Integrand size = 106, antiderivative size = 27 \[ \int \frac {e^{16} \left (12 x-9 x^2\right )+e^{2 x} \left (12 x-9 x^2\right )+e^8 \left (-12 x+9 x^2\right )+e^x \left (-12 x+3 x^2+3 x^3+e^8 \left (24 x-18 x^2\right )\right )}{4-8 e^8+4 e^{16}+4 e^{2 x}+e^x \left (-8+8 e^8\right )} \, dx=3-\frac {3 (2-x) x^2}{-4+\frac {4}{e^8+e^x}} \]
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Leaf count is larger than twice the leaf count of optimal. \(57\) vs. \(2(27)=54\).
Time = 1.18 (sec) , antiderivative size = 57, normalized size of antiderivative = 2.11, number of steps used = 49, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.132, Rules used = {6820, 12, 6874, 2216, 2215, 2221, 2611, 2320, 6724, 2222, 2317, 2438, 6744, 45} \[ \int \frac {e^{16} \left (12 x-9 x^2\right )+e^{2 x} \left (12 x-9 x^2\right )+e^8 \left (-12 x+9 x^2\right )+e^x \left (-12 x+3 x^2+3 x^3+e^8 \left (24 x-18 x^2\right )\right )}{4-8 e^8+4 e^{16}+4 e^{2 x}+e^x \left (-8+8 e^8\right )} \, dx=\frac {3 x^3}{4 \left (-e^x+1-e^8\right )}-\frac {3 x^3}{4}-\frac {3 x^2}{2 \left (-e^x+1-e^8\right )}+\frac {3 x^2}{2} \]
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Rule 12
Rule 45
Rule 2215
Rule 2216
Rule 2221
Rule 2222
Rule 2317
Rule 2320
Rule 2438
Rule 2611
Rule 6724
Rule 6744
Rule 6820
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \int \frac {3 x \left (e^{8+x} (8-6 x)+\left (1-\frac {1}{e^8}\right ) e^{16} (4-3 x)+e^{2 x} (4-3 x)+e^x \left (-4+x+x^2\right )\right )}{4 \left (1-e^8-e^x\right )^2} \, dx \\ & = \frac {3}{4} \int \frac {x \left (e^{8+x} (8-6 x)+\left (1-\frac {1}{e^8}\right ) e^{16} (4-3 x)+e^{2 x} (4-3 x)+e^x \left (-4+x+x^2\right )\right )}{\left (1-e^8-e^x\right )^2} \, dx \\ & = \frac {3}{4} \int \left (-\frac {\left (-1+e^8\right ) (-2+x) x^2}{\left (-1+e^8+e^x\right )^2}-x (-4+3 x)+\frac {x \left (4-5 x+x^2\right )}{-1+e^8+e^x}\right ) \, dx \\ & = -\left (\frac {3}{4} \int x (-4+3 x) \, dx\right )+\frac {3}{4} \int \frac {x \left (4-5 x+x^2\right )}{-1+e^8+e^x} \, dx+\frac {1}{4} \left (3 \left (1-e^8\right )\right ) \int \frac {(-2+x) x^2}{\left (-1+e^8+e^x\right )^2} \, dx \\ & = -\left (\frac {3}{4} \int \left (-4 x+3 x^2\right ) \, dx\right )+\frac {3}{4} \int \left (\frac {4 x}{-1+e^8+e^x}-\frac {5 x^2}{-1+e^8+e^x}+\frac {x^3}{-1+e^8+e^x}\right ) \, dx+\frac {1}{4} \left (3 \left (1-e^8\right )\right ) \int \left (-\frac {2 x^2}{\left (-1+e^8+e^x\right )^2}+\frac {x^3}{\left (-1+e^8+e^x\right )^2}\right ) \, dx \\ & = \frac {3 x^2}{2}-\frac {3 x^3}{4}+\frac {3}{4} \int \frac {x^3}{-1+e^8+e^x} \, dx+3 \int \frac {x}{-1+e^8+e^x} \, dx-\frac {15}{4} \int \frac {x^2}{-1+e^8+e^x} \, dx+\frac {1}{4} \left (3 \left (1-e^8\right )\right ) \int \frac {x^3}{\left (-1+e^8+e^x\right )^2} \, dx-\frac {1}{2} \left (3 \left (1-e^8\right )\right ) \int \frac {x^2}{\left (-1+e^8+e^x\right )^2} \, dx \\ & = \frac {3 x^2}{2}-\frac {3 x^2}{2 \left (1-e^8\right )}-\frac {3 x^3}{4}+\frac {5 x^3}{4 \left (1-e^8\right )}-\frac {3 x^4}{16 \left (1-e^8\right )}+\frac {3}{4} \int \frac {e^x x^3}{\left (-1+e^8+e^x\right )^2} \, dx-\frac {3}{4} \int \frac {x^3}{-1+e^8+e^x} \, dx-\frac {3}{2} \int \frac {e^x x^2}{\left (-1+e^8+e^x\right )^2} \, dx+\frac {3}{2} \int \frac {x^2}{-1+e^8+e^x} \, dx+\frac {3 \int \frac {e^x x^3}{-1+e^8+e^x} \, dx}{4 \left (1-e^8\right )}+\frac {3 \int \frac {e^x x}{-1+e^8+e^x} \, dx}{1-e^8}-\frac {15 \int \frac {e^x x^2}{-1+e^8+e^x} \, dx}{4 \left (1-e^8\right )} \\ & = \frac {3 x^2}{2}-\frac {3 x^2}{2 \left (1-e^8\right )}-\frac {3 x^2}{2 \left (1-e^8-e^x\right )}-\frac {3 x^3}{4}+\frac {3 x^3}{4 \left (1-e^8\right )}+\frac {3 x^3}{4 \left (1-e^8-e^x\right )}+\frac {3 x \log \left (1-\frac {e^x}{1-e^8}\right )}{1-e^8}-\frac {15 x^2 \log \left (1-\frac {e^x}{1-e^8}\right )}{4 \left (1-e^8\right )}+\frac {3 x^3 \log \left (1-\frac {e^x}{1-e^8}\right )}{4 \left (1-e^8\right )}+\frac {9}{4} \int \frac {x^2}{-1+e^8+e^x} \, dx-3 \int \frac {x}{-1+e^8+e^x} \, dx-\frac {3 \int \frac {e^x x^3}{-1+e^8+e^x} \, dx}{4 \left (1-e^8\right )}+\frac {3 \int \frac {e^x x^2}{-1+e^8+e^x} \, dx}{2 \left (1-e^8\right )}-\frac {9 \int x^2 \log \left (1+\frac {e^x}{-1+e^8}\right ) \, dx}{4 \left (1-e^8\right )}-\frac {3 \int \log \left (1+\frac {e^x}{-1+e^8}\right ) \, dx}{1-e^8}+\frac {15 \int x \log \left (1+\frac {e^x}{-1+e^8}\right ) \, dx}{2 \left (1-e^8\right )} \\ & = \frac {3 x^2}{2}-\frac {3 x^2}{2 \left (1-e^8-e^x\right )}-\frac {3 x^3}{4}+\frac {3 x^3}{4 \left (1-e^8-e^x\right )}+\frac {3 x \log \left (1-\frac {e^x}{1-e^8}\right )}{1-e^8}-\frac {9 x^2 \log \left (1-\frac {e^x}{1-e^8}\right )}{4 \left (1-e^8\right )}-\frac {15 x \text {Li}_2\left (\frac {e^x}{1-e^8}\right )}{2 \left (1-e^8\right )}+\frac {9 x^2 \text {Li}_2\left (\frac {e^x}{1-e^8}\right )}{4 \left (1-e^8\right )}+\frac {9 \int \frac {e^x x^2}{-1+e^8+e^x} \, dx}{4 \left (1-e^8\right )}+\frac {9 \int x^2 \log \left (1+\frac {e^x}{-1+e^8}\right ) \, dx}{4 \left (1-e^8\right )}-\frac {3 \int \frac {e^x x}{-1+e^8+e^x} \, dx}{1-e^8}-\frac {3 \int x \log \left (1+\frac {e^x}{-1+e^8}\right ) \, dx}{1-e^8}-\frac {3 \text {Subst}\left (\int \frac {\log \left (1+\frac {x}{-1+e^8}\right )}{x} \, dx,x,e^x\right )}{1-e^8}-\frac {9 \int x \text {Li}_2\left (-\frac {e^x}{-1+e^8}\right ) \, dx}{2 \left (1-e^8\right )}+\frac {15 \int \text {Li}_2\left (-\frac {e^x}{-1+e^8}\right ) \, dx}{2 \left (1-e^8\right )} \\ & = \frac {3 x^2}{2}-\frac {3 x^2}{2 \left (1-e^8-e^x\right )}-\frac {3 x^3}{4}+\frac {3 x^3}{4 \left (1-e^8-e^x\right )}+\frac {3 \text {Li}_2\left (\frac {e^x}{1-e^8}\right )}{1-e^8}-\frac {9 x \text {Li}_2\left (\frac {e^x}{1-e^8}\right )}{2 \left (1-e^8\right )}-\frac {9 x \text {Li}_3\left (\frac {e^x}{1-e^8}\right )}{2 \left (1-e^8\right )}+\frac {3 \int \log \left (1+\frac {e^x}{-1+e^8}\right ) \, dx}{1-e^8}-\frac {3 \int \text {Li}_2\left (-\frac {e^x}{-1+e^8}\right ) \, dx}{1-e^8}-\frac {9 \int x \log \left (1+\frac {e^x}{-1+e^8}\right ) \, dx}{2 \left (1-e^8\right )}+\frac {9 \int x \text {Li}_2\left (-\frac {e^x}{-1+e^8}\right ) \, dx}{2 \left (1-e^8\right )}+\frac {9 \int \text {Li}_3\left (-\frac {e^x}{-1+e^8}\right ) \, dx}{2 \left (1-e^8\right )}+\frac {15 \text {Subst}\left (\int \frac {\text {Li}_2\left (-\frac {x}{-1+e^8}\right )}{x} \, dx,x,e^x\right )}{2 \left (1-e^8\right )} \\ & = \frac {3 x^2}{2}-\frac {3 x^2}{2 \left (1-e^8-e^x\right )}-\frac {3 x^3}{4}+\frac {3 x^3}{4 \left (1-e^8-e^x\right )}+\frac {3 \text {Li}_2\left (\frac {e^x}{1-e^8}\right )}{1-e^8}+\frac {15 \text {Li}_3\left (\frac {e^x}{1-e^8}\right )}{2 \left (1-e^8\right )}+\frac {3 \text {Subst}\left (\int \frac {\log \left (1+\frac {x}{-1+e^8}\right )}{x} \, dx,x,e^x\right )}{1-e^8}-\frac {3 \text {Subst}\left (\int \frac {\text {Li}_2\left (-\frac {x}{-1+e^8}\right )}{x} \, dx,x,e^x\right )}{1-e^8}-\frac {9 \int \text {Li}_2\left (-\frac {e^x}{-1+e^8}\right ) \, dx}{2 \left (1-e^8\right )}-\frac {9 \int \text {Li}_3\left (-\frac {e^x}{-1+e^8}\right ) \, dx}{2 \left (1-e^8\right )}+\frac {9 \text {Subst}\left (\int \frac {\text {Li}_3\left (-\frac {x}{-1+e^8}\right )}{x} \, dx,x,e^x\right )}{2 \left (1-e^8\right )} \\ & = \frac {3 x^2}{2}-\frac {3 x^2}{2 \left (1-e^8-e^x\right )}-\frac {3 x^3}{4}+\frac {3 x^3}{4 \left (1-e^8-e^x\right )}+\frac {9 \text {Li}_3\left (\frac {e^x}{1-e^8}\right )}{2 \left (1-e^8\right )}+\frac {9 \text {Li}_4\left (\frac {e^x}{1-e^8}\right )}{2 \left (1-e^8\right )}-\frac {9 \text {Subst}\left (\int \frac {\text {Li}_2\left (-\frac {x}{-1+e^8}\right )}{x} \, dx,x,e^x\right )}{2 \left (1-e^8\right )}-\frac {9 \text {Subst}\left (\int \frac {\text {Li}_3\left (-\frac {x}{-1+e^8}\right )}{x} \, dx,x,e^x\right )}{2 \left (1-e^8\right )} \\ & = \frac {3 x^2}{2}-\frac {3 x^2}{2 \left (1-e^8-e^x\right )}-\frac {3 x^3}{4}+\frac {3 x^3}{4 \left (1-e^8-e^x\right )} \\ \end{align*}
Time = 1.70 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {e^{16} \left (12 x-9 x^2\right )+e^{2 x} \left (12 x-9 x^2\right )+e^8 \left (-12 x+9 x^2\right )+e^x \left (-12 x+3 x^2+3 x^3+e^8 \left (24 x-18 x^2\right )\right )}{4-8 e^8+4 e^{16}+4 e^{2 x}+e^x \left (-8+8 e^8\right )} \, dx=-\frac {3 \left (e^8+e^x\right ) (-2+x) x^2}{4 \left (-1+e^8+e^x\right )} \]
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Time = 0.33 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.04
method | result | size |
risch | \(-\frac {3 x^{3}}{4}+\frac {3 x^{2}}{2}-\frac {3 \left (-2+x \right ) x^{2}}{4 \left ({\mathrm e}^{x}+{\mathrm e}^{8}-1\right )}\) | \(28\) |
norman | \(\frac {\frac {3 x^{2} {\mathrm e}^{8}}{2}-\frac {3 \,{\mathrm e}^{8} x^{3}}{4}+\frac {3 \,{\mathrm e}^{x} x^{2}}{2}-\frac {3 \,{\mathrm e}^{x} x^{3}}{4}}{{\mathrm e}^{x}+{\mathrm e}^{8}-1}\) | \(39\) |
parallelrisch | \(-\frac {3 \,{\mathrm e}^{8} x^{3}+3 \,{\mathrm e}^{x} x^{3}-6 x^{2} {\mathrm e}^{8}-6 \,{\mathrm e}^{x} x^{2}}{4 \left ({\mathrm e}^{x}+{\mathrm e}^{8}-1\right )}\) | \(40\) |
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Time = 0.24 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.30 \[ \int \frac {e^{16} \left (12 x-9 x^2\right )+e^{2 x} \left (12 x-9 x^2\right )+e^8 \left (-12 x+9 x^2\right )+e^x \left (-12 x+3 x^2+3 x^3+e^8 \left (24 x-18 x^2\right )\right )}{4-8 e^8+4 e^{16}+4 e^{2 x}+e^x \left (-8+8 e^8\right )} \, dx=-\frac {3 \, {\left ({\left (x^{3} - 2 \, x^{2}\right )} e^{8} + {\left (x^{3} - 2 \, x^{2}\right )} e^{x}\right )}}{4 \, {\left (e^{8} + e^{x} - 1\right )}} \]
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Time = 0.08 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.26 \[ \int \frac {e^{16} \left (12 x-9 x^2\right )+e^{2 x} \left (12 x-9 x^2\right )+e^8 \left (-12 x+9 x^2\right )+e^x \left (-12 x+3 x^2+3 x^3+e^8 \left (24 x-18 x^2\right )\right )}{4-8 e^8+4 e^{16}+4 e^{2 x}+e^x \left (-8+8 e^8\right )} \, dx=- \frac {3 x^{3}}{4} + \frac {3 x^{2}}{2} + \frac {- 3 x^{3} + 6 x^{2}}{4 e^{x} - 4 + 4 e^{8}} \]
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Time = 0.22 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.33 \[ \int \frac {e^{16} \left (12 x-9 x^2\right )+e^{2 x} \left (12 x-9 x^2\right )+e^8 \left (-12 x+9 x^2\right )+e^x \left (-12 x+3 x^2+3 x^3+e^8 \left (24 x-18 x^2\right )\right )}{4-8 e^8+4 e^{16}+4 e^{2 x}+e^x \left (-8+8 e^8\right )} \, dx=-\frac {3 \, {\left (x^{3} e^{8} - 2 \, x^{2} e^{8} + {\left (x^{3} - 2 \, x^{2}\right )} e^{x}\right )}}{4 \, {\left (e^{8} + e^{x} - 1\right )}} \]
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Time = 0.27 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.37 \[ \int \frac {e^{16} \left (12 x-9 x^2\right )+e^{2 x} \left (12 x-9 x^2\right )+e^8 \left (-12 x+9 x^2\right )+e^x \left (-12 x+3 x^2+3 x^3+e^8 \left (24 x-18 x^2\right )\right )}{4-8 e^8+4 e^{16}+4 e^{2 x}+e^x \left (-8+8 e^8\right )} \, dx=-\frac {3 \, {\left (x^{3} e^{8} + x^{3} e^{x} - 2 \, x^{2} e^{8} - 2 \, x^{2} e^{x}\right )}}{4 \, {\left (e^{8} + e^{x} - 1\right )}} \]
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Time = 12.49 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93 \[ \int \frac {e^{16} \left (12 x-9 x^2\right )+e^{2 x} \left (12 x-9 x^2\right )+e^8 \left (-12 x+9 x^2\right )+e^x \left (-12 x+3 x^2+3 x^3+e^8 \left (24 x-18 x^2\right )\right )}{4-8 e^8+4 e^{16}+4 e^{2 x}+e^x \left (-8+8 e^8\right )} \, dx=-\frac {3\,x^2\,\left ({\mathrm {e}}^8+{\mathrm {e}}^x\right )\,\left (x-2\right )}{4\,\left ({\mathrm {e}}^8+{\mathrm {e}}^x-1\right )} \]
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