Optimal. Leaf size=212 \[ \frac {\log \left (-\sqrt [3]{d} \sqrt [3]{k^2 x^4+\left (-k^2-1\right ) x^2+1}+x+1\right )}{d^{2/3}}-\frac {\log \left (d^{2/3} \left (k^2 x^4+\left (-k^2-1\right ) x^2+1\right )^{2/3}+\left (\sqrt [3]{d} x+\sqrt [3]{d}\right ) \sqrt [3]{k^2 x^4+\left (-k^2-1\right ) x^2+1}+x^2+2 x+1\right )}{2 d^{2/3}}+\frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{d} \sqrt [3]{k^2 x^4+\left (-k^2-1\right ) x^2+1}}{\sqrt [3]{d} \sqrt [3]{k^2 x^4+\left (-k^2-1\right ) x^2+1}+2 x+2}\right )}{d^{2/3}} \]
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Rubi [F] time = 6.32, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \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 \end {gather*}
Verification is not applicable to the result.
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\begin {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 &=\frac {\left (\sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2}\right ) \int \frac {-3+\left (1-2 k^2\right ) x+3 k^2 x^2+k^2 x^3}{\sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2} \left (-1+d-(2+d) x-\left (1+d k^2\right ) x^2+d k^2 x^3\right )} \, dx}{\sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}}\\ &=\frac {\left (\sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2}\right ) \int \left (\frac {1}{d \sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2}}+\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 \sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2} \left (-1+d-(2+d) x-\left (1+d k^2\right ) x^2+d k^2 x^3\right )}\right ) \, dx}{\sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}}\\ &=\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]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}}+\frac {\left (\sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2}\right ) \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}{\sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2} \left (-1+d-(2+d) x-\left (1+d k^2\right ) x^2+d k^2 x^3\right )} \, dx}{d \sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}}\\ &=\frac {x \sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2} F_1\left (\frac {1}{2};\frac {1}{3},\frac {1}{3};\frac {3}{2};x^2,k^2 x^2\right )}{d \sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}}+\frac {\left (\sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2}\right ) \int \frac {-1+4 d-2 \left (1+d-d k^2\right ) x-\left (1+4 d k^2\right ) x^2}{\sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2} \left (1-d+(2+d) x+\left (1+d k^2\right ) x^2-d k^2 x^3\right )} \, dx}{d \sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}}\\ &=\frac {x \sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2} F_1\left (\frac {1}{2};\frac {1}{3},\frac {1}{3};\frac {3}{2};x^2,k^2 x^2\right )}{d \sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}}+\frac {\left (\sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2}\right ) \int \left (\frac {4 \left (1-\frac {1}{4 d}\right ) d}{\sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2} \left (1-d+(2+d) x+\left (1+d k^2\right ) x^2-d k^2 x^3\right )}+\frac {2 \left (-1-d \left (1-k^2\right )\right ) x}{\sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2} \left (1-d+(2+d) x+\left (1+d k^2\right ) x^2-d k^2 x^3\right )}+\frac {\left (-1-4 d k^2\right ) x^2}{\sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2} \left (1-d+(2+d) x+\left (1+d k^2\right ) x^2-d k^2 x^3\right )}\right ) \, dx}{d \sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}}\\ &=\frac {x \sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2} F_1\left (\frac {1}{2};\frac {1}{3},\frac {1}{3};\frac {3}{2};x^2,k^2 x^2\right )}{d \sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}}+\frac {\left ((-1+4 d) \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} \left (1-d+(2+d) x+\left (1+d k^2\right ) x^2-d k^2 x^3\right )} \, dx}{d \sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}}+\frac {\left (\left (-1-4 d k^2\right ) \sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2}\right ) \int \frac {x^2}{\sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2} \left (1-d+(2+d) x+\left (1+d k^2\right ) x^2-d k^2 x^3\right )} \, dx}{d \sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}}+\frac {\left (2 \left (-1-d \left (1-k^2\right )\right ) \sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2}\right ) \int \frac {x}{\sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2} \left (1-d+(2+d) x+\left (1+d k^2\right ) x^2-d k^2 x^3\right )} \, dx}{d \sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}}\\ \end {align*}
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Mathematica [F] time = 4.04, size = 0, normalized size = 0.00 \begin {gather*} \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 \end {gather*}
Verification is not applicable to the result.
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IntegrateAlgebraic [A] time = 5.63, size = 212, normalized size = 1.00 \begin {gather*} \frac {\sqrt {3} \tan ^{-1}\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}} \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^{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} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.03, size = 0, normalized size = 0.00 \[\int \frac {-3+\left (-2 k^{2}+1\right ) x +3 k^{2} x^{2}+k^{2} x^{3}}{\left (\left (-x^{2}+1\right ) \left (-k^{2} x^{2}+1\right )\right )^{\frac {1}{3}} \left (-1+d -\left (2+d \right ) x -\left (d \,k^{2}+1\right ) x^{2}+d \,k^{2} x^{3}\right )}\, dx\]
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^{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} \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 {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 \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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