Integrand size = 49, antiderivative size = 27 \[ \int \frac {x^2+e^{e^{\frac {-1-e-x+\log (5)}{x}}+\frac {-1-e-x+\log (5)}{x}} (1+e-\log (5))}{x^2} \, dx=e^{e^{-x+\frac {-1-e-x+x^2+\log (5)}{x}}}+x \]
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\[ \int \frac {x^2+e^{e^{\frac {-1-e-x+\log (5)}{x}}+\frac {-1-e-x+\log (5)}{x}} (1+e-\log (5))}{x^2} \, dx=\int \frac {x^2+e^{e^{\frac {-1-e-x+\log (5)}{x}}+\frac {-1-e-x+\log (5)}{x}} (1+e-\log (5))}{x^2} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \left (1+\frac {5^{\frac {1}{x}} e^{-1+5^{\frac {1}{x}} e^{-1-\frac {1}{x}-\frac {e}{x}}-\frac {1+e}{x}} (1+e-\log (5))}{x^2}\right ) \, dx \\ & = x+(1+e-\log (5)) \int \frac {5^{\frac {1}{x}} e^{-1+5^{\frac {1}{x}} e^{-1-\frac {1}{x}-\frac {e}{x}}-\frac {1+e}{x}}}{x^2} \, dx \\ & = x+(-1-e+\log (5)) \text {Subst}\left (\int 5^x e^{-1+5^x e^{-1-x-e x}+(-1-e) x} \, dx,x,\frac {1}{x}\right ) \\ \end{align*}
Time = 1.42 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.78 \[ \int \frac {x^2+e^{e^{\frac {-1-e-x+\log (5)}{x}}+\frac {-1-e-x+\log (5)}{x}} (1+e-\log (5))}{x^2} \, dx=e^{5^{\frac {1}{x}} e^{-\frac {1+e+x}{x}}}+x \]
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Time = 0.24 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.74
method | result | size |
parallelrisch | \(x +{\mathrm e}^{{\mathrm e}^{\frac {\ln \left (5\right )-{\mathrm e}-x -1}{x}}}\) | \(20\) |
risch | \(x +{\mathrm e}^{5^{\frac {1}{x}} {\mathrm e}^{-\frac {{\mathrm e}+x +1}{x}}}\) | \(21\) |
norman | \(\frac {x^{2}+x \,{\mathrm e}^{{\mathrm e}^{\frac {\ln \left (5\right )-{\mathrm e}-x -1}{x}}}}{x}\) | \(28\) |
parts | \(x -\frac {\left (-\ln \left (5\right )+1+{\mathrm e}\right ) {\mathrm e}^{{\mathrm e}^{\frac {\ln \left (5\right )-{\mathrm e}-x -1}{x}}}}{\ln \left (5\right )-{\mathrm e}-1}\) | \(40\) |
default | \(x -\frac {{\mathrm e}^{5^{\frac {1}{x}} {\mathrm e}^{-\frac {{\mathrm e}}{x}} {\mathrm e}^{-1} {\mathrm e}^{-\frac {1}{x}}}}{\ln \left (5\right )-{\mathrm e}-1}-\frac {{\mathrm e} \,{\mathrm e}^{5^{\frac {1}{x}} {\mathrm e}^{-\frac {{\mathrm e}}{x}} {\mathrm e}^{-1} {\mathrm e}^{-\frac {1}{x}}}}{\ln \left (5\right )-{\mathrm e}-1}+\frac {\ln \left (5\right ) {\mathrm e}^{5^{\frac {1}{x}} {\mathrm e}^{-\frac {{\mathrm e}}{x}} {\mathrm e}^{-1} {\mathrm e}^{-\frac {1}{x}}}}{\ln \left (5\right )-{\mathrm e}-1}\) | \(114\) |
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Leaf count of result is larger than twice the leaf count of optimal. 66 vs. \(2 (26) = 52\).
Time = 0.25 (sec) , antiderivative size = 66, normalized size of antiderivative = 2.44 \[ \int \frac {x^2+e^{e^{\frac {-1-e-x+\log (5)}{x}}+\frac {-1-e-x+\log (5)}{x}} (1+e-\log (5))}{x^2} \, dx={\left (x e^{\left (-\frac {x + e - \log \left (5\right ) + 1}{x}\right )} + e^{\left (\frac {x e^{\left (-\frac {x + e - \log \left (5\right ) + 1}{x}\right )} - x - e + \log \left (5\right ) - 1}{x}\right )}\right )} e^{\left (\frac {x + e - \log \left (5\right ) + 1}{x}\right )} \]
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Time = 0.13 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.56 \[ \int \frac {x^2+e^{e^{\frac {-1-e-x+\log (5)}{x}}+\frac {-1-e-x+\log (5)}{x}} (1+e-\log (5))}{x^2} \, dx=x + e^{e^{\frac {- x - e - 1 + \log {\left (5 \right )}}{x}}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 106 vs. \(2 (26) = 52\).
Time = 0.33 (sec) , antiderivative size = 106, normalized size of antiderivative = 3.93 \[ \int \frac {x^2+e^{e^{\frac {-1-e-x+\log (5)}{x}}+\frac {-1-e-x+\log (5)}{x}} (1+e-\log (5))}{x^2} \, dx=x - \frac {e^{\left (e^{\left (-\frac {e}{x} + \frac {\log \left (5\right )}{x} - \frac {1}{x} - 1\right )}\right )} \log \left (5\right )}{e - \log \left (5\right ) + 1} + \frac {e^{\left (e^{\left (-\frac {e}{x} + \frac {\log \left (5\right )}{x} - \frac {1}{x} - 1\right )} + 1\right )}}{e - \log \left (5\right ) + 1} + \frac {e^{\left (e^{\left (-\frac {e}{x} + \frac {\log \left (5\right )}{x} - \frac {1}{x} - 1\right )}\right )}}{e - \log \left (5\right ) + 1} \]
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Leaf count of result is larger than twice the leaf count of optimal. 219 vs. \(2 (26) = 52\).
Time = 0.32 (sec) , antiderivative size = 219, normalized size of antiderivative = 8.11 \[ \int \frac {x^2+e^{e^{\frac {-1-e-x+\log (5)}{x}}+\frac {-1-e-x+\log (5)}{x}} (1+e-\log (5))}{x^2} \, dx=\frac {x e^{\left (-\frac {x + e - \log \left (5\right ) + 1}{x}\right )} \log \left (5\right ) - x e^{\left (-\frac {x + e - \log \left (5\right ) + 1}{x} + 1\right )} - x e^{\left (-\frac {x + e - \log \left (5\right ) + 1}{x}\right )} + e^{\left (\frac {x e^{\left (-\frac {x + e - \log \left (5\right ) + 1}{x}\right )} - x - e + \log \left (5\right ) - 1}{x}\right )} \log \left (5\right ) - e^{\left (\frac {x e^{\left (-\frac {x + e - \log \left (5\right ) + 1}{x}\right )} - x - e + \log \left (5\right ) - 1}{x}\right )} - e^{\left (\frac {x e^{\left (-\frac {x + e - \log \left (5\right ) + 1}{x}\right )} - e + \log \left (5\right ) - 1}{x}\right )}}{e^{\left (-\frac {x + e - \log \left (5\right ) + 1}{x}\right )} \log \left (5\right ) - e^{\left (-\frac {x + e - \log \left (5\right ) + 1}{x} + 1\right )} - e^{\left (-\frac {x + e - \log \left (5\right ) + 1}{x}\right )}} \]
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Time = 18.21 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93 \[ \int \frac {x^2+e^{e^{\frac {-1-e-x+\log (5)}{x}}+\frac {-1-e-x+\log (5)}{x}} (1+e-\log (5))}{x^2} \, dx=x+{\mathrm {e}}^{5^{1/x}\,{\mathrm {e}}^{-\frac {\mathrm {e}}{x}}\,{\mathrm {e}}^{-1}\,{\mathrm {e}}^{-\frac {1}{x}}} \]
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