Integrand size = 66, antiderivative size = 22 \[ \int \frac {e^{\frac {1}{2} \left (-1-\left (-10 x^2+2 x^3\right ) \log (\log (\log (x)))\right )} \left (5 x-x^2+\left (10 x-3 x^2\right ) \log (x) \log (\log (x)) \log (\log (\log (x)))\right )}{20 \log (x) \log (\log (x))} \, dx=\frac {1}{20} e^{-\frac {1}{2}-(-5+x) x^2 \log (\log (\log (x)))} \]
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\[ \int \frac {e^{\frac {1}{2} \left (-1-\left (-10 x^2+2 x^3\right ) \log (\log (\log (x)))\right )} \left (5 x-x^2+\left (10 x-3 x^2\right ) \log (x) \log (\log (x)) \log (\log (\log (x)))\right )}{20 \log (x) \log (\log (x))} \, dx=\int \frac {\exp \left (\frac {1}{2} \left (-1-\left (-10 x^2+2 x^3\right ) \log (\log (\log (x)))\right )\right ) \left (5 x-x^2+\left (10 x-3 x^2\right ) \log (x) \log (\log (x)) \log (\log (\log (x)))\right )}{20 \log (x) \log (\log (x))} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \frac {1}{20} \int \frac {\exp \left (\frac {1}{2} \left (-1-\left (-10 x^2+2 x^3\right ) \log (\log (\log (x)))\right )\right ) \left (5 x-x^2+\left (10 x-3 x^2\right ) \log (x) \log (\log (x)) \log (\log (\log (x)))\right )}{\log (x) \log (\log (x))} \, dx \\ & = \frac {1}{20} \int \frac {x \log ^{-1+5 x^2-x^3}(\log (x)) (5-x-(-10+3 x) \log (x) \log (\log (x)) \log (\log (\log (x))))}{\sqrt {e} \log (x)} \, dx \\ & = \frac {\int \frac {x \log ^{-1+5 x^2-x^3}(\log (x)) (5-x-(-10+3 x) \log (x) \log (\log (x)) \log (\log (\log (x))))}{\log (x)} \, dx}{20 \sqrt {e}} \\ & = \frac {\int \left (-\frac {(-5+x) x \log ^{-1+5 x^2-x^3}(\log (x))}{\log (x)}-x (-10+3 x) \log ^{5 x^2-x^3}(\log (x)) \log (\log (\log (x)))\right ) \, dx}{20 \sqrt {e}} \\ & = -\frac {\int \frac {(-5+x) x \log ^{-1+5 x^2-x^3}(\log (x))}{\log (x)} \, dx}{20 \sqrt {e}}-\frac {\int x (-10+3 x) \log ^{5 x^2-x^3}(\log (x)) \log (\log (\log (x))) \, dx}{20 \sqrt {e}} \\ & = -\frac {\int \left (-\frac {5 x \log ^{-1+5 x^2-x^3}(\log (x))}{\log (x)}+\frac {x^2 \log ^{-1+5 x^2-x^3}(\log (x))}{\log (x)}\right ) \, dx}{20 \sqrt {e}}-\frac {\int x (-10+3 x) \log ^{(5-x) x^2}(\log (x)) \log (\log (\log (x))) \, dx}{20 \sqrt {e}} \\ & = -\frac {\int \frac {x^2 \log ^{-1+5 x^2-x^3}(\log (x))}{\log (x)} \, dx}{20 \sqrt {e}}-\frac {\int \left (-10 x \log ^{(5-x) x^2}(\log (x)) \log (\log (\log (x)))+3 x^2 \log ^{(5-x) x^2}(\log (x)) \log (\log (\log (x)))\right ) \, dx}{20 \sqrt {e}}+\frac {\int \frac {x \log ^{-1+5 x^2-x^3}(\log (x))}{\log (x)} \, dx}{4 \sqrt {e}} \\ & = -\frac {\int \frac {x^2 \log ^{-1+5 x^2-x^3}(\log (x))}{\log (x)} \, dx}{20 \sqrt {e}}-\frac {3 \int x^2 \log ^{(5-x) x^2}(\log (x)) \log (\log (\log (x))) \, dx}{20 \sqrt {e}}+\frac {\int \frac {x \log ^{-1+5 x^2-x^3}(\log (x))}{\log (x)} \, dx}{4 \sqrt {e}}+\frac {\int x \log ^{(5-x) x^2}(\log (x)) \log (\log (\log (x))) \, dx}{2 \sqrt {e}} \\ \end{align*}
Time = 0.17 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95 \[ \int \frac {e^{\frac {1}{2} \left (-1-\left (-10 x^2+2 x^3\right ) \log (\log (\log (x)))\right )} \left (5 x-x^2+\left (10 x-3 x^2\right ) \log (x) \log (\log (x)) \log (\log (\log (x)))\right )}{20 \log (x) \log (\log (x))} \, dx=\frac {\log ^{-\left ((-5+x) x^2\right )}(\log (x))}{20 \sqrt {e}} \]
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Time = 4.69 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.82
method | result | size |
risch | \(\frac {\ln \left (\ln \left (x \right )\right )^{-x^{2} \left (-5+x \right )} {\mathrm e}^{-\frac {1}{2}}}{20}\) | \(18\) |
parallelrisch | \(\frac {{\mathrm e}^{\left (-x^{3}+5 x^{2}\right ) \ln \left (\ln \left (\ln \left (x \right )\right )\right )-\frac {1}{2}}}{20}\) | \(25\) |
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Time = 0.28 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {e^{\frac {1}{2} \left (-1-\left (-10 x^2+2 x^3\right ) \log (\log (\log (x)))\right )} \left (5 x-x^2+\left (10 x-3 x^2\right ) \log (x) \log (\log (x)) \log (\log (\log (x)))\right )}{20 \log (x) \log (\log (x))} \, dx=\frac {1}{20} \, e^{\left (-{\left (x^{3} - 5 \, x^{2}\right )} \log \left (\log \left (\log \left (x\right )\right )\right ) - \frac {1}{2}\right )} \]
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Time = 0.72 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {1}{2} \left (-1-\left (-10 x^2+2 x^3\right ) \log (\log (\log (x)))\right )} \left (5 x-x^2+\left (10 x-3 x^2\right ) \log (x) \log (\log (x)) \log (\log (\log (x)))\right )}{20 \log (x) \log (\log (x))} \, dx=\frac {e^{- \left (x^{3} - 5 x^{2}\right ) \log {\left (\log {\left (\log {\left (x \right )} \right )} \right )} - \frac {1}{2}}}{20} \]
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Time = 0.33 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.05 \[ \int \frac {e^{\frac {1}{2} \left (-1-\left (-10 x^2+2 x^3\right ) \log (\log (\log (x)))\right )} \left (5 x-x^2+\left (10 x-3 x^2\right ) \log (x) \log (\log (x)) \log (\log (\log (x)))\right )}{20 \log (x) \log (\log (x))} \, dx=\frac {1}{20} \, e^{\left (-x^{3} \log \left (\log \left (\log \left (x\right )\right )\right ) + 5 \, x^{2} \log \left (\log \left (\log \left (x\right )\right )\right ) - \frac {1}{2}\right )} \]
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Time = 1.71 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.05 \[ \int \frac {e^{\frac {1}{2} \left (-1-\left (-10 x^2+2 x^3\right ) \log (\log (\log (x)))\right )} \left (5 x-x^2+\left (10 x-3 x^2\right ) \log (x) \log (\log (x)) \log (\log (\log (x)))\right )}{20 \log (x) \log (\log (x))} \, dx=\frac {1}{20} \, e^{\left (-x^{3} \log \left (\log \left (\log \left (x\right )\right )\right ) + 5 \, x^{2} \log \left (\log \left (\log \left (x\right )\right )\right ) - \frac {1}{2}\right )} \]
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Time = 9.39 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.86 \[ \int \frac {e^{\frac {1}{2} \left (-1-\left (-10 x^2+2 x^3\right ) \log (\log (\log (x)))\right )} \left (5 x-x^2+\left (10 x-3 x^2\right ) \log (x) \log (\log (x)) \log (\log (\log (x)))\right )}{20 \log (x) \log (\log (x))} \, dx=\frac {{\ln \left (\ln \left (x\right )\right )}^{5\,x^2-x^3}\,{\mathrm {e}}^{-\frac {1}{2}}}{20} \]
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