Integrand size = 39, antiderivative size = 25 \[ \int \frac {4+2 x^3+8 \log \left (\frac {5}{x}\right )+e^{x^{\frac {1}{x}}} x^{\frac {1}{x}} (x-x \log (x))}{x^3} \, dx=\frac {3}{4}+e^{x^{\frac {1}{x}}}+2 x-\frac {4 \log \left (\frac {5}{x}\right )}{x^2} \]
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\[ \int \frac {4+2 x^3+8 \log \left (\frac {5}{x}\right )+e^{x^{\frac {1}{x}}} x^{\frac {1}{x}} (x-x \log (x))}{x^3} \, dx=\int \frac {4+2 x^3+8 \log \left (\frac {5}{x}\right )+e^{x^{\frac {1}{x}}} x^{\frac {1}{x}} (x-x \log (x))}{x^3} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {2 \left (2+x^3+4 \log \left (\frac {5}{x}\right )\right )}{x^3}-e^{x^{\frac {1}{x}}} x^{-2+\frac {1}{x}} (-1+\log (x))\right ) \, dx \\ & = 2 \int \frac {2+x^3+4 \log \left (\frac {5}{x}\right )}{x^3} \, dx-\int e^{x^{\frac {1}{x}}} x^{-2+\frac {1}{x}} (-1+\log (x)) \, dx \\ & = 2 \int \left (\frac {2+x^3}{x^3}+\frac {4 \log \left (\frac {5}{x}\right )}{x^3}\right ) \, dx-\int \left (-e^{x^{\frac {1}{x}}} x^{-2+\frac {1}{x}}+e^{x^{\frac {1}{x}}} x^{-2+\frac {1}{x}} \log (x)\right ) \, dx \\ & = 2 \int \frac {2+x^3}{x^3} \, dx+8 \int \frac {\log \left (\frac {5}{x}\right )}{x^3} \, dx+\int e^{x^{\frac {1}{x}}} x^{-2+\frac {1}{x}} \, dx-\int e^{x^{\frac {1}{x}}} x^{-2+\frac {1}{x}} \log (x) \, dx \\ & = \frac {2}{x^2}-\frac {4 \log \left (\frac {5}{x}\right )}{x^2}+2 \int \left (1+\frac {2}{x^3}\right ) \, dx-\log (x) \int e^{x^{\frac {1}{x}}} x^{-2+\frac {1}{x}} \, dx+\int e^{x^{\frac {1}{x}}} x^{-2+\frac {1}{x}} \, dx+\int \frac {\int e^{x^{\frac {1}{x}}} x^{-2+\frac {1}{x}} \, dx}{x} \, dx \\ & = 2 x-\frac {4 \log \left (\frac {5}{x}\right )}{x^2}-\log (x) \int e^{x^{\frac {1}{x}}} x^{-2+\frac {1}{x}} \, dx+\int e^{x^{\frac {1}{x}}} x^{-2+\frac {1}{x}} \, dx+\int \frac {\int e^{x^{\frac {1}{x}}} x^{-2+\frac {1}{x}} \, dx}{x} \, dx \\ \end{align*}
Time = 0.11 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {4+2 x^3+8 \log \left (\frac {5}{x}\right )+e^{x^{\frac {1}{x}}} x^{\frac {1}{x}} (x-x \log (x))}{x^3} \, dx=e^{x^{\frac {1}{x}}}+2 x-\frac {4 \log \left (\frac {5}{x}\right )}{x^2} \]
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Time = 2.72 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.20
method | result | size |
risch | \(\frac {4 \ln \left (x \right )}{x^{2}}-\frac {2 \left (-x^{3}+2 \ln \left (5\right )\right )}{x^{2}}+{\mathrm e}^{x^{\frac {1}{x}}}\) | \(30\) |
parallelrisch | \(-\frac {-4 x^{3}-2 \,{\mathrm e}^{{\mathrm e}^{\frac {\ln \left (x \right )}{x}}} x^{2}+8 \ln \left (\frac {5}{x}\right )}{2 x^{2}}\) | \(33\) |
default | \({\mathrm e}^{{\mathrm e}^{\frac {\ln \left (x \right )}{x}}}-\frac {4 \ln \left (\frac {1}{x}\right )}{x^{2}}+\frac {2}{x^{2}}-\frac {4 \ln \left (5\right )+2}{x^{2}}+2 x\) | \(38\) |
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Time = 0.26 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.56 \[ \int \frac {4+2 x^3+8 \log \left (\frac {5}{x}\right )+e^{x^{\frac {1}{x}}} x^{\frac {1}{x}} (x-x \log (x))}{x^3} \, dx=\frac {2 \, x^{3} + x^{2} e^{\left (e^{\left (\frac {\log \left (5\right ) - \log \left (\frac {5}{x}\right )}{x}\right )}\right )} - 4 \, \log \left (\frac {5}{x}\right )}{x^{2}} \]
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Time = 0.27 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08 \[ \int \frac {4+2 x^3+8 \log \left (\frac {5}{x}\right )+e^{x^{\frac {1}{x}}} x^{\frac {1}{x}} (x-x \log (x))}{x^3} \, dx=2 x + e^{e^{\frac {\log {\left (x \right )}}{x}}} + \frac {4 \log {\left (x \right )}}{x^{2}} - \frac {4 \log {\left (5 \right )}}{x^{2}} \]
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Time = 0.25 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84 \[ \int \frac {4+2 x^3+8 \log \left (\frac {5}{x}\right )+e^{x^{\frac {1}{x}}} x^{\frac {1}{x}} (x-x \log (x))}{x^3} \, dx=2 \, x - \frac {4 \, \log \left (\frac {5}{x}\right )}{x^{2}} + e^{\left (x^{\left (\frac {1}{x}\right )}\right )} \]
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Time = 0.30 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int \frac {4+2 x^3+8 \log \left (\frac {5}{x}\right )+e^{x^{\frac {1}{x}}} x^{\frac {1}{x}} (x-x \log (x))}{x^3} \, dx=2 \, x - \frac {4 \, \log \left (5\right )}{x^{2}} + \frac {4 \, \log \left (x\right )}{x^{2}} + e^{\left (x^{\left (\frac {1}{x}\right )}\right )} \]
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Time = 9.53 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.04 \[ \int \frac {4+2 x^3+8 \log \left (\frac {5}{x}\right )+e^{x^{\frac {1}{x}}} x^{\frac {1}{x}} (x-x \log (x))}{x^3} \, dx=2\,x+{\mathrm {e}}^{x^{1/x}}-\frac {4\,\ln \left (\frac {1}{x}\right )}{x^2}-\frac {4\,\ln \left (5\right )}{x^2} \]
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