Integrand size = 70, antiderivative size = 29 \[ \int \frac {e^{-1-e^x} \left (-16 e^{2 x} x-15 x^2-15 e^2 x^2+e^x \left (-16+16 x+15 x^3+15 e^2 x^3\right )\right )}{16 x^2+16 e^2 x^2} \, dx=e^{-1-e^x} \left (-\frac {15 x}{16}+\frac {e^x}{x+e^2 x}\right ) \]
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\[ \int \frac {e^{-1-e^x} \left (-16 e^{2 x} x-15 x^2-15 e^2 x^2+e^x \left (-16+16 x+15 x^3+15 e^2 x^3\right )\right )}{16 x^2+16 e^2 x^2} \, dx=\int \frac {e^{-1-e^x} \left (-16 e^{2 x} x-15 x^2-15 e^2 x^2+e^x \left (-16+16 x+15 x^3+15 e^2 x^3\right )\right )}{16 x^2+16 e^2 x^2} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{-1-e^x} \left (-16 e^{2 x} x-15 x^2-15 e^2 x^2+e^x \left (-16+16 x+15 x^3+15 e^2 x^3\right )\right )}{\left (16+16 e^2\right ) x^2} \, dx \\ & = \int \frac {e^{-1-e^x} \left (-16 e^{2 x} x+\left (-15-15 e^2\right ) x^2+e^x \left (-16+16 x+15 x^3+15 e^2 x^3\right )\right )}{\left (16+16 e^2\right ) x^2} \, dx \\ & = \frac {\int \frac {e^{-1-e^x} \left (-16 e^{2 x} x+\left (-15-15 e^2\right ) x^2+e^x \left (-16+16 x+15 x^3+15 e^2 x^3\right )\right )}{x^2} \, dx}{16 \left (1+e^2\right )} \\ & = \frac {\int \left (-15 e^{-1-e^x} \left (1+e^2\right )-\frac {16 e^{-1-e^x+2 x}}{x}+\frac {e^{-1-e^x+x} \left (-16+16 x+15 \left (1+e^2\right ) x^3\right )}{x^2}\right ) \, dx}{16 \left (1+e^2\right )} \\ & = -\left (\frac {15}{16} \int e^{-1-e^x} \, dx\right )+\frac {\int \frac {e^{-1-e^x+x} \left (-16+16 x+15 \left (1+e^2\right ) x^3\right )}{x^2} \, dx}{16 \left (1+e^2\right )}-\frac {\int \frac {e^{-1-e^x+2 x}}{x} \, dx}{1+e^2} \\ & = -\left (\frac {15}{16} \text {Subst}\left (\int \frac {e^{-1-x}}{x} \, dx,x,e^x\right )\right )+\frac {\int \left (-\frac {16 e^{-1-e^x+x}}{x^2}+\frac {16 e^{-1-e^x+x}}{x}+15 e^{-1-e^x+x} \left (1+e^2\right ) x\right ) \, dx}{16 \left (1+e^2\right )}-\frac {\int \frac {e^{-1-e^x+2 x}}{x} \, dx}{1+e^2} \\ & = -\frac {15 \operatorname {ExpIntegralEi}\left (-e^x\right )}{16 e}+\frac {15}{16} \int e^{-1-e^x+x} x \, dx-\frac {\int \frac {e^{-1-e^x+x}}{x^2} \, dx}{1+e^2}+\frac {\int \frac {e^{-1-e^x+x}}{x} \, dx}{1+e^2}-\frac {\int \frac {e^{-1-e^x+2 x}}{x} \, dx}{1+e^2} \\ \end{align*}
Time = 0.54 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.45 \[ \int \frac {e^{-1-e^x} \left (-16 e^{2 x} x-15 x^2-15 e^2 x^2+e^x \left (-16+16 x+15 x^3+15 e^2 x^3\right )\right )}{16 x^2+16 e^2 x^2} \, dx=\frac {e^{-1-e^x} \left (16 e^x-15 x^2-15 e^2 x^2\right )}{16 \left (1+e^2\right ) x} \]
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Time = 0.84 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.93
method | result | size |
norman | \(\frac {\left (\frac {{\mathrm e}^{x}}{{\mathrm e}^{2}+1}-\frac {15 x^{2}}{16}\right ) {\mathrm e}^{-{\mathrm e}^{x}-1}}{x}\) | \(27\) |
risch | \(-\frac {\left (15 x^{2} {\mathrm e}^{2}+15 x^{2}-16 \,{\mathrm e}^{x}\right ) {\mathrm e}^{-{\mathrm e}^{x}-1}}{16 x \left ({\mathrm e}^{2}+1\right )}\) | \(36\) |
parallelrisch | \(\frac {\left (-15 x^{2} {\mathrm e}^{2}-15 x^{2}+16 \,{\mathrm e}^{x}\right ) {\mathrm e}^{-{\mathrm e}^{x}-1}}{16 \left ({\mathrm e}^{2}+1\right ) x}\) | \(36\) |
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Time = 0.27 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.17 \[ \int \frac {e^{-1-e^x} \left (-16 e^{2 x} x-15 x^2-15 e^2 x^2+e^x \left (-16+16 x+15 x^3+15 e^2 x^3\right )\right )}{16 x^2+16 e^2 x^2} \, dx=-\frac {{\left (15 \, x^{2} e^{2} + 15 \, x^{2} - 16 \, e^{x}\right )} e^{\left (-e^{x} - 1\right )}}{16 \, {\left (x e^{2} + x\right )}} \]
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Time = 0.10 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.24 \[ \int \frac {e^{-1-e^x} \left (-16 e^{2 x} x-15 x^2-15 e^2 x^2+e^x \left (-16+16 x+15 x^3+15 e^2 x^3\right )\right )}{16 x^2+16 e^2 x^2} \, dx=\frac {\left (- 15 x^{2} e^{2} - 15 x^{2} + 16 e^{x}\right ) e^{- e^{x} - 1}}{16 x + 16 x e^{2}} \]
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Time = 0.34 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.07 \[ \int \frac {e^{-1-e^x} \left (-16 e^{2 x} x-15 x^2-15 e^2 x^2+e^x \left (-16+16 x+15 x^3+15 e^2 x^3\right )\right )}{16 x^2+16 e^2 x^2} \, dx=-\frac {{\left (15 \, x^{2} {\left (e^{2} + 1\right )} - 16 \, e^{x}\right )} e^{\left (-e^{x}\right )}}{16 \, x {\left (e^{3} + e\right )}} \]
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\[ \int \frac {e^{-1-e^x} \left (-16 e^{2 x} x-15 x^2-15 e^2 x^2+e^x \left (-16+16 x+15 x^3+15 e^2 x^3\right )\right )}{16 x^2+16 e^2 x^2} \, dx=\int { -\frac {{\left (15 \, x^{2} e^{2} + 15 \, x^{2} + 16 \, x e^{\left (2 \, x\right )} - {\left (15 \, x^{3} e^{2} + 15 \, x^{3} + 16 \, x - 16\right )} e^{x}\right )} e^{\left (-e^{x} - 1\right )}}{16 \, {\left (x^{2} e^{2} + x^{2}\right )}} \,d x } \]
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Time = 13.13 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.28 \[ \int \frac {e^{-1-e^x} \left (-16 e^{2 x} x-15 x^2-15 e^2 x^2+e^x \left (-16+16 x+15 x^3+15 e^2 x^3\right )\right )}{16 x^2+16 e^2 x^2} \, dx=\frac {{\mathrm {e}}^{x-{\mathrm {e}}^x-1}-\frac {x^2\,{\mathrm {e}}^{-{\mathrm {e}}^x-1}\,\left (15\,{\mathrm {e}}^2+15\right )}{16}}{x\,\left ({\mathrm {e}}^2+1\right )} \]
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