Optimal. Leaf size=85 \[ \frac {\tan ^{-1}\left (\frac {(k+1) x}{\sqrt {k^2 x^3+\left (-k^2-1\right ) x^2+x}}\right )}{2 k (k+1)}-\frac {\tan ^{-1}\left (\frac {(k-1) x}{\sqrt {k^2 x^3+\left (-k^2-1\right ) x^2+x}}\right )}{2 (k-1) k} \]
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Rubi [C] time = 1.69, antiderivative size = 203, normalized size of antiderivative = 2.39, number of steps used = 13, number of rules used = 7, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.219, Rules used = {6718, 6725, 175, 115, 168, 538, 537} \begin {gather*} \frac {\sqrt {1-x} \sqrt {x} \sqrt {\frac {k^2 (1-x)}{1-k^2}+1} \Pi \left (-\frac {k}{1-k};\sin ^{-1}\left (\sqrt {1-x}\right )|-\frac {k^2}{1-k^2}\right )}{(1-k) k \sqrt {(1-x) x \left (1-k^2 x\right )}}-\frac {\sqrt {1-x} \sqrt {x} \sqrt {\frac {k^2 (1-x)}{1-k^2}+1} \Pi \left (\frac {k}{k+1};\sin ^{-1}\left (\sqrt {1-x}\right )|-\frac {k^2}{1-k^2}\right )}{k (k+1) \sqrt {(1-x) x \left (1-k^2 x\right )}} \end {gather*}
Antiderivative was successfully verified.
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Rule 115
Rule 168
Rule 175
Rule 537
Rule 538
Rule 6718
Rule 6725
Rubi steps
\begin {align*} \int \frac {x}{\sqrt {(1-x) x \left (1-k^2 x\right )} \left (-1+k^2 x^2\right )} \, dx &=\frac {\left (\sqrt {1-x} \sqrt {x} \sqrt {1-k^2 x}\right ) \int \frac {\sqrt {x}}{\sqrt {1-x} \sqrt {1-k^2 x} \left (-1+k^2 x^2\right )} \, dx}{\sqrt {(1-x) x \left (1-k^2 x\right )}}\\ &=\frac {\left (\sqrt {1-x} \sqrt {x} \sqrt {1-k^2 x}\right ) \int \left (-\frac {\sqrt {x}}{2 \sqrt {1-x} (1-k x) \sqrt {1-k^2 x}}-\frac {\sqrt {x}}{2 \sqrt {1-x} (1+k x) \sqrt {1-k^2 x}}\right ) \, dx}{\sqrt {(1-x) x \left (1-k^2 x\right )}}\\ &=-\frac {\left (\sqrt {1-x} \sqrt {x} \sqrt {1-k^2 x}\right ) \int \frac {\sqrt {x}}{\sqrt {1-x} (1-k x) \sqrt {1-k^2 x}} \, dx}{2 \sqrt {(1-x) x \left (1-k^2 x\right )}}-\frac {\left (\sqrt {1-x} \sqrt {x} \sqrt {1-k^2 x}\right ) \int \frac {\sqrt {x}}{\sqrt {1-x} (1+k x) \sqrt {1-k^2 x}} \, dx}{2 \sqrt {(1-x) x \left (1-k^2 x\right )}}\\ &=-\frac {\left (\sqrt {1-x} \sqrt {x} \sqrt {1-k^2 x}\right ) \int \frac {1}{\sqrt {1-x} \sqrt {x} (1-k x) \sqrt {1-k^2 x}} \, dx}{2 k \sqrt {(1-x) x \left (1-k^2 x\right )}}+\frac {\left (\sqrt {1-x} \sqrt {x} \sqrt {1-k^2 x}\right ) \int \frac {1}{\sqrt {1-x} \sqrt {x} (1+k x) \sqrt {1-k^2 x}} \, dx}{2 k \sqrt {(1-x) x \left (1-k^2 x\right )}}\\ &=-\frac {\left (\sqrt {1-x} \sqrt {x} \sqrt {1-k^2 x}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-x^2} \left (1+k-k x^2\right ) \sqrt {1-k^2+k^2 x^2}} \, dx,x,\sqrt {1-x}\right )}{k \sqrt {(1-x) x \left (1-k^2 x\right )}}+\frac {\left (\sqrt {1-x} \sqrt {x} \sqrt {1-k^2 x}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-x^2} \left (1-k+k x^2\right ) \sqrt {1-k^2+k^2 x^2}} \, dx,x,\sqrt {1-x}\right )}{k \sqrt {(1-x) x \left (1-k^2 x\right )}}\\ &=-\frac {\left (\sqrt {1+\frac {k^2 (-1+x)}{-1+k^2}} \sqrt {1-x} \sqrt {x}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-x^2} \left (1+k-k x^2\right ) \sqrt {1+\frac {k^2 x^2}{1-k^2}}} \, dx,x,\sqrt {1-x}\right )}{k \sqrt {(1-x) x \left (1-k^2 x\right )}}+\frac {\left (\sqrt {1+\frac {k^2 (-1+x)}{-1+k^2}} \sqrt {1-x} \sqrt {x}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-x^2} \left (1-k+k x^2\right ) \sqrt {1+\frac {k^2 x^2}{1-k^2}}} \, dx,x,\sqrt {1-x}\right )}{k \sqrt {(1-x) x \left (1-k^2 x\right )}}\\ &=\frac {\sqrt {1+\frac {k^2 (1-x)}{1-k^2}} \sqrt {1-x} \sqrt {x} \Pi \left (-\frac {k}{1-k};\sin ^{-1}\left (\sqrt {1-x}\right )|-\frac {k^2}{1-k^2}\right )}{(1-k) k \sqrt {(1-x) x \left (1-k^2 x\right )}}-\frac {\sqrt {1+\frac {k^2 (1-x)}{1-k^2}} \sqrt {1-x} \sqrt {x} \Pi \left (\frac {k}{1+k};\sin ^{-1}\left (\sqrt {1-x}\right )|-\frac {k^2}{1-k^2}\right )}{k (1+k) \sqrt {(1-x) x \left (1-k^2 x\right )}}\\ \end {align*}
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Mathematica [C] time = 1.39, size = 154, normalized size = 1.81 \begin {gather*} \frac {i \sqrt {x-1} x \sqrt {\frac {1-\frac {1}{k^2}}{x-1}+1} \left (2 k F\left (i \sinh ^{-1}\left (\frac {1}{\sqrt {x-1}}\right )|1-\frac {1}{k^2}\right )-(k-1) \Pi \left (1+\frac {1}{k};i \sinh ^{-1}\left (\frac {1}{\sqrt {x-1}}\right )|1-\frac {1}{k^2}\right )-(k+1) \Pi \left (\frac {k-1}{k};i \sinh ^{-1}\left (\frac {1}{\sqrt {x-1}}\right )|1-\frac {1}{k^2}\right )\right )}{k \left (k^2-1\right ) \sqrt {\frac {1}{x-1}+1} \sqrt {(x-1) x \left (k^2 x-1\right )}} \end {gather*}
Antiderivative was successfully verified.
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IntegrateAlgebraic [A] time = 0.19, size = 85, normalized size = 1.00 \begin {gather*} \frac {\tan ^{-1}\left (\frac {(k+1) x}{\sqrt {k^2 x^3+\left (-k^2-1\right ) x^2+x}}\right )}{2 k (k+1)}-\frac {\tan ^{-1}\left (\frac {(k-1) x}{\sqrt {k^2 x^3+\left (-k^2-1\right ) x^2+x}}\right )}{2 (k-1) k} \end {gather*}
Antiderivative was successfully verified.
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fricas [B] time = 0.47, size = 177, normalized size = 2.08 \begin {gather*} -\frac {{\left (k - 1\right )} \arctan \left (\frac {\sqrt {k^{2} x^{3} - {\left (k^{2} + 1\right )} x^{2} + x} {\left (k^{2} x^{2} - 2 \, {\left (k^{2} + k + 1\right )} x + 1\right )}}{2 \, {\left ({\left (k^{3} + k^{2}\right )} x^{3} - {\left (k^{3} + k^{2} + k + 1\right )} x^{2} + {\left (k + 1\right )} x\right )}}\right ) - {\left (k + 1\right )} \arctan \left (\frac {\sqrt {k^{2} x^{3} - {\left (k^{2} + 1\right )} x^{2} + x} {\left (k^{2} x^{2} - 2 \, {\left (k^{2} - k + 1\right )} x + 1\right )}}{2 \, {\left ({\left (k^{3} - k^{2}\right )} x^{3} - {\left (k^{3} - k^{2} + k - 1\right )} x^{2} + {\left (k - 1\right )} x\right )}}\right )}{4 \, {\left (k^{3} - k\right )}} \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 {x}{{\left (k^{2} x^{2} - 1\right )} \sqrt {{\left (k^{2} x - 1\right )} {\left (x - 1\right )} x}}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 0.03, size = 232, normalized size = 2.73 \begin {gather*} -\frac {\sqrt {-\left (x -\frac {1}{k^{2}}\right ) k^{2}}\, \sqrt {\frac {-1+x}{\frac {1}{k^{2}}-1}}\, \sqrt {k^{2} x}\, \EllipticPi \left (\sqrt {-\left (x -\frac {1}{k^{2}}\right ) k^{2}}, \frac {1}{k^{2} \left (\frac {1}{k^{2}}-\frac {1}{k}\right )}, \sqrt {\frac {1}{k^{2} \left (\frac {1}{k^{2}}-1\right )}}\right )}{k^{4} \sqrt {k^{2} x^{3}-k^{2} x^{2}-x^{2}+x}\, \left (\frac {1}{k^{2}}-\frac {1}{k}\right )}-\frac {\sqrt {-\left (x -\frac {1}{k^{2}}\right ) k^{2}}\, \sqrt {\frac {-1+x}{\frac {1}{k^{2}}-1}}\, \sqrt {k^{2} x}\, \EllipticPi \left (\sqrt {-\left (x -\frac {1}{k^{2}}\right ) k^{2}}, \frac {1}{k^{2} \left (\frac {1}{k^{2}}+\frac {1}{k}\right )}, \sqrt {\frac {1}{k^{2} \left (\frac {1}{k^{2}}-1\right )}}\right )}{k^{4} \sqrt {k^{2} x^{3}-k^{2} x^{2}-x^{2}+x}\, \left (\frac {1}{k^{2}}+\frac {1}{k}\right )} \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 {x}{{\left (k^{2} x^{2} - 1\right )} \sqrt {{\left (k^{2} x - 1\right )} {\left (x - 1\right )} x}}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F(-1)] time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \text {Hanged} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x}{\sqrt {x \left (x - 1\right ) \left (k^{2} x - 1\right )} \left (k x - 1\right ) \left (k x + 1\right )}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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