Integrand size = 95, antiderivative size = 22 \[ \int \frac {e^{\frac {x+\log \left (1600-800 e^{10/x}+100 e^{20/x}\right )}{x}} \left (-20 e^{10/x}+\left (4 x-e^{10/x} x\right ) \log \left (1600-800 e^{10/x}+100 e^{20/x}\right )\right )}{-4 x^3+e^{10/x} x^3} \, dx=e^{\frac {x+\log \left (100 \left (-4+e^{10/x}\right )^2\right )}{x}} \]
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Time = 0.25 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.18, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.011, Rules used = {6838} \[ \int \frac {e^{\frac {x+\log \left (1600-800 e^{10/x}+100 e^{20/x}\right )}{x}} \left (-20 e^{10/x}+\left (4 x-e^{10/x} x\right ) \log \left (1600-800 e^{10/x}+100 e^{20/x}\right )\right )}{-4 x^3+e^{10/x} x^3} \, dx=e \left (-800 e^{10/x}+100 e^{20/x}+1600\right )^{\frac {1}{x}} \]
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Rule 6838
Rubi steps \begin{align*} \text {integral}& = e \left (1600-800 e^{10/x}+100 e^{20/x}\right )^{\frac {1}{x}} \\ \end{align*}
\[ \int \frac {e^{\frac {x+\log \left (1600-800 e^{10/x}+100 e^{20/x}\right )}{x}} \left (-20 e^{10/x}+\left (4 x-e^{10/x} x\right ) \log \left (1600-800 e^{10/x}+100 e^{20/x}\right )\right )}{-4 x^3+e^{10/x} x^3} \, dx=\int \frac {e^{\frac {x+\log \left (1600-800 e^{10/x}+100 e^{20/x}\right )}{x}} \left (-20 e^{10/x}+\left (4 x-e^{10/x} x\right ) \log \left (1600-800 e^{10/x}+100 e^{20/x}\right )\right )}{-4 x^3+e^{10/x} x^3} \, dx \]
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Time = 3.00 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.32
method | result | size |
parallelrisch | \({\mathrm e}^{\frac {\ln \left (100 \,{\mathrm e}^{\frac {20}{x}}-800 \,{\mathrm e}^{\frac {10}{x}}+1600\right )+x}{x}}\) | \(29\) |
risch | \(4^{\frac {1}{x}} 25^{\frac {1}{x}} \left ({\mathrm e}^{\frac {10}{x}}-4\right )^{\frac {2}{x}} {\mathrm e}^{\frac {-i \pi {\operatorname {csgn}\left (i \left ({\mathrm e}^{\frac {10}{x}}-4\right )\right )}^{2} \operatorname {csgn}\left (i \left ({\mathrm e}^{\frac {10}{x}}-4\right )^{2}\right )+2 i \pi \,\operatorname {csgn}\left (i \left ({\mathrm e}^{\frac {10}{x}}-4\right )\right ) {\operatorname {csgn}\left (i \left ({\mathrm e}^{\frac {10}{x}}-4\right )^{2}\right )}^{2}-i \pi {\operatorname {csgn}\left (i \left ({\mathrm e}^{\frac {10}{x}}-4\right )^{2}\right )}^{3}+2 x}{2 x}}\) | \(120\) |
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none
Time = 0.25 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.18 \[ \int \frac {e^{\frac {x+\log \left (1600-800 e^{10/x}+100 e^{20/x}\right )}{x}} \left (-20 e^{10/x}+\left (4 x-e^{10/x} x\right ) \log \left (1600-800 e^{10/x}+100 e^{20/x}\right )\right )}{-4 x^3+e^{10/x} x^3} \, dx=e^{\left (\frac {x + \log \left (100 \, e^{\frac {20}{x}} - 800 \, e^{\frac {10}{x}} + 1600\right )}{x}\right )} \]
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Timed out. \[ \int \frac {e^{\frac {x+\log \left (1600-800 e^{10/x}+100 e^{20/x}\right )}{x}} \left (-20 e^{10/x}+\left (4 x-e^{10/x} x\right ) \log \left (1600-800 e^{10/x}+100 e^{20/x}\right )\right )}{-4 x^3+e^{10/x} x^3} \, dx=\text {Timed out} \]
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Leaf count of result is larger than twice the leaf count of optimal. 45 vs. \(2 (20) = 40\).
Time = 0.38 (sec) , antiderivative size = 45, normalized size of antiderivative = 2.05 \[ \int \frac {e^{\frac {x+\log \left (1600-800 e^{10/x}+100 e^{20/x}\right )}{x}} \left (-20 e^{10/x}+\left (4 x-e^{10/x} x\right ) \log \left (1600-800 e^{10/x}+100 e^{20/x}\right )\right )}{-4 x^3+e^{10/x} x^3} \, dx=e^{\left (\frac {2 \, \log \left (5\right )}{x} + \frac {2 \, \log \left (2\right )}{x} + \frac {2 \, \log \left (e^{\frac {5}{x}} + 2\right )}{x} + \frac {2 \, \log \left (e^{\frac {5}{x}} - 2\right )}{x} + 1\right )} \]
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\[ \int \frac {e^{\frac {x+\log \left (1600-800 e^{10/x}+100 e^{20/x}\right )}{x}} \left (-20 e^{10/x}+\left (4 x-e^{10/x} x\right ) \log \left (1600-800 e^{10/x}+100 e^{20/x}\right )\right )}{-4 x^3+e^{10/x} x^3} \, dx=\int { -\frac {{\left ({\left (x e^{\frac {10}{x}} - 4 \, x\right )} \log \left (100 \, e^{\frac {20}{x}} - 800 \, e^{\frac {10}{x}} + 1600\right ) + 20 \, e^{\frac {10}{x}}\right )} e^{\left (\frac {x + \log \left (100 \, e^{\frac {20}{x}} - 800 \, e^{\frac {10}{x}} + 1600\right )}{x}\right )}}{x^{3} e^{\frac {10}{x}} - 4 \, x^{3}} \,d x } \]
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Time = 8.55 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.14 \[ \int \frac {e^{\frac {x+\log \left (1600-800 e^{10/x}+100 e^{20/x}\right )}{x}} \left (-20 e^{10/x}+\left (4 x-e^{10/x} x\right ) \log \left (1600-800 e^{10/x}+100 e^{20/x}\right )\right )}{-4 x^3+e^{10/x} x^3} \, dx=\mathrm {e}\,{\left (100\,{\mathrm {e}}^{20/x}-800\,{\mathrm {e}}^{10/x}+1600\right )}^{1/x} \]
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