Integrand size = 58, antiderivative size = 21 \[ \int \frac {10+e^2 (2-2 x)-10 x+x^2-x^3-2 x^4+2 x^3 \log (x)}{-5 x-e^2 x-x^4+x^3 \log (x)} \, dx=-5+2 x+\log \left (\frac {5+e^2}{x^2}+x-\log (x)\right ) \]
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\[ \int \frac {10+e^2 (2-2 x)-10 x+x^2-x^3-2 x^4+2 x^3 \log (x)}{-5 x-e^2 x-x^4+x^3 \log (x)} \, dx=\int \frac {10+e^2 (2-2 x)-10 x+x^2-x^3-2 x^4+2 x^3 \log (x)}{-5 x-e^2 x-x^4+x^3 \log (x)} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {10+e^2 (2-2 x)-10 x+x^2-x^3-2 x^4+2 x^3 \log (x)}{\left (-5-e^2\right ) x-x^4+x^3 \log (x)} \, dx \\ & = \int \left (2+\frac {-2 \left (5+e^2\right )-x^2+x^3}{x \left (5 \left (1+\frac {e^2}{5}\right )+x^3-x^2 \log (x)\right )}\right ) \, dx \\ & = 2 x+\int \frac {-2 \left (5+e^2\right )-x^2+x^3}{x \left (5 \left (1+\frac {e^2}{5}\right )+x^3-x^2 \log (x)\right )} \, dx \\ & = 2 x+\int \left (\frac {2 \left (-5-e^2\right )}{x \left (5 \left (1+\frac {e^2}{5}\right )+x^3-x^2 \log (x)\right )}+\frac {x^2}{5 \left (1+\frac {e^2}{5}\right )+x^3-x^2 \log (x)}+\frac {x}{-5 \left (1+\frac {e^2}{5}\right )-x^3+x^2 \log (x)}\right ) \, dx \\ & = 2 x-\left (2 \left (5+e^2\right )\right ) \int \frac {1}{x \left (5 \left (1+\frac {e^2}{5}\right )+x^3-x^2 \log (x)\right )} \, dx+\int \frac {x^2}{5 \left (1+\frac {e^2}{5}\right )+x^3-x^2 \log (x)} \, dx+\int \frac {x}{-5 \left (1+\frac {e^2}{5}\right )-x^3+x^2 \log (x)} \, dx \\ \end{align*}
Time = 0.28 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.14 \[ \int \frac {10+e^2 (2-2 x)-10 x+x^2-x^3-2 x^4+2 x^3 \log (x)}{-5 x-e^2 x-x^4+x^3 \log (x)} \, dx=2 x-2 \log (x)+\log \left (5+e^2+x^3-x^2 \log (x)\right ) \]
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Time = 0.51 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00
method | result | size |
risch | \(2 x +\ln \left (\ln \left (x \right )-\frac {x^{3}+{\mathrm e}^{2}+5}{x^{2}}\right )\) | \(21\) |
default | \(-2 \ln \left (x \right )+2 x +\ln \left (-x^{2} \ln \left (x \right )+x^{3}+{\mathrm e}^{2}+5\right )\) | \(26\) |
norman | \(-2 \ln \left (x \right )+2 x +\ln \left (-x^{2} \ln \left (x \right )+x^{3}+{\mathrm e}^{2}+5\right )\) | \(26\) |
parallelrisch | \(-2 \ln \left (x \right )+2 x +\ln \left (-x^{2} \ln \left (x \right )+x^{3}+{\mathrm e}^{2}+5\right )\) | \(26\) |
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Time = 0.25 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.14 \[ \int \frac {10+e^2 (2-2 x)-10 x+x^2-x^3-2 x^4+2 x^3 \log (x)}{-5 x-e^2 x-x^4+x^3 \log (x)} \, dx=2 \, x + \log \left (-\frac {x^{3} - x^{2} \log \left (x\right ) + e^{2} + 5}{x^{2}}\right ) \]
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Time = 0.10 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95 \[ \int \frac {10+e^2 (2-2 x)-10 x+x^2-x^3-2 x^4+2 x^3 \log (x)}{-5 x-e^2 x-x^4+x^3 \log (x)} \, dx=2 x + \log {\left (\log {\left (x \right )} + \frac {- x^{3} - e^{2} - 5}{x^{2}} \right )} \]
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Time = 0.22 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.14 \[ \int \frac {10+e^2 (2-2 x)-10 x+x^2-x^3-2 x^4+2 x^3 \log (x)}{-5 x-e^2 x-x^4+x^3 \log (x)} \, dx=2 \, x + \log \left (-\frac {x^{3} - x^{2} \log \left (x\right ) + e^{2} + 5}{x^{2}}\right ) \]
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Time = 0.30 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.24 \[ \int \frac {10+e^2 (2-2 x)-10 x+x^2-x^3-2 x^4+2 x^3 \log (x)}{-5 x-e^2 x-x^4+x^3 \log (x)} \, dx=2 \, x + \log \left (-x^{3} + x^{2} \log \left (x\right ) - e^{2} - 5\right ) - 2 \, \log \left (x\right ) \]
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Time = 8.42 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.62 \[ \int \frac {10+e^2 (2-2 x)-10 x+x^2-x^3-2 x^4+2 x^3 \log (x)}{-5 x-e^2 x-x^4+x^3 \log (x)} \, dx=\ln \left ({\mathrm {e}}^2-x^2\,\ln \left (x\right )+x^3+5\right )-\frac {2\,x^2\,\ln \left (x\right )-2\,x^3}{x^2} \]
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