Optimal. Leaf size=226 \[ -\frac {\log \left (\sqrt [3]{d} \left (k^2 x^4+\left (-k^2-1\right ) x^2+1\right )^{2/3}+x^2-1\right )}{2 d^{2/3}}+\frac {\log \left (d^{2/3} \left (k^2 x^4+\left (-k^2-1\right ) x^2+1\right )^{4/3}+\left (\sqrt [3]{d}-\sqrt [3]{d} x^2\right ) \left (k^2 x^4+\left (-k^2-1\right ) x^2+1\right )^{2/3}+x^4-2 x^2+1\right )}{4 d^{2/3}}-\frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{d} \left (k^2 x^4+\left (-k^2-1\right ) x^2+1\right )^{2/3}}{\sqrt [3]{d} \left (k^2 x^4+\left (-k^2-1\right ) x^2+1\right )^{2/3}-2 x^2+2}\right )}{2 d^{2/3}} \]
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Rubi [C] time = 2.91, antiderivative size = 423, normalized size of antiderivative = 1.87, number of steps used = 10, number of rules used = 7, integrand size = 73, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.096, Rules used = {1594, 6715, 6719, 1586, 6728, 137, 136} \begin {gather*} \frac {3 k^2 \left (1-x^2\right ) \left (1-\sqrt {1-4 d k^2 \left (1-k^2\right )}\right ) \left (1-k^2 x^2\right ) F_1\left (\frac {1}{3};-\frac {1}{3},1;\frac {4}{3};-\frac {k^2 \left (1-x^2\right )}{1-k^2},\frac {2 d k^4 \left (1-x^2\right )}{-2 d \left (1-k^2\right ) k^2-\sqrt {1-4 d k^2 \left (1-k^2\right )}+1}\right )}{2 \left (-2 d \left (1-k^2\right ) k^2-\sqrt {1-4 d k^2 \left (1-k^2\right )}+1\right ) \sqrt [3]{\frac {1-k^2 x^2}{1-k^2}} \left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3}}+\frac {3 k^2 \left (1-x^2\right ) \left (\sqrt {1-4 d k^2 \left (1-k^2\right )}+1\right ) \left (1-k^2 x^2\right ) F_1\left (\frac {1}{3};-\frac {1}{3},1;\frac {4}{3};-\frac {k^2 \left (1-x^2\right )}{1-k^2},\frac {2 d k^4 \left (1-x^2\right )}{-2 d \left (1-k^2\right ) k^2+\sqrt {1-4 d k^2 \left (1-k^2\right )}+1}\right )}{2 \left (-2 d \left (1-k^2\right ) k^2+\sqrt {1-4 d k^2 \left (1-k^2\right )}+1\right ) \sqrt [3]{\frac {1-k^2 x^2}{1-k^2}} \left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3}} \end {gather*}
Warning: Unable to verify antiderivative.
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Rule 136
Rule 137
Rule 1586
Rule 1594
Rule 6715
Rule 6719
Rule 6728
Rubi steps
\begin {align*} \int \frac {\left (-1+2 k^2\right ) x-2 k^4 x^3+k^4 x^5}{\left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3} \left (-1+d+\left (1-2 d k^2\right ) x^2+d k^4 x^4\right )} \, dx &=\int \frac {x \left (-1+2 k^2-2 k^4 x^2+k^4 x^4\right )}{\left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3} \left (-1+d+\left (1-2 d k^2\right ) x^2+d k^4 x^4\right )} \, dx\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {-1+2 k^2-2 k^4 x+k^4 x^2}{\left ((1-x) \left (1-k^2 x\right )\right )^{2/3} \left (-1+d+\left (1-2 d k^2\right ) x+d k^4 x^2\right )} \, dx,x,x^2\right )\\ &=\frac {\left (\left (1-x^2\right )^{2/3} \left (1-k^2 x^2\right )^{2/3}\right ) \operatorname {Subst}\left (\int \frac {-1+2 k^2-2 k^4 x+k^4 x^2}{(1-x)^{2/3} \left (1-k^2 x\right )^{2/3} \left (-1+d+\left (1-2 d k^2\right ) x+d k^4 x^2\right )} \, dx,x,x^2\right )}{2 \left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3}}\\ &=\frac {\left (\left (1-x^2\right )^{2/3} \left (1-k^2 x^2\right )^{2/3}\right ) \operatorname {Subst}\left (\int \frac {\sqrt [3]{1-k^2 x} \left (-1+2 k^2-k^2 x\right )}{(1-x)^{2/3} \left (-1+d+\left (1-2 d k^2\right ) x+d k^4 x^2\right )} \, dx,x,x^2\right )}{2 \left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3}}\\ &=\frac {\left (\left (1-x^2\right )^{2/3} \left (1-k^2 x^2\right )^{2/3}\right ) \operatorname {Subst}\left (\int \left (\frac {\left (-k^2+k^2 \sqrt {1-4 d k^2+4 d k^4}\right ) \sqrt [3]{1-k^2 x}}{(1-x)^{2/3} \left (1-2 d k^2-\sqrt {1-4 d k^2+4 d k^4}+2 d k^4 x\right )}+\frac {\left (-k^2-k^2 \sqrt {1-4 d k^2+4 d k^4}\right ) \sqrt [3]{1-k^2 x}}{(1-x)^{2/3} \left (1-2 d k^2+\sqrt {1-4 d k^2+4 d k^4}+2 d k^4 x\right )}\right ) \, dx,x,x^2\right )}{2 \left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3}}\\ &=-\frac {\left (k^2 \left (1-\sqrt {1-4 d k^2 \left (1-k^2\right )}\right ) \left (1-x^2\right )^{2/3} \left (1-k^2 x^2\right )^{2/3}\right ) \operatorname {Subst}\left (\int \frac {\sqrt [3]{1-k^2 x}}{(1-x)^{2/3} \left (1-2 d k^2-\sqrt {1-4 d k^2+4 d k^4}+2 d k^4 x\right )} \, dx,x,x^2\right )}{2 \left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3}}-\frac {\left (k^2 \left (1+\sqrt {1-4 d k^2 \left (1-k^2\right )}\right ) \left (1-x^2\right )^{2/3} \left (1-k^2 x^2\right )^{2/3}\right ) \operatorname {Subst}\left (\int \frac {\sqrt [3]{1-k^2 x}}{(1-x)^{2/3} \left (1-2 d k^2+\sqrt {1-4 d k^2+4 d k^4}+2 d k^4 x\right )} \, dx,x,x^2\right )}{2 \left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3}}\\ &=-\frac {\left (k^2 \left (1-\sqrt {1-4 d k^2 \left (1-k^2\right )}\right ) \left (1-x^2\right )^{2/3} \left (1-k^2 x^2\right )\right ) \operatorname {Subst}\left (\int \frac {\sqrt [3]{-\frac {1}{-1+k^2}+\frac {k^2 x}{-1+k^2}}}{(1-x)^{2/3} \left (1-2 d k^2-\sqrt {1-4 d k^2+4 d k^4}+2 d k^4 x\right )} \, dx,x,x^2\right )}{2 \left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3} \sqrt [3]{\frac {-1+k^2 x^2}{-1+k^2}}}-\frac {\left (k^2 \left (1+\sqrt {1-4 d k^2 \left (1-k^2\right )}\right ) \left (1-x^2\right )^{2/3} \left (1-k^2 x^2\right )\right ) \operatorname {Subst}\left (\int \frac {\sqrt [3]{-\frac {1}{-1+k^2}+\frac {k^2 x}{-1+k^2}}}{(1-x)^{2/3} \left (1-2 d k^2+\sqrt {1-4 d k^2+4 d k^4}+2 d k^4 x\right )} \, dx,x,x^2\right )}{2 \left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3} \sqrt [3]{\frac {-1+k^2 x^2}{-1+k^2}}}\\ &=\frac {3 k^2 \left (1-\sqrt {1-4 d k^2 \left (1-k^2\right )}\right ) \left (1-x^2\right ) \left (1-k^2 x^2\right ) F_1\left (\frac {1}{3};-\frac {1}{3},1;\frac {4}{3};-\frac {k^2 \left (1-x^2\right )}{1-k^2},\frac {2 d k^4 \left (1-x^2\right )}{1-2 d k^2 \left (1-k^2\right )-\sqrt {1-4 d k^2 \left (1-k^2\right )}}\right )}{2 \left (1-2 d k^2 \left (1-k^2\right )-\sqrt {1-4 d k^2 \left (1-k^2\right )}\right ) \sqrt [3]{\frac {1-k^2 x^2}{1-k^2}} \left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3}}+\frac {3 k^2 \left (1+\sqrt {1-4 d k^2 \left (1-k^2\right )}\right ) \left (1-x^2\right ) \left (1-k^2 x^2\right ) F_1\left (\frac {1}{3};-\frac {1}{3},1;\frac {4}{3};-\frac {k^2 \left (1-x^2\right )}{1-k^2},\frac {2 d k^4 \left (1-x^2\right )}{1-2 d k^2 \left (1-k^2\right )+\sqrt {1-4 d k^2 \left (1-k^2\right )}}\right )}{2 \left (1-2 d k^2 \left (1-k^2\right )+\sqrt {1-4 d k^2 \left (1-k^2\right )}\right ) \sqrt [3]{\frac {1-k^2 x^2}{1-k^2}} \left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3}}\\ \end {align*}
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Mathematica [F] time = 1.83, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (-1+2 k^2\right ) x-2 k^4 x^3+k^4 x^5}{\left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3} \left (-1+d+\left (1-2 d k^2\right ) x^2+d k^4 x^4\right )} \, dx \end {gather*}
Verification is not applicable to the result.
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IntegrateAlgebraic [A] time = 5.63, size = 226, normalized size = 1.00 \begin {gather*} -\frac {\log \left (\sqrt [3]{d} \left (k^2 x^4+\left (-k^2-1\right ) x^2+1\right )^{2/3}+x^2-1\right )}{2 d^{2/3}}+\frac {\log \left (d^{2/3} \left (k^2 x^4+\left (-k^2-1\right ) x^2+1\right )^{4/3}+\left (\sqrt [3]{d}-\sqrt [3]{d} x^2\right ) \left (k^2 x^4+\left (-k^2-1\right ) x^2+1\right )^{2/3}+x^4-2 x^2+1\right )}{4 d^{2/3}}-\frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{d} \left (k^2 x^4+\left (-k^2-1\right ) x^2+1\right )^{2/3}}{\sqrt [3]{d} \left (k^2 x^4+\left (-k^2-1\right ) x^2+1\right )^{2/3}-2 x^2+2}\right )}{2 d^{2/3}} \end {gather*}
Antiderivative was successfully verified.
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fricas [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {k^{4} x^{5} - 2 \, k^{4} x^{3} + {\left (2 \, k^{2} - 1\right )} x}{{\left (d k^{4} x^{4} - {\left (2 \, 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 {2}{3}}}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.08, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (2 k^{2}-1\right ) x -2 k^{4} x^{3}+k^{4} x^{5}}{\left (\left (-x^{2}+1\right ) \left (-k^{2} x^{2}+1\right )\right )^{\frac {2}{3}} \left (-1+d +\left (-2 d \,k^{2}+1\right ) x^{2}+d \,k^{4} x^{4}\right )}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {k^{4} x^{5} - 2 \, k^{4} x^{3} + {\left (2 \, k^{2} - 1\right )} x}{{\left (d k^{4} x^{4} - {\left (2 \, 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 {2}{3}}}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {k^4\,x^5-2\,k^4\,x^3+x\,\left (2\,k^2-1\right )}{{\left (\left (x^2-1\right )\,\left (k^2\,x^2-1\right )\right )}^{2/3}\,\left (d-x^2\,\left (2\,d\,k^2-1\right )+d\,k^4\,x^4-1\right )} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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