Integrand size = 21, antiderivative size = 46 \[ \int \frac {x}{(1+x) (1+2 x)^2 \left (1+x^2\right )} \, dx=\frac {2}{5 (1+2 x)}+\frac {\arctan (x)}{50}-\frac {1}{2} \log (1+x)+\frac {16}{25} \log (1+2 x)-\frac {7}{100} \log \left (1+x^2\right ) \]
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Time = 0.19 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {6857, 649, 209, 266} \[ \int \frac {x}{(1+x) (1+2 x)^2 \left (1+x^2\right )} \, dx=\frac {\arctan (x)}{50}-\frac {7}{100} \log \left (x^2+1\right )+\frac {2}{5 (2 x+1)}-\frac {1}{2} \log (x+1)+\frac {16}{25} \log (2 x+1) \]
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Rule 209
Rule 266
Rule 649
Rule 6857
Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {1}{2 (1+x)}-\frac {4}{5 (1+2 x)^2}+\frac {32}{25 (1+2 x)}+\frac {1-7 x}{50 \left (1+x^2\right )}\right ) \, dx \\ & = \frac {2}{5 (1+2 x)}-\frac {1}{2} \log (1+x)+\frac {16}{25} \log (1+2 x)+\frac {1}{50} \int \frac {1-7 x}{1+x^2} \, dx \\ & = \frac {2}{5 (1+2 x)}-\frac {1}{2} \log (1+x)+\frac {16}{25} \log (1+2 x)+\frac {1}{50} \int \frac {1}{1+x^2} \, dx-\frac {7}{50} \int \frac {x}{1+x^2} \, dx \\ & = \frac {2}{5 (1+2 x)}+\frac {\arctan (x)}{50}-\frac {1}{2} \log (1+x)+\frac {16}{25} \log (1+2 x)-\frac {7}{100} \log \left (1+x^2\right ) \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.87 \[ \int \frac {x}{(1+x) (1+2 x)^2 \left (1+x^2\right )} \, dx=\frac {1}{100} \left (\frac {40}{1+2 x}+2 \arctan (x)-50 \log (1+x)+64 \log (1+2 x)-7 \log \left (1+x^2\right )\right ) \]
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Time = 0.28 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.76
method | result | size |
risch | \(\frac {1}{5 x +\frac {5}{2}}-\frac {\ln \left (1+x \right )}{2}-\frac {7 \ln \left (x^{2}+1\right )}{100}+\frac {\arctan \left (x \right )}{50}+\frac {16 \ln \left (1+2 x \right )}{25}\) | \(35\) |
default | \(\frac {2}{5 \left (1+2 x \right )}+\frac {\arctan \left (x \right )}{50}-\frac {\ln \left (1+x \right )}{2}+\frac {16 \ln \left (1+2 x \right )}{25}-\frac {7 \ln \left (x^{2}+1\right )}{100}\) | \(37\) |
parallelrisch | \(-\frac {2 i \ln \left (x -i\right ) x -2 i \ln \left (x +i\right ) x +i \ln \left (x -i\right )-i \ln \left (x +i\right )+100 \ln \left (1+x \right ) x -128 \ln \left (x +\frac {1}{2}\right ) x +14 \ln \left (x -i\right ) x +14 \ln \left (x +i\right ) x -40+50 \ln \left (1+x \right )-64 \ln \left (x +\frac {1}{2}\right )+7 \ln \left (x -i\right )+7 \ln \left (x +i\right )}{100 \left (1+2 x \right )}\) | \(102\) |
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Time = 0.24 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.24 \[ \int \frac {x}{(1+x) (1+2 x)^2 \left (1+x^2\right )} \, dx=\frac {2 \, {\left (2 \, x + 1\right )} \arctan \left (x\right ) - 7 \, {\left (2 \, x + 1\right )} \log \left (x^{2} + 1\right ) + 64 \, {\left (2 \, x + 1\right )} \log \left (2 \, x + 1\right ) - 50 \, {\left (2 \, x + 1\right )} \log \left (x + 1\right ) + 40}{100 \, {\left (2 \, x + 1\right )}} \]
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Time = 0.09 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.80 \[ \int \frac {x}{(1+x) (1+2 x)^2 \left (1+x^2\right )} \, dx=\frac {16 \log {\left (x + \frac {1}{2} \right )}}{25} - \frac {\log {\left (x + 1 \right )}}{2} - \frac {7 \log {\left (x^{2} + 1 \right )}}{100} + \frac {\operatorname {atan}{\left (x \right )}}{50} + \frac {2}{10 x + 5} \]
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Time = 0.33 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.78 \[ \int \frac {x}{(1+x) (1+2 x)^2 \left (1+x^2\right )} \, dx=\frac {2}{5 \, {\left (2 \, x + 1\right )}} + \frac {1}{50} \, \arctan \left (x\right ) - \frac {7}{100} \, \log \left (x^{2} + 1\right ) + \frac {16}{25} \, \log \left (2 \, x + 1\right ) - \frac {1}{2} \, \log \left (x + 1\right ) \]
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Time = 0.29 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.35 \[ \int \frac {x}{(1+x) (1+2 x)^2 \left (1+x^2\right )} \, dx=\frac {2}{5 \, {\left (2 \, x + 1\right )}} + \frac {1}{50} \, \arctan \left (-\frac {5}{2 \, {\left (2 \, x + 1\right )}} + \frac {1}{2}\right ) - \frac {7}{100} \, \log \left (-\frac {2}{2 \, x + 1} + \frac {5}{{\left (2 \, x + 1\right )}^{2}} + 1\right ) - \frac {1}{2} \, \log \left ({\left | -\frac {1}{2 \, x + 1} - 1 \right |}\right ) \]
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Time = 0.22 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.83 \[ \int \frac {x}{(1+x) (1+2 x)^2 \left (1+x^2\right )} \, dx=\frac {16\,\ln \left (x+\frac {1}{2}\right )}{25}-\frac {\ln \left (x+1\right )}{2}+\frac {1}{5\,\left (x+\frac {1}{2}\right )}+\ln \left (x-\mathrm {i}\right )\,\left (-\frac {7}{100}-\frac {1}{100}{}\mathrm {i}\right )+\ln \left (x+1{}\mathrm {i}\right )\,\left (-\frac {7}{100}+\frac {1}{100}{}\mathrm {i}\right ) \]
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