Integrand size = 47, antiderivative size = 33 \[ \int \frac {5 x+e^{e^{\frac {-3 x+3 \log (x)}{x}}+\frac {-3 x+3 \log (x)}{x}} (6-6 \log (x))}{2 x^2} \, dx=e^{e^{\frac {3 (-x+\log (x))}{x}}}+\frac {5}{2} \log \left (-x+\frac {e^4 x}{5}\right ) \]
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\[ \int \frac {5 x+e^{e^{\frac {-3 x+3 \log (x)}{x}}+\frac {-3 x+3 \log (x)}{x}} (6-6 \log (x))}{2 x^2} \, dx=\int \frac {5 x+\exp \left (e^{\frac {-3 x+3 \log (x)}{x}}+\frac {-3 x+3 \log (x)}{x}\right ) (6-6 \log (x))}{2 x^2} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \int \frac {5 x+\exp \left (e^{\frac {-3 x+3 \log (x)}{x}}+\frac {-3 x+3 \log (x)}{x}\right ) (6-6 \log (x))}{x^2} \, dx \\ & = \frac {1}{2} \int \left (\frac {5}{x}-6 e^{-3+\frac {x^{3/x}}{e^3}} x^{-2+\frac {3}{x}} (-1+\log (x))\right ) \, dx \\ & = \frac {5 \log (x)}{2}-3 \int e^{-3+\frac {x^{3/x}}{e^3}} x^{-2+\frac {3}{x}} (-1+\log (x)) \, dx \\ & = \frac {5 \log (x)}{2}-3 \int \left (-e^{-3+\frac {x^{3/x}}{e^3}} x^{-2+\frac {3}{x}}+e^{-3+\frac {x^{3/x}}{e^3}} x^{-2+\frac {3}{x}} \log (x)\right ) \, dx \\ & = \frac {5 \log (x)}{2}+3 \int e^{-3+\frac {x^{3/x}}{e^3}} x^{-2+\frac {3}{x}} \, dx-3 \int e^{-3+\frac {x^{3/x}}{e^3}} x^{-2+\frac {3}{x}} \log (x) \, dx \\ & = \frac {5 \log (x)}{2}+3 \int e^{-3+\frac {x^{3/x}}{e^3}} x^{-2+\frac {3}{x}} \, dx+3 \int \frac {\int e^{-3+\frac {x^{3/x}}{e^3}} x^{-2+\frac {3}{x}} \, dx}{x} \, dx-(3 \log (x)) \int e^{-3+\frac {x^{3/x}}{e^3}} x^{-2+\frac {3}{x}} \, dx \\ \end{align*}
Time = 0.23 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.73 \[ \int \frac {5 x+e^{e^{\frac {-3 x+3 \log (x)}{x}}+\frac {-3 x+3 \log (x)}{x}} (6-6 \log (x))}{2 x^2} \, dx=\frac {1}{2} \left (2 e^{\frac {x^{3/x}}{e^3}}+5 \log (x)\right ) \]
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Time = 0.12 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.52
method | result | size |
risch | \(\frac {5 \ln \left (x \right )}{2}+{\mathrm e}^{x^{\frac {3}{x}} {\mathrm e}^{-3}}\) | \(17\) |
parallelrisch | \(\frac {5 \ln \left (x \right )}{2}+{\mathrm e}^{{\mathrm e}^{\frac {3 \ln \left (x \right )-3 x}{x}}}\) | \(19\) |
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Leaf count of result is larger than twice the leaf count of optimal. 60 vs. \(2 (26) = 52\).
Time = 0.26 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.82 \[ \int \frac {5 x+e^{e^{\frac {-3 x+3 \log (x)}{x}}+\frac {-3 x+3 \log (x)}{x}} (6-6 \log (x))}{2 x^2} \, dx=\frac {1}{2} \, {\left (5 \, e^{\left (-\frac {3 \, {\left (x - \log \left (x\right )\right )}}{x}\right )} \log \left (x\right ) + 2 \, e^{\left (\frac {x e^{\left (-\frac {3 \, {\left (x - \log \left (x\right )\right )}}{x}\right )} - 3 \, x + 3 \, \log \left (x\right )}{x}\right )}\right )} e^{\left (\frac {3 \, {\left (x - \log \left (x\right )\right )}}{x}\right )} \]
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Time = 0.24 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.58 \[ \int \frac {5 x+e^{e^{\frac {-3 x+3 \log (x)}{x}}+\frac {-3 x+3 \log (x)}{x}} (6-6 \log (x))}{2 x^2} \, dx=e^{e^{\frac {- 3 x + 3 \log {\left (x \right )}}{x}}} + \frac {5 \log {\left (x \right )}}{2} \]
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Time = 0.32 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.48 \[ \int \frac {5 x+e^{e^{\frac {-3 x+3 \log (x)}{x}}+\frac {-3 x+3 \log (x)}{x}} (6-6 \log (x))}{2 x^2} \, dx=e^{\left (e^{\left (\frac {3 \, \log \left (x\right )}{x} - 3\right )}\right )} + \frac {5}{2} \, \log \left (x\right ) \]
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Leaf count of result is larger than twice the leaf count of optimal. 60 vs. \(2 (26) = 52\).
Time = 0.29 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.82 \[ \int \frac {5 x+e^{e^{\frac {-3 x+3 \log (x)}{x}}+\frac {-3 x+3 \log (x)}{x}} (6-6 \log (x))}{2 x^2} \, dx=\frac {1}{2} \, {\left (5 \, e^{\left (-\frac {3 \, {\left (x - \log \left (x\right )\right )}}{x}\right )} \log \left (x\right ) + 2 \, e^{\left (\frac {x e^{\left (-\frac {3 \, {\left (x - \log \left (x\right )\right )}}{x}\right )} - 3 \, x + 3 \, \log \left (x\right )}{x}\right )}\right )} e^{\left (\frac {3 \, {\left (x - \log \left (x\right )\right )}}{x}\right )} \]
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Time = 12.95 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.48 \[ \int \frac {5 x+e^{e^{\frac {-3 x+3 \log (x)}{x}}+\frac {-3 x+3 \log (x)}{x}} (6-6 \log (x))}{2 x^2} \, dx={\mathrm {e}}^{x^{3/x}\,{\mathrm {e}}^{-3}}+\frac {5\,\ln \left (x\right )}{2} \]
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