Integrand size = 32, antiderivative size = 21 \[ \int \frac {-1-2 x^2+e^x \left (-x+2 x^3\right )}{-x+2 x^3} \, dx=e^3+e^x-\log \left (\frac {1}{2 x}-x\right ) \]
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Time = 0.12 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.76, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.156, Rules used = {1607, 6820, 2225, 457, 78} \[ \int \frac {-1-2 x^2+e^x \left (-x+2 x^3\right )}{-x+2 x^3} \, dx=-\log \left (1-2 x^2\right )+e^x+\log (x) \]
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Rule 78
Rule 457
Rule 1607
Rule 2225
Rule 6820
Rubi steps \begin{align*} \text {integral}& = \int \frac {-1-2 x^2+e^x \left (-x+2 x^3\right )}{x \left (-1+2 x^2\right )} \, dx \\ & = \int \left (e^x+\frac {1+2 x^2}{x \left (1-2 x^2\right )}\right ) \, dx \\ & = \int e^x \, dx+\int \frac {1+2 x^2}{x \left (1-2 x^2\right )} \, dx \\ & = e^x+\frac {1}{2} \text {Subst}\left (\int \frac {1+2 x}{(1-2 x) x} \, dx,x,x^2\right ) \\ & = e^x+\frac {1}{2} \text {Subst}\left (\int \left (\frac {1}{x}-\frac {4}{-1+2 x}\right ) \, dx,x,x^2\right ) \\ & = e^x+\log (x)-\log \left (1-2 x^2\right ) \\ \end{align*}
Time = 0.24 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.76 \[ \int \frac {-1-2 x^2+e^x \left (-x+2 x^3\right )}{-x+2 x^3} \, dx=e^x+\log (x)-\log \left (1-2 x^2\right ) \]
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Time = 0.18 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.67
method | result | size |
parallelrisch | \({\mathrm e}^{x}+\ln \left (x \right )-\ln \left (x^{2}-\frac {1}{2}\right )\) | \(14\) |
default | \(-\ln \left (2 x^{2}-1\right )+\ln \left (x \right )+{\mathrm e}^{x}\) | \(16\) |
norman | \(-\ln \left (2 x^{2}-1\right )+\ln \left (x \right )+{\mathrm e}^{x}\) | \(16\) |
risch | \(-\ln \left (2 x^{2}-1\right )+\ln \left (x \right )+{\mathrm e}^{x}\) | \(16\) |
parts | \(-\ln \left (2 x^{2}-1\right )+\ln \left (x \right )+{\mathrm e}^{x}\) | \(16\) |
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Time = 0.24 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.71 \[ \int \frac {-1-2 x^2+e^x \left (-x+2 x^3\right )}{-x+2 x^3} \, dx=e^{x} - \log \left (2 \, x^{2} - 1\right ) + \log \left (x\right ) \]
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Time = 0.07 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.67 \[ \int \frac {-1-2 x^2+e^x \left (-x+2 x^3\right )}{-x+2 x^3} \, dx=e^{x} + \log {\left (x \right )} - \log {\left (2 x^{2} - 1 \right )} \]
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Time = 0.21 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.71 \[ \int \frac {-1-2 x^2+e^x \left (-x+2 x^3\right )}{-x+2 x^3} \, dx=e^{x} - \log \left (2 \, x^{2} - 1\right ) + \log \left (x\right ) \]
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Time = 0.28 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.71 \[ \int \frac {-1-2 x^2+e^x \left (-x+2 x^3\right )}{-x+2 x^3} \, dx=e^{x} - \log \left (2 \, x^{2} - 1\right ) + \log \left (x\right ) \]
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Time = 0.15 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.62 \[ \int \frac {-1-2 x^2+e^x \left (-x+2 x^3\right )}{-x+2 x^3} \, dx={\mathrm {e}}^x-\ln \left (x^2-\frac {1}{2}\right )+\ln \left (x\right ) \]
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