Integrand size = 30, antiderivative size = 41 \[ \int \frac {x}{\left (1+x^2\right ) \left (2+x^2\right ) \left (3+x^2\right ) \left (4+x^2\right )} \, dx=\frac {1}{12} \log \left (1+x^2\right )-\frac {1}{4} \log \left (2+x^2\right )+\frac {1}{4} \log \left (3+x^2\right )-\frac {1}{12} \log \left (4+x^2\right ) \]
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Time = 0.26 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {6826, 186} \[ \int \frac {x}{\left (1+x^2\right ) \left (2+x^2\right ) \left (3+x^2\right ) \left (4+x^2\right )} \, dx=\frac {1}{12} \log \left (x^2+1\right )-\frac {1}{4} \log \left (x^2+2\right )+\frac {1}{4} \log \left (x^2+3\right )-\frac {1}{12} \log \left (x^2+4\right ) \]
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Rule 186
Rule 6826
Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {1}{(1+x) (2+x) (3+x) (4+x)} \, dx,x,x^2\right ) \\ & = \frac {1}{2} \text {Subst}\left (\int \left (\frac {1}{6 (1+x)}-\frac {1}{2 (2+x)}+\frac {1}{2 (3+x)}-\frac {1}{6 (4+x)}\right ) \, dx,x,x^2\right ) \\ & = \frac {1}{12} \log \left (1+x^2\right )-\frac {1}{4} \log \left (2+x^2\right )+\frac {1}{4} \log \left (3+x^2\right )-\frac {1}{12} \log \left (4+x^2\right ) \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.00 \[ \int \frac {x}{\left (1+x^2\right ) \left (2+x^2\right ) \left (3+x^2\right ) \left (4+x^2\right )} \, dx=\frac {1}{12} \log \left (1+x^2\right )-\frac {1}{4} \log \left (2+x^2\right )+\frac {1}{4} \log \left (3+x^2\right )-\frac {1}{12} \log \left (4+x^2\right ) \]
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Time = 0.24 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.83
method | result | size |
default | \(\frac {\ln \left (x^{2}+1\right )}{12}-\frac {\ln \left (x^{2}+2\right )}{4}+\frac {\ln \left (x^{2}+3\right )}{4}-\frac {\ln \left (x^{2}+4\right )}{12}\) | \(34\) |
norman | \(\frac {\ln \left (x^{2}+1\right )}{12}-\frac {\ln \left (x^{2}+2\right )}{4}+\frac {\ln \left (x^{2}+3\right )}{4}-\frac {\ln \left (x^{2}+4\right )}{12}\) | \(34\) |
risch | \(\frac {\ln \left (x^{2}+1\right )}{12}-\frac {\ln \left (x^{2}+2\right )}{4}+\frac {\ln \left (x^{2}+3\right )}{4}-\frac {\ln \left (x^{2}+4\right )}{12}\) | \(34\) |
parallelrisch | \(\frac {\ln \left (x^{2}+1\right )}{12}-\frac {\ln \left (x^{2}+2\right )}{4}+\frac {\ln \left (x^{2}+3\right )}{4}-\frac {\ln \left (x^{2}+4\right )}{12}\) | \(34\) |
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Time = 0.25 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.80 \[ \int \frac {x}{\left (1+x^2\right ) \left (2+x^2\right ) \left (3+x^2\right ) \left (4+x^2\right )} \, dx=-\frac {1}{12} \, \log \left (x^{2} + 4\right ) + \frac {1}{4} \, \log \left (x^{2} + 3\right ) - \frac {1}{4} \, \log \left (x^{2} + 2\right ) + \frac {1}{12} \, \log \left (x^{2} + 1\right ) \]
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Time = 0.08 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.78 \[ \int \frac {x}{\left (1+x^2\right ) \left (2+x^2\right ) \left (3+x^2\right ) \left (4+x^2\right )} \, dx=\frac {\log {\left (x^{2} + 1 \right )}}{12} - \frac {\log {\left (x^{2} + 2 \right )}}{4} + \frac {\log {\left (x^{2} + 3 \right )}}{4} - \frac {\log {\left (x^{2} + 4 \right )}}{12} \]
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Time = 0.20 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.80 \[ \int \frac {x}{\left (1+x^2\right ) \left (2+x^2\right ) \left (3+x^2\right ) \left (4+x^2\right )} \, dx=-\frac {1}{12} \, \log \left (x^{2} + 4\right ) + \frac {1}{4} \, \log \left (x^{2} + 3\right ) - \frac {1}{4} \, \log \left (x^{2} + 2\right ) + \frac {1}{12} \, \log \left (x^{2} + 1\right ) \]
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Time = 0.28 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.80 \[ \int \frac {x}{\left (1+x^2\right ) \left (2+x^2\right ) \left (3+x^2\right ) \left (4+x^2\right )} \, dx=-\frac {1}{12} \, \log \left (x^{2} + 4\right ) + \frac {1}{4} \, \log \left (x^{2} + 3\right ) - \frac {1}{4} \, \log \left (x^{2} + 2\right ) + \frac {1}{12} \, \log \left (x^{2} + 1\right ) \]
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Time = 0.06 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.80 \[ \int \frac {x}{\left (1+x^2\right ) \left (2+x^2\right ) \left (3+x^2\right ) \left (4+x^2\right )} \, dx=\frac {\mathrm {atanh}\left (\frac {3072}{5\,\left (1280\,x^2+3072\right )}-\frac {1}{5}\right )}{2}-\frac {\mathrm {atanh}\left (\frac {1024}{405\,\left (\frac {640\,x^2}{243}+\frac {1024}{243}\right )}-\frac {3}{5}\right )}{6} \]
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