Integrand size = 105, antiderivative size = 31 \[ \int \frac {162 x^5+9 x^6+e^{2 e^x} \left (135 x^4-108 e^x x^5\right )}{27 e^{6 e^x}+216 x^3+108 x^4+18 x^5+x^6+e^{4 e^x} \left (162 x+27 x^2\right )+e^{2 e^x} \left (324 x^2+108 x^3+9 x^4\right )} \, dx=\frac {x^3}{\left (\frac {x}{3}+\left (\frac {e^{2 e^x}}{x^2}+\frac {2}{x}\right ) x\right )^2} \]
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\[ \int \frac {162 x^5+9 x^6+e^{2 e^x} \left (135 x^4-108 e^x x^5\right )}{27 e^{6 e^x}+216 x^3+108 x^4+18 x^5+x^6+e^{4 e^x} \left (162 x+27 x^2\right )+e^{2 e^x} \left (324 x^2+108 x^3+9 x^4\right )} \, dx=\int \frac {162 x^5+9 x^6+e^{2 e^x} \left (135 x^4-108 e^x x^5\right )}{27 e^{6 e^x}+216 x^3+108 x^4+18 x^5+x^6+e^{4 e^x} \left (162 x+27 x^2\right )+e^{2 e^x} \left (324 x^2+108 x^3+9 x^4\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {9 x^4 \left (15 e^{2 e^x}-12 e^{2 e^x+x} x+x (18+x)\right )}{\left (3 e^{2 e^x}+x (6+x)\right )^3} \, dx \\ & = 9 \int \frac {x^4 \left (15 e^{2 e^x}-12 e^{2 e^x+x} x+x (18+x)\right )}{\left (3 e^{2 e^x}+x (6+x)\right )^3} \, dx \\ & = 9 \int \left (-\frac {12 e^{2 e^x+x} x^5}{\left (3 e^{2 e^x}+6 x+x^2\right )^3}+\frac {x^4 \left (15 e^{2 e^x}+18 x+x^2\right )}{\left (3 e^{2 e^x}+6 x+x^2\right )^3}\right ) \, dx \\ & = 9 \int \frac {x^4 \left (15 e^{2 e^x}+18 x+x^2\right )}{\left (3 e^{2 e^x}+6 x+x^2\right )^3} \, dx-108 \int \frac {e^{2 e^x+x} x^5}{\left (3 e^{2 e^x}+6 x+x^2\right )^3} \, dx \\ & = 9 \int \left (-\frac {4 x^5 (3+x)}{\left (3 e^{2 e^x}+6 x+x^2\right )^3}+\frac {5 x^4}{\left (3 e^{2 e^x}+6 x+x^2\right )^2}\right ) \, dx-108 \int \frac {e^{2 e^x+x} x^5}{\left (3 e^{2 e^x}+6 x+x^2\right )^3} \, dx \\ & = -\left (36 \int \frac {x^5 (3+x)}{\left (3 e^{2 e^x}+6 x+x^2\right )^3} \, dx\right )+45 \int \frac {x^4}{\left (3 e^{2 e^x}+6 x+x^2\right )^2} \, dx-108 \int \frac {e^{2 e^x+x} x^5}{\left (3 e^{2 e^x}+6 x+x^2\right )^3} \, dx \\ & = -\left (36 \int \left (\frac {3 x^5}{\left (3 e^{2 e^x}+6 x+x^2\right )^3}+\frac {x^6}{\left (3 e^{2 e^x}+6 x+x^2\right )^3}\right ) \, dx\right )+45 \int \frac {x^4}{\left (3 e^{2 e^x}+6 x+x^2\right )^2} \, dx-108 \int \frac {e^{2 e^x+x} x^5}{\left (3 e^{2 e^x}+6 x+x^2\right )^3} \, dx \\ & = -\left (36 \int \frac {x^6}{\left (3 e^{2 e^x}+6 x+x^2\right )^3} \, dx\right )+45 \int \frac {x^4}{\left (3 e^{2 e^x}+6 x+x^2\right )^2} \, dx-108 \int \frac {x^5}{\left (3 e^{2 e^x}+6 x+x^2\right )^3} \, dx-108 \int \frac {e^{2 e^x+x} x^5}{\left (3 e^{2 e^x}+6 x+x^2\right )^3} \, dx \\ \end{align*}
Time = 0.59 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.74 \[ \int \frac {162 x^5+9 x^6+e^{2 e^x} \left (135 x^4-108 e^x x^5\right )}{27 e^{6 e^x}+216 x^3+108 x^4+18 x^5+x^6+e^{4 e^x} \left (162 x+27 x^2\right )+e^{2 e^x} \left (324 x^2+108 x^3+9 x^4\right )} \, dx=\frac {9 x^5}{\left (3 e^{2 e^x}+6 x+x^2\right )^2} \]
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Time = 0.60 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.71
method | result | size |
risch | \(\frac {9 x^{5}}{\left (3 \,{\mathrm e}^{2 \,{\mathrm e}^{x}}+x^{2}+6 x \right )^{2}}\) | \(22\) |
parallelrisch | \(\frac {9 x^{5}}{x^{4}+6 \,{\mathrm e}^{2 \,{\mathrm e}^{x}} x^{2}+9 \,{\mathrm e}^{4 \,{\mathrm e}^{x}}+12 x^{3}+36 \,{\mathrm e}^{2 \,{\mathrm e}^{x}} x +36 x^{2}}\) | \(47\) |
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Time = 0.25 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.35 \[ \int \frac {162 x^5+9 x^6+e^{2 e^x} \left (135 x^4-108 e^x x^5\right )}{27 e^{6 e^x}+216 x^3+108 x^4+18 x^5+x^6+e^{4 e^x} \left (162 x+27 x^2\right )+e^{2 e^x} \left (324 x^2+108 x^3+9 x^4\right )} \, dx=\frac {9 \, x^{5}}{x^{4} + 12 \, x^{3} + 36 \, x^{2} + 6 \, {\left (x^{2} + 6 \, x\right )} e^{\left (2 \, e^{x}\right )} + 9 \, e^{\left (4 \, e^{x}\right )}} \]
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Time = 0.11 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.35 \[ \int \frac {162 x^5+9 x^6+e^{2 e^x} \left (135 x^4-108 e^x x^5\right )}{27 e^{6 e^x}+216 x^3+108 x^4+18 x^5+x^6+e^{4 e^x} \left (162 x+27 x^2\right )+e^{2 e^x} \left (324 x^2+108 x^3+9 x^4\right )} \, dx=\frac {x^{5}}{\frac {x^{4}}{9} + \frac {4 x^{3}}{3} + 4 x^{2} + \left (\frac {2 x^{2}}{3} + 4 x\right ) e^{2 e^{x}} + e^{4 e^{x}}} \]
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Time = 0.27 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.35 \[ \int \frac {162 x^5+9 x^6+e^{2 e^x} \left (135 x^4-108 e^x x^5\right )}{27 e^{6 e^x}+216 x^3+108 x^4+18 x^5+x^6+e^{4 e^x} \left (162 x+27 x^2\right )+e^{2 e^x} \left (324 x^2+108 x^3+9 x^4\right )} \, dx=\frac {9 \, x^{5}}{x^{4} + 12 \, x^{3} + 36 \, x^{2} + 6 \, {\left (x^{2} + 6 \, x\right )} e^{\left (2 \, e^{x}\right )} + 9 \, e^{\left (4 \, e^{x}\right )}} \]
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Time = 0.31 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.48 \[ \int \frac {162 x^5+9 x^6+e^{2 e^x} \left (135 x^4-108 e^x x^5\right )}{27 e^{6 e^x}+216 x^3+108 x^4+18 x^5+x^6+e^{4 e^x} \left (162 x+27 x^2\right )+e^{2 e^x} \left (324 x^2+108 x^3+9 x^4\right )} \, dx=\frac {9 \, x^{5}}{x^{4} + 12 \, x^{3} + 6 \, x^{2} e^{\left (2 \, e^{x}\right )} + 36 \, x^{2} + 36 \, x e^{\left (2 \, e^{x}\right )} + 9 \, e^{\left (4 \, e^{x}\right )}} \]
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Timed out. \[ \int \frac {162 x^5+9 x^6+e^{2 e^x} \left (135 x^4-108 e^x x^5\right )}{27 e^{6 e^x}+216 x^3+108 x^4+18 x^5+x^6+e^{4 e^x} \left (162 x+27 x^2\right )+e^{2 e^x} \left (324 x^2+108 x^3+9 x^4\right )} \, dx=\int \frac {162\,x^5-{\mathrm {e}}^{2\,{\mathrm {e}}^x}\,\left (108\,x^5\,{\mathrm {e}}^x-135\,x^4\right )+9\,x^6}{27\,{\mathrm {e}}^{6\,{\mathrm {e}}^x}+{\mathrm {e}}^{2\,{\mathrm {e}}^x}\,\left (9\,x^4+108\,x^3+324\,x^2\right )+{\mathrm {e}}^{4\,{\mathrm {e}}^x}\,\left (27\,x^2+162\,x\right )+216\,x^3+108\,x^4+18\,x^5+x^6} \,d x \]
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