Integrand size = 18, antiderivative size = 41 \[ \int \frac {x^3}{1-2 x^2+3 x^4} \, dx=-\frac {\arctan \left (\frac {1-3 x^2}{\sqrt {2}}\right )}{6 \sqrt {2}}+\frac {1}{12} \log \left (1-2 x^2+3 x^4\right ) \]
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Time = 0.03 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {1128, 648, 632, 210, 642} \[ \int \frac {x^3}{1-2 x^2+3 x^4} \, dx=\frac {1}{12} \log \left (3 x^4-2 x^2+1\right )-\frac {\arctan \left (\frac {1-3 x^2}{\sqrt {2}}\right )}{6 \sqrt {2}} \]
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Rule 210
Rule 632
Rule 642
Rule 648
Rule 1128
Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {x}{1-2 x+3 x^2} \, dx,x,x^2\right ) \\ & = \frac {1}{12} \text {Subst}\left (\int \frac {-2+6 x}{1-2 x+3 x^2} \, dx,x,x^2\right )+\frac {1}{6} \text {Subst}\left (\int \frac {1}{1-2 x+3 x^2} \, dx,x,x^2\right ) \\ & = \frac {1}{12} \log \left (1-2 x^2+3 x^4\right )-\frac {1}{3} \text {Subst}\left (\int \frac {1}{-8-x^2} \, dx,x,2 \left (-1+3 x^2\right )\right ) \\ & = -\frac {\arctan \left (\frac {1-3 x^2}{\sqrt {2}}\right )}{6 \sqrt {2}}+\frac {1}{12} \log \left (1-2 x^2+3 x^4\right ) \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.93 \[ \int \frac {x^3}{1-2 x^2+3 x^4} \, dx=\frac {1}{12} \left (\sqrt {2} \arctan \left (\frac {-1+3 x^2}{\sqrt {2}}\right )+\log \left (1-2 x^2+3 x^4\right )\right ) \]
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Time = 0.05 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.85
method | result | size |
default | \(\frac {\ln \left (3 x^{4}-2 x^{2}+1\right )}{12}+\frac {\sqrt {2}\, \arctan \left (\frac {\left (6 x^{2}-2\right ) \sqrt {2}}{4}\right )}{12}\) | \(35\) |
risch | \(\frac {\ln \left (9 x^{4}-6 x^{2}+3\right )}{12}+\frac {\sqrt {2}\, \arctan \left (\frac {\left (3 x^{2}-1\right ) \sqrt {2}}{2}\right )}{12}\) | \(35\) |
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Time = 0.25 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.83 \[ \int \frac {x^3}{1-2 x^2+3 x^4} \, dx=\frac {1}{12} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (3 \, x^{2} - 1\right )}\right ) + \frac {1}{12} \, \log \left (3 \, x^{4} - 2 \, x^{2} + 1\right ) \]
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Time = 0.05 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.02 \[ \int \frac {x^3}{1-2 x^2+3 x^4} \, dx=\frac {\log {\left (x^{4} - \frac {2 x^{2}}{3} + \frac {1}{3} \right )}}{12} + \frac {\sqrt {2} \operatorname {atan}{\left (\frac {3 \sqrt {2} x^{2}}{2} - \frac {\sqrt {2}}{2} \right )}}{12} \]
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Time = 0.28 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.83 \[ \int \frac {x^3}{1-2 x^2+3 x^4} \, dx=\frac {1}{12} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (3 \, x^{2} - 1\right )}\right ) + \frac {1}{12} \, \log \left (3 \, x^{4} - 2 \, x^{2} + 1\right ) \]
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Time = 0.43 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.83 \[ \int \frac {x^3}{1-2 x^2+3 x^4} \, dx=\frac {1}{12} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (3 \, x^{2} - 1\right )}\right ) + \frac {1}{12} \, \log \left (3 \, x^{4} - 2 \, x^{2} + 1\right ) \]
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Time = 0.23 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.83 \[ \int \frac {x^3}{1-2 x^2+3 x^4} \, dx=\frac {\ln \left (x^4-\frac {2\,x^2}{3}+\frac {1}{3}\right )}{12}-\frac {\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}}{2}-\frac {3\,\sqrt {2}\,x^2}{2}\right )}{12} \]
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