Optimal. Leaf size=94 \[ \frac {\tan ^{-1}\left (\frac {(k+1) x}{\sqrt {k^2 x^4+\left (-k^2-1\right ) x^2+1}+k x^2+1}\right )}{2 k (k+1)}-\frac {\tan ^{-1}\left (\frac {(k-1) x}{\sqrt {k^2 x^4+\left (-k^2-1\right ) x^2+1}}\right )}{4 (k-1) k} \]
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Rubi [A] time = 1.10, antiderivative size = 87, normalized size of antiderivative = 0.93, number of steps used = 11, number of rules used = 7, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.189, Rules used = {1992, 6725, 1210, 1103, 1698, 204, 203} \begin {gather*} \frac {\tan ^{-1}\left (\frac {(k+1) x}{\sqrt {k^2 x^4-\left (k^2+1\right ) x^2+1}}\right )}{4 k (k+1)}-\frac {\tan ^{-1}\left (\frac {(1-k) x}{\sqrt {k^2 x^4-\left (k^2+1\right ) x^2+1}}\right )}{4 (1-k) k} \end {gather*}
Antiderivative was successfully verified.
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Rule 203
Rule 204
Rule 1103
Rule 1210
Rule 1698
Rule 1992
Rule 6725
Rubi steps
\begin {align*} \int \frac {x^2}{\sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (-1+k^2 x^4\right )} \, dx &=\int \frac {x^2}{\left (-1+k^2 x^4\right ) \sqrt {1-\left (1+k^2\right ) x^2+k^2 x^4}} \, dx\\ &=\int \left (\frac {1}{2 k \left (-1+k x^2\right ) \sqrt {1-\left (1+k^2\right ) x^2+k^2 x^4}}+\frac {1}{2 k \left (1+k x^2\right ) \sqrt {1-\left (1+k^2\right ) x^2+k^2 x^4}}\right ) \, dx\\ &=\frac {\int \frac {1}{\left (-1+k x^2\right ) \sqrt {1+\left (-1-k^2\right ) x^2+k^2 x^4}} \, dx}{2 k}+\frac {\int \frac {1}{\left (1+k x^2\right ) \sqrt {1+\left (-1-k^2\right ) x^2+k^2 x^4}} \, dx}{2 k}\\ &=-\frac {\int \frac {-1-k x^2}{\left (-1+k x^2\right ) \sqrt {1+\left (-1-k^2\right ) x^2+k^2 x^4}} \, dx}{4 k}+\frac {\int \frac {1-k x^2}{\left (1+k x^2\right ) \sqrt {1+\left (-1-k^2\right ) x^2+k^2 x^4}} \, dx}{4 k}\\ &=\frac {\operatorname {Subst}\left (\int \frac {1}{1-\left (-1-2 k-k^2\right ) x^2} \, dx,x,\frac {x}{\sqrt {1+\left (-1-k^2\right ) x^2+k^2 x^4}}\right )}{4 k}+\frac {\operatorname {Subst}\left (\int \frac {1}{-1-\left (1-2 k+k^2\right ) x^2} \, dx,x,\frac {x}{\sqrt {1+\left (-1-k^2\right ) x^2+k^2 x^4}}\right )}{4 k}\\ &=-\frac {\tan ^{-1}\left (\frac {(1-k) x}{\sqrt {1-\left (1+k^2\right ) x^2+k^2 x^4}}\right )}{4 (1-k) k}+\frac {\tan ^{-1}\left (\frac {(1+k) x}{\sqrt {1-\left (1+k^2\right ) x^2+k^2 x^4}}\right )}{4 k (1+k)}\\ \end {align*}
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Mathematica [C] time = 0.39, size = 70, normalized size = 0.74 \begin {gather*} \frac {\sqrt {1-x^2} \sqrt {1-k^2 x^2} \left (\Pi \left (-k;\sin ^{-1}(x)|k^2\right )-\Pi \left (k;\sin ^{-1}(x)|k^2\right )\right )}{2 k \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 = 0.98, size = 94, normalized size = 1.00 \begin {gather*} -\frac {\tan ^{-1}\left (\frac {(-1+k) x}{\sqrt {1+\left (-1-k^2\right ) x^2+k^2 x^4}}\right )}{4 (-1+k) k}+\frac {\tan ^{-1}\left (\frac {(1+k) x}{1+k x^2+\sqrt {1+\left (-1-k^2\right ) x^2+k^2 x^4}}\right )}{2 k (1+k)} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.64, size = 83, normalized size = 0.88 \begin {gather*} -\frac {{\left (k - 1\right )} \arctan \left (\frac {\sqrt {k^{2} x^{4} - {\left (k^{2} + 1\right )} x^{2} + 1}}{{\left (k + 1\right )} x}\right ) - {\left (k + 1\right )} \arctan \left (\frac {\sqrt {k^{2} x^{4} - {\left (k^{2} + 1\right )} x^{2} + 1}}{{\left (k - 1\right )} x}\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^{2}}{{\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 [A] time = 0.12, size = 93, normalized size = 0.99
method | result | size |
elliptic | \(\frac {\left (-\frac {\sqrt {2}\, \arctan \left (\frac {\sqrt {\left (-x^{2}+1\right ) \left (-k^{2} x^{2}+1\right )}}{x \left (1+k \right )}\right )}{4 k \left (1+k \right )}+\frac {\sqrt {2}\, \arctan \left (\frac {\sqrt {\left (-x^{2}+1\right ) \left (-k^{2} x^{2}+1\right )}}{x \left (-1+k \right )}\right )}{4 k \left (-1+k \right )}\right ) \sqrt {2}}{2}\) | \(93\) |
default | \(\frac {\sqrt {-x^{2}+1}\, \sqrt {-k^{2} x^{2}+1}\, \EllipticPi \left (x , -k , k\right )}{2 k \sqrt {k^{2} x^{4}-k^{2} x^{2}-x^{2}+1}}-\frac {\sqrt {-x^{2}+1}\, \sqrt {-k^{2} x^{2}+1}\, \EllipticPi \left (x , k , k\right )}{2 k \sqrt {k^{2} x^{4}-k^{2} x^{2}-x^{2}+1}}\) | \(112\) |
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^{2}}{{\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 = -0.01 \begin {gather*} \int \frac {x^2}{\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 {x^{2}}{\sqrt {\left (x - 1\right ) \left (x + 1\right ) \left (k x - 1\right ) \left (k x + 1\right )} \left (k x^{2} - 1\right ) \left (k x^{2} + 1\right )}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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