Integrand size = 9, antiderivative size = 41 \[ \int \frac {x}{1+x^3} \, dx=-\frac {\arctan \left (\frac {1-2 x}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {1}{3} \log (1+x)+\frac {1}{6} \log \left (1-x+x^2\right ) \]
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Time = 0.02 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {298, 31, 648, 632, 210, 642} \[ \int \frac {x}{1+x^3} \, dx=-\frac {\arctan \left (\frac {1-2 x}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {1}{6} \log \left (x^2-x+1\right )-\frac {1}{3} \log (x+1) \]
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Rule 31
Rule 210
Rule 298
Rule 632
Rule 642
Rule 648
Rubi steps \begin{align*} \text {integral}& = -\left (\frac {1}{3} \int \frac {1}{1+x} \, dx\right )+\frac {1}{3} \int \frac {1+x}{1-x+x^2} \, dx \\ & = -\frac {1}{3} \log (1+x)+\frac {1}{6} \int \frac {-1+2 x}{1-x+x^2} \, dx+\frac {1}{2} \int \frac {1}{1-x+x^2} \, dx \\ & = -\frac {1}{3} \log (1+x)+\frac {1}{6} \log \left (1-x+x^2\right )-\text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,-1+2 x\right ) \\ & = \frac {\arctan \left (\frac {-1+2 x}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {1}{3} \log (1+x)+\frac {1}{6} \log \left (1-x+x^2\right ) \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.98 \[ \int \frac {x}{1+x^3} \, dx=\frac {\arctan \left (\frac {-1+2 x}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {1}{3} \log (1+x)+\frac {1}{6} \log \left (1-x+x^2\right ) \]
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Time = 0.07 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.80
method | result | size |
risch | \(-\frac {\ln \left (1+x \right )}{3}+\frac {\ln \left (x^{2}-x +1\right )}{6}+\frac {\sqrt {3}\, \arctan \left (\frac {2 \left (x -\frac {1}{2}\right ) \sqrt {3}}{3}\right )}{3}\) | \(33\) |
default | \(\frac {\ln \left (x^{2}-x +1\right )}{6}+\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 x -1\right ) \sqrt {3}}{3}\right )}{3}-\frac {\ln \left (1+x \right )}{3}\) | \(35\) |
meijerg | \(-\frac {x^{2} \ln \left (1+\left (x^{3}\right )^{\frac {1}{3}}\right )}{3 \left (x^{3}\right )^{\frac {2}{3}}}+\frac {x^{2} \ln \left (1-\left (x^{3}\right )^{\frac {1}{3}}+\left (x^{3}\right )^{\frac {2}{3}}\right )}{6 \left (x^{3}\right )^{\frac {2}{3}}}+\frac {x^{2} \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (x^{3}\right )^{\frac {1}{3}}}{2-\left (x^{3}\right )^{\frac {1}{3}}}\right )}{3 \left (x^{3}\right )^{\frac {2}{3}}}\) | \(80\) |
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Time = 0.27 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.83 \[ \int \frac {x}{1+x^3} \, dx=\frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) + \frac {1}{6} \, \log \left (x^{2} - x + 1\right ) - \frac {1}{3} \, \log \left (x + 1\right ) \]
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Time = 0.06 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.00 \[ \int \frac {x}{1+x^3} \, dx=- \frac {\log {\left (x + 1 \right )}}{3} + \frac {\log {\left (x^{2} - x + 1 \right )}}{6} + \frac {\sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} x}{3} - \frac {\sqrt {3}}{3} \right )}}{3} \]
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Time = 0.27 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.83 \[ \int \frac {x}{1+x^3} \, dx=\frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) + \frac {1}{6} \, \log \left (x^{2} - x + 1\right ) - \frac {1}{3} \, \log \left (x + 1\right ) \]
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Time = 0.28 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.85 \[ \int \frac {x}{1+x^3} \, dx=\frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) + \frac {1}{6} \, \log \left (x^{2} - x + 1\right ) - \frac {1}{3} \, \log \left ({\left | x + 1 \right |}\right ) \]
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Time = 0.20 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.12 \[ \int \frac {x}{1+x^3} \, dx=-\frac {\ln \left (x+1\right )}{3}-\ln \left (x-\frac {1}{2}-\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (-\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )+\ln \left (x-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right ) \]
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