Integrand size = 29, antiderivative size = 58 \[ \int \frac {x \left (4+x^2+3 x^4+5 x^6\right )}{\left (3+2 x^2+x^4\right )^2} \, dx=\frac {25 \left (1+x^2\right )}{8 \left (3+2 x^2+x^4\right )}-\frac {23 \arctan \left (\frac {1+x^2}{\sqrt {2}}\right )}{8 \sqrt {2}}+\frac {5}{4} \log \left (3+2 x^2+x^4\right ) \]
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Time = 0.04 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {1677, 1674, 648, 632, 210, 642} \[ \int \frac {x \left (4+x^2+3 x^4+5 x^6\right )}{\left (3+2 x^2+x^4\right )^2} \, dx=-\frac {23 \arctan \left (\frac {x^2+1}{\sqrt {2}}\right )}{8 \sqrt {2}}+\frac {25 \left (x^2+1\right )}{8 \left (x^4+2 x^2+3\right )}+\frac {5}{4} \log \left (x^4+2 x^2+3\right ) \]
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Rule 210
Rule 632
Rule 642
Rule 648
Rule 1674
Rule 1677
Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {4+x+3 x^2+5 x^3}{\left (3+2 x+x^2\right )^2} \, dx,x,x^2\right ) \\ & = \frac {25 \left (1+x^2\right )}{8 \left (3+2 x^2+x^4\right )}+\frac {1}{16} \text {Subst}\left (\int \frac {-6+40 x}{3+2 x+x^2} \, dx,x,x^2\right ) \\ & = \frac {25 \left (1+x^2\right )}{8 \left (3+2 x^2+x^4\right )}+\frac {5}{4} \text {Subst}\left (\int \frac {2+2 x}{3+2 x+x^2} \, dx,x,x^2\right )-\frac {23}{8} \text {Subst}\left (\int \frac {1}{3+2 x+x^2} \, dx,x,x^2\right ) \\ & = \frac {25 \left (1+x^2\right )}{8 \left (3+2 x^2+x^4\right )}+\frac {5}{4} \log \left (3+2 x^2+x^4\right )+\frac {23}{4} \text {Subst}\left (\int \frac {1}{-8-x^2} \, dx,x,2 \left (1+x^2\right )\right ) \\ & = \frac {25 \left (1+x^2\right )}{8 \left (3+2 x^2+x^4\right )}-\frac {23 \tan ^{-1}\left (\frac {1+x^2}{\sqrt {2}}\right )}{8 \sqrt {2}}+\frac {5}{4} \log \left (3+2 x^2+x^4\right ) \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.00 \[ \int \frac {x \left (4+x^2+3 x^4+5 x^6\right )}{\left (3+2 x^2+x^4\right )^2} \, dx=\frac {25 \left (1+x^2\right )}{8 \left (3+2 x^2+x^4\right )}-\frac {23 \arctan \left (\frac {1+x^2}{\sqrt {2}}\right )}{8 \sqrt {2}}+\frac {5}{4} \log \left (3+2 x^2+x^4\right ) \]
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Time = 0.05 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.88
method | result | size |
risch | \(\frac {\frac {25 x^{2}}{8}+\frac {25}{8}}{x^{4}+2 x^{2}+3}+\frac {5 \ln \left (x^{4}+2 x^{2}+3\right )}{4}-\frac {23 \arctan \left (\frac {\left (x^{2}+1\right ) \sqrt {2}}{2}\right ) \sqrt {2}}{16}\) | \(51\) |
default | \(\frac {\frac {25 x^{2}}{4}+\frac {25}{4}}{2 x^{4}+4 x^{2}+6}+\frac {5 \ln \left (x^{4}+2 x^{2}+3\right )}{4}-\frac {23 \sqrt {2}\, \arctan \left (\frac {\left (2 x^{2}+2\right ) \sqrt {2}}{4}\right )}{16}\) | \(54\) |
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Time = 0.25 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.21 \[ \int \frac {x \left (4+x^2+3 x^4+5 x^6\right )}{\left (3+2 x^2+x^4\right )^2} \, dx=-\frac {23 \, \sqrt {2} {\left (x^{4} + 2 \, x^{2} + 3\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (x^{2} + 1\right )}\right ) - 50 \, x^{2} - 20 \, {\left (x^{4} + 2 \, x^{2} + 3\right )} \log \left (x^{4} + 2 \, x^{2} + 3\right ) - 50}{16 \, {\left (x^{4} + 2 \, x^{2} + 3\right )}} \]
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Time = 0.08 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.03 \[ \int \frac {x \left (4+x^2+3 x^4+5 x^6\right )}{\left (3+2 x^2+x^4\right )^2} \, dx=\frac {25 x^{2} + 25}{8 x^{4} + 16 x^{2} + 24} + \frac {5 \log {\left (x^{4} + 2 x^{2} + 3 \right )}}{4} - \frac {23 \sqrt {2} \operatorname {atan}{\left (\frac {\sqrt {2} x^{2}}{2} + \frac {\sqrt {2}}{2} \right )}}{16} \]
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Time = 0.27 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.84 \[ \int \frac {x \left (4+x^2+3 x^4+5 x^6\right )}{\left (3+2 x^2+x^4\right )^2} \, dx=-\frac {23}{16} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (x^{2} + 1\right )}\right ) + \frac {25 \, {\left (x^{2} + 1\right )}}{8 \, {\left (x^{4} + 2 \, x^{2} + 3\right )}} + \frac {5}{4} \, \log \left (x^{4} + 2 \, x^{2} + 3\right ) \]
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Time = 0.46 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.84 \[ \int \frac {x \left (4+x^2+3 x^4+5 x^6\right )}{\left (3+2 x^2+x^4\right )^2} \, dx=-\frac {23}{16} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (x^{2} + 1\right )}\right ) + \frac {25 \, {\left (x^{2} + 1\right )}}{8 \, {\left (x^{4} + 2 \, x^{2} + 3\right )}} + \frac {5}{4} \, \log \left (x^{4} + 2 \, x^{2} + 3\right ) \]
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Time = 0.03 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.19 \[ \int \frac {x \left (4+x^2+3 x^4+5 x^6\right )}{\left (3+2 x^2+x^4\right )^2} \, dx=\frac {5\,\ln \left (x^4+2\,x^2+3\right )}{4}+\frac {25\,x^2}{8\,\left (x^4+2\,x^2+3\right )}+\frac {25}{8\,\left (x^4+2\,x^2+3\right )}-\frac {23\,\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,x^2}{2}+\frac {\sqrt {2}}{2}\right )}{16} \]
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