Integrand size = 93, antiderivative size = 26 \[ \int \frac {e^{\frac {2 \left (3 e^x-3 x\right )}{-32+2 e^x+30 x-8 x^2}} \left (48-12 x^2+e^x \left (-96+72 x-12 x^2\right )\right )}{256+e^{2 x}-480 x+353 x^2-120 x^3+16 x^4+e^x \left (-32+30 x-8 x^2\right )} \, dx=e^{\frac {6}{2-\frac {8 (2-x)^2}{e^x-x}}} \]
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\[ \int \frac {e^{\frac {2 \left (3 e^x-3 x\right )}{-32+2 e^x+30 x-8 x^2}} \left (48-12 x^2+e^x \left (-96+72 x-12 x^2\right )\right )}{256+e^{2 x}-480 x+353 x^2-120 x^3+16 x^4+e^x \left (-32+30 x-8 x^2\right )} \, dx=\int \frac {\exp \left (\frac {2 \left (3 e^x-3 x\right )}{-32+2 e^x+30 x-8 x^2}\right ) \left (48-12 x^2+e^x \left (-96+72 x-12 x^2\right )\right )}{256+e^{2 x}-480 x+353 x^2-120 x^3+16 x^4+e^x \left (-32+30 x-8 x^2\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {12 e^{\frac {3 \left (e^x-x\right )}{-16+e^x+15 x-4 x^2}} (2-x) \left (2+e^x (-4+x)+x\right )}{\left (16-e^x-15 x+4 x^2\right )^2} \, dx \\ & = 12 \int \frac {e^{\frac {3 \left (e^x-x\right )}{-16+e^x+15 x-4 x^2}} (2-x) \left (2+e^x (-4+x)+x\right )}{\left (16-e^x-15 x+4 x^2\right )^2} \, dx \\ & = 12 \int \left (-\frac {e^{\frac {3 \left (e^x-x\right )}{-16+e^x+15 x-4 x^2}} (-2+x)^2 \left (31-23 x+4 x^2\right )}{\left (16-e^x-15 x+4 x^2\right )^2}+\frac {e^{\frac {3 \left (e^x-x\right )}{-16+e^x+15 x-4 x^2}} \left (8-6 x+x^2\right )}{16-e^x-15 x+4 x^2}\right ) \, dx \\ & = -\left (12 \int \frac {e^{\frac {3 \left (e^x-x\right )}{-16+e^x+15 x-4 x^2}} (-2+x)^2 \left (31-23 x+4 x^2\right )}{\left (16-e^x-15 x+4 x^2\right )^2} \, dx\right )+12 \int \frac {e^{\frac {3 \left (e^x-x\right )}{-16+e^x+15 x-4 x^2}} \left (8-6 x+x^2\right )}{16-e^x-15 x+4 x^2} \, dx \\ & = -\left (12 \int \left (\frac {124 e^{\frac {3 \left (e^x-x\right )}{-16+e^x+15 x-4 x^2}}}{\left (-16+e^x+15 x-4 x^2\right )^2}-\frac {216 e^{\frac {3 \left (e^x-x\right )}{-16+e^x+15 x-4 x^2}} x}{\left (16-e^x-15 x+4 x^2\right )^2}+\frac {139 e^{\frac {3 \left (e^x-x\right )}{-16+e^x+15 x-4 x^2}} x^2}{\left (16-e^x-15 x+4 x^2\right )^2}-\frac {39 e^{\frac {3 \left (e^x-x\right )}{-16+e^x+15 x-4 x^2}} x^3}{\left (16-e^x-15 x+4 x^2\right )^2}+\frac {4 e^{\frac {3 \left (e^x-x\right )}{-16+e^x+15 x-4 x^2}} x^4}{\left (16-e^x-15 x+4 x^2\right )^2}\right ) \, dx\right )+12 \int \left (-\frac {8 e^{\frac {3 \left (e^x-x\right )}{-16+e^x+15 x-4 x^2}}}{-16+e^x+15 x-4 x^2}-\frac {6 e^{\frac {3 \left (e^x-x\right )}{-16+e^x+15 x-4 x^2}} x}{16-e^x-15 x+4 x^2}+\frac {e^{\frac {3 \left (e^x-x\right )}{-16+e^x+15 x-4 x^2}} x^2}{16-e^x-15 x+4 x^2}\right ) \, dx \\ & = 12 \int \frac {e^{\frac {3 \left (e^x-x\right )}{-16+e^x+15 x-4 x^2}} x^2}{16-e^x-15 x+4 x^2} \, dx-48 \int \frac {e^{\frac {3 \left (e^x-x\right )}{-16+e^x+15 x-4 x^2}} x^4}{\left (16-e^x-15 x+4 x^2\right )^2} \, dx-72 \int \frac {e^{\frac {3 \left (e^x-x\right )}{-16+e^x+15 x-4 x^2}} x}{16-e^x-15 x+4 x^2} \, dx-96 \int \frac {e^{\frac {3 \left (e^x-x\right )}{-16+e^x+15 x-4 x^2}}}{-16+e^x+15 x-4 x^2} \, dx+468 \int \frac {e^{\frac {3 \left (e^x-x\right )}{-16+e^x+15 x-4 x^2}} x^3}{\left (16-e^x-15 x+4 x^2\right )^2} \, dx-1488 \int \frac {e^{\frac {3 \left (e^x-x\right )}{-16+e^x+15 x-4 x^2}}}{\left (-16+e^x+15 x-4 x^2\right )^2} \, dx-1668 \int \frac {e^{\frac {3 \left (e^x-x\right )}{-16+e^x+15 x-4 x^2}} x^2}{\left (16-e^x-15 x+4 x^2\right )^2} \, dx+2592 \int \frac {e^{\frac {3 \left (e^x-x\right )}{-16+e^x+15 x-4 x^2}} x}{\left (16-e^x-15 x+4 x^2\right )^2} \, dx \\ \end{align*}
Time = 2.64 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {2 \left (3 e^x-3 x\right )}{-32+2 e^x+30 x-8 x^2}} \left (48-12 x^2+e^x \left (-96+72 x-12 x^2\right )\right )}{256+e^{2 x}-480 x+353 x^2-120 x^3+16 x^4+e^x \left (-32+30 x-8 x^2\right )} \, dx=e^{\frac {3 \left (e^x-x\right )}{-16+e^x+15 x-4 x^2}} \]
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Time = 0.54 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92
method | result | size |
risch | \({\mathrm e}^{\frac {3 \,{\mathrm e}^{x}-3 x}{-4 x^{2}+{\mathrm e}^{x}+15 x -16}}\) | \(24\) |
parallelrisch | \({\mathrm e}^{\frac {3 \,{\mathrm e}^{x}-3 x}{-4 x^{2}+{\mathrm e}^{x}+15 x -16}}\) | \(26\) |
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Time = 0.25 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96 \[ \int \frac {e^{\frac {2 \left (3 e^x-3 x\right )}{-32+2 e^x+30 x-8 x^2}} \left (48-12 x^2+e^x \left (-96+72 x-12 x^2\right )\right )}{256+e^{2 x}-480 x+353 x^2-120 x^3+16 x^4+e^x \left (-32+30 x-8 x^2\right )} \, dx=e^{\left (\frac {3 \, {\left (x - e^{x}\right )}}{4 \, x^{2} - 15 \, x - e^{x} + 16}\right )} \]
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Time = 0.26 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int \frac {e^{\frac {2 \left (3 e^x-3 x\right )}{-32+2 e^x+30 x-8 x^2}} \left (48-12 x^2+e^x \left (-96+72 x-12 x^2\right )\right )}{256+e^{2 x}-480 x+353 x^2-120 x^3+16 x^4+e^x \left (-32+30 x-8 x^2\right )} \, dx=e^{\frac {2 \left (- 3 x + 3 e^{x}\right )}{- 8 x^{2} + 30 x + 2 e^{x} - 32}} \]
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Time = 0.30 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.58 \[ \int \frac {e^{\frac {2 \left (3 e^x-3 x\right )}{-32+2 e^x+30 x-8 x^2}} \left (48-12 x^2+e^x \left (-96+72 x-12 x^2\right )\right )}{256+e^{2 x}-480 x+353 x^2-120 x^3+16 x^4+e^x \left (-32+30 x-8 x^2\right )} \, dx=e^{\left (\frac {3 \, x}{4 \, x^{2} - 15 \, x - e^{x} + 16} - \frac {3 \, e^{x}}{4 \, x^{2} - 15 \, x - e^{x} + 16}\right )} \]
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Time = 0.34 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.58 \[ \int \frac {e^{\frac {2 \left (3 e^x-3 x\right )}{-32+2 e^x+30 x-8 x^2}} \left (48-12 x^2+e^x \left (-96+72 x-12 x^2\right )\right )}{256+e^{2 x}-480 x+353 x^2-120 x^3+16 x^4+e^x \left (-32+30 x-8 x^2\right )} \, dx=e^{\left (\frac {3 \, x}{4 \, x^{2} - 15 \, x - e^{x} + 16} - \frac {3 \, e^{x}}{4 \, x^{2} - 15 \, x - e^{x} + 16}\right )} \]
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Time = 10.95 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.46 \[ \int \frac {e^{\frac {2 \left (3 e^x-3 x\right )}{-32+2 e^x+30 x-8 x^2}} \left (48-12 x^2+e^x \left (-96+72 x-12 x^2\right )\right )}{256+e^{2 x}-480 x+353 x^2-120 x^3+16 x^4+e^x \left (-32+30 x-8 x^2\right )} \, dx={\mathrm {e}}^{\frac {3\,{\mathrm {e}}^x}{15\,x+{\mathrm {e}}^x-4\,x^2-16}}\,{\mathrm {e}}^{-\frac {3\,x}{15\,x+{\mathrm {e}}^x-4\,x^2-16}} \]
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