Integrand size = 41, antiderivative size = 17 \[ \int \frac {e^{\frac {1-e^8 \log \left (e^2 \left (x+e x^2\right )\right )}{e^8}} (-1-2 e x)}{x+e x^2} \, dx=\frac {e^{-2+\frac {1}{e^8}}}{x+e x^2} \]
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Time = 0.23 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.06, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.122, Rules used = {1607, 2306, 12, 1598, 75} \[ \int \frac {e^{\frac {1-e^8 \log \left (e^2 \left (x+e x^2\right )\right )}{e^8}} (-1-2 e x)}{x+e x^2} \, dx=\frac {e^{\frac {1}{e^8}-2}}{x (e x+1)} \]
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Rule 12
Rule 75
Rule 1598
Rule 1607
Rule 2306
Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{\frac {1-e^8 \log \left (e^2 \left (x+e x^2\right )\right )}{e^8}} (-1-2 e x)}{x (1+e x)} \, dx \\ & = \int \frac {e^{-2+\frac {1}{e^8}} (-1-2 e x)}{x (1+e x) \left (x+e x^2\right )} \, dx \\ & = e^{-2+\frac {1}{e^8}} \int \frac {-1-2 e x}{x (1+e x) \left (x+e x^2\right )} \, dx \\ & = e^{-2+\frac {1}{e^8}} \int \frac {-1-2 e x}{x^2 (1+e x)^2} \, dx \\ & = \frac {e^{-2+\frac {1}{e^8}}}{x (1+e x)} \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {1-e^8 \log \left (e^2 \left (x+e x^2\right )\right )}{e^8}} (-1-2 e x)}{x+e x^2} \, dx=\frac {e^{-2+\frac {1}{e^8}}}{x+e x^2} \]
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Time = 0.20 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.06
method | result | size |
risch | \(\frac {{\mathrm e}^{-2+{\mathrm e}^{-8}}}{x \left (x \,{\mathrm e}+1\right )}\) | \(18\) |
norman | \(\frac {{\mathrm e}^{{\mathrm e}^{-8}} {\mathrm e}^{-2}}{x \left (x \,{\mathrm e}+1\right )}\) | \(22\) |
default | \({\mathrm e}^{-2+{\mathrm e}^{-8}} \left (-\frac {{\mathrm e}}{x \,{\mathrm e}+1}+\frac {1}{x}\right )\) | \(23\) |
gosper | \({\mathrm e}^{-\left ({\mathrm e}^{8} \ln \left (x \left (x \,{\mathrm e}+1\right ) {\mathrm e}^{2}\right )-1\right ) {\mathrm e}^{-8}}\) | \(26\) |
parallelrisch | \({\mathrm e}^{\left (-{\mathrm e}^{8} \ln \left (\left (x^{2} {\mathrm e}+x \right ) {\mathrm e}^{2}\right )+1\right ) {\mathrm e}^{-8}}\) | \(27\) |
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Time = 0.25 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {1-e^8 \log \left (e^2 \left (x+e x^2\right )\right )}{e^8}} (-1-2 e x)}{x+e x^2} \, dx=\frac {e^{\left (e^{\left (-8\right )}\right )}}{x^{2} e^{3} + x e^{2}} \]
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Time = 0.13 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {1-e^8 \log \left (e^2 \left (x+e x^2\right )\right )}{e^8}} (-1-2 e x)}{x+e x^2} \, dx=\frac {e^{e^{-8}}}{x^{2} e^{3} + x e^{2}} \]
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Time = 0.19 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.29 \[ \int \frac {e^{\frac {1-e^8 \log \left (e^2 \left (x+e x^2\right )\right )}{e^8}} (-1-2 e x)}{x+e x^2} \, dx=e^{\left (-{\left (e^{8} \log \left ({\left (x^{2} e + x\right )} e^{2}\right ) - 1\right )} e^{\left (-8\right )}\right )} \]
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Time = 0.26 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.94 \[ \int \frac {e^{\frac {1-e^8 \log \left (e^2 \left (x+e x^2\right )\right )}{e^8}} (-1-2 e x)}{x+e x^2} \, dx=\frac {e^{\left (e^{\left (-8\right )} - 2\right )}}{x^{2} e + x} \]
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Time = 0.16 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {1-e^8 \log \left (e^2 \left (x+e x^2\right )\right )}{e^8}} (-1-2 e x)}{x+e x^2} \, dx=\frac {{\mathrm {e}}^{{\mathrm {e}}^{-8}-2}}{x\,\left (x\,\mathrm {e}+1\right )} \]
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