Integrand size = 37, antiderivative size = 94 \[ \int \frac {x^2}{\sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (-1+k^2 x^4\right )} \, dx=-\frac {\arctan \left (\frac {(-1+k) x}{\sqrt {1+\left (-1-k^2\right ) x^2+k^2 x^4}}\right )}{4 (-1+k) k}+\frac {\arctan \left (\frac {(1+k) x}{1+k x^2+\sqrt {1+\left (-1-k^2\right ) x^2+k^2 x^4}}\right )}{2 k (1+k)} \]
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Time = 0.55 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.93, number of steps used = 11, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.189, Rules used = {1976, 6857, 1224, 1117, 1712, 210, 209} \[ \int \frac {x^2}{\sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (-1+k^2 x^4\right )} \, dx=\frac {\arctan \left (\frac {(k+1) x}{\sqrt {k^2 x^4-\left (k^2+1\right ) x^2+1}}\right )}{4 k (k+1)}-\frac {\arctan \left (\frac {(1-k) x}{\sqrt {k^2 x^4-\left (k^2+1\right ) x^2+1}}\right )}{4 (1-k) k} \]
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Rule 209
Rule 210
Rule 1117
Rule 1224
Rule 1712
Rule 1976
Rule 6857
Rubi steps \begin{align*} \text {integral}& = \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 {\text {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 {\text {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 {\arctan \left (\frac {(1-k) x}{\sqrt {1-\left (1+k^2\right ) x^2+k^2 x^4}}\right )}{4 (1-k) k}+\frac {\arctan \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*}
Time = 5.09 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.86 \[ \int \frac {x^2}{\sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (-1+k^2 x^4\right )} \, dx=\frac {-\frac {\arctan \left (\frac {(-1+k) x}{\sqrt {\left (-1+x^2\right ) \left (-1+k^2 x^2\right )}}\right )}{-1+k}+\frac {2 \arctan \left (\frac {(1+k) x}{1+k x^2+\sqrt {\left (-1+x^2\right ) \left (-1+k^2 x^2\right )}}\right )}{1+k}}{4 k} \]
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Time = 3.52 (sec) , antiderivative size = 93, normalized size of antiderivative = 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 | \(\frac {-\frac {\sqrt {-\left (1+k \right )^{2}}\, \ln \left (\frac {\sqrt {-\left (-1+k \right )^{2}}\, \sqrt {\left (x^{2}-1\right ) \left (k^{2} x^{2}-1\right )}-2 k^{\frac {3}{2}} x^{2}-2 \sqrt {k}+\left (-k^{2}-2 k -1\right ) x}{1+2 \sqrt {k}\, x +k \,x^{2}}\right )}{2}-\frac {\sqrt {-\left (1+k \right )^{2}}\, \ln \left (\frac {\sqrt {-\left (-1+k \right )^{2}}\, \sqrt {\left (x^{2}-1\right ) \left (k^{2} x^{2}-1\right )}+2 k^{\frac {3}{2}} x^{2}+2 \sqrt {k}+\left (-k^{2}-2 k -1\right ) x}{1-2 \sqrt {k}\, x +k \,x^{2}}\right )}{2}+\sqrt {-\left (-1+k \right )^{2}}\, \ln \left (\frac {\sqrt {-\left (1+k \right )^{2}}\, \sqrt {\left (x^{2}-1\right ) \left (k^{2} x^{2}-1\right )}-x \left (1+k \right )^{2}}{k \,x^{2}+1}\right )+\ln \left (2\right ) \left (\sqrt {-\left (-1+k \right )^{2}}-\sqrt {-\left (1+k \right )^{2}}\right )}{4 \sqrt {-\left (1+k \right )^{2}}\, \sqrt {-\left (-1+k \right )^{2}}\, k}\) | \(268\) |
pseudoelliptic | \(\frac {-\frac {\sqrt {-\left (1+k \right )^{2}}\, \ln \left (\frac {\sqrt {-\left (-1+k \right )^{2}}\, \sqrt {\left (x^{2}-1\right ) \left (k^{2} x^{2}-1\right )}-2 k^{\frac {3}{2}} x^{2}-2 \sqrt {k}+\left (-k^{2}-2 k -1\right ) x}{1+2 \sqrt {k}\, x +k \,x^{2}}\right )}{2}-\frac {\sqrt {-\left (1+k \right )^{2}}\, \ln \left (\frac {\sqrt {-\left (-1+k \right )^{2}}\, \sqrt {\left (x^{2}-1\right ) \left (k^{2} x^{2}-1\right )}+2 k^{\frac {3}{2}} x^{2}+2 \sqrt {k}+\left (-k^{2}-2 k -1\right ) x}{1-2 \sqrt {k}\, x +k \,x^{2}}\right )}{2}+\sqrt {-\left (-1+k \right )^{2}}\, \ln \left (\frac {\sqrt {-\left (1+k \right )^{2}}\, \sqrt {\left (x^{2}-1\right ) \left (k^{2} x^{2}-1\right )}-x \left (1+k \right )^{2}}{k \,x^{2}+1}\right )+\ln \left (2\right ) \left (\sqrt {-\left (-1+k \right )^{2}}-\sqrt {-\left (1+k \right )^{2}}\right )}{4 \sqrt {-\left (1+k \right )^{2}}\, \sqrt {-\left (-1+k \right )^{2}}\, k}\) | \(268\) |
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Time = 0.36 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.88 \[ \int \frac {x^2}{\sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (-1+k^2 x^4\right )} \, dx=-\frac {{\left (k - 1\right )} \arctan \left (\frac {\sqrt {k^{2} x^{4} - {\left (k^{2} + 1\right )} x^{2} + 1}}{{\left (k + 1\right )} x}\right ) - {\left (k + 1\right )} \arctan \left (\frac {\sqrt {k^{2} x^{4} - {\left (k^{2} + 1\right )} x^{2} + 1}}{{\left (k - 1\right )} x}\right )}{4 \, {\left (k^{3} - k\right )}} \]
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\[ \int \frac {x^2}{\sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (-1+k^2 x^4\right )} \, dx=\int \frac {x^{2}}{\sqrt {\left (x - 1\right ) \left (x + 1\right ) \left (k x - 1\right ) \left (k x + 1\right )} \left (k x^{2} - 1\right ) \left (k x^{2} + 1\right )}\, dx \]
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\[ \int \frac {x^2}{\sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (-1+k^2 x^4\right )} \, dx=\int { \frac {x^{2}}{{\left (k^{2} x^{4} - 1\right )} \sqrt {{\left (k^{2} x^{2} - 1\right )} {\left (x^{2} - 1\right )}}} \,d x } \]
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\[ \int \frac {x^2}{\sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (-1+k^2 x^4\right )} \, dx=\int { \frac {x^{2}}{{\left (k^{2} x^{4} - 1\right )} \sqrt {{\left (k^{2} x^{2} - 1\right )} {\left (x^{2} - 1\right )}}} \,d x } \]
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Timed out. \[ \int \frac {x^2}{\sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (-1+k^2 x^4\right )} \, dx=\int \frac {x^2}{\left (k^2\,x^4-1\right )\,\sqrt {\left (x^2-1\right )\,\left (k^2\,x^2-1\right )}} \,d x \]
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