Integrand size = 109, antiderivative size = 22 \[ \int \frac {2^{\frac {10 x}{6 \log (x)+2 e^{\frac {x^2}{e^5}} \log (x)}} \left (-15 \log (2)+15 \log (2) \log (x)+e^{\frac {x^2}{e^5}} \left (-5 \log (2)+5 \log (2) \log (x)-\frac {10 x^2 \log (2) \log (x)}{e^5}\right )\right )}{9 \log ^2(x)+6 e^{\frac {x^2}{e^5}} \log ^2(x)+e^{\frac {2 x^2}{e^5}} \log ^2(x)} \, dx=2^{\frac {5 x}{\left (3+e^{\frac {x^2}{e^5}}\right ) \log (x)}} \]
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\[ \int \frac {2^{\frac {10 x}{6 \log (x)+2 e^{\frac {x^2}{e^5}} \log (x)}} \left (-15 \log (2)+15 \log (2) \log (x)+e^{\frac {x^2}{e^5}} \left (-5 \log (2)+5 \log (2) \log (x)-\frac {10 x^2 \log (2) \log (x)}{e^5}\right )\right )}{9 \log ^2(x)+6 e^{\frac {x^2}{e^5}} \log ^2(x)+e^{\frac {2 x^2}{e^5}} \log ^2(x)} \, dx=\int \frac {2^{\frac {10 x}{6 \log (x)+2 e^{\frac {x^2}{e^5}} \log (x)}} \left (-15 \log (2)+15 \log (2) \log (x)+e^{\frac {x^2}{e^5}} \left (-5 \log (2)+5 \log (2) \log (x)-\frac {10 x^2 \log (2) \log (x)}{e^5}\right )\right )}{9 \log ^2(x)+6 e^{\frac {x^2}{e^5}} \log ^2(x)+e^{\frac {2 x^2}{e^5}} \log ^2(x)} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {2^{\frac {5 x}{\left (3+e^{\frac {x^2}{e^5}}\right ) \log (x)}} \left (-15 \log (2)+15 \log (2) \log (x)+e^{\frac {x^2}{e^5}} \left (-5 \log (2)+5 \log (2) \log (x)-\frac {10 x^2 \log (2) \log (x)}{e^5}\right )\right )}{\left (3+e^{\frac {x^2}{e^5}}\right )^2 \log ^2(x)} \, dx \\ & = \int \left (\frac {15\ 2^{1+\frac {5 x}{\left (3+e^{\frac {x^2}{e^5}}\right ) \log (x)}} x^2 \log (2)}{e^5 \left (3+e^{\frac {x^2}{e^5}}\right )^2 \log (x)}+\frac {5\ 2^{\frac {5 x}{\left (3+e^{\frac {x^2}{e^5}}\right ) \log (x)}} \log (2) \left (-e^5+e^5 \log (x)-2 x^2 \log (x)\right )}{e^5 \left (3+e^{\frac {x^2}{e^5}}\right ) \log ^2(x)}\right ) \, dx \\ & = \frac {(5 \log (2)) \int \frac {2^{\frac {5 x}{\left (3+e^{\frac {x^2}{e^5}}\right ) \log (x)}} \left (-e^5+e^5 \log (x)-2 x^2 \log (x)\right )}{\left (3+e^{\frac {x^2}{e^5}}\right ) \log ^2(x)} \, dx}{e^5}+\frac {(15 \log (2)) \int \frac {2^{1+\frac {5 x}{\left (3+e^{\frac {x^2}{e^5}}\right ) \log (x)}} x^2}{\left (3+e^{\frac {x^2}{e^5}}\right )^2 \log (x)} \, dx}{e^5} \\ & = \frac {(5 \log (2)) \int \left (-\frac {2^{\frac {5 x}{\left (3+e^{\frac {x^2}{e^5}}\right ) \log (x)}} e^5}{\left (3+e^{\frac {x^2}{e^5}}\right ) \log ^2(x)}+\frac {2^{\frac {5 x}{\left (3+e^{\frac {x^2}{e^5}}\right ) \log (x)}} e^5}{\left (3+e^{\frac {x^2}{e^5}}\right ) \log (x)}-\frac {2^{1+\frac {5 x}{\left (3+e^{\frac {x^2}{e^5}}\right ) \log (x)}} x^2}{\left (3+e^{\frac {x^2}{e^5}}\right ) \log (x)}\right ) \, dx}{e^5}+\frac {(15 \log (2)) \int \frac {2^{1+\frac {5 x}{\left (3+e^{\frac {x^2}{e^5}}\right ) \log (x)}} x^2}{\left (3+e^{\frac {x^2}{e^5}}\right )^2 \log (x)} \, dx}{e^5} \\ & = -\left ((5 \log (2)) \int \frac {2^{\frac {5 x}{\left (3+e^{\frac {x^2}{e^5}}\right ) \log (x)}}}{\left (3+e^{\frac {x^2}{e^5}}\right ) \log ^2(x)} \, dx\right )+(5 \log (2)) \int \frac {2^{\frac {5 x}{\left (3+e^{\frac {x^2}{e^5}}\right ) \log (x)}}}{\left (3+e^{\frac {x^2}{e^5}}\right ) \log (x)} \, dx-\frac {(5 \log (2)) \int \frac {2^{1+\frac {5 x}{\left (3+e^{\frac {x^2}{e^5}}\right ) \log (x)}} x^2}{\left (3+e^{\frac {x^2}{e^5}}\right ) \log (x)} \, dx}{e^5}+\frac {(15 \log (2)) \int \frac {2^{1+\frac {5 x}{\left (3+e^{\frac {x^2}{e^5}}\right ) \log (x)}} x^2}{\left (3+e^{\frac {x^2}{e^5}}\right )^2 \log (x)} \, dx}{e^5} \\ \end{align*}
Time = 0.11 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.36 \[ \int \frac {2^{\frac {10 x}{6 \log (x)+2 e^{\frac {x^2}{e^5}} \log (x)}} \left (-15 \log (2)+15 \log (2) \log (x)+e^{\frac {x^2}{e^5}} \left (-5 \log (2)+5 \log (2) \log (x)-\frac {10 x^2 \log (2) \log (x)}{e^5}\right )\right )}{9 \log ^2(x)+6 e^{\frac {x^2}{e^5}} \log ^2(x)+e^{\frac {2 x^2}{e^5}} \log ^2(x)} \, dx=\frac {5\ 2^{\frac {5 x}{\left (3+e^{\frac {x^2}{e^5}}\right ) \log (x)}} \log (2)}{\log (32)} \]
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Time = 0.26 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.05
\[2^{\frac {5 x}{\ln \left (x \right ) \left ({\mathrm e}^{x^{2} {\mathrm e}^{-5}}+3\right )}}\]
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Time = 0.28 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {2^{\frac {10 x}{6 \log (x)+2 e^{\frac {x^2}{e^5}} \log (x)}} \left (-15 \log (2)+15 \log (2) \log (x)+e^{\frac {x^2}{e^5}} \left (-5 \log (2)+5 \log (2) \log (x)-\frac {10 x^2 \log (2) \log (x)}{e^5}\right )\right )}{9 \log ^2(x)+6 e^{\frac {x^2}{e^5}} \log ^2(x)+e^{\frac {2 x^2}{e^5}} \log ^2(x)} \, dx=2^{\frac {5 \, x}{{\left (e^{\left (x^{2} e^{\left (-5\right )}\right )} + 3\right )} \log \left (x\right )}} \]
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Time = 0.59 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.18 \[ \int \frac {2^{\frac {10 x}{6 \log (x)+2 e^{\frac {x^2}{e^5}} \log (x)}} \left (-15 \log (2)+15 \log (2) \log (x)+e^{\frac {x^2}{e^5}} \left (-5 \log (2)+5 \log (2) \log (x)-\frac {10 x^2 \log (2) \log (x)}{e^5}\right )\right )}{9 \log ^2(x)+6 e^{\frac {x^2}{e^5}} \log ^2(x)+e^{\frac {2 x^2}{e^5}} \log ^2(x)} \, dx=e^{\frac {10 x \log {\left (2 \right )}}{2 e^{\frac {x^{2}}{e^{5}}} \log {\left (x \right )} + 6 \log {\left (x \right )}}} \]
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Time = 0.32 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {2^{\frac {10 x}{6 \log (x)+2 e^{\frac {x^2}{e^5}} \log (x)}} \left (-15 \log (2)+15 \log (2) \log (x)+e^{\frac {x^2}{e^5}} \left (-5 \log (2)+5 \log (2) \log (x)-\frac {10 x^2 \log (2) \log (x)}{e^5}\right )\right )}{9 \log ^2(x)+6 e^{\frac {x^2}{e^5}} \log ^2(x)+e^{\frac {2 x^2}{e^5}} \log ^2(x)} \, dx=2^{\frac {5 \, x}{{\left (e^{\left (x^{2} e^{\left (-5\right )}\right )} + 3\right )} \log \left (x\right )}} \]
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Time = 0.27 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {2^{\frac {10 x}{6 \log (x)+2 e^{\frac {x^2}{e^5}} \log (x)}} \left (-15 \log (2)+15 \log (2) \log (x)+e^{\frac {x^2}{e^5}} \left (-5 \log (2)+5 \log (2) \log (x)-\frac {10 x^2 \log (2) \log (x)}{e^5}\right )\right )}{9 \log ^2(x)+6 e^{\frac {x^2}{e^5}} \log ^2(x)+e^{\frac {2 x^2}{e^5}} \log ^2(x)} \, dx=2^{\frac {5 \, x}{e^{\left (x^{2} e^{\left (-5\right )}\right )} \log \left (x\right ) + 3 \, \log \left (x\right )}} \]
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Time = 11.82 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.05 \[ \int \frac {2^{\frac {10 x}{6 \log (x)+2 e^{\frac {x^2}{e^5}} \log (x)}} \left (-15 \log (2)+15 \log (2) \log (x)+e^{\frac {x^2}{e^5}} \left (-5 \log (2)+5 \log (2) \log (x)-\frac {10 x^2 \log (2) \log (x)}{e^5}\right )\right )}{9 \log ^2(x)+6 e^{\frac {x^2}{e^5}} \log ^2(x)+e^{\frac {2 x^2}{e^5}} \log ^2(x)} \, dx={\mathrm {e}}^{\frac {5\,x\,\ln \left (2\right )}{3\,\ln \left (x\right )+{\mathrm {e}}^{x^2\,{\mathrm {e}}^{-5}}\,\ln \left (x\right )}} \]
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