Integrand size = 76, antiderivative size = 13 \[ \int \frac {-25-50 e^x-25 x}{16 x-8 x^2+x^3+e^x \left (16-8 x+x^2\right )+\left (-8 x+2 x^2+e^x (-8+2 x)\right ) \log \left (e^x+x\right )+\left (e^x+x\right ) \log ^2\left (e^x+x\right )} \, dx=\frac {25}{-4+x+\log \left (e^x+x\right )} \]
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Time = 0.14 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.31, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.039, Rules used = {6820, 12, 6818} \[ \int \frac {-25-50 e^x-25 x}{16 x-8 x^2+x^3+e^x \left (16-8 x+x^2\right )+\left (-8 x+2 x^2+e^x (-8+2 x)\right ) \log \left (e^x+x\right )+\left (e^x+x\right ) \log ^2\left (e^x+x\right )} \, dx=-\frac {25}{-x-\log \left (x+e^x\right )+4} \]
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Rule 12
Rule 6818
Rule 6820
Rubi steps \begin{align*} \text {integral}& = \int \frac {25 \left (-1-2 e^x-x\right )}{\left (e^x+x\right ) \left (4-x-\log \left (e^x+x\right )\right )^2} \, dx \\ & = 25 \int \frac {-1-2 e^x-x}{\left (e^x+x\right ) \left (4-x-\log \left (e^x+x\right )\right )^2} \, dx \\ & = -\frac {25}{4-x-\log \left (e^x+x\right )} \\ \end{align*}
Time = 0.04 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00 \[ \int \frac {-25-50 e^x-25 x}{16 x-8 x^2+x^3+e^x \left (16-8 x+x^2\right )+\left (-8 x+2 x^2+e^x (-8+2 x)\right ) \log \left (e^x+x\right )+\left (e^x+x\right ) \log ^2\left (e^x+x\right )} \, dx=\frac {25}{-4+x+\log \left (e^x+x\right )} \]
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Time = 0.11 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00
method | result | size |
risch | \(\frac {25}{\ln \left ({\mathrm e}^{x}+x \right )-4+x}\) | \(13\) |
parallelrisch | \(\frac {25}{\ln \left ({\mathrm e}^{x}+x \right )-4+x}\) | \(13\) |
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Time = 0.25 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.92 \[ \int \frac {-25-50 e^x-25 x}{16 x-8 x^2+x^3+e^x \left (16-8 x+x^2\right )+\left (-8 x+2 x^2+e^x (-8+2 x)\right ) \log \left (e^x+x\right )+\left (e^x+x\right ) \log ^2\left (e^x+x\right )} \, dx=\frac {25}{x + \log \left (x + e^{x}\right ) - 4} \]
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Time = 0.11 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.77 \[ \int \frac {-25-50 e^x-25 x}{16 x-8 x^2+x^3+e^x \left (16-8 x+x^2\right )+\left (-8 x+2 x^2+e^x (-8+2 x)\right ) \log \left (e^x+x\right )+\left (e^x+x\right ) \log ^2\left (e^x+x\right )} \, dx=\frac {25}{x + \log {\left (x + e^{x} \right )} - 4} \]
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Time = 0.28 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.92 \[ \int \frac {-25-50 e^x-25 x}{16 x-8 x^2+x^3+e^x \left (16-8 x+x^2\right )+\left (-8 x+2 x^2+e^x (-8+2 x)\right ) \log \left (e^x+x\right )+\left (e^x+x\right ) \log ^2\left (e^x+x\right )} \, dx=\frac {25}{x + \log \left (x + e^{x}\right ) - 4} \]
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Time = 0.30 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.92 \[ \int \frac {-25-50 e^x-25 x}{16 x-8 x^2+x^3+e^x \left (16-8 x+x^2\right )+\left (-8 x+2 x^2+e^x (-8+2 x)\right ) \log \left (e^x+x\right )+\left (e^x+x\right ) \log ^2\left (e^x+x\right )} \, dx=\frac {25}{x + \log \left (x + e^{x}\right ) - 4} \]
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Time = 15.30 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.92 \[ \int \frac {-25-50 e^x-25 x}{16 x-8 x^2+x^3+e^x \left (16-8 x+x^2\right )+\left (-8 x+2 x^2+e^x (-8+2 x)\right ) \log \left (e^x+x\right )+\left (e^x+x\right ) \log ^2\left (e^x+x\right )} \, dx=\frac {25}{x+\ln \left (x+{\mathrm {e}}^x\right )-4} \]
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