Integrand size = 25, antiderivative size = 25 \[ \int \frac {-1+2 x+x^2}{-2 x+3 x^2+2 x^3} \, dx=\frac {1}{10} \log (1-2 x)+\frac {\log (x)}{2}-\frac {1}{10} \log (2+x) \]
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Time = 0.03 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {1608, 1642} \[ \int \frac {-1+2 x+x^2}{-2 x+3 x^2+2 x^3} \, dx=\frac {1}{10} \log (1-2 x)+\frac {\log (x)}{2}-\frac {1}{10} \log (x+2) \]
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Rule 1608
Rule 1642
Rubi steps \begin{align*} \text {integral}& = \int \frac {-1+2 x+x^2}{x \left (-2+3 x+2 x^2\right )} \, dx \\ & = \int \left (\frac {1}{2 x}-\frac {1}{10 (2+x)}+\frac {1}{5 (-1+2 x)}\right ) \, dx \\ & = \frac {1}{10} \log (1-2 x)+\frac {\log (x)}{2}-\frac {1}{10} \log (2+x) \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {-1+2 x+x^2}{-2 x+3 x^2+2 x^3} \, dx=\frac {1}{10} \log (1-2 x)+\frac {\log (x)}{2}-\frac {1}{10} \log (2+x) \]
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Time = 0.04 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.72
method | result | size |
parallelrisch | \(\frac {\ln \left (x \right )}{2}-\frac {\ln \left (2+x \right )}{10}+\frac {\ln \left (x -\frac {1}{2}\right )}{10}\) | \(18\) |
default | \(-\frac {\ln \left (2+x \right )}{10}+\frac {\ln \left (x \right )}{2}+\frac {\ln \left (2 x -1\right )}{10}\) | \(20\) |
norman | \(-\frac {\ln \left (2+x \right )}{10}+\frac {\ln \left (x \right )}{2}+\frac {\ln \left (2 x -1\right )}{10}\) | \(20\) |
risch | \(-\frac {\ln \left (2+x \right )}{10}+\frac {\ln \left (x \right )}{2}+\frac {\ln \left (2 x -1\right )}{10}\) | \(20\) |
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Time = 0.24 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.76 \[ \int \frac {-1+2 x+x^2}{-2 x+3 x^2+2 x^3} \, dx=\frac {1}{10} \, \log \left (2 \, x - 1\right ) - \frac {1}{10} \, \log \left (x + 2\right ) + \frac {1}{2} \, \log \left (x\right ) \]
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Time = 0.08 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.76 \[ \int \frac {-1+2 x+x^2}{-2 x+3 x^2+2 x^3} \, dx=\frac {\log {\left (x \right )}}{2} + \frac {\log {\left (x - \frac {1}{2} \right )}}{10} - \frac {\log {\left (x + 2 \right )}}{10} \]
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Time = 0.21 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.76 \[ \int \frac {-1+2 x+x^2}{-2 x+3 x^2+2 x^3} \, dx=\frac {1}{10} \, \log \left (2 \, x - 1\right ) - \frac {1}{10} \, \log \left (x + 2\right ) + \frac {1}{2} \, \log \left (x\right ) \]
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Time = 0.29 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {-1+2 x+x^2}{-2 x+3 x^2+2 x^3} \, dx=\frac {1}{10} \, \log \left ({\left | 2 \, x - 1 \right |}\right ) - \frac {1}{10} \, \log \left ({\left | x + 2 \right |}\right ) + \frac {1}{2} \, \log \left ({\left | x \right |}\right ) \]
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Time = 0.25 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.76 \[ \int \frac {-1+2 x+x^2}{-2 x+3 x^2+2 x^3} \, dx=\frac {\mathrm {atanh}\left (\frac {24}{145\,\left (\frac {29\,x}{100}-\frac {11}{50}\right )}+\frac {35}{29}\right )}{5}+\frac {\ln \left (x\right )}{2} \]
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