Integrand size = 172, antiderivative size = 37 \[ \int \frac {5 e^{8+18 x} x^2+5 x^4+e^{\left .\frac {2}{5}\right /x} \left (-9 x+15 x^2\right )+e^{4+9 x} \left (10 x^3+e^{\left .\frac {2}{5}\right /x} \left (6+135 x^2\right )\right )+\left (-6 e^{\left .\frac {2}{5}\right /x}-10 e^{4+9 x} x^2-10 x^3\right ) \log (x)+5 x^2 \log ^2(x)}{60 e^{8+18 x} x^2+120 e^{4+9 x} x^3+60 x^4+\left (-120 e^{4+9 x} x^2-120 x^3\right ) \log (x)+60 x^2 \log ^2(x)} \, dx=\frac {1}{4} \left (\frac {x}{3}+\frac {e^{\left .\frac {2}{5}\right /x}}{-e^{4+9 x}-x+\log (x)}\right ) \]
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\[ \int \frac {5 e^{8+18 x} x^2+5 x^4+e^{\left .\frac {2}{5}\right /x} \left (-9 x+15 x^2\right )+e^{4+9 x} \left (10 x^3+e^{\left .\frac {2}{5}\right /x} \left (6+135 x^2\right )\right )+\left (-6 e^{\left .\frac {2}{5}\right /x}-10 e^{4+9 x} x^2-10 x^3\right ) \log (x)+5 x^2 \log ^2(x)}{60 e^{8+18 x} x^2+120 e^{4+9 x} x^3+60 x^4+\left (-120 e^{4+9 x} x^2-120 x^3\right ) \log (x)+60 x^2 \log ^2(x)} \, dx=\int \frac {5 e^{8+18 x} x^2+5 x^4+e^{\left .\frac {2}{5}\right /x} \left (-9 x+15 x^2\right )+e^{4+9 x} \left (10 x^3+e^{\left .\frac {2}{5}\right /x} \left (6+135 x^2\right )\right )+\left (-6 e^{\left .\frac {2}{5}\right /x}-10 e^{4+9 x} x^2-10 x^3\right ) \log (x)+5 x^2 \log ^2(x)}{60 e^{8+18 x} x^2+120 e^{4+9 x} x^3+60 x^4+\left (-120 e^{4+9 x} x^2-120 x^3\right ) \log (x)+60 x^2 \log ^2(x)} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {5 e^{8+18 x} x^2+5 x^4+e^{\left .\frac {2}{5}\right /x} \left (-9 x+15 x^2\right )+e^{4+9 x} \left (10 x^3+e^{\left .\frac {2}{5}\right /x} \left (6+135 x^2\right )\right )+\left (-6 e^{\left .\frac {2}{5}\right /x}-10 e^{4+9 x} x^2-10 x^3\right ) \log (x)+5 x^2 \log ^2(x)}{60 x^2 \left (e^{4+9 x}+x-\log (x)\right )^2} \, dx \\ & = \frac {1}{60} \int \frac {5 e^{8+18 x} x^2+5 x^4+e^{\left .\frac {2}{5}\right /x} \left (-9 x+15 x^2\right )+e^{4+9 x} \left (10 x^3+e^{\left .\frac {2}{5}\right /x} \left (6+135 x^2\right )\right )+\left (-6 e^{\left .\frac {2}{5}\right /x}-10 e^{4+9 x} x^2-10 x^3\right ) \log (x)+5 x^2 \log ^2(x)}{x^2 \left (e^{4+9 x}+x-\log (x)\right )^2} \, dx \\ & = \frac {1}{60} \int \left (5+\frac {3 e^{\left .\frac {2}{5}\right /x} \left (2+45 x^2\right )}{x^2 \left (e^{4+9 x}+x-\log (x)\right )}-\frac {15 e^{\left .\frac {2}{5}\right /x} \left (1-x+9 x^2-9 x \log (x)\right )}{x \left (e^{4+9 x}+x-\log (x)\right )^2}\right ) \, dx \\ & = \frac {x}{12}+\frac {1}{20} \int \frac {e^{\left .\frac {2}{5}\right /x} \left (2+45 x^2\right )}{x^2 \left (e^{4+9 x}+x-\log (x)\right )} \, dx-\frac {1}{4} \int \frac {e^{\left .\frac {2}{5}\right /x} \left (1-x+9 x^2-9 x \log (x)\right )}{x \left (e^{4+9 x}+x-\log (x)\right )^2} \, dx \\ & = \frac {x}{12}+\frac {1}{20} \int \left (\frac {45 e^{\left .\frac {2}{5}\right /x}}{e^{4+9 x}+x-\log (x)}+\frac {2 e^{\left .\frac {2}{5}\right /x}}{x^2 \left (e^{4+9 x}+x-\log (x)\right )}\right ) \, dx-\frac {1}{4} \int \left (-\frac {e^{\left .\frac {2}{5}\right /x}}{\left (e^{4+9 x}+x-\log (x)\right )^2}+\frac {e^{\left .\frac {2}{5}\right /x}}{x \left (e^{4+9 x}+x-\log (x)\right )^2}+\frac {9 e^{\left .\frac {2}{5}\right /x} x}{\left (e^{4+9 x}+x-\log (x)\right )^2}-\frac {9 e^{\left .\frac {2}{5}\right /x} \log (x)}{\left (e^{4+9 x}+x-\log (x)\right )^2}\right ) \, dx \\ & = \frac {x}{12}+\frac {1}{10} \int \frac {e^{\left .\frac {2}{5}\right /x}}{x^2 \left (e^{4+9 x}+x-\log (x)\right )} \, dx+\frac {1}{4} \int \frac {e^{\left .\frac {2}{5}\right /x}}{\left (e^{4+9 x}+x-\log (x)\right )^2} \, dx-\frac {1}{4} \int \frac {e^{\left .\frac {2}{5}\right /x}}{x \left (e^{4+9 x}+x-\log (x)\right )^2} \, dx-\frac {9}{4} \int \frac {e^{\left .\frac {2}{5}\right /x} x}{\left (e^{4+9 x}+x-\log (x)\right )^2} \, dx+\frac {9}{4} \int \frac {e^{\left .\frac {2}{5}\right /x}}{e^{4+9 x}+x-\log (x)} \, dx+\frac {9}{4} \int \frac {e^{\left .\frac {2}{5}\right /x} \log (x)}{\left (e^{4+9 x}+x-\log (x)\right )^2} \, dx \\ \end{align*}
Time = 0.11 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.30 \[ \int \frac {5 e^{8+18 x} x^2+5 x^4+e^{\left .\frac {2}{5}\right /x} \left (-9 x+15 x^2\right )+e^{4+9 x} \left (10 x^3+e^{\left .\frac {2}{5}\right /x} \left (6+135 x^2\right )\right )+\left (-6 e^{\left .\frac {2}{5}\right /x}-10 e^{4+9 x} x^2-10 x^3\right ) \log (x)+5 x^2 \log ^2(x)}{60 e^{8+18 x} x^2+120 e^{4+9 x} x^3+60 x^4+\left (-120 e^{4+9 x} x^2-120 x^3\right ) \log (x)+60 x^2 \log ^2(x)} \, dx=\frac {-3 e^{\left .\frac {2}{5}\right /x}+e^{4+9 x} x+x^2-x \log (x)}{12 \left (e^{4+9 x}+x-\log (x)\right )} \]
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Time = 0.68 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.73
method | result | size |
risch | \(\frac {x}{12}-\frac {{\mathrm e}^{\frac {2}{5 x}}}{4 \left (x +{\mathrm e}^{9 x +4}-\ln \left (x \right )\right )}\) | \(27\) |
parallelrisch | \(\frac {5 x^{2}-5 x \ln \left (x \right )+5 x \,{\mathrm e}^{9 x +4}-15 \,{\mathrm e}^{\frac {2}{5 x}}}{60 x +60 \,{\mathrm e}^{9 x +4}-60 \ln \left (x \right )}\) | \(45\) |
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Time = 0.25 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.11 \[ \int \frac {5 e^{8+18 x} x^2+5 x^4+e^{\left .\frac {2}{5}\right /x} \left (-9 x+15 x^2\right )+e^{4+9 x} \left (10 x^3+e^{\left .\frac {2}{5}\right /x} \left (6+135 x^2\right )\right )+\left (-6 e^{\left .\frac {2}{5}\right /x}-10 e^{4+9 x} x^2-10 x^3\right ) \log (x)+5 x^2 \log ^2(x)}{60 e^{8+18 x} x^2+120 e^{4+9 x} x^3+60 x^4+\left (-120 e^{4+9 x} x^2-120 x^3\right ) \log (x)+60 x^2 \log ^2(x)} \, dx=\frac {x^{2} + x e^{\left (9 \, x + 4\right )} - x \log \left (x\right ) - 3 \, e^{\left (\frac {2}{5 \, x}\right )}}{12 \, {\left (x + e^{\left (9 \, x + 4\right )} - \log \left (x\right )\right )}} \]
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Time = 0.11 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.70 \[ \int \frac {5 e^{8+18 x} x^2+5 x^4+e^{\left .\frac {2}{5}\right /x} \left (-9 x+15 x^2\right )+e^{4+9 x} \left (10 x^3+e^{\left .\frac {2}{5}\right /x} \left (6+135 x^2\right )\right )+\left (-6 e^{\left .\frac {2}{5}\right /x}-10 e^{4+9 x} x^2-10 x^3\right ) \log (x)+5 x^2 \log ^2(x)}{60 e^{8+18 x} x^2+120 e^{4+9 x} x^3+60 x^4+\left (-120 e^{4+9 x} x^2-120 x^3\right ) \log (x)+60 x^2 \log ^2(x)} \, dx=\frac {x}{12} - \frac {e^{\frac {2}{5 x}}}{4 x + 4 e^{9 x + 4} - 4 \log {\left (x \right )}} \]
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Time = 0.24 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.11 \[ \int \frac {5 e^{8+18 x} x^2+5 x^4+e^{\left .\frac {2}{5}\right /x} \left (-9 x+15 x^2\right )+e^{4+9 x} \left (10 x^3+e^{\left .\frac {2}{5}\right /x} \left (6+135 x^2\right )\right )+\left (-6 e^{\left .\frac {2}{5}\right /x}-10 e^{4+9 x} x^2-10 x^3\right ) \log (x)+5 x^2 \log ^2(x)}{60 e^{8+18 x} x^2+120 e^{4+9 x} x^3+60 x^4+\left (-120 e^{4+9 x} x^2-120 x^3\right ) \log (x)+60 x^2 \log ^2(x)} \, dx=\frac {x^{2} + x e^{\left (9 \, x + 4\right )} - x \log \left (x\right ) - 3 \, e^{\left (\frac {2}{5 \, x}\right )}}{12 \, {\left (x + e^{\left (9 \, x + 4\right )} - \log \left (x\right )\right )}} \]
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Exception generated. \[ \int \frac {5 e^{8+18 x} x^2+5 x^4+e^{\left .\frac {2}{5}\right /x} \left (-9 x+15 x^2\right )+e^{4+9 x} \left (10 x^3+e^{\left .\frac {2}{5}\right /x} \left (6+135 x^2\right )\right )+\left (-6 e^{\left .\frac {2}{5}\right /x}-10 e^{4+9 x} x^2-10 x^3\right ) \log (x)+5 x^2 \log ^2(x)}{60 e^{8+18 x} x^2+120 e^{4+9 x} x^3+60 x^4+\left (-120 e^{4+9 x} x^2-120 x^3\right ) \log (x)+60 x^2 \log ^2(x)} \, dx=\text {Exception raised: TypeError} \]
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Time = 12.55 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.81 \[ \int \frac {5 e^{8+18 x} x^2+5 x^4+e^{\left .\frac {2}{5}\right /x} \left (-9 x+15 x^2\right )+e^{4+9 x} \left (10 x^3+e^{\left .\frac {2}{5}\right /x} \left (6+135 x^2\right )\right )+\left (-6 e^{\left .\frac {2}{5}\right /x}-10 e^{4+9 x} x^2-10 x^3\right ) \log (x)+5 x^2 \log ^2(x)}{60 e^{8+18 x} x^2+120 e^{4+9 x} x^3+60 x^4+\left (-120 e^{4+9 x} x^2-120 x^3\right ) \log (x)+60 x^2 \log ^2(x)} \, dx=\frac {x}{12}-\frac {{\mathrm {e}}^{\frac {2}{5\,x}}}{4\,\left (x+{\mathrm {e}}^{9\,x+4}-\ln \left (x\right )\right )} \]
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