Integrand size = 100, antiderivative size = 33 \[ \int \frac {-6 x^2-4 x^3+e^x \left (6 x^2+4 x^3-x^4\right )+\left (-18 x-18 x^2+e^x \left (18 x+18 x^2-6 x^3\right )\right ) \log (x)+\left (-18 x+e^x \left (18 x-9 x^2\right )\right ) \log ^2(x)}{9-18 e^x+9 e^{2 x}} \, dx=\frac {e^{-x} x \left (\frac {x^2}{3}+x \log (x)\right )^2}{x-e^{-x} x} \]
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\[ \int \frac {-6 x^2-4 x^3+e^x \left (6 x^2+4 x^3-x^4\right )+\left (-18 x-18 x^2+e^x \left (18 x+18 x^2-6 x^3\right )\right ) \log (x)+\left (-18 x+e^x \left (18 x-9 x^2\right )\right ) \log ^2(x)}{9-18 e^x+9 e^{2 x}} \, dx=\int \frac {-6 x^2-4 x^3+e^x \left (6 x^2+4 x^3-x^4\right )+\left (-18 x-18 x^2+e^x \left (18 x+18 x^2-6 x^3\right )\right ) \log (x)+\left (-18 x+e^x \left (18 x-9 x^2\right )\right ) \log ^2(x)}{9-18 e^x+9 e^{2 x}} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {x (x+3 \log (x)) \left (-6-4 x-e^x \left (-6-4 x+x^2\right )-3 \left (2+e^x (-2+x)\right ) \log (x)\right )}{9 \left (1-e^x\right )^2} \, dx \\ & = \frac {1}{9} \int \frac {x (x+3 \log (x)) \left (-6-4 x-e^x \left (-6-4 x+x^2\right )-3 \left (2+e^x (-2+x)\right ) \log (x)\right )}{\left (1-e^x\right )^2} \, dx \\ & = \frac {1}{9} \int \left (-\frac {x^2 (x+3 \log (x))^2}{\left (-1+e^x\right )^2}-\frac {x \left (-6 x-4 x^2+x^3-18 \log (x)-18 x \log (x)+6 x^2 \log (x)-18 \log ^2(x)+9 x \log ^2(x)\right )}{-1+e^x}\right ) \, dx \\ & = -\left (\frac {1}{9} \int \frac {x^2 (x+3 \log (x))^2}{\left (-1+e^x\right )^2} \, dx\right )-\frac {1}{9} \int \frac {x \left (-6 x-4 x^2+x^3-18 \log (x)-18 x \log (x)+6 x^2 \log (x)-18 \log ^2(x)+9 x \log ^2(x)\right )}{-1+e^x} \, dx \\ & = -\left (\frac {1}{9} \int \left (\frac {x^4}{\left (-1+e^x\right )^2}+\frac {6 x^3 \log (x)}{\left (-1+e^x\right )^2}+\frac {9 x^2 \log ^2(x)}{\left (-1+e^x\right )^2}\right ) \, dx\right )-\frac {1}{9} \int \left (-\frac {6 x^2}{-1+e^x}-\frac {4 x^3}{-1+e^x}+\frac {x^4}{-1+e^x}-\frac {18 x \log (x)}{-1+e^x}-\frac {18 x^2 \log (x)}{-1+e^x}+\frac {6 x^3 \log (x)}{-1+e^x}-\frac {18 x \log ^2(x)}{-1+e^x}+\frac {9 x^2 \log ^2(x)}{-1+e^x}\right ) \, dx \\ & = -\left (\frac {1}{9} \int \frac {x^4}{\left (-1+e^x\right )^2} \, dx\right )-\frac {1}{9} \int \frac {x^4}{-1+e^x} \, dx+\frac {4}{9} \int \frac {x^3}{-1+e^x} \, dx+\frac {2}{3} \int \frac {x^2}{-1+e^x} \, dx-\frac {2}{3} \int \frac {x^3 \log (x)}{\left (-1+e^x\right )^2} \, dx-\frac {2}{3} \int \frac {x^3 \log (x)}{-1+e^x} \, dx+2 \int \frac {x \log (x)}{-1+e^x} \, dx+2 \int \frac {x^2 \log (x)}{-1+e^x} \, dx+2 \int \frac {x \log ^2(x)}{-1+e^x} \, dx-\int \frac {x^2 \log ^2(x)}{\left (-1+e^x\right )^2} \, dx-\int \frac {x^2 \log ^2(x)}{-1+e^x} \, dx \\ & = -\frac {2 x^3}{9}-\frac {x^4}{9}+\frac {x^5}{45}-\frac {1}{9} \int \frac {e^x x^4}{\left (-1+e^x\right )^2} \, dx+\frac {1}{9} \int \frac {x^4}{-1+e^x} \, dx-\frac {1}{9} \int \frac {e^x x^4}{-1+e^x} \, dx+\frac {4}{9} \int \frac {e^x x^3}{-1+e^x} \, dx+\frac {2}{3} \int \frac {e^x x^2}{-1+e^x} \, dx-\frac {2}{3} \int \frac {x^3 \log (x)}{\left (-1+e^x\right )^2} \, dx-\frac {2}{3} \int \frac {x^3 \log (x)}{-1+e^x} \, dx+2 \int \frac {x \log (x)}{-1+e^x} \, dx+2 \int \frac {x^2 \log (x)}{-1+e^x} \, dx+2 \int \frac {x \log ^2(x)}{-1+e^x} \, dx-\int \frac {x^2 \log ^2(x)}{\left (-1+e^x\right )^2} \, dx-\int \frac {x^2 \log ^2(x)}{-1+e^x} \, dx \\ & = -\frac {2 x^3}{9}-\frac {x^4}{9}-\frac {x^4}{9 \left (1-e^x\right )}+\frac {2}{3} x^2 \log \left (1-e^x\right )+\frac {4}{9} x^3 \log \left (1-e^x\right )-\frac {1}{9} x^4 \log \left (1-e^x\right )+\frac {1}{9} \int \frac {e^x x^4}{-1+e^x} \, dx-\frac {4}{9} \int \frac {x^3}{-1+e^x} \, dx+\frac {4}{9} \int x^3 \log \left (1-e^x\right ) \, dx-\frac {2}{3} \int \frac {x^3 \log (x)}{\left (-1+e^x\right )^2} \, dx-\frac {2}{3} \int \frac {x^3 \log (x)}{-1+e^x} \, dx-\frac {4}{3} \int x \log \left (1-e^x\right ) \, dx-\frac {4}{3} \int x^2 \log \left (1-e^x\right ) \, dx+2 \int \frac {x \log (x)}{-1+e^x} \, dx+2 \int \frac {x^2 \log (x)}{-1+e^x} \, dx+2 \int \frac {x \log ^2(x)}{-1+e^x} \, dx-\int \frac {x^2 \log ^2(x)}{\left (-1+e^x\right )^2} \, dx-\int \frac {x^2 \log ^2(x)}{-1+e^x} \, dx \\ & = -\frac {2 x^3}{9}-\frac {x^4}{9 \left (1-e^x\right )}+\frac {2}{3} x^2 \log \left (1-e^x\right )+\frac {4}{9} x^3 \log \left (1-e^x\right )+\frac {4 x \operatorname {PolyLog}\left (2,e^x\right )}{3}+\frac {4}{3} x^2 \operatorname {PolyLog}\left (2,e^x\right )-\frac {4}{9} x^3 \operatorname {PolyLog}\left (2,e^x\right )-\frac {4}{9} \int \frac {e^x x^3}{-1+e^x} \, dx-\frac {4}{9} \int x^3 \log \left (1-e^x\right ) \, dx-\frac {2}{3} \int \frac {x^3 \log (x)}{\left (-1+e^x\right )^2} \, dx-\frac {2}{3} \int \frac {x^3 \log (x)}{-1+e^x} \, dx-\frac {4}{3} \int \operatorname {PolyLog}\left (2,e^x\right ) \, dx+\frac {4}{3} \int x^2 \operatorname {PolyLog}\left (2,e^x\right ) \, dx+2 \int \frac {x \log (x)}{-1+e^x} \, dx+2 \int \frac {x^2 \log (x)}{-1+e^x} \, dx+2 \int \frac {x \log ^2(x)}{-1+e^x} \, dx-\frac {8}{3} \int x \operatorname {PolyLog}\left (2,e^x\right ) \, dx-\int \frac {x^2 \log ^2(x)}{\left (-1+e^x\right )^2} \, dx-\int \frac {x^2 \log ^2(x)}{-1+e^x} \, dx \\ & = -\frac {2 x^3}{9}-\frac {x^4}{9 \left (1-e^x\right )}+\frac {2}{3} x^2 \log \left (1-e^x\right )+\frac {4 x \operatorname {PolyLog}\left (2,e^x\right )}{3}+\frac {4}{3} x^2 \operatorname {PolyLog}\left (2,e^x\right )-\frac {8 x \operatorname {PolyLog}\left (3,e^x\right )}{3}+\frac {4}{3} x^2 \operatorname {PolyLog}\left (3,e^x\right )-\frac {2}{3} \int \frac {x^3 \log (x)}{\left (-1+e^x\right )^2} \, dx-\frac {2}{3} \int \frac {x^3 \log (x)}{-1+e^x} \, dx+\frac {4}{3} \int x^2 \log \left (1-e^x\right ) \, dx-\frac {4}{3} \int x^2 \operatorname {PolyLog}\left (2,e^x\right ) \, dx-\frac {4}{3} \text {Subst}\left (\int \frac {\operatorname {PolyLog}(2,x)}{x} \, dx,x,e^x\right )+2 \int \frac {x \log (x)}{-1+e^x} \, dx+2 \int \frac {x^2 \log (x)}{-1+e^x} \, dx+2 \int \frac {x \log ^2(x)}{-1+e^x} \, dx+\frac {8}{3} \int \operatorname {PolyLog}\left (3,e^x\right ) \, dx-\frac {8}{3} \int x \operatorname {PolyLog}\left (3,e^x\right ) \, dx-\int \frac {x^2 \log ^2(x)}{\left (-1+e^x\right )^2} \, dx-\int \frac {x^2 \log ^2(x)}{-1+e^x} \, dx \\ & = -\frac {2 x^3}{9}-\frac {x^4}{9 \left (1-e^x\right )}+\frac {2}{3} x^2 \log \left (1-e^x\right )+\frac {4 x \operatorname {PolyLog}\left (2,e^x\right )}{3}-\frac {4 \operatorname {PolyLog}\left (3,e^x\right )}{3}-\frac {8 x \operatorname {PolyLog}\left (3,e^x\right )}{3}-\frac {8 x \operatorname {PolyLog}\left (4,e^x\right )}{3}-\frac {2}{3} \int \frac {x^3 \log (x)}{\left (-1+e^x\right )^2} \, dx-\frac {2}{3} \int \frac {x^3 \log (x)}{-1+e^x} \, dx+2 \int \frac {x \log (x)}{-1+e^x} \, dx+2 \int \frac {x^2 \log (x)}{-1+e^x} \, dx+2 \int \frac {x \log ^2(x)}{-1+e^x} \, dx+\frac {8}{3} \int x \operatorname {PolyLog}\left (2,e^x\right ) \, dx+\frac {8}{3} \int x \operatorname {PolyLog}\left (3,e^x\right ) \, dx+\frac {8}{3} \int \operatorname {PolyLog}\left (4,e^x\right ) \, dx+\frac {8}{3} \text {Subst}\left (\int \frac {\operatorname {PolyLog}(3,x)}{x} \, dx,x,e^x\right )-\int \frac {x^2 \log ^2(x)}{\left (-1+e^x\right )^2} \, dx-\int \frac {x^2 \log ^2(x)}{-1+e^x} \, dx \\ & = -\frac {2 x^3}{9}-\frac {x^4}{9 \left (1-e^x\right )}+\frac {2}{3} x^2 \log \left (1-e^x\right )+\frac {4 x \operatorname {PolyLog}\left (2,e^x\right )}{3}-\frac {4 \operatorname {PolyLog}\left (3,e^x\right )}{3}+\frac {8 \operatorname {PolyLog}\left (4,e^x\right )}{3}-\frac {2}{3} \int \frac {x^3 \log (x)}{\left (-1+e^x\right )^2} \, dx-\frac {2}{3} \int \frac {x^3 \log (x)}{-1+e^x} \, dx+2 \int \frac {x \log (x)}{-1+e^x} \, dx+2 \int \frac {x^2 \log (x)}{-1+e^x} \, dx+2 \int \frac {x \log ^2(x)}{-1+e^x} \, dx-\frac {8}{3} \int \operatorname {PolyLog}\left (3,e^x\right ) \, dx-\frac {8}{3} \int \operatorname {PolyLog}\left (4,e^x\right ) \, dx+\frac {8}{3} \text {Subst}\left (\int \frac {\operatorname {PolyLog}(4,x)}{x} \, dx,x,e^x\right )-\int \frac {x^2 \log ^2(x)}{\left (-1+e^x\right )^2} \, dx-\int \frac {x^2 \log ^2(x)}{-1+e^x} \, dx \\ & = -\frac {2 x^3}{9}-\frac {x^4}{9 \left (1-e^x\right )}+\frac {2}{3} x^2 \log \left (1-e^x\right )+\frac {4 x \operatorname {PolyLog}\left (2,e^x\right )}{3}-\frac {4 \operatorname {PolyLog}\left (3,e^x\right )}{3}+\frac {8 \operatorname {PolyLog}\left (4,e^x\right )}{3}+\frac {8 \operatorname {PolyLog}\left (5,e^x\right )}{3}-\frac {2}{3} \int \frac {x^3 \log (x)}{\left (-1+e^x\right )^2} \, dx-\frac {2}{3} \int \frac {x^3 \log (x)}{-1+e^x} \, dx+2 \int \frac {x \log (x)}{-1+e^x} \, dx+2 \int \frac {x^2 \log (x)}{-1+e^x} \, dx+2 \int \frac {x \log ^2(x)}{-1+e^x} \, dx-\frac {8}{3} \text {Subst}\left (\int \frac {\operatorname {PolyLog}(3,x)}{x} \, dx,x,e^x\right )-\frac {8}{3} \text {Subst}\left (\int \frac {\operatorname {PolyLog}(4,x)}{x} \, dx,x,e^x\right )-\int \frac {x^2 \log ^2(x)}{\left (-1+e^x\right )^2} \, dx-\int \frac {x^2 \log ^2(x)}{-1+e^x} \, dx \\ & = -\frac {2 x^3}{9}-\frac {x^4}{9 \left (1-e^x\right )}+\frac {2}{3} x^2 \log \left (1-e^x\right )+\frac {4 x \operatorname {PolyLog}\left (2,e^x\right )}{3}-\frac {4 \operatorname {PolyLog}\left (3,e^x\right )}{3}-\frac {2}{3} \int \frac {x^3 \log (x)}{\left (-1+e^x\right )^2} \, dx-\frac {2}{3} \int \frac {x^3 \log (x)}{-1+e^x} \, dx+2 \int \frac {x \log (x)}{-1+e^x} \, dx+2 \int \frac {x^2 \log (x)}{-1+e^x} \, dx+2 \int \frac {x \log ^2(x)}{-1+e^x} \, dx-\int \frac {x^2 \log ^2(x)}{\left (-1+e^x\right )^2} \, dx-\int \frac {x^2 \log ^2(x)}{-1+e^x} \, dx \\ \end{align*}
Time = 0.30 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.67 \[ \int \frac {-6 x^2-4 x^3+e^x \left (6 x^2+4 x^3-x^4\right )+\left (-18 x-18 x^2+e^x \left (18 x+18 x^2-6 x^3\right )\right ) \log (x)+\left (-18 x+e^x \left (18 x-9 x^2\right )\right ) \log ^2(x)}{9-18 e^x+9 e^{2 x}} \, dx=\frac {x^2 (x+3 \log (x))^2}{9 \left (-1+e^x\right )} \]
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Time = 0.13 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.94
method | result | size |
parallelrisch | \(-\frac {-x^{4}-6 x^{3} \ln \left (x \right )-9 x^{2} \ln \left (x \right )^{2}}{9 \left ({\mathrm e}^{x}-1\right )}\) | \(31\) |
risch | \(\frac {x^{2} \ln \left (x \right )^{2}}{{\mathrm e}^{x}-1}+\frac {2 x^{3} \ln \left (x \right )}{3 \left ({\mathrm e}^{x}-1\right )}+\frac {x^{4}}{9 \,{\mathrm e}^{x}-9}\) | \(40\) |
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Time = 0.25 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.85 \[ \int \frac {-6 x^2-4 x^3+e^x \left (6 x^2+4 x^3-x^4\right )+\left (-18 x-18 x^2+e^x \left (18 x+18 x^2-6 x^3\right )\right ) \log (x)+\left (-18 x+e^x \left (18 x-9 x^2\right )\right ) \log ^2(x)}{9-18 e^x+9 e^{2 x}} \, dx=\frac {x^{4} + 6 \, x^{3} \log \left (x\right ) + 9 \, x^{2} \log \left (x\right )^{2}}{9 \, {\left (e^{x} - 1\right )}} \]
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Time = 0.07 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.82 \[ \int \frac {-6 x^2-4 x^3+e^x \left (6 x^2+4 x^3-x^4\right )+\left (-18 x-18 x^2+e^x \left (18 x+18 x^2-6 x^3\right )\right ) \log (x)+\left (-18 x+e^x \left (18 x-9 x^2\right )\right ) \log ^2(x)}{9-18 e^x+9 e^{2 x}} \, dx=\frac {x^{4} + 6 x^{3} \log {\left (x \right )} + 9 x^{2} \log {\left (x \right )}^{2}}{9 e^{x} - 9} \]
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Time = 0.23 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.85 \[ \int \frac {-6 x^2-4 x^3+e^x \left (6 x^2+4 x^3-x^4\right )+\left (-18 x-18 x^2+e^x \left (18 x+18 x^2-6 x^3\right )\right ) \log (x)+\left (-18 x+e^x \left (18 x-9 x^2\right )\right ) \log ^2(x)}{9-18 e^x+9 e^{2 x}} \, dx=\frac {x^{4} + 6 \, x^{3} \log \left (x\right ) + 9 \, x^{2} \log \left (x\right )^{2}}{9 \, {\left (e^{x} - 1\right )}} \]
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Time = 0.29 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.85 \[ \int \frac {-6 x^2-4 x^3+e^x \left (6 x^2+4 x^3-x^4\right )+\left (-18 x-18 x^2+e^x \left (18 x+18 x^2-6 x^3\right )\right ) \log (x)+\left (-18 x+e^x \left (18 x-9 x^2\right )\right ) \log ^2(x)}{9-18 e^x+9 e^{2 x}} \, dx=\frac {x^{4} + 6 \, x^{3} \log \left (x\right ) + 9 \, x^{2} \log \left (x\right )^{2}}{9 \, {\left (e^{x} - 1\right )}} \]
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Time = 12.04 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.61 \[ \int \frac {-6 x^2-4 x^3+e^x \left (6 x^2+4 x^3-x^4\right )+\left (-18 x-18 x^2+e^x \left (18 x+18 x^2-6 x^3\right )\right ) \log (x)+\left (-18 x+e^x \left (18 x-9 x^2\right )\right ) \log ^2(x)}{9-18 e^x+9 e^{2 x}} \, dx=\frac {x^2\,{\left (x+3\,\ln \left (x\right )\right )}^2}{9\,\left ({\mathrm {e}}^x-1\right )} \]
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