-d k^4\right ) \left (1-x^2\right )^{3/4} \left (1-k^2 x^2\right )^{3/4}\right ) \int \frac {1}{\left (1-x^2\right )^{3/4} \left (1-k^2 x^2\right )^{3/4}} \, dx}{\left (1-d k^2\right )^2 \left (1+\left (-1-k^2\right ) x^2+k^2 x^4\right )^{3/4}} \\ & = -\frac {k \left (5+(1+3 d) k^2-d k^4\right ) x \left (\frac {1-x^2}{1-k^2 x^2}\right )^{3/4} \left (1-k^2 x^2\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {3}{2},\frac {\left (1-k^2\right ) x^2}{1-k^2 x^2}\right )}{\left (1-d k^2\right )^2 \left (1-\left (1+k^2\right ) x^2+k^2 x^4\right )^{3/4}}-\frac {\left (k \left (5+3 (1+8 d) k^2+3 d^2 k^4-3 d^2 k^6\right )\right ) \int \frac {x^2}{\left (1-d-(1+3 d) k x-\left (1+3 d k^2\right ) x^2+k \left (1-d k^2\right ) x^3\right ) \left (1+\left (-1-k^2\right ) x^2+k^2 x^4\right )^{3/4}} \, dx}{\left (1-d k^2\right )^2}+\frac {\left (1-k^4-6 d^2 k^4 \left (1-k^2\right )-d k^2 \left (19+14 k^2-k^4\right )\right ) \int \frac {x}{\left (1-d-(1+3 d) k x-\left (1+3 d k^2\right ) x^2+k \left (1-d k^2\right ) x^3\right ) \left (1+\left (-1-k^2\right ) x^2+k^2 x^4\right )^{3/4}} \, dx}{\left (1-d k^2\right )^2}+\frac {\left (k \left (7+k^2-3 d^2 k^2 \left (1-k^2\right )-d \left (5+2 k^2+k^4\right )\right )\right ) \int \frac {1}{\left (1-d-(1+3 d) k x-\left (1+3 d k^2\right ) x^2+k \left (1-d k^2\right ) x^3\right ) \left (1+\left (-1-k^2\right ) x^2+k^2 x^4\right )^{3/4}} \, dx}{\left (1-d k^2\right )^2}-\frac {\left (4 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^4}} \, dx,x,\sqrt [4]{\left (-1+x^2\right ) \left (-1+k^2 x^2\right )}\right )}{\left (1-d k^2\right ) \left (-1-k^2+2 k^2 x^2\right )} \\ & = \frac {\sqrt {2} k^{3/2} \sqrt {-1+k^2} \sqrt {\left (-1-k^2+2 k^2 x^2\right )^2} \sqrt {\frac {\left (1+k^2 \left (1-2 x^2\right )\right )^2}{\left (1-k^2\right )^2 \left (1-\frac {2 k \sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )}}{1-k^2}\right )^2}} \left (1-\frac {2 k \sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )}}{1-k^2}\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt {2} \sqrt {k} \sqrt [4]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}}{\sqrt {-1+k^2}}\right ),\frac {1}{2}\right )}{\left (1-d k^2\right ) \left (1+k^2-2 k^2 x^2\right ) \sqrt {\left (-1-k^2 \left (1-2 x^2\right )\right )^2}}-\frac {k \left (5+(1+3 d) k^2-d k^4\right ) x \left (\frac {1-x^2}{1-k^2 x^2}\right )^{3/4} \left (1-k^2 x^2\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {3}{2},\frac {\left (1-k^2\right ) x^2}{1-k^2 x^2}\right )}{\left (1-d k^2\right )^2 \left (1-\left (1+k^2\right ) x^2+k^2 x^4\right )^{3/4}}-\frac {\left (k \left (5+3 (1+8 d) k^2+3 d^2 k^4-3 d^2 k^6\right )\right ) \int \frac {x^2}{\left (1-d-(1+3 d) k x-\left (1+3 d k^2\right ) x^2+k \left (1-d k^2\right ) x^3\right ) \left (1+\left (-1-k^2\right ) x^2+k^2 x^4\right )^{3/4}} \, dx}{\left (1-d k^2\right )^2}+\frac {\left (1-k^4-6 d^2 k^4 \left (1-k^2\right )-d k^2 \left (19+14 k^2-k^4\right )\right ) \int \frac {x}{\left (1-d-(1+3 d) k x-\left (1+3 d k^2\right ) x^2+k \left (1-d k^2\right ) x^3\right ) \left (1+\left (-1-k^2\right ) x^2+k^2 x^4\right )^{3/4}} \, dx}{\left (1-d k^2\right )^2}+\frac {\left (k \left (7+k^2-3 d^2 k^2 \left (1-k^2\right )-d \left (5+2 k^2+k^4\right )\right )\right ) \int \frac {1}{\left (1-d-(1+3 d) k x-\left (1+3 d k^2\right ) x^2+k \left (1-d k^2\right ) x^3\right ) \left (1+\left (-1-k^2\right ) x^2+k^2 x^4\right )^{3/4}} \, dx}{\left (1-d k^2\right )^2} \\ \end{align*}
\[ \int \frac {\left (-2 k+(-1+k) (1+k) x+2 k x^2\right ) \left (1+2 k x+k^2 x^2\right )}{\left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{3/4} \left (-1+d+(1+3 d) k x+\left (1+3 d k^2\right ) x^2+k \left (-1+d k^2\right ) x^3\right )} \, dx=\int \frac {\left (-2 k+(-1+k) (1+k) x+2 k x^2\right ) \left (1+2 k x+k^2 x^2\right )}{\left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{3/4} \left (-1+d+(1+3 d) k x+\left (1+3 d k^2\right ) x^2+k \left (-1+d k^2\right ) x^3\right )} \, dx \]
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\[\int \frac {\left (-2 k +\left (-1+k \right ) \left (1+k \right ) x +2 k \,x^{2}\right ) \left (k^{2} x^{2}+2 k x +1\right )}{{\left (\left (-x^{2}+1\right ) \left (-k^{2} x^{2}+1\right )\right )}^{\frac {3}{4}} \left (-1+d +\left (1+3 d \right ) k x +\left (3 d \,k^{2}+1\right ) x^{2}+k \left (d \,k^{2}-1\right ) x^{3}\right )}d x\]
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Timed out. \[ \int \frac {\left (-2 k+(-1+k) (1+k) x+2 k x^2\right ) \left (1+2 k x+k^2 x^2\right )}{\left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{3/4} \left (-1+d+(1+3 d) k x+\left (1+3 d k^2\right ) x^2+k \left (-1+d k^2\right ) x^3\right )} \, dx=\text {Timed out} \]
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Timed out. \[ \int \frac {\left (-2 k+(-1+k) (1+k) x+2 k x^2\right ) \left (1+2 k x+k^2 x^2\right )}{\left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{3/4} \left (-1+d+(1+3 d) k x+\left (1+3 d k^2\right ) x^2+k \left (-1+d k^2\right ) x^3\right )} \, dx=\text {Timed out} \]
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\[ \int \frac {\left (-2 k+(-1+k) (1+k) x+2 k x^2\right ) \left (1+2 k x+k^2 x^2\right )}{\left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{3/4} \left (-1+d+(1+3 d) k x+\left (1+3 d k^2\right ) x^2+k \left (-1+d k^2\right ) x^3\right )} \, dx=\int { \frac {{\left (k^{2} x^{2} + 2 \, k x + 1\right )} {\left ({\left (k + 1\right )} {\left (k - 1\right )} x + 2 \, k x^{2} - 2 \, k\right )}}{{\left ({\left (d k^{2} - 1\right )} k x^{3} + {\left (3 \, d + 1\right )} k x + {\left (3 \, d k^{2} + 1\right )} x^{2} + d - 1\right )} \left ({\left (k^{2} x^{2} - 1\right )} {\left (x^{2} - 1\right )}\right )^{\frac {3}{4}}} \,d x } \]
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\[ \int \frac {\left (-2 k+(-1+k) (1+k) x+2 k x^2\right ) \left (1+2 k x+k^2 x^2\right )}{\left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{3/4} \left (-1+d+(1+3 d) k x+\left (1+3 d k^2\right ) x^2+k \left (-1+d k^2\right ) x^3\right )} \, dx=\int { \frac {{\left (k^{2} x^{2} + 2 \, k x + 1\right )} {\left ({\left (k + 1\right )} {\left (k - 1\right )} x + 2 \, k x^{2} - 2 \, k\right )}}{{\left ({\left (d k^{2} - 1\right )} k x^{3} + {\left (3 \, d + 1\right )} k x + {\left (3 \, d k^{2} + 1\right )} x^{2} + d - 1\right )} \left ({\left (k^{2} x^{2} - 1\right )} {\left (x^{2} - 1\right )}\right )^{\frac {3}{4}}} \,d x } \]
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Timed out. \[ \int \frac {\left (-2 k+(-1+k) (1+k) x+2 k x^2\right ) \left (1+2 k x+k^2 x^2\right )}{\left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{3/4} \left (-1+d+(1+3 d) k x+\left (1+3 d k^2\right ) x^2+k \left (-1+d k^2\right ) x^3\right )} \, dx=\int \frac {\left (2\,k\,x^2+\left (k-1\right )\,\left (k+1\right )\,x-2\,k\right )\,\left (k^2\,x^2+2\,k\,x+1\right )}{{\left (\left (x^2-1\right )\,\left (k^2\,x^2-1\right )\right )}^{3/4}\,\left (k\,\left (d\,k^2-1\right )\,x^3+\left (3\,d\,k^2+1\right )\,x^2+k\,\left (3\,d+1\right )\,x+d-1\right )} \,d x \]
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