Integrand size = 194, antiderivative size = 28 \[ \int \frac {\left (\frac {-e^{2 x}-8 e^x x^4-16 x^8-2 x^9-x^{10}-x^8 \log (5)}{x^8}\right )^x \left (2 x^9+2 x^{10}+e^{2 x} (-8+2 x)+e^x \left (-32 x^4+8 x^5\right )+\left (e^{2 x}+8 e^x x^4+16 x^8+2 x^9+x^{10}+x^8 \log (5)\right ) \log \left (\frac {-e^{2 x}-8 e^x x^4-16 x^8-2 x^9-x^{10}-x^8 \log (5)}{x^8}\right )\right )}{e^{2 x}+8 e^x x^4+16 x^8+2 x^9+x^{10}+x^8 \log (5)} \, dx=\left (1-\left (4+\frac {e^x}{x^4}\right )^2-(1+x)^2-\log (5)\right )^x \]
[Out]
\[ \int \frac {\left (\frac {-e^{2 x}-8 e^x x^4-16 x^8-2 x^9-x^{10}-x^8 \log (5)}{x^8}\right )^x \left (2 x^9+2 x^{10}+e^{2 x} (-8+2 x)+e^x \left (-32 x^4+8 x^5\right )+\left (e^{2 x}+8 e^x x^4+16 x^8+2 x^9+x^{10}+x^8 \log (5)\right ) \log \left (\frac {-e^{2 x}-8 e^x x^4-16 x^8-2 x^9-x^{10}-x^8 \log (5)}{x^8}\right )\right )}{e^{2 x}+8 e^x x^4+16 x^8+2 x^9+x^{10}+x^8 \log (5)} \, dx=\int \frac {\left (\frac {-e^{2 x}-8 e^x x^4-16 x^8-2 x^9-x^{10}-x^8 \log (5)}{x^8}\right )^x \left (2 x^9+2 x^{10}+e^{2 x} (-8+2 x)+e^x \left (-32 x^4+8 x^5\right )+\left (e^{2 x}+8 e^x x^4+16 x^8+2 x^9+x^{10}+x^8 \log (5)\right ) \log \left (\frac {-e^{2 x}-8 e^x x^4-16 x^8-2 x^9-x^{10}-x^8 \log (5)}{x^8}\right )\right )}{e^{2 x}+8 e^x x^4+16 x^8+2 x^9+x^{10}+x^8 \log (5)} \, dx \]
[In]
[Out]
Rubi steps \begin{align*} \text {integral}& = \int \frac {\left (\frac {-e^{2 x}-8 e^x x^4-16 x^8-2 x^9-x^{10}-x^8 \log (5)}{x^8}\right )^x \left (2 x^9+2 x^{10}+e^{2 x} (-8+2 x)+e^x \left (-32 x^4+8 x^5\right )+\left (e^{2 x}+8 e^x x^4+16 x^8+2 x^9+x^{10}+x^8 \log (5)\right ) \log \left (\frac {-e^{2 x}-8 e^x x^4-16 x^8-2 x^9-x^{10}-x^8 \log (5)}{x^8}\right )\right )}{e^{2 x}+8 e^x x^4+2 x^9+x^{10}+x^8 (16+\log (5))} \, dx \\ & = \int \frac {\left (-\frac {e^{2 x}}{x^8}-\frac {8 e^x}{x^4}-2 x-x^2-16 \left (1+\frac {\log (5)}{16}\right )\right )^{-1+x} \left (-2 e^{2 x} (-4+x)-8 e^x (-4+x) x^4-2 x^9-2 x^{10}-\left (e^{2 x}+8 e^x x^4+x^8 \left (16+2 x+x^2+\log (5)\right )\right ) \log \left (-\frac {e^{2 x}+8 e^x x^4+x^8 \left (16+2 x+x^2+\log (5)\right )}{x^8}\right )\right )}{x^8} \, dx \\ & = \int \left (-2 x \left (-\frac {e^{2 x}}{x^8}-\frac {8 e^x}{x^4}-2 x-x^2-16 \left (1+\frac {\log (5)}{16}\right )\right )^{-1+x}-2 x^2 \left (-\frac {e^{2 x}}{x^8}-\frac {8 e^x}{x^4}-2 x-x^2-16 \left (1+\frac {\log (5)}{16}\right )\right )^{-1+x}-2 x \left (-\frac {e^{2 x}}{x^8}-\frac {8 e^x}{x^4}-2 x-x^2-16 \left (1+\frac {\log (5)}{16}\right )\right )^{-1+x} \log \left (-\frac {e^{2 x}}{x^8}-\frac {8 e^x}{x^4}-2 x-x^2-16 \left (1+\frac {\log (5)}{16}\right )\right )-x^2 \left (-\frac {e^{2 x}}{x^8}-\frac {8 e^x}{x^4}-2 x-x^2-16 \left (1+\frac {\log (5)}{16}\right )\right )^{-1+x} \log \left (-\frac {e^{2 x}}{x^8}-\frac {8 e^x}{x^4}-2 x-x^2-16 \left (1+\frac {\log (5)}{16}\right )\right )-16 \left (-\frac {e^{2 x}}{x^8}-\frac {8 e^x}{x^4}-2 x-x^2-16 \left (1+\frac {\log (5)}{16}\right )\right )^{-1+x} \left (1+\frac {\log (5)}{16}\right ) \log \left (-\frac {e^{2 x}}{x^8}-\frac {8 e^x}{x^4}-2 x-x^2-16 \left (1+\frac {\log (5)}{16}\right )\right )-\frac {8 e^x \left (-\frac {e^{2 x}}{x^8}-\frac {8 e^x}{x^4}-2 x-x^2-16 \left (1+\frac {\log (5)}{16}\right )\right )^{-1+x} \left (-4+x+\log \left (-\frac {e^{2 x}+8 e^x x^4+x^8 \left (16+2 x+x^2+\log (5)\right )}{x^8}\right )\right )}{x^4}-\frac {e^{2 x} \left (-\frac {e^{2 x}}{x^8}-\frac {8 e^x}{x^4}-2 x-x^2-16 \left (1+\frac {\log (5)}{16}\right )\right )^{-1+x} \left (-8+2 x+\log \left (-\frac {e^{2 x}+8 e^x x^4+x^8 \left (16+2 x+x^2+\log (5)\right )}{x^8}\right )\right )}{x^8}\right ) \, dx \\ & = -\left (2 \int x \left (-\frac {e^{2 x}}{x^8}-\frac {8 e^x}{x^4}-2 x-x^2-16 \left (1+\frac {\log (5)}{16}\right )\right )^{-1+x} \, dx\right )-2 \int x^2 \left (-\frac {e^{2 x}}{x^8}-\frac {8 e^x}{x^4}-2 x-x^2-16 \left (1+\frac {\log (5)}{16}\right )\right )^{-1+x} \, dx-2 \int x \left (-\frac {e^{2 x}}{x^8}-\frac {8 e^x}{x^4}-2 x-x^2-16 \left (1+\frac {\log (5)}{16}\right )\right )^{-1+x} \log \left (-\frac {e^{2 x}}{x^8}-\frac {8 e^x}{x^4}-2 x-x^2-16 \left (1+\frac {\log (5)}{16}\right )\right ) \, dx-8 \int \frac {e^x \left (-\frac {e^{2 x}}{x^8}-\frac {8 e^x}{x^4}-2 x-x^2-16 \left (1+\frac {\log (5)}{16}\right )\right )^{-1+x} \left (-4+x+\log \left (-\frac {e^{2 x}+8 e^x x^4+x^8 \left (16+2 x+x^2+\log (5)\right )}{x^8}\right )\right )}{x^4} \, dx-(16+\log (5)) \int \left (-\frac {e^{2 x}}{x^8}-\frac {8 e^x}{x^4}-2 x-x^2-16 \left (1+\frac {\log (5)}{16}\right )\right )^{-1+x} \log \left (-\frac {e^{2 x}}{x^8}-\frac {8 e^x}{x^4}-2 x-x^2-16 \left (1+\frac {\log (5)}{16}\right )\right ) \, dx-\int x^2 \left (-\frac {e^{2 x}}{x^8}-\frac {8 e^x}{x^4}-2 x-x^2-16 \left (1+\frac {\log (5)}{16}\right )\right )^{-1+x} \log \left (-\frac {e^{2 x}}{x^8}-\frac {8 e^x}{x^4}-2 x-x^2-16 \left (1+\frac {\log (5)}{16}\right )\right ) \, dx-\int \frac {e^{2 x} \left (-\frac {e^{2 x}}{x^8}-\frac {8 e^x}{x^4}-2 x-x^2-16 \left (1+\frac {\log (5)}{16}\right )\right )^{-1+x} \left (-8+2 x+\log \left (-\frac {e^{2 x}+8 e^x x^4+x^8 \left (16+2 x+x^2+\log (5)\right )}{x^8}\right )\right )}{x^8} \, dx \\ & = -\left (2 \int x \left (-\frac {e^{2 x}}{x^8}-\frac {8 e^x}{x^4}-2 x-x^2-16 \left (1+\frac {\log (5)}{16}\right )\right )^{-1+x} \, dx\right )-2 \int x^2 \left (-\frac {e^{2 x}}{x^8}-\frac {8 e^x}{x^4}-2 x-x^2-16 \left (1+\frac {\log (5)}{16}\right )\right )^{-1+x} \, dx+2 \int \frac {2 \left (e^{2 x} (-4+x)+4 e^x (-4+x) x^4+x^9 (1+x)\right ) \int x \left (-\frac {e^{2 x}}{x^8}-\frac {8 e^x}{x^4}-2 x-x^2-16 \left (1+\frac {\log (5)}{16}\right )\right )^{-1+x} \, dx}{x \left (e^{2 x}+8 e^x x^4+x^8 \left (16+2 x+x^2+\log (5)\right )\right )} \, dx-8 \int \left (\frac {e^x (-4+x) \left (-\frac {e^{2 x}}{x^8}-\frac {8 e^x}{x^4}-2 x-x^2-16 \left (1+\frac {\log (5)}{16}\right )\right )^{-1+x}}{x^4}+\frac {e^x \left (-\frac {e^{2 x}}{x^8}-\frac {8 e^x}{x^4}-2 x-x^2-16 \left (1+\frac {\log (5)}{16}\right )\right )^{-1+x} \log \left (-\frac {e^{2 x}}{x^8}-\frac {8 e^x}{x^4}-2 x-x^2-16 \left (1+\frac {\log (5)}{16}\right )\right )}{x^4}\right ) \, dx-(-16-\log (5)) \int \frac {2 \left (e^{2 x} (-4+x)+4 e^x (-4+x) x^4+x^9 (1+x)\right ) \int \left (-\frac {e^{2 x}}{x^8}-\frac {8 e^x}{x^4}-2 x-x^2-16 \left (1+\frac {\log (5)}{16}\right )\right )^{-1+x} \, dx}{x \left (e^{2 x}+8 e^x x^4+x^8 \left (16+2 x+x^2+\log (5)\right )\right )} \, dx-\log \left (-\frac {e^{2 x}}{x^8}-\frac {8 e^x}{x^4}-2 x-x^2-16 \left (1+\frac {\log (5)}{16}\right )\right ) \int x^2 \left (-\frac {e^{2 x}}{x^8}-\frac {8 e^x}{x^4}-2 x-x^2-16 \left (1+\frac {\log (5)}{16}\right )\right )^{-1+x} \, dx-\left (2 \log \left (-\frac {e^{2 x}}{x^8}-\frac {8 e^x}{x^4}-2 x-x^2-16 \left (1+\frac {\log (5)}{16}\right )\right )\right ) \int x \left (-\frac {e^{2 x}}{x^8}-\frac {8 e^x}{x^4}-2 x-x^2-16 \left (1+\frac {\log (5)}{16}\right )\right )^{-1+x} \, dx-\left ((16+\log (5)) \log \left (-\frac {e^{2 x}}{x^8}-\frac {8 e^x}{x^4}-2 x-x^2-16 \left (1+\frac {\log (5)}{16}\right )\right )\right ) \int \left (-\frac {e^{2 x}}{x^8}-\frac {8 e^x}{x^4}-2 x-x^2-16 \left (1+\frac {\log (5)}{16}\right )\right )^{-1+x} \, dx-\int \left (\frac {2 e^{2 x} (-4+x) \left (-\frac {e^{2 x}}{x^8}-\frac {8 e^x}{x^4}-2 x-x^2-16 \left (1+\frac {\log (5)}{16}\right )\right )^{-1+x}}{x^8}+\frac {e^{2 x} \left (-\frac {e^{2 x}}{x^8}-\frac {8 e^x}{x^4}-2 x-x^2-16 \left (1+\frac {\log (5)}{16}\right )\right )^{-1+x} \log \left (-\frac {e^{2 x}}{x^8}-\frac {8 e^x}{x^4}-2 x-x^2-16 \left (1+\frac {\log (5)}{16}\right )\right )}{x^8}\right ) \, dx+\int \frac {2 \left (e^{2 x} (-4+x)+4 e^x (-4+x) x^4+x^9 (1+x)\right ) \int x^2 \left (-\frac {e^{2 x}}{x^8}-\frac {8 e^x}{x^4}-2 x-x^2-16 \left (1+\frac {\log (5)}{16}\right )\right )^{-1+x} \, dx}{x \left (e^{2 x}+8 e^x x^4+x^8 \left (16+2 x+x^2+\log (5)\right )\right )} \, dx \\ & = -\left (2 \int \frac {e^{2 x} (-4+x) \left (-\frac {e^{2 x}}{x^8}-\frac {8 e^x}{x^4}-2 x-x^2-16 \left (1+\frac {\log (5)}{16}\right )\right )^{-1+x}}{x^8} \, dx\right )-2 \int x \left (-\frac {e^{2 x}}{x^8}-\frac {8 e^x}{x^4}-2 x-x^2-16 \left (1+\frac {\log (5)}{16}\right )\right )^{-1+x} \, dx-2 \int x^2 \left (-\frac {e^{2 x}}{x^8}-\frac {8 e^x}{x^4}-2 x-x^2-16 \left (1+\frac {\log (5)}{16}\right )\right )^{-1+x} \, dx+2 \int \frac {\left (e^{2 x} (-4+x)+4 e^x (-4+x) x^4+x^9 (1+x)\right ) \int x^2 \left (-\frac {e^{2 x}}{x^8}-\frac {8 e^x}{x^4}-2 x-x^2-16 \left (1+\frac {\log (5)}{16}\right )\right )^{-1+x} \, dx}{x \left (e^{2 x}+8 e^x x^4+x^8 \left (16+2 x+x^2+\log (5)\right )\right )} \, dx+4 \int \frac {\left (e^{2 x} (-4+x)+4 e^x (-4+x) x^4+x^9 (1+x)\right ) \int x \left (-\frac {e^{2 x}}{x^8}-\frac {8 e^x}{x^4}-2 x-x^2-16 \left (1+\frac {\log (5)}{16}\right )\right )^{-1+x} \, dx}{x \left (e^{2 x}+8 e^x x^4+x^8 \left (16+2 x+x^2+\log (5)\right )\right )} \, dx-8 \int \frac {e^x (-4+x) \left (-\frac {e^{2 x}}{x^8}-\frac {8 e^x}{x^4}-2 x-x^2-16 \left (1+\frac {\log (5)}{16}\right )\right )^{-1+x}}{x^4} \, dx-8 \int \frac {e^x \left (-\frac {e^{2 x}}{x^8}-\frac {8 e^x}{x^4}-2 x-x^2-16 \left (1+\frac {\log (5)}{16}\right )\right )^{-1+x} \log \left (-\frac {e^{2 x}}{x^8}-\frac {8 e^x}{x^4}-2 x-x^2-16 \left (1+\frac {\log (5)}{16}\right )\right )}{x^4} \, dx+(2 (16+\log (5))) \int \frac {\left (e^{2 x} (-4+x)+4 e^x (-4+x) x^4+x^9 (1+x)\right ) \int \left (-\frac {e^{2 x}}{x^8}-\frac {8 e^x}{x^4}-2 x-x^2-16 \left (1+\frac {\log (5)}{16}\right )\right )^{-1+x} \, dx}{x \left (e^{2 x}+8 e^x x^4+x^8 \left (16+2 x+x^2+\log (5)\right )\right )} \, dx-\log \left (-\frac {e^{2 x}}{x^8}-\frac {8 e^x}{x^4}-2 x-x^2-16 \left (1+\frac {\log (5)}{16}\right )\right ) \int x^2 \left (-\frac {e^{2 x}}{x^8}-\frac {8 e^x}{x^4}-2 x-x^2-16 \left (1+\frac {\log (5)}{16}\right )\right )^{-1+x} \, dx-\left (2 \log \left (-\frac {e^{2 x}}{x^8}-\frac {8 e^x}{x^4}-2 x-x^2-16 \left (1+\frac {\log (5)}{16}\right )\right )\right ) \int x \left (-\frac {e^{2 x}}{x^8}-\frac {8 e^x}{x^4}-2 x-x^2-16 \left (1+\frac {\log (5)}{16}\right )\right )^{-1+x} \, dx-\left ((16+\log (5)) \log \left (-\frac {e^{2 x}}{x^8}-\frac {8 e^x}{x^4}-2 x-x^2-16 \left (1+\frac {\log (5)}{16}\right )\right )\right ) \int \left (-\frac {e^{2 x}}{x^8}-\frac {8 e^x}{x^4}-2 x-x^2-16 \left (1+\frac {\log (5)}{16}\right )\right )^{-1+x} \, dx-\int \frac {e^{2 x} \left (-\frac {e^{2 x}}{x^8}-\frac {8 e^x}{x^4}-2 x-x^2-16 \left (1+\frac {\log (5)}{16}\right )\right )^{-1+x} \log \left (-\frac {e^{2 x}}{x^8}-\frac {8 e^x}{x^4}-2 x-x^2-16 \left (1+\frac {\log (5)}{16}\right )\right )}{x^8} \, dx \\ & = \text {Too large to display} \\ \end{align*}
Time = 0.17 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.25 \[ \int \frac {\left (\frac {-e^{2 x}-8 e^x x^4-16 x^8-2 x^9-x^{10}-x^8 \log (5)}{x^8}\right )^x \left (2 x^9+2 x^{10}+e^{2 x} (-8+2 x)+e^x \left (-32 x^4+8 x^5\right )+\left (e^{2 x}+8 e^x x^4+16 x^8+2 x^9+x^{10}+x^8 \log (5)\right ) \log \left (\frac {-e^{2 x}-8 e^x x^4-16 x^8-2 x^9-x^{10}-x^8 \log (5)}{x^8}\right )\right )}{e^{2 x}+8 e^x x^4+16 x^8+2 x^9+x^{10}+x^8 \log (5)} \, dx=\left (-\frac {e^{2 x}+8 e^x x^4+x^8 \left (16+2 x+x^2+\log (5)\right )}{x^8}\right )^x \]
[In]
[Out]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.14 (sec) , antiderivative size = 754, normalized size of antiderivative = 26.93
\[\text {Expression too large to display}\]
[In]
[Out]
none
Time = 0.26 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.36 \[ \int \frac {\left (\frac {-e^{2 x}-8 e^x x^4-16 x^8-2 x^9-x^{10}-x^8 \log (5)}{x^8}\right )^x \left (2 x^9+2 x^{10}+e^{2 x} (-8+2 x)+e^x \left (-32 x^4+8 x^5\right )+\left (e^{2 x}+8 e^x x^4+16 x^8+2 x^9+x^{10}+x^8 \log (5)\right ) \log \left (\frac {-e^{2 x}-8 e^x x^4-16 x^8-2 x^9-x^{10}-x^8 \log (5)}{x^8}\right )\right )}{e^{2 x}+8 e^x x^4+16 x^8+2 x^9+x^{10}+x^8 \log (5)} \, dx=\left (-\frac {x^{10} + 2 \, x^{9} + x^{8} \log \left (5\right ) + 16 \, x^{8} + 8 \, x^{4} e^{x} + e^{\left (2 \, x\right )}}{x^{8}}\right )^{x} \]
[In]
[Out]
Timed out. \[ \int \frac {\left (\frac {-e^{2 x}-8 e^x x^4-16 x^8-2 x^9-x^{10}-x^8 \log (5)}{x^8}\right )^x \left (2 x^9+2 x^{10}+e^{2 x} (-8+2 x)+e^x \left (-32 x^4+8 x^5\right )+\left (e^{2 x}+8 e^x x^4+16 x^8+2 x^9+x^{10}+x^8 \log (5)\right ) \log \left (\frac {-e^{2 x}-8 e^x x^4-16 x^8-2 x^9-x^{10}-x^8 \log (5)}{x^8}\right )\right )}{e^{2 x}+8 e^x x^4+16 x^8+2 x^9+x^{10}+x^8 \log (5)} \, dx=\text {Timed out} \]
[In]
[Out]
none
Time = 0.37 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.54 \[ \int \frac {\left (\frac {-e^{2 x}-8 e^x x^4-16 x^8-2 x^9-x^{10}-x^8 \log (5)}{x^8}\right )^x \left (2 x^9+2 x^{10}+e^{2 x} (-8+2 x)+e^x \left (-32 x^4+8 x^5\right )+\left (e^{2 x}+8 e^x x^4+16 x^8+2 x^9+x^{10}+x^8 \log (5)\right ) \log \left (\frac {-e^{2 x}-8 e^x x^4-16 x^8-2 x^9-x^{10}-x^8 \log (5)}{x^8}\right )\right )}{e^{2 x}+8 e^x x^4+16 x^8+2 x^9+x^{10}+x^8 \log (5)} \, dx=e^{\left (x \log \left (-x^{10} - 2 \, x^{9} - x^{8} {\left (\log \left (5\right ) + 16\right )} - 8 \, x^{4} e^{x} - e^{\left (2 \, x\right )}\right ) - 8 \, x \log \left (x\right )\right )} \]
[In]
[Out]
\[ \int \frac {\left (\frac {-e^{2 x}-8 e^x x^4-16 x^8-2 x^9-x^{10}-x^8 \log (5)}{x^8}\right )^x \left (2 x^9+2 x^{10}+e^{2 x} (-8+2 x)+e^x \left (-32 x^4+8 x^5\right )+\left (e^{2 x}+8 e^x x^4+16 x^8+2 x^9+x^{10}+x^8 \log (5)\right ) \log \left (\frac {-e^{2 x}-8 e^x x^4-16 x^8-2 x^9-x^{10}-x^8 \log (5)}{x^8}\right )\right )}{e^{2 x}+8 e^x x^4+16 x^8+2 x^9+x^{10}+x^8 \log (5)} \, dx=\int { \frac {{\left (2 \, x^{10} + 2 \, x^{9} + 2 \, {\left (x - 4\right )} e^{\left (2 \, x\right )} + 8 \, {\left (x^{5} - 4 \, x^{4}\right )} e^{x} + {\left (x^{10} + 2 \, x^{9} + x^{8} \log \left (5\right ) + 16 \, x^{8} + 8 \, x^{4} e^{x} + e^{\left (2 \, x\right )}\right )} \log \left (-\frac {x^{10} + 2 \, x^{9} + x^{8} \log \left (5\right ) + 16 \, x^{8} + 8 \, x^{4} e^{x} + e^{\left (2 \, x\right )}}{x^{8}}\right )\right )} \left (-\frac {x^{10} + 2 \, x^{9} + x^{8} \log \left (5\right ) + 16 \, x^{8} + 8 \, x^{4} e^{x} + e^{\left (2 \, x\right )}}{x^{8}}\right )^{x}}{x^{10} + 2 \, x^{9} + x^{8} \log \left (5\right ) + 16 \, x^{8} + 8 \, x^{4} e^{x} + e^{\left (2 \, x\right )}} \,d x } \]
[In]
[Out]
Time = 12.75 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.57 \[ \int \frac {\left (\frac {-e^{2 x}-8 e^x x^4-16 x^8-2 x^9-x^{10}-x^8 \log (5)}{x^8}\right )^x \left (2 x^9+2 x^{10}+e^{2 x} (-8+2 x)+e^x \left (-32 x^4+8 x^5\right )+\left (e^{2 x}+8 e^x x^4+16 x^8+2 x^9+x^{10}+x^8 \log (5)\right ) \log \left (\frac {-e^{2 x}-8 e^x x^4-16 x^8-2 x^9-x^{10}-x^8 \log (5)}{x^8}\right )\right )}{e^{2 x}+8 e^x x^4+16 x^8+2 x^9+x^{10}+x^8 \log (5)} \, dx={\left (\frac {1}{x^8}\right )}^x\,{\left (-{\mathrm {e}}^{2\,x}-8\,x^4\,{\mathrm {e}}^x-x^8\,\ln \left (5\right )-16\,x^8-2\,x^9-x^{10}\right )}^x \]
[In]
[Out]