Integrand size = 16, antiderivative size = 42 \[ \int \frac {1-x}{x \left (1+x^3\right )} \, dx=\frac {\arctan \left (\frac {1-2 x}{\sqrt {3}}\right )}{\sqrt {3}}+\log (x)-\frac {2}{3} \log (1+x)-\frac {1}{6} \log \left (1-x+x^2\right ) \]
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Time = 0.03 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {1848, 648, 632, 210, 642} \[ \int \frac {1-x}{x \left (1+x^3\right )} \, dx=\frac {\arctan \left (\frac {1-2 x}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {1}{6} \log \left (x^2-x+1\right )+\log (x)-\frac {2}{3} \log (x+1) \]
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Rule 210
Rule 632
Rule 642
Rule 648
Rule 1848
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {1}{x}-\frac {2}{3 (1+x)}+\frac {-1-x}{3 \left (1-x+x^2\right )}\right ) \, dx \\ & = \log (x)-\frac {2}{3} \log (1+x)+\frac {1}{3} \int \frac {-1-x}{1-x+x^2} \, dx \\ & = \log (x)-\frac {2}{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 \\ & = \log (x)-\frac {2}{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 {\tan ^{-1}\left (\frac {-1+2 x}{\sqrt {3}}\right )}{\sqrt {3}}+\log (x)-\frac {2}{3} \log (1+x)-\frac {1}{6} \log \left (1-x+x^2\right ) \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.26 \[ \int \frac {1-x}{x \left (1+x^3\right )} \, dx=-\frac {\arctan \left (\frac {-1+2 x}{\sqrt {3}}\right )}{\sqrt {3}}+\log (x)-\frac {1}{3} \log (1+x)+\frac {1}{6} \log \left (1-x+x^2\right )-\frac {1}{3} \log \left (1+x^3\right ) \]
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Time = 1.53 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.83
method | result | size |
risch | \(-\frac {2 \ln \left (1+x \right )}{3}+\ln \left (x \right )-\frac {\ln \left (x^{2}-x +1\right )}{6}-\frac {\sqrt {3}\, \arctan \left (\frac {2 \left (x -\frac {1}{2}\right ) \sqrt {3}}{3}\right )}{3}\) | \(35\) |
default | \(\ln \left (x \right )-\frac {2 \ln \left (1+x \right )}{3}-\frac {\ln \left (x^{2}-x +1\right )}{6}-\frac {\sqrt {3}\, \arctan \left (\frac {\left (-1+2 x \right ) \sqrt {3}}{3}\right )}{3}\) | \(37\) |
meijerg | \(\ln \left (x \right )-\frac {\ln \left (x^{3}+1\right )}{3}-\frac {x \ln \left (1+\left (x^{3}\right )^{\frac {1}{3}}\right )}{3 \left (x^{3}\right )^{\frac {1}{3}}}+\frac {x \ln \left (1-\left (x^{3}\right )^{\frac {1}{3}}+\left (x^{3}\right )^{\frac {2}{3}}\right )}{6 \left (x^{3}\right )^{\frac {1}{3}}}-\frac {x \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (x^{3}\right )^{\frac {1}{3}}}{2-\left (x^{3}\right )^{\frac {1}{3}}}\right )}{3 \left (x^{3}\right )^{\frac {1}{3}}}\) | \(84\) |
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Time = 0.39 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.86 \[ \int \frac {1-x}{x \left (1+x^3\right )} \, dx=-\frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) - \frac {1}{6} \, \log \left (x^{2} - x + 1\right ) - \frac {2}{3} \, \log \left (x + 1\right ) + \log \left (x\right ) \]
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Time = 0.09 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.10 \[ \int \frac {1-x}{x \left (1+x^3\right )} \, dx=\log {\left (x \right )} - \frac {2 \log {\left (x + 1 \right )}}{3} - \frac {\log {\left (x^{2} - x + 1 \right )}}{6} - \frac {\sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} x}{3} - \frac {\sqrt {3}}{3} \right )}}{3} \]
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Time = 0.30 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.86 \[ \int \frac {1-x}{x \left (1+x^3\right )} \, dx=-\frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) - \frac {1}{6} \, \log \left (x^{2} - x + 1\right ) - \frac {2}{3} \, \log \left (x + 1\right ) + \log \left (x\right ) \]
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Time = 0.26 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.90 \[ \int \frac {1-x}{x \left (1+x^3\right )} \, dx=-\frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) - \frac {1}{6} \, \log \left (x^{2} - x + 1\right ) - \frac {2}{3} \, \log \left ({\left | x + 1 \right |}\right ) + \log \left ({\left | x \right |}\right ) \]
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Time = 8.96 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.14 \[ \int \frac {1-x}{x \left (1+x^3\right )} \, dx=\ln \left (x\right )-\frac {2\,\ln \left (x+1\right )}{3}+\ln \left (x-\frac {1}{2}-\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (-\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )-\ln \left (x-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right ) \]
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