Integrand size = 54, antiderivative size = 30 \[ \int \frac {-1+2 x^2+2 x^3-2 e^{-4+e x} x^3+\left (1-x+e^{-4+e x} x\right ) \log (x)}{-2 x^3+x \log (x)} \, dx=-x+\log \left (\frac {e^{-5+e^{-5+e x}} x}{2 x^2-\log (x)}\right ) \]
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Time = 0.60 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.87, number of steps used = 9, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.093, Rules used = {2641, 6874, 2225, 45, 6816} \[ \int \frac {-1+2 x^2+2 x^3-2 e^{-4+e x} x^3+\left (1-x+e^{-4+e x} x\right ) \log (x)}{-2 x^3+x \log (x)} \, dx=-\log \left (2 x^2-\log (x)\right )-x+e^{e x-5}+\log (x) \]
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Rule 45
Rule 2225
Rule 2641
Rule 6816
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \int \frac {-1+2 x^2+2 x^3-2 e^{-4+e x} x^3+\left (1-x+e^{-4+e x} x\right ) \log (x)}{x \left (-2 x^2+\log (x)\right )} \, dx \\ & = \int \left (e^{-4+e x}+\frac {1-2 x^2-2 x^3-\log (x)+x \log (x)}{x \left (2 x^2-\log (x)\right )}\right ) \, dx \\ & = \int e^{-4+e x} \, dx+\int \frac {1-2 x^2-2 x^3-\log (x)+x \log (x)}{x \left (2 x^2-\log (x)\right )} \, dx \\ & = e^{-5+e x}+\int \left (\frac {1-x}{x}+\frac {1-4 x^2}{x \left (2 x^2-\log (x)\right )}\right ) \, dx \\ & = e^{-5+e x}+\int \frac {1-x}{x} \, dx+\int \frac {1-4 x^2}{x \left (2 x^2-\log (x)\right )} \, dx \\ & = e^{-5+e x}-\log \left (2 x^2-\log (x)\right )+\int \left (-1+\frac {1}{x}\right ) \, dx \\ & = e^{-5+e x}-x+\log (x)-\log \left (2 x^2-\log (x)\right ) \\ \end{align*}
Time = 1.08 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.87 \[ \int \frac {-1+2 x^2+2 x^3-2 e^{-4+e x} x^3+\left (1-x+e^{-4+e x} x\right ) \log (x)}{-2 x^3+x \log (x)} \, dx=e^{-5+e x}-x+\log (x)-\log \left (2 x^2-\log (x)\right ) \]
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Time = 0.56 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.83
method | result | size |
default | \(\ln \left (x \right )-x -\ln \left (-2 x^{2}+\ln \left (x \right )\right )+{\mathrm e}^{x \,{\mathrm e}-5}\) | \(25\) |
risch | \(\ln \left (x \right )-x -\ln \left (-2 x^{2}+\ln \left (x \right )\right )+{\mathrm e}^{x \,{\mathrm e}-5}\) | \(25\) |
parallelrisch | \(-\ln \left (x^{2}-\frac {\ln \left (x \right )}{2}\right )-x +\ln \left (x \right )+{\mathrm e}^{x \,{\mathrm e}-5}\) | \(25\) |
parts | \(\ln \left (x \right )-x -\ln \left (-2 x^{2}+\ln \left (x \right )\right )+{\mathrm e}^{x \,{\mathrm e}-5}\) | \(25\) |
norman | \(\ln \left (x \right )-x +{\mathrm e}^{x \,{\mathrm e}-5}-\ln \left (2 x^{2}-\ln \left (x \right )\right )\) | \(27\) |
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Time = 0.28 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.20 \[ \int \frac {-1+2 x^2+2 x^3-2 e^{-4+e x} x^3+\left (1-x+e^{-4+e x} x\right ) \log (x)}{-2 x^3+x \log (x)} \, dx=-{\left (x e + e \log \left (-2 \, x^{2} + \log \left (x\right )\right ) - e \log \left (x\right ) - e^{\left (x e - 4\right )}\right )} e^{\left (-1\right )} \]
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Time = 0.14 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.73 \[ \int \frac {-1+2 x^2+2 x^3-2 e^{-4+e x} x^3+\left (1-x+e^{-4+e x} x\right ) \log (x)}{-2 x^3+x \log (x)} \, dx=- x + e^{e x - 5} + \log {\left (x \right )} - \log {\left (- 2 x^{2} + \log {\left (x \right )} \right )} \]
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Time = 0.25 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00 \[ \int \frac {-1+2 x^2+2 x^3-2 e^{-4+e x} x^3+\left (1-x+e^{-4+e x} x\right ) \log (x)}{-2 x^3+x \log (x)} \, dx=-{\left (x e^{5} - e^{\left (x e\right )}\right )} e^{\left (-5\right )} - \log \left (-2 \, x^{2} + \log \left (x\right )\right ) + \log \left (x\right ) \]
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Time = 0.30 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.20 \[ \int \frac {-1+2 x^2+2 x^3-2 e^{-4+e x} x^3+\left (1-x+e^{-4+e x} x\right ) \log (x)}{-2 x^3+x \log (x)} \, dx=-{\left (x e^{5} + e^{5} \log \left (2 \, x^{2} - \log \left (x\right )\right ) - e^{5} \log \left (x\right ) - e^{\left (x e\right )}\right )} e^{\left (-5\right )} \]
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Time = 13.11 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \[ \int \frac {-1+2 x^2+2 x^3-2 e^{-4+e x} x^3+\left (1-x+e^{-4+e x} x\right ) \log (x)}{-2 x^3+x \log (x)} \, dx={\mathrm {e}}^{x\,\mathrm {e}-5}-x+\ln \left (x\right )-\ln \left (\ln \left (x\right )-2\,x^2\right ) \]
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