Integrand size = 70, antiderivative size = 28 \[ \int \frac {-2-2 x-6 x^3-16 x^4-10 x^5+e^x \left (-1-3 x^3-8 x^4-5 x^5\right )+\left (-2 x-e^x x\right ) \log \left (\frac {16 x}{25}\right )}{2 x+e^x x} \, dx=\log \left (\frac {2+e^x}{x}\right )-x \left (\left (x+x^2\right )^2+\log \left (\frac {16 x}{25}\right )\right ) \]
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Time = 0.25 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.25, number of steps used = 11, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {6874, 2320, 36, 29, 31, 14, 2332} \[ \int \frac {-2-2 x-6 x^3-16 x^4-10 x^5+e^x \left (-1-3 x^3-8 x^4-5 x^5\right )+\left (-2 x-e^x x\right ) \log \left (\frac {16 x}{25}\right )}{2 x+e^x x} \, dx=-x^5-2 x^4-x^3-x \log \left (\frac {16 x}{25}\right )+\log \left (e^x+2\right )-\log (x) \]
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Rule 14
Rule 29
Rule 31
Rule 36
Rule 2320
Rule 2332
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {2}{2+e^x}+\frac {-1-3 x^3-8 x^4-5 x^5-x \log \left (\frac {16 x}{25}\right )}{x}\right ) \, dx \\ & = -\left (2 \int \frac {1}{2+e^x} \, dx\right )+\int \frac {-1-3 x^3-8 x^4-5 x^5-x \log \left (\frac {16 x}{25}\right )}{x} \, dx \\ & = -\left (2 \text {Subst}\left (\int \frac {1}{x (2+x)} \, dx,x,e^x\right )\right )+\int \left (\frac {-1-3 x^3-8 x^4-5 x^5}{x}-\log \left (\frac {16 x}{25}\right )\right ) \, dx \\ & = \int \frac {-1-3 x^3-8 x^4-5 x^5}{x} \, dx-\int \log \left (\frac {16 x}{25}\right ) \, dx-\text {Subst}\left (\int \frac {1}{x} \, dx,x,e^x\right )+\text {Subst}\left (\int \frac {1}{2+x} \, dx,x,e^x\right ) \\ & = \log \left (2+e^x\right )-x \log \left (\frac {16 x}{25}\right )+\int \left (-\frac {1}{x}-3 x^2-8 x^3-5 x^4\right ) \, dx \\ & = -x^3-2 x^4-x^5+\log \left (2+e^x\right )-x \log \left (\frac {16 x}{25}\right )-\log (x) \\ \end{align*}
Time = 0.12 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.25 \[ \int \frac {-2-2 x-6 x^3-16 x^4-10 x^5+e^x \left (-1-3 x^3-8 x^4-5 x^5\right )+\left (-2 x-e^x x\right ) \log \left (\frac {16 x}{25}\right )}{2 x+e^x x} \, dx=-x^3-2 x^4-x^5+\log \left (2+e^x\right )-x \log \left (\frac {16 x}{25}\right )-\log (x) \]
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Time = 0.07 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.18
method | result | size |
risch | \(-\ln \left (\frac {16 x}{25}\right ) x -x^{5}-2 x^{4}-x^{3}-\ln \left (x \right )+\ln \left ({\mathrm e}^{x}+2\right )\) | \(33\) |
parallelrisch | \(-\ln \left (\frac {16 x}{25}\right ) x -x^{5}-2 x^{4}-x^{3}-\ln \left (x \right )+\ln \left ({\mathrm e}^{x}+2\right )\) | \(33\) |
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Time = 0.23 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.14 \[ \int \frac {-2-2 x-6 x^3-16 x^4-10 x^5+e^x \left (-1-3 x^3-8 x^4-5 x^5\right )+\left (-2 x-e^x x\right ) \log \left (\frac {16 x}{25}\right )}{2 x+e^x x} \, dx=-x^{5} - 2 \, x^{4} - x^{3} - x \log \left (\frac {16}{25} \, x\right ) - \log \left (x\right ) + \log \left (e^{x} + 2\right ) \]
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Time = 0.10 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.04 \[ \int \frac {-2-2 x-6 x^3-16 x^4-10 x^5+e^x \left (-1-3 x^3-8 x^4-5 x^5\right )+\left (-2 x-e^x x\right ) \log \left (\frac {16 x}{25}\right )}{2 x+e^x x} \, dx=- x^{5} - 2 x^{4} - x^{3} - x \log {\left (\frac {16 x}{25} \right )} - \log {\left (x \right )} + \log {\left (e^{x} + 2 \right )} \]
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Time = 0.30 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.36 \[ \int \frac {-2-2 x-6 x^3-16 x^4-10 x^5+e^x \left (-1-3 x^3-8 x^4-5 x^5\right )+\left (-2 x-e^x x\right ) \log \left (\frac {16 x}{25}\right )}{2 x+e^x x} \, dx=-x^{5} - 2 \, x^{4} - x^{3} + 2 \, x {\left (\log \left (5\right ) - 2 \, \log \left (2\right )\right )} - {\left (x + 1\right )} \log \left (x\right ) + \log \left (e^{x} + 2\right ) \]
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Time = 0.25 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.21 \[ \int \frac {-2-2 x-6 x^3-16 x^4-10 x^5+e^x \left (-1-3 x^3-8 x^4-5 x^5\right )+\left (-2 x-e^x x\right ) \log \left (\frac {16 x}{25}\right )}{2 x+e^x x} \, dx=-x^{5} - 2 \, x^{4} - x^{3} - x \log \left (\frac {16}{25} \, x\right ) - \log \left (\frac {1}{25} \, x\right ) + \log \left (e^{x} + 2\right ) \]
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Time = 8.38 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.14 \[ \int \frac {-2-2 x-6 x^3-16 x^4-10 x^5+e^x \left (-1-3 x^3-8 x^4-5 x^5\right )+\left (-2 x-e^x x\right ) \log \left (\frac {16 x}{25}\right )}{2 x+e^x x} \, dx=\ln \left ({\mathrm {e}}^x+2\right )-\ln \left (x\right )-x\,\ln \left (\frac {16\,x}{25}\right )-x^3-2\,x^4-x^5 \]
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