Integrand size = 63, antiderivative size = 28 \[ \int \frac {1-4 x^3-x^4+e^{e^5} \left (2 x+x^2\right )+\left (-2 x-x^2\right ) \log (x)}{-x-e^{e^5} x^2+x^4+x^2 \log (x)} \, dx=2-x-\log \left (x-x^2 \left (-e^{e^5}+x^2+\log (x)\right )\right ) \]
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\[ \int \frac {1-4 x^3-x^4+e^{e^5} \left (2 x+x^2\right )+\left (-2 x-x^2\right ) \log (x)}{-x-e^{e^5} x^2+x^4+x^2 \log (x)} \, dx=\int \frac {1-4 x^3-x^4+e^{e^5} \left (2 x+x^2\right )+\left (-2 x-x^2\right ) \log (x)}{-x-e^{e^5} x^2+x^4+x^2 \log (x)} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {-2-x}{x}-\frac {1+x+2 x^3}{x \left (-1-e^{e^5} x+x^3+x \log (x)\right )}\right ) \, dx \\ & = \int \frac {-2-x}{x} \, dx-\int \frac {1+x+2 x^3}{x \left (-1-e^{e^5} x+x^3+x \log (x)\right )} \, dx \\ & = \int \left (-1-\frac {2}{x}\right ) \, dx-\int \left (-\frac {1}{1+e^{e^5} x-x^3-x \log (x)}+\frac {1}{x \left (-1-e^{e^5} x+x^3+x \log (x)\right )}+\frac {2 x^2}{-1-e^{e^5} x+x^3+x \log (x)}\right ) \, dx \\ & = -x-2 \log (x)-2 \int \frac {x^2}{-1-e^{e^5} x+x^3+x \log (x)} \, dx+\int \frac {1}{1+e^{e^5} x-x^3-x \log (x)} \, dx-\int \frac {1}{x \left (-1-e^{e^5} x+x^3+x \log (x)\right )} \, dx \\ \end{align*}
Time = 0.33 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.07 \[ \int \frac {1-4 x^3-x^4+e^{e^5} \left (2 x+x^2\right )+\left (-2 x-x^2\right ) \log (x)}{-x-e^{e^5} x^2+x^4+x^2 \log (x)} \, dx=-x-\log (x)-\log \left (1+e^{e^5} x-x^3-x \log (x)\right ) \]
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Time = 0.19 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.96
method | result | size |
parallelrisch | \(-\ln \left (x^{3}+x \ln \left (x \right )-x \,{\mathrm e}^{{\mathrm e}^{5}}-1\right )-x -\ln \left (x \right )\) | \(27\) |
default | \(-\ln \left (x \right )-x -\ln \left (-x^{3}+x \,{\mathrm e}^{{\mathrm e}^{5}}-x \ln \left (x \right )+1\right )\) | \(29\) |
norman | \(-\ln \left (x \right )-x -\ln \left (-x^{3}+x \,{\mathrm e}^{{\mathrm e}^{5}}-x \ln \left (x \right )+1\right )\) | \(29\) |
risch | \(-x -2 \ln \left (x \right )-\ln \left (\ln \left (x \right )-\frac {-x^{3}+x \,{\mathrm e}^{{\mathrm e}^{5}}+1}{x}\right )\) | \(32\) |
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Time = 0.27 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.07 \[ \int \frac {1-4 x^3-x^4+e^{e^5} \left (2 x+x^2\right )+\left (-2 x-x^2\right ) \log (x)}{-x-e^{e^5} x^2+x^4+x^2 \log (x)} \, dx=-x - 2 \, \log \left (x\right ) - \log \left (\frac {x^{3} - x e^{\left (e^{5}\right )} + x \log \left (x\right ) - 1}{x}\right ) \]
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Time = 0.10 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93 \[ \int \frac {1-4 x^3-x^4+e^{e^5} \left (2 x+x^2\right )+\left (-2 x-x^2\right ) \log (x)}{-x-e^{e^5} x^2+x^4+x^2 \log (x)} \, dx=- x - 2 \log {\left (x \right )} - \log {\left (\log {\left (x \right )} + \frac {x^{3} - x e^{e^{5}} - 1}{x} \right )} \]
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Time = 0.21 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.07 \[ \int \frac {1-4 x^3-x^4+e^{e^5} \left (2 x+x^2\right )+\left (-2 x-x^2\right ) \log (x)}{-x-e^{e^5} x^2+x^4+x^2 \log (x)} \, dx=-x - 2 \, \log \left (x\right ) - \log \left (\frac {x^{3} - x e^{\left (e^{5}\right )} + x \log \left (x\right ) - 1}{x}\right ) \]
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Time = 0.28 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93 \[ \int \frac {1-4 x^3-x^4+e^{e^5} \left (2 x+x^2\right )+\left (-2 x-x^2\right ) \log (x)}{-x-e^{e^5} x^2+x^4+x^2 \log (x)} \, dx=-x - \log \left (x^{3} - x e^{\left (e^{5}\right )} + x \log \left (x\right ) - 1\right ) - \log \left (x\right ) \]
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Time = 8.14 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.21 \[ \int \frac {1-4 x^3-x^4+e^{e^5} \left (2 x+x^2\right )+\left (-2 x-x^2\right ) \log (x)}{-x-e^{e^5} x^2+x^4+x^2 \log (x)} \, dx=-\ln \left (x\,\ln \left (x\right )-x\,{\mathrm {e}}^{{\mathrm {e}}^5}+x^3-1\right )-\frac {x^2\,\ln \left (x\right )+x^3}{x^2} \]
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