Optimal. Leaf size=497 \[ \frac {i \left (i \sqrt {k^2+2 i \sqrt {2} \sqrt {k^2+1} \sqrt {k}-2 k+1} k^2+\sqrt {2} \sqrt {k^2+1} \sqrt {k^2+2 i \sqrt {2} \sqrt {k^2+1} \sqrt {k}-2 k+1} \sqrt {k}+i \sqrt {k^2+2 i \sqrt {2} \sqrt {k^2+1} \sqrt {k}-2 k+1}\right ) \tan ^{-1}\left (\frac {\sqrt {k^2+2 i \sqrt {2} \sqrt {k^2+1} \sqrt {k}-2 k+1} x}{\sqrt {k^2 x^4+\left (-k^2-1\right ) x^2+1}+k x^2+1}\right )}{\left (\sqrt {k}-i\right )^2 \left (\sqrt {k}+i\right )^2 (k-i) (k+i)}-\frac {i \left (-i \sqrt {k^2-2 i \sqrt {2} \sqrt {k^2+1} \sqrt {k}-2 k+1} k^2+\sqrt {2} \sqrt {k^2+1} \sqrt {k^2-2 i \sqrt {2} \sqrt {k^2+1} \sqrt {k}-2 k+1} \sqrt {k}-i \sqrt {k^2-2 i \sqrt {2} \sqrt {k^2+1} \sqrt {k}-2 k+1}\right ) \tan ^{-1}\left (\frac {\sqrt {k^2-2 i \sqrt {2} \sqrt {k^2+1} \sqrt {k}-2 k+1} x}{\sqrt {k^2 x^4+\left (-k^2-1\right ) x^2+1}+k x^2+1}\right )}{\left (\sqrt {k}-i\right )^2 \left (\sqrt {k}+i\right )^2 (k-i) (k+i)} \]
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Rubi [A] time = 0.16, antiderivative size = 45, normalized size of antiderivative = 0.09, number of steps used = 2, number of rules used = 2, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.047, Rules used = {2112, 203} \begin {gather*} -\frac {\tan ^{-1}\left (\frac {\sqrt {k^2+1} x}{\sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )}}\right )}{\sqrt {k^2+1}} \end {gather*}
Antiderivative was successfully verified.
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Rule 203
Rule 2112
Rubi steps
\begin {align*} \int \frac {-1+k^2 x^4}{\sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (1+k^2 x^4\right )} \, dx &=-\operatorname {Subst}\left (\int \frac {1}{1-\left (-1-k^2\right ) x^2} \, dx,x,\frac {x}{\sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )}}\right )\\ &=-\frac {\tan ^{-1}\left (\frac {\sqrt {1+k^2} x}{\sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )}}\right )}{\sqrt {1+k^2}}\\ \end {align*}
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Mathematica [C] time = 0.47, size = 78, normalized size = 0.16 \begin {gather*} \frac {\sqrt {1-x^2} \sqrt {1-k^2 x^2} \left (F\left (\sin ^{-1}(x)|k^2\right )-\Pi \left (-i k;\sin ^{-1}(x)|k^2\right )-\Pi \left (i k;\sin ^{-1}(x)|k^2\right )\right )}{\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 = 2.89, size = 497, normalized size = 1.00 \begin {gather*} \frac {i \left (i \sqrt {k^2+2 i \sqrt {2} \sqrt {k^2+1} \sqrt {k}-2 k+1} k^2+\sqrt {2} \sqrt {k^2+1} \sqrt {k^2+2 i \sqrt {2} \sqrt {k^2+1} \sqrt {k}-2 k+1} \sqrt {k}+i \sqrt {k^2+2 i \sqrt {2} \sqrt {k^2+1} \sqrt {k}-2 k+1}\right ) \tan ^{-1}\left (\frac {\sqrt {k^2+2 i \sqrt {2} \sqrt {k^2+1} \sqrt {k}-2 k+1} x}{\sqrt {k^2 x^4+\left (-k^2-1\right ) x^2+1}+k x^2+1}\right )}{\left (\sqrt {k}-i\right )^2 \left (\sqrt {k}+i\right )^2 (k-i) (k+i)}-\frac {i \left (-i \sqrt {k^2-2 i \sqrt {2} \sqrt {k^2+1} \sqrt {k}-2 k+1} k^2+\sqrt {2} \sqrt {k^2+1} \sqrt {k^2-2 i \sqrt {2} \sqrt {k^2+1} \sqrt {k}-2 k+1} \sqrt {k}-i \sqrt {k^2-2 i \sqrt {2} \sqrt {k^2+1} \sqrt {k}-2 k+1}\right ) \tan ^{-1}\left (\frac {\sqrt {k^2-2 i \sqrt {2} \sqrt {k^2+1} \sqrt {k}-2 k+1} x}{\sqrt {k^2 x^4+\left (-k^2-1\right ) x^2+1}+k x^2+1}\right )}{\left (\sqrt {k}-i\right )^2 \left (\sqrt {k}+i\right )^2 (k-i) (k+i)} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.46, size = 62, normalized size = 0.12 \begin {gather*} -\frac {\arctan \left (\frac {2 \, \sqrt {k^{2} x^{4} - {\left (k^{2} + 1\right )} x^{2} + 1} \sqrt {k^{2} + 1} x}{k^{2} x^{4} - 2 \, {\left (k^{2} + 1\right )} x^{2} + 1}\right )}{2 \, \sqrt {k^{2} + 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^{2} x^{4} - 1}{{\left (k^{2} x^{4} + 1\right )} \sqrt {{\left (k^{2} x^{2} - 1\right )} {\left (x^{2} - 1\right )}}}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 0.07, size = 269, normalized size = 0.54 \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 {\munderset {\underline {\hspace {1.25 ex}}\alpha =\RootOf \left (k^{2} \textit {\_Z}^{4}+1\right )}{\sum }\frac {-\frac {\arctanh \left (\frac {\left (2 k^{2} \underline {\hspace {1.25 ex}}\alpha ^{2}-k^{2}-1\right ) \left (\underline {\hspace {1.25 ex}}\alpha ^{2} k^{4}+k^{4} x^{2}-2 k^{2} \underline {\hspace {1.25 ex}}\alpha ^{2}+6 k^{2} x^{2}+\underline {\hspace {1.25 ex}}\alpha ^{2}-4 k^{2}+x^{2}-4\right )}{2 \left (k^{4}+6 k^{2}+1\right ) \sqrt {-k^{2} \underline {\hspace {1.25 ex}}\alpha ^{2}-\underline {\hspace {1.25 ex}}\alpha ^{2}}\, \sqrt {k^{2} x^{4}-k^{2} x^{2}-x^{2}+1}}\right )}{\sqrt {-k^{2} \underline {\hspace {1.25 ex}}\alpha ^{2}-\underline {\hspace {1.25 ex}}\alpha ^{2}}}+\frac {2 \sqrt {-x^{2}+1}\, \sqrt {-k^{2} x^{2}+1}\, \underline {\hspace {1.25 ex}}\alpha ^{3} k^{2} \EllipticPi \left (x , -k^{2} \underline {\hspace {1.25 ex}}\alpha ^{2}, k\right )}{\sqrt {k^{2} x^{4}-k^{2} x^{2}-x^{2}+1}}}{\underline {\hspace {1.25 ex}}\alpha ^{3}}}{4 k^{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 {k^{2} x^{4} - 1}{{\left (k^{2} x^{4} + 1\right )} \sqrt {{\left (k^{2} x^{2} - 1\right )} {\left (x^{2} - 1\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 = -0.00 \begin {gather*} \int \frac {k^2\,x^4-1}{\left (k^2\,x^4+1\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 {\left (k x^{2} - 1\right ) \left (k x^{2} + 1\right )}{\sqrt {\left (x - 1\right ) \left (x + 1\right ) \left (k x - 1\right ) \left (k x + 1\right )} \left (k^{2} x^{4} + 1\right )}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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