Optimal. Leaf size=187 \[ \frac {\tan ^{-1}\left (\frac {\left (-k-2 \sqrt {k}-1\right ) x}{\sqrt {k^2 x^4+\left (-k^2-1\right ) x^2+1}+k x^2-1}\right )}{k+1}+\frac {\tan ^{-1}\left (\frac {\left (-k+2 \sqrt {k}-1\right ) x}{\sqrt {k^2 x^4+\left (-k^2-1\right ) x^2+1}+k x^2-1}\right )}{k+1}+\frac {i \tanh ^{-1}\left (\frac {\left (2 k^{3/2}+2 \sqrt {k}\right ) x^2}{k^2 x^4+\left (k x^2-1\right ) \sqrt {k^2 x^4+\left (-k^2-1\right ) x^2+1}+2 k x^2+1}\right )}{k+1} \]
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Rubi [C] time = 1.45, antiderivative size = 201, normalized size of antiderivative = 1.07, number of steps used = 9, number of rules used = 8, integrand size = 47, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.170, Rules used = {6719, 6742, 419, 2113, 537, 571, 93, 208} \begin {gather*} \frac {2 i \sqrt {1-x^2} \sqrt {1-k^2 x^2} \tanh ^{-1}\left (\frac {\sqrt {k} \sqrt {1-x^2}}{\sqrt {1-k^2 x^2}}\right )}{(k+1) \sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )}}+\frac {\sqrt {1-x^2} \sqrt {1-k^2 x^2} F\left (\sin ^{-1}(x)|k^2\right )}{\sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )}}-\frac {2 \sqrt {1-x^2} \sqrt {1-k^2 x^2} \Pi \left (-k;\sin ^{-1}(x)|k^2\right )}{\sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )}} \end {gather*}
Antiderivative was successfully verified.
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Rule 93
Rule 208
Rule 419
Rule 537
Rule 571
Rule 2113
Rule 6719
Rule 6742
Rubi steps
\begin {align*} \int \frac {-i+\sqrt {k} x}{\left (i+\sqrt {k} x\right ) \sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )}} \, dx &=\frac {\left (\sqrt {1-x^2} \sqrt {1-k^2 x^2}\right ) \int \frac {-i+\sqrt {k} x}{\left (i+\sqrt {k} x\right ) \sqrt {1-x^2} \sqrt {1-k^2 x^2}} \, dx}{\sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )}}\\ &=\frac {\left (\sqrt {1-x^2} \sqrt {1-k^2 x^2}\right ) \int \left (\frac {1}{\sqrt {1-x^2} \sqrt {1-k^2 x^2}}-\frac {2 i}{\left (i+\sqrt {k} x\right ) \sqrt {1-x^2} \sqrt {1-k^2 x^2}}\right ) \, dx}{\sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )}}\\ &=-\frac {\left (2 i \sqrt {1-x^2} \sqrt {1-k^2 x^2}\right ) \int \frac {1}{\left (i+\sqrt {k} x\right ) \sqrt {1-x^2} \sqrt {1-k^2 x^2}} \, dx}{\sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )}}+\frac {\left (\sqrt {1-x^2} \sqrt {1-k^2 x^2}\right ) \int \frac {1}{\sqrt {1-x^2} \sqrt {1-k^2 x^2}} \, dx}{\sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )}}\\ &=\frac {\sqrt {1-x^2} \sqrt {1-k^2 x^2} F\left (\sin ^{-1}(x)|k^2\right )}{\sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )}}+\frac {\left (2 \sqrt {1-x^2} \sqrt {1-k^2 x^2}\right ) \int \frac {1}{\sqrt {1-x^2} \left (-1-k x^2\right ) \sqrt {1-k^2 x^2}} \, dx}{\sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )}}+\frac {\left (2 i \sqrt {k} \sqrt {1-x^2} \sqrt {1-k^2 x^2}\right ) \int \frac {x}{\sqrt {1-x^2} \left (-1-k x^2\right ) \sqrt {1-k^2 x^2}} \, dx}{\sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )}}\\ &=\frac {\sqrt {1-x^2} \sqrt {1-k^2 x^2} F\left (\sin ^{-1}(x)|k^2\right )}{\sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )}}-\frac {2 \sqrt {1-x^2} \sqrt {1-k^2 x^2} \Pi \left (-k;\sin ^{-1}(x)|k^2\right )}{\sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )}}+\frac {\left (i \sqrt {k} \sqrt {1-x^2} \sqrt {1-k^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-x} (-1-k x) \sqrt {1-k^2 x}} \, dx,x,x^2\right )}{\sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )}}\\ &=\frac {\sqrt {1-x^2} \sqrt {1-k^2 x^2} F\left (\sin ^{-1}(x)|k^2\right )}{\sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )}}-\frac {2 \sqrt {1-x^2} \sqrt {1-k^2 x^2} \Pi \left (-k;\sin ^{-1}(x)|k^2\right )}{\sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )}}+\frac {\left (2 i \sqrt {k} \sqrt {1-x^2} \sqrt {1-k^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{1+k-\left (k+k^2\right ) x^2} \, dx,x,\frac {\sqrt {1-x^2}}{\sqrt {1-k^2 x^2}}\right )}{\sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )}}\\ &=\frac {2 i \sqrt {1-x^2} \sqrt {1-k^2 x^2} \tanh ^{-1}\left (\frac {\sqrt {k} \sqrt {1-x^2}}{\sqrt {1-k^2 x^2}}\right )}{(1+k) \sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )}}+\frac {\sqrt {1-x^2} \sqrt {1-k^2 x^2} F\left (\sin ^{-1}(x)|k^2\right )}{\sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )}}-\frac {2 \sqrt {1-x^2} \sqrt {1-k^2 x^2} \Pi \left (-k;\sin ^{-1}(x)|k^2\right )}{\sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )}}\\ \end {align*}
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Mathematica [C] time = 0.36, size = 208, normalized size = 1.11 \begin {gather*} \frac {-2 i \sqrt {k} \sqrt {x^2-1} \sqrt {k^2 x^2-1} \tanh ^{-1}\left (\frac {\sqrt {k (k+1)} \sqrt {x^2-1}}{\sqrt {k+1} \sqrt {k^2 x^2-1}}\right )+\sqrt {k+1} \sqrt {k (k+1)} \sqrt {1-x^2} \sqrt {1-k^2 x^2} F\left (\sin ^{-1}(x)|k^2\right )-2 \sqrt {k+1} \sqrt {k (k+1)} \sqrt {1-x^2} \sqrt {1-k^2 x^2} \Pi \left (-k;\sin ^{-1}(x)|k^2\right )}{\sqrt {k+1} \sqrt {k (k+1)} \sqrt {\left (x^2-1\right ) \left (k^2 x^2-1\right )}} \end {gather*}
Antiderivative was successfully verified.
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IntegrateAlgebraic [A] time = 3.66, size = 56, normalized size = 0.30 \begin {gather*} -\frac {2 \tan ^{-1}\left (\frac {(k+1) x}{\sqrt {k^2 x^4+\left (-k^2-1\right ) x^2+1}+k x^2+2 i \sqrt {k} x-1}\right )}{k+1} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 1.04, size = 249, normalized size = 1.33 \begin {gather*} \frac {i \, \log \left (\frac {{\left (-i \, k^{6} - 5 i \, k^{5} - 10 i \, k^{4} - 10 i \, k^{3} - 5 i \, k^{2} - i \, k\right )} x^{3} + {\left (i \, k^{5} + 5 i \, k^{4} + 10 i \, k^{3} + 10 i \, k^{2} + 5 i \, k + i\right )} x + \sqrt {k^{2} x^{4} - {\left (k^{2} + 1\right )} x^{2} + 1} {\left (k^{4} + 4 \, k^{3} - {\left (k^{5} + 4 \, k^{4} + 6 \, k^{3} + 4 \, k^{2} + k\right )} x^{2} + {\left (2 i \, k^{4} + 8 i \, k^{3} + 12 i \, k^{2} + 8 i \, k + 2 i\right )} \sqrt {k} x + 6 \, k^{2} + 4 \, k + 1\right )} + 2 \, {\left ({\left (k^{5} + 3 \, k^{4} + 3 \, k^{3} + k^{2}\right )} x^{4} + k^{3} - {\left (k^{5} + 3 \, k^{4} + 4 \, k^{3} + 4 \, k^{2} + 3 \, k + 1\right )} x^{2} + 3 \, k^{2} + 3 \, k + 1\right )} \sqrt {k}}{4 \, {\left (k^{5} x^{4} + 2 \, k^{4} x^{2} + k^{3}\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 {\sqrt {k} x - i}{\sqrt {{\left (k^{2} x^{2} - 1\right )} {\left (x^{2} - 1\right )}} {\left (\sqrt {k} x + i\right )}}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 0.17, size = 211, normalized size = 1.13 \begin {gather*} \frac {\sqrt {-x^{2}+1}\, \sqrt {-k^{2} x^{2}+1}\, \EllipticF \left (x , k\right )}{\sqrt {k^{2} x^{4}-k^{2} x^{2}-x^{2}+1}}-\frac {2 \sqrt {-x^{2}+1}\, \sqrt {-k^{2} x^{2}+1}\, \EllipticPi \left (x , -k , k\right )}{\sqrt {k^{2} x^{4}-k^{2} x^{2}-x^{2}+1}}+\frac {i \ln \left (\frac {2 k^{2}+4 k +2+\left (-k^{3}-2 k^{2}-k \right ) \left (x^{2}+\frac {1}{k}\right )+2 \sqrt {\left (1+k \right )^{2}}\, \sqrt {k^{3} \left (x^{2}+\frac {1}{k}\right )^{2}+\left (-k^{3}-2 k^{2}-k \right ) \left (x^{2}+\frac {1}{k}\right )+k^{2}+2 k +1}}{x^{2}+\frac {1}{k}}\right )}{\sqrt {\left (1+k \right )^{2}}} \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 {\sqrt {k} x - i}{\sqrt {{\left (k^{2} x^{2} - 1\right )} {\left (x^{2} - 1\right )}} {\left (\sqrt {k} x + i\right )}}\,{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 {\sqrt {k}\,x-\mathrm {i}}{\left (\sqrt {k}\,x+1{}\mathrm {i}\right )\,\sqrt {\left (x^2-1\right )\,\left (k^2\,x^2-1\right )}} \,d x \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 {\sqrt {k} x - i}{\sqrt {\left (x - 1\right ) \left (x + 1\right ) \left (k x - 1\right ) \left (k x + 1\right )} \left (\sqrt {k} x + i\right )}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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