Optimal. Leaf size=37 \[ -\frac {2 \tan ^{-1}\left (\frac {(k+1) x}{\sqrt {k^2 x^3+\left (-k^2-1\right ) x^2+x}}\right )}{k+1} \]
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Rubi [C] time = 0.64, antiderivative size = 154, normalized size of antiderivative = 4.16, number of steps used = 6, number of rules used = 6, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {6718, 1607, 168, 538, 537, 115} \begin {gather*} \frac {2 \sqrt {1-x} \sqrt {x} \sqrt {1-k^2 x} F\left (\sin ^{-1}\left (\sqrt {x}\right )|k^2\right )}{\sqrt {(1-x) x \left (1-k^2 x\right )}}+\frac {4 \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+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 537
Rule 538
Rule 1607
Rule 6718
Rubi steps
\begin {align*} \int \frac {-1+k x}{(1+k x) \sqrt {(1-x) x \left (1-k^2 x\right )}} \, dx &=\frac {\left (\sqrt {1-x} \sqrt {x} \sqrt {1-k^2 x}\right ) \int \frac {-1+k x}{\sqrt {1-x} \sqrt {x} (1+k x) \sqrt {1-k^2 x}} \, 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 {1}{\sqrt {1-x} \sqrt {x} \sqrt {1-k^2 x}} \, dx}{\sqrt {(1-x) x \left (1-k^2 x\right )}}-\frac {\left (2 \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}{\sqrt {(1-x) x \left (1-k^2 x\right )}}\\ &=\frac {2 \sqrt {1-x} \sqrt {x} \sqrt {1-k^2 x} F\left (\sin ^{-1}\left (\sqrt {x}\right )|k^2\right )}{\sqrt {(1-x) x \left (1-k^2 x\right )}}+\frac {\left (4 \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 )}{\sqrt {(1-x) x \left (1-k^2 x\right )}}\\ &=\frac {2 \sqrt {1-x} \sqrt {x} \sqrt {1-k^2 x} F\left (\sin ^{-1}\left (\sqrt {x}\right )|k^2\right )}{\sqrt {(1-x) x \left (1-k^2 x\right )}}+\frac {\left (4 \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 )}{\sqrt {(1-x) x \left (1-k^2 x\right )}}\\ &=\frac {2 \sqrt {1-x} \sqrt {x} \sqrt {1-k^2 x} F\left (\sin ^{-1}\left (\sqrt {x}\right )|k^2\right )}{\sqrt {(1-x) x \left (1-k^2 x\right )}}+\frac {4 \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) \sqrt {(1-x) x \left (1-k^2 x\right )}}\\ \end {align*}
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Mathematica [C] time = 0.85, size = 114, normalized size = 3.08 \begin {gather*} \frac {2 i \sqrt {\frac {1}{x-1}+1} (x-1)^{3/2} \sqrt {\frac {1-\frac {1}{k^2}}{x-1}+1} \left ((k-1) F\left (i \sinh ^{-1}\left (\frac {1}{\sqrt {x-1}}\right )|1-\frac {1}{k^2}\right )+2 \Pi \left (1+\frac {1}{k};i \sinh ^{-1}\left (\frac {1}{\sqrt {x-1}}\right )|1-\frac {1}{k^2}\right )\right )}{(k+1) \sqrt {(x-1) x \left (k^2 x-1\right )}} \end {gather*}
Antiderivative was successfully verified.
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IntegrateAlgebraic [A] time = 0.18, size = 37, normalized size = 1.00 \begin {gather*} -\frac {2 \tan ^{-1}\left (\frac {(k+1) x}{\sqrt {k^2 x^3+\left (-k^2-1\right ) x^2+x}}\right )}{k+1} \end {gather*}
Antiderivative was successfully verified.
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fricas [B] time = 0.46, size = 81, normalized size = 2.19 \begin {gather*} \frac {\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 )}{k + 1} \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 x - 1}{\sqrt {{\left (k^{2} x - 1\right )} {\left (x - 1\right )} x} {\left (k x + 1\right )}}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 0.06, size = 206, normalized size = 5.57 \begin {gather*} -\frac {2 \sqrt {-\left (x -\frac {1}{k^{2}}\right ) k^{2}}\, \sqrt {\frac {-1+x}{\frac {1}{k^{2}}-1}}\, \sqrt {k^{2} x}\, \EllipticF \left (\sqrt {-\left (x -\frac {1}{k^{2}}\right ) k^{2}}, \sqrt {\frac {1}{k^{2} \left (\frac {1}{k^{2}}-1\right )}}\right )}{k^{2} \sqrt {k^{2} x^{3}-k^{2} x^{2}-x^{2}+x}}+\frac {4 \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^{3} \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 {k x - 1}{\sqrt {{\left (k^{2} x - 1\right )} {\left (x - 1\right )} x} {\left (k x + 1\right )}}\,{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 = -0.03 \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 {k x - 1}{\sqrt {x \left (x - 1\right ) \left (k^{2} 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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