Integrand size = 55, antiderivative size = 26 \[ \int \frac {-24+16 x-56 x^2+12 x^3+e^x \left (-3-13 x+3 x^2\right )+\left (e^x+4 x\right ) \log \left (e^x+4 x\right )}{e^x+4 x} \, dx=x \left (-3+(-6+x) \left (x+\frac {-x+\log \left (e^x+4 x\right )}{x}\right )\right ) \]
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\[ \int \frac {-24+16 x-56 x^2+12 x^3+e^x \left (-3-13 x+3 x^2\right )+\left (e^x+4 x\right ) \log \left (e^x+4 x\right )}{e^x+4 x} \, dx=\int \frac {-24+16 x-56 x^2+12 x^3+e^x \left (-3-13 x+3 x^2\right )+\left (e^x+4 x\right ) \log \left (e^x+4 x\right )}{e^x+4 x} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \left (-3-13 x+3 x^2-\frac {4 \left (6-7 x+x^2\right )}{e^x+4 x}+\log \left (e^x+4 x\right )\right ) \, dx \\ & = -3 x-\frac {13 x^2}{2}+x^3-4 \int \frac {6-7 x+x^2}{e^x+4 x} \, dx+\int \log \left (e^x+4 x\right ) \, dx \\ & = -3 x-\frac {13 x^2}{2}+x^3+x \log \left (e^x+4 x\right )-4 \int \left (\frac {6}{e^x+4 x}-\frac {7 x}{e^x+4 x}+\frac {x^2}{e^x+4 x}\right ) \, dx-\int \frac {\left (4+e^x\right ) x}{e^x+4 x} \, dx \\ & = -3 x-\frac {13 x^2}{2}+x^3+x \log \left (e^x+4 x\right )-4 \int \frac {x^2}{e^x+4 x} \, dx-24 \int \frac {1}{e^x+4 x} \, dx+28 \int \frac {x}{e^x+4 x} \, dx-\int \left (x-\frac {4 (-1+x) x}{e^x+4 x}\right ) \, dx \\ & = -3 x-7 x^2+x^3+x \log \left (e^x+4 x\right )+4 \int \frac {(-1+x) x}{e^x+4 x} \, dx-4 \int \frac {x^2}{e^x+4 x} \, dx-24 \int \frac {1}{e^x+4 x} \, dx+28 \int \frac {x}{e^x+4 x} \, dx \\ & = -3 x-7 x^2+x^3+x \log \left (e^x+4 x\right )-4 \int \frac {x^2}{e^x+4 x} \, dx+4 \int \left (-\frac {x}{e^x+4 x}+\frac {x^2}{e^x+4 x}\right ) \, dx-24 \int \frac {1}{e^x+4 x} \, dx+28 \int \frac {x}{e^x+4 x} \, dx \\ & = -3 x-7 x^2+x^3+x \log \left (e^x+4 x\right )-4 \int \frac {x}{e^x+4 x} \, dx-24 \int \frac {1}{e^x+4 x} \, dx+28 \int \frac {x}{e^x+4 x} \, dx \\ \end{align*}
Time = 0.18 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.23 \[ \int \frac {-24+16 x-56 x^2+12 x^3+e^x \left (-3-13 x+3 x^2\right )+\left (e^x+4 x\right ) \log \left (e^x+4 x\right )}{e^x+4 x} \, dx=3 x-7 x^2+x^3-6 \log \left (e^x+4 x\right )+x \log \left (e^x+4 x\right ) \]
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Time = 0.04 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.19
method | result | size |
norman | \(x^{3}-6 \ln \left (4 x +{\mathrm e}^{x}\right )+\ln \left (4 x +{\mathrm e}^{x}\right ) x +3 x -7 x^{2}\) | \(31\) |
risch | \(x^{3}-6 \ln \left (4 x +{\mathrm e}^{x}\right )+\ln \left (4 x +{\mathrm e}^{x}\right ) x +3 x -7 x^{2}\) | \(31\) |
parallelrisch | \(x^{3}-6 \ln \left (4 x +{\mathrm e}^{x}\right )+\ln \left (4 x +{\mathrm e}^{x}\right ) x +3 x -7 x^{2}\) | \(31\) |
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Time = 0.26 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.88 \[ \int \frac {-24+16 x-56 x^2+12 x^3+e^x \left (-3-13 x+3 x^2\right )+\left (e^x+4 x\right ) \log \left (e^x+4 x\right )}{e^x+4 x} \, dx=x^{3} - 7 \, x^{2} + {\left (x - 6\right )} \log \left (4 \, x + e^{x}\right ) + 3 \, x \]
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Time = 0.13 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.19 \[ \int \frac {-24+16 x-56 x^2+12 x^3+e^x \left (-3-13 x+3 x^2\right )+\left (e^x+4 x\right ) \log \left (e^x+4 x\right )}{e^x+4 x} \, dx=x^{3} - 7 x^{2} + x \log {\left (4 x + e^{x} \right )} + 3 x - 6 \log {\left (4 x + e^{x} \right )} \]
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Time = 0.22 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.88 \[ \int \frac {-24+16 x-56 x^2+12 x^3+e^x \left (-3-13 x+3 x^2\right )+\left (e^x+4 x\right ) \log \left (e^x+4 x\right )}{e^x+4 x} \, dx=x^{3} - 7 \, x^{2} + {\left (x - 6\right )} \log \left (4 \, x + e^{x}\right ) + 3 \, x \]
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Time = 0.29 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.23 \[ \int \frac {-24+16 x-56 x^2+12 x^3+e^x \left (-3-13 x+3 x^2\right )+\left (e^x+4 x\right ) \log \left (e^x+4 x\right )}{e^x+4 x} \, dx=x^{3} - 7 \, x^{2} + x \log \left (4 \, x + e^{x}\right ) + 3 \, x - 6 \, \log \left (-4 \, x - e^{x}\right ) \]
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Time = 12.50 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.15 \[ \int \frac {-24+16 x-56 x^2+12 x^3+e^x \left (-3-13 x+3 x^2\right )+\left (e^x+4 x\right ) \log \left (e^x+4 x\right )}{e^x+4 x} \, dx=3\,x-6\,\ln \left (4\,x+{\mathrm {e}}^x\right )+x\,\ln \left (4\,x+{\mathrm {e}}^x\right )-7\,x^2+x^3 \]
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