Optimal. Leaf size=90 \[ \frac {\tanh ^{-1}\left (\frac {\sqrt [4]{d} \sqrt [4]{k^2 x^4+\left (-k^2-1\right ) x^2+1}}{k x+1}\right )}{d^{3/4}}-\frac {\tan ^{-1}\left (\frac {\sqrt [4]{d} \sqrt [4]{k^2 x^4+\left (-k^2-1\right ) x^2+1}}{k x+1}\right )}{d^{3/4}} \]
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Rubi [F] time = 5.81, 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 {-2 k+(-1+k) (1+k) x+2 k x^2}{\sqrt [4]{\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (1-d+(3+d) k x+\left (d+3 k^2\right ) x^2+k \left (-d+k^2\right ) x^3\right )} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {align*} \int \frac {-2 k+(-1+k) (1+k) x+2 k x^2}{\sqrt [4]{\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (1-d+(3+d) k x+\left (d+3 k^2\right ) x^2+k \left (-d+k^2\right ) x^3\right )} \, dx &=\frac {\left (\sqrt [4]{1-x^2} \sqrt [4]{1-k^2 x^2}\right ) \int \frac {-2 k+(-1+k) (1+k) x+2 k x^2}{\sqrt [4]{1-x^2} \sqrt [4]{1-k^2 x^2} \left (1-d+(3+d) k x+\left (d+3 k^2\right ) x^2+k \left (-d+k^2\right ) x^3\right )} \, dx}{\sqrt [4]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}}\\ &=\frac {\left (\sqrt [4]{1-x^2} \sqrt [4]{1-k^2 x^2}\right ) \int \frac {-2 k-(1-k) (1+k) x+2 k x^2}{\sqrt [4]{1-x^2} \sqrt [4]{1-k^2 x^2} \left (1-d+(3+d) k x+\left (d+3 k^2\right ) x^2-k \left (d-k^2\right ) x^3\right )} \, dx}{\sqrt [4]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}}\\ &=\frac {\left (\sqrt [4]{1-x^2} \sqrt [4]{1-k^2 x^2}\right ) \int \left (\frac {(-1-k) (1-k) x}{\sqrt [4]{1-x^2} \sqrt [4]{1-k^2 x^2} \left (1-d+(3+d) k x+\left (d+3 k^2\right ) x^2-k \left (d-k^2\right ) x^3\right )}+\frac {2 k x^2}{\sqrt [4]{1-x^2} \sqrt [4]{1-k^2 x^2} \left (1-d+(3+d) k x+\left (d+3 k^2\right ) x^2-k \left (d-k^2\right ) x^3\right )}+\frac {2 k}{\sqrt [4]{1-x^2} \sqrt [4]{1-k^2 x^2} \left (-1+d-(3+d) k x-\left (d+3 k^2\right ) x^2+k \left (d-k^2\right ) x^3\right )}\right ) \, dx}{\sqrt [4]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}}\\ &=\frac {\left ((-1-k) (1-k) \sqrt [4]{1-x^2} \sqrt [4]{1-k^2 x^2}\right ) \int \frac {x}{\sqrt [4]{1-x^2} \sqrt [4]{1-k^2 x^2} \left (1-d+(3+d) k x+\left (d+3 k^2\right ) x^2-k \left (d-k^2\right ) x^3\right )} \, dx}{\sqrt [4]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}}+\frac {\left (2 k \sqrt [4]{1-x^2} \sqrt [4]{1-k^2 x^2}\right ) \int \frac {x^2}{\sqrt [4]{1-x^2} \sqrt [4]{1-k^2 x^2} \left (1-d+(3+d) k x+\left (d+3 k^2\right ) x^2-k \left (d-k^2\right ) x^3\right )} \, dx}{\sqrt [4]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}}+\frac {\left (2 k \sqrt [4]{1-x^2} \sqrt [4]{1-k^2 x^2}\right ) \int \frac {1}{\sqrt [4]{1-x^2} \sqrt [4]{1-k^2 x^2} \left (-1+d-(3+d) k x-\left (d+3 k^2\right ) x^2+k \left (d-k^2\right ) x^3\right )} \, dx}{\sqrt [4]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}}\\ \end {align*}
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Mathematica [F] time = 1.17, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {-2 k+(-1+k) (1+k) x+2 k x^2}{\sqrt [4]{\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (1-d+(3+d) k x+\left (d+3 k^2\right ) x^2+k \left (-d+k^2\right ) x^3\right )} \, dx \end {gather*}
Verification is not applicable to the result.
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IntegrateAlgebraic [A] time = 12.98, size = 90, normalized size = 1.00 \begin {gather*} -\frac {\tan ^{-1}\left (\frac {\sqrt [4]{d} \sqrt [4]{1+\left (-1-k^2\right ) x^2+k^2 x^4}}{1+k x}\right )}{d^{3/4}}+\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{d} \sqrt [4]{1+\left (-1-k^2\right ) x^2+k^2 x^4}}{1+k x}\right )}{d^{3/4}} \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 {{\left (k + 1\right )} {\left (k - 1\right )} x + 2 \, k x^{2} - 2 \, k}{{\left ({\left (k^{2} - d\right )} k x^{3} + {\left (d + 3\right )} k x + {\left (3 \, k^{2} + d\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} \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 \[\int \frac {-2 k +\left (-1+k \right ) \left (1+k \right ) x +2 k \,x^{2}}{\left (\left (-x^{2}+1\right ) \left (-k^{2} x^{2}+1\right )\right )^{\frac {1}{4}} \left (1-d +\left (3+d \right ) k x +\left (3 k^{2}+d \right ) x^{2}+k \left (k^{2}-d \right ) x^{3}\right )}\, dx\]
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 {{\left (k + 1\right )} {\left (k - 1\right )} x + 2 \, k x^{2} - 2 \, k}{{\left ({\left (k^{2} - d\right )} k x^{3} + {\left (d + 3\right )} k x + {\left (3 \, k^{2} + d\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} \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 = -0.01 \begin {gather*} \int \frac {2\,k\,x^2+\left (k-1\right )\,\left (k+1\right )\,x-2\,k}{{\left (\left (x^2-1\right )\,\left (k^2\,x^2-1\right )\right )}^{1/4}\,\left (-k\,\left (d-k^2\right )\,x^3+\left (3\,k^2+d\right )\,x^2+k\,\left (d+3\right )\,x-d+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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