Integrand size = 336, antiderivative size = 25 \[ \int \frac {e^x \left (10 x+14 x^2+4 x^3+\left (16 x+8 x^2\right ) \log (3)+4 x \log ^2(3)\right )+\left (e^x \left (2 x+5 x^2+5 x^3+2 x^4+\left (4 x+8 x^2+4 x^3\right ) \log (3)+\left (2 x+2 x^2\right ) \log ^2(3)\right )+e^x \left (2+5 x+5 x^2+2 x^3+\left (4+8 x+4 x^2\right ) \log (3)+(2+2 x) \log ^2(3)\right ) \log \left (-2-3 x-2 x^2+(-4-4 x) \log (3)-2 \log ^2(3)\right )\right ) \log \left (x+\log \left (-2-3 x-2 x^2+(-4-4 x) \log (3)-2 \log ^2(3)\right )\right ) \log \left (\log ^2\left (x+\log \left (-2-3 x-2 x^2+(-4-4 x) \log (3)-2 \log ^2(3)\right )\right )\right )}{\left (2 x+3 x^2+2 x^3+\left (4 x+4 x^2\right ) \log (3)+2 x \log ^2(3)+\left (2+3 x+2 x^2+(4+4 x) \log (3)+2 \log ^2(3)\right ) \log \left (-2-3 x-2 x^2+(-4-4 x) \log (3)-2 \log ^2(3)\right )\right ) \log \left (x+\log \left (-2-3 x-2 x^2+(-4-4 x) \log (3)-2 \log ^2(3)\right )\right )} \, dx=-3+e^x x \log \left (\log ^2\left (x+\log \left (x-2 (1+x+\log (3))^2\right )\right )\right ) \]
[Out]
\[ \int \frac {e^x \left (10 x+14 x^2+4 x^3+\left (16 x+8 x^2\right ) \log (3)+4 x \log ^2(3)\right )+\left (e^x \left (2 x+5 x^2+5 x^3+2 x^4+\left (4 x+8 x^2+4 x^3\right ) \log (3)+\left (2 x+2 x^2\right ) \log ^2(3)\right )+e^x \left (2+5 x+5 x^2+2 x^3+\left (4+8 x+4 x^2\right ) \log (3)+(2+2 x) \log ^2(3)\right ) \log \left (-2-3 x-2 x^2+(-4-4 x) \log (3)-2 \log ^2(3)\right )\right ) \log \left (x+\log \left (-2-3 x-2 x^2+(-4-4 x) \log (3)-2 \log ^2(3)\right )\right ) \log \left (\log ^2\left (x+\log \left (-2-3 x-2 x^2+(-4-4 x) \log (3)-2 \log ^2(3)\right )\right )\right )}{\left (2 x+3 x^2+2 x^3+\left (4 x+4 x^2\right ) \log (3)+2 x \log ^2(3)+\left (2+3 x+2 x^2+(4+4 x) \log (3)+2 \log ^2(3)\right ) \log \left (-2-3 x-2 x^2+(-4-4 x) \log (3)-2 \log ^2(3)\right )\right ) \log \left (x+\log \left (-2-3 x-2 x^2+(-4-4 x) \log (3)-2 \log ^2(3)\right )\right )} \, dx=\int \frac {e^x \left (10 x+14 x^2+4 x^3+\left (16 x+8 x^2\right ) \log (3)+4 x \log ^2(3)\right )+\left (e^x \left (2 x+5 x^2+5 x^3+2 x^4+\left (4 x+8 x^2+4 x^3\right ) \log (3)+\left (2 x+2 x^2\right ) \log ^2(3)\right )+e^x \left (2+5 x+5 x^2+2 x^3+\left (4+8 x+4 x^2\right ) \log (3)+(2+2 x) \log ^2(3)\right ) \log \left (-2-3 x-2 x^2+(-4-4 x) \log (3)-2 \log ^2(3)\right )\right ) \log \left (x+\log \left (-2-3 x-2 x^2+(-4-4 x) \log (3)-2 \log ^2(3)\right )\right ) \log \left (\log ^2\left (x+\log \left (-2-3 x-2 x^2+(-4-4 x) \log (3)-2 \log ^2(3)\right )\right )\right )}{\left (2 x+3 x^2+2 x^3+\left (4 x+4 x^2\right ) \log (3)+2 x \log ^2(3)+\left (2+3 x+2 x^2+(4+4 x) \log (3)+2 \log ^2(3)\right ) \log \left (-2-3 x-2 x^2+(-4-4 x) \log (3)-2 \log ^2(3)\right )\right ) \log \left (x+\log \left (-2-3 x-2 x^2+(-4-4 x) \log (3)-2 \log ^2(3)\right )\right )} \, dx \]
[In]
[Out]
Rubi steps \begin{align*} \text {integral}& = \int \frac {e^x \left (10 x+14 x^2+4 x^3+\left (16 x+8 x^2\right ) \log (3)+4 x \log ^2(3)\right )+\left (e^x \left (2 x+5 x^2+5 x^3+2 x^4+\left (4 x+8 x^2+4 x^3\right ) \log (3)+\left (2 x+2 x^2\right ) \log ^2(3)\right )+e^x \left (2+5 x+5 x^2+2 x^3+\left (4+8 x+4 x^2\right ) \log (3)+(2+2 x) \log ^2(3)\right ) \log \left (-2-3 x-2 x^2+(-4-4 x) \log (3)-2 \log ^2(3)\right )\right ) \log \left (x+\log \left (-2-3 x-2 x^2+(-4-4 x) \log (3)-2 \log ^2(3)\right )\right ) \log \left (\log ^2\left (x+\log \left (-2-3 x-2 x^2+(-4-4 x) \log (3)-2 \log ^2(3)\right )\right )\right )}{\left (3 x^2+2 x^3+\left (4 x+4 x^2\right ) \log (3)+x \left (2+2 \log ^2(3)\right )+\left (2+3 x+2 x^2+(4+4 x) \log (3)+2 \log ^2(3)\right ) \log \left (-2-3 x-2 x^2+(-4-4 x) \log (3)-2 \log ^2(3)\right )\right ) \log \left (x+\log \left (-2-3 x-2 x^2+(-4-4 x) \log (3)-2 \log ^2(3)\right )\right )} \, dx \\ & = \int e^x \left (\frac {2 x \left (5+2 x^2+8 \log (3)+2 \log ^2(3)+x (7+\log (81))\right )}{\left (2+2 x^2+2 \log ^2(3)+\log (81)+x (3+\log (81))\right ) \left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right ) \log \left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right )}+(1+x) \log \left (\log ^2\left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right )\right )\right ) \, dx \\ & = \int \left (\frac {2 e^x x \left (5+2 x^2+8 \log (3)+2 \log ^2(3)+x (7+\log (81))\right )}{\left (2+2 x^2+2 \log ^2(3)+\log (81)+x (3+\log (81))\right ) \left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right ) \log \left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right )}+e^x (1+x) \log \left (\log ^2\left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right )\right )\right ) \, dx \\ & = 2 \int \frac {e^x x \left (5+2 x^2+8 \log (3)+2 \log ^2(3)+x (7+\log (81))\right )}{\left (2+2 x^2+2 \log ^2(3)+\log (81)+x (3+\log (81))\right ) \left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right ) \log \left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right )} \, dx+\int e^x (1+x) \log \left (\log ^2\left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right )\right ) \, dx \\ & = 2 \int \left (\frac {2 e^x}{\left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right ) \log \left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right )}+\frac {e^x x}{\left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right ) \log \left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right )}+\frac {e^x \left (-4-4 \log ^2(3)-x (3+\log (81))-\log (6561)\right )}{\left (2+2 x^2+2 \log ^2(3)+\log (81)+x (3+\log (81))\right ) \left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right ) \log \left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right )}\right ) \, dx+\int \left (e^x \log \left (\log ^2\left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right )\right )+e^x x \log \left (\log ^2\left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right )\right )\right ) \, dx \\ & = 2 \int \frac {e^x x}{\left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right ) \log \left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right )} \, dx+2 \int \frac {e^x \left (-4-4 \log ^2(3)-x (3+\log (81))-\log (6561)\right )}{\left (2+2 x^2+2 \log ^2(3)+\log (81)+x (3+\log (81))\right ) \left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right ) \log \left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right )} \, dx+4 \int \frac {e^x}{\left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right ) \log \left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right )} \, dx+\int e^x \log \left (\log ^2\left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right )\right ) \, dx+\int e^x x \log \left (\log ^2\left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right )\right ) \, dx \\ & = 2 \int \left (\frac {4 e^x \left (1+\log ^2(3)+\log (9)\right )}{\left (-2-2 x^2-2 \log ^2(3)-\log (81)-x (3+\log (81))\right ) \left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right ) \log \left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right )}+\frac {e^x x (-3-\log (81))}{\left (2+2 x^2+2 \log ^2(3)+\log (81)+x (3+\log (81))\right ) \left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right ) \log \left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right )}\right ) \, dx+2 \int \frac {e^x x}{\left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right ) \log \left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right )} \, dx+4 \int \frac {e^x}{\left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right ) \log \left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right )} \, dx+\int e^x \log \left (\log ^2\left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right )\right ) \, dx+\int e^x x \log \left (\log ^2\left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right )\right ) \, dx \\ & = 2 \int \frac {e^x x}{\left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right ) \log \left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right )} \, dx+4 \int \frac {e^x}{\left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right ) \log \left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right )} \, dx+\left (8 \left (1+\log ^2(3)+\log (9)\right )\right ) \int \frac {e^x}{\left (-2-2 x^2-2 \log ^2(3)-\log (81)-x (3+\log (81))\right ) \left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right ) \log \left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right )} \, dx-(2 (3+\log (81))) \int \frac {e^x x}{\left (2+2 x^2+2 \log ^2(3)+\log (81)+x (3+\log (81))\right ) \left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right ) \log \left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right )} \, dx+\int e^x \log \left (\log ^2\left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right )\right ) \, dx+\int e^x x \log \left (\log ^2\left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right )\right ) \, dx \\ & = 2 \int \frac {e^x x}{\left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right ) \log \left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right )} \, dx+4 \int \frac {e^x}{\left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right ) \log \left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right )} \, dx+\left (8 \left (1+\log ^2(3)+\log (9)\right )\right ) \int \left (-\frac {4 i e^x}{\sqrt {7+16 \log ^2(3)+2 \log (81)-\log ^2(81)} \left (-3-4 x-\log (81)+i \sqrt {7+16 \log ^2(3)+2 \log (81)-\log ^2(81)}\right ) \left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right ) \log \left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right )}-\frac {4 i e^x}{\sqrt {7+16 \log ^2(3)+2 \log (81)-\log ^2(81)} \left (3+4 x+\log (81)+i \sqrt {7+16 \log ^2(3)+2 \log (81)-\log ^2(81)}\right ) \left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right ) \log \left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right )}\right ) \, dx-(2 (3+\log (81))) \int \left (\frac {e^x \left (1-\frac {i (3+\log (81))}{\sqrt {7+16 \log ^2(3)-\log ^2(81)+\log (6561)}}\right )}{\left (3+4 x+\log (81)+i \sqrt {7+16 \log ^2(3)+2 \log (81)-\log ^2(81)}\right ) \left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right ) \log \left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right )}+\frac {e^x \left (1+\frac {i (3+\log (81))}{\sqrt {7+16 \log ^2(3)-\log ^2(81)+\log (6561)}}\right )}{\left (3+4 x+\log (81)-i \sqrt {7+16 \log ^2(3)+2 \log (81)-\log ^2(81)}\right ) \left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right ) \log \left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right )}\right ) \, dx+\int e^x \log \left (\log ^2\left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right )\right ) \, dx+\int e^x x \log \left (\log ^2\left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right )\right ) \, dx \\ & = 2 \int \frac {e^x x}{\left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right ) \log \left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right )} \, dx+4 \int \frac {e^x}{\left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right ) \log \left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right )} \, dx-\frac {\left (32 i \left (1+\log ^2(3)+\log (9)\right )\right ) \int \frac {e^x}{\left (-3-4 x-\log (81)+i \sqrt {7+16 \log ^2(3)+2 \log (81)-\log ^2(81)}\right ) \left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right ) \log \left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right )} \, dx}{\sqrt {7+16 \log ^2(3)-\log ^2(81)+\log (6561)}}-\frac {\left (32 i \left (1+\log ^2(3)+\log (9)\right )\right ) \int \frac {e^x}{\left (3+4 x+\log (81)+i \sqrt {7+16 \log ^2(3)+2 \log (81)-\log ^2(81)}\right ) \left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right ) \log \left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right )} \, dx}{\sqrt {7+16 \log ^2(3)-\log ^2(81)+\log (6561)}}-\left (2 (3+\log (81)) \left (1-\frac {i (3+\log (81))}{\sqrt {7+16 \log ^2(3)-\log ^2(81)+\log (6561)}}\right )\right ) \int \frac {e^x}{\left (3+4 x+\log (81)+i \sqrt {7+16 \log ^2(3)+2 \log (81)-\log ^2(81)}\right ) \left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right ) \log \left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right )} \, dx-\left (2 (3+\log (81)) \left (1+\frac {i (3+\log (81))}{\sqrt {7+16 \log ^2(3)-\log ^2(81)+\log (6561)}}\right )\right ) \int \frac {e^x}{\left (3+4 x+\log (81)-i \sqrt {7+16 \log ^2(3)+2 \log (81)-\log ^2(81)}\right ) \left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right ) \log \left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right )} \, dx+\int e^x \log \left (\log ^2\left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right )\right ) \, dx+\int e^x x \log \left (\log ^2\left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right )\right ) \, dx \\ \end{align*}
Time = 0.23 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.32 \[ \int \frac {e^x \left (10 x+14 x^2+4 x^3+\left (16 x+8 x^2\right ) \log (3)+4 x \log ^2(3)\right )+\left (e^x \left (2 x+5 x^2+5 x^3+2 x^4+\left (4 x+8 x^2+4 x^3\right ) \log (3)+\left (2 x+2 x^2\right ) \log ^2(3)\right )+e^x \left (2+5 x+5 x^2+2 x^3+\left (4+8 x+4 x^2\right ) \log (3)+(2+2 x) \log ^2(3)\right ) \log \left (-2-3 x-2 x^2+(-4-4 x) \log (3)-2 \log ^2(3)\right )\right ) \log \left (x+\log \left (-2-3 x-2 x^2+(-4-4 x) \log (3)-2 \log ^2(3)\right )\right ) \log \left (\log ^2\left (x+\log \left (-2-3 x-2 x^2+(-4-4 x) \log (3)-2 \log ^2(3)\right )\right )\right )}{\left (2 x+3 x^2+2 x^3+\left (4 x+4 x^2\right ) \log (3)+2 x \log ^2(3)+\left (2+3 x+2 x^2+(4+4 x) \log (3)+2 \log ^2(3)\right ) \log \left (-2-3 x-2 x^2+(-4-4 x) \log (3)-2 \log ^2(3)\right )\right ) \log \left (x+\log \left (-2-3 x-2 x^2+(-4-4 x) \log (3)-2 \log ^2(3)\right )\right )} \, dx=e^x x \log \left (\log ^2\left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right )\right ) \]
[In]
[Out]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.42 (sec) , antiderivative size = 216, normalized size of antiderivative = 8.64
\[2 x \,{\mathrm e}^{x} \ln \left (\ln \left (\ln \left (-2 \ln \left (3\right )^{2}+\left (-4-4 x \right ) \ln \left (3\right )-2 x^{2}-3 x -2\right )+x \right )\right )-\frac {i \pi \,\operatorname {csgn}\left (i {\ln \left (\ln \left (-2 \ln \left (3\right )^{2}+\left (-4-4 x \right ) \ln \left (3\right )-2 x^{2}-3 x -2\right )+x \right )}^{2}\right ) \left ({\operatorname {csgn}\left (i \ln \left (\ln \left (-2 \ln \left (3\right )^{2}+\left (-4-4 x \right ) \ln \left (3\right )-2 x^{2}-3 x -2\right )+x \right )\right )}^{2}-2 \,\operatorname {csgn}\left (i {\ln \left (\ln \left (-2 \ln \left (3\right )^{2}+\left (-4-4 x \right ) \ln \left (3\right )-2 x^{2}-3 x -2\right )+x \right )}^{2}\right ) \operatorname {csgn}\left (i \ln \left (\ln \left (-2 \ln \left (3\right )^{2}+\left (-4-4 x \right ) \ln \left (3\right )-2 x^{2}-3 x -2\right )+x \right )\right )+{\operatorname {csgn}\left (i {\ln \left (\ln \left (-2 \ln \left (3\right )^{2}+\left (-4-4 x \right ) \ln \left (3\right )-2 x^{2}-3 x -2\right )+x \right )}^{2}\right )}^{2}\right ) x \,{\mathrm e}^{x}}{2}\]
[In]
[Out]
none
Time = 0.26 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.36 \[ \int \frac {e^x \left (10 x+14 x^2+4 x^3+\left (16 x+8 x^2\right ) \log (3)+4 x \log ^2(3)\right )+\left (e^x \left (2 x+5 x^2+5 x^3+2 x^4+\left (4 x+8 x^2+4 x^3\right ) \log (3)+\left (2 x+2 x^2\right ) \log ^2(3)\right )+e^x \left (2+5 x+5 x^2+2 x^3+\left (4+8 x+4 x^2\right ) \log (3)+(2+2 x) \log ^2(3)\right ) \log \left (-2-3 x-2 x^2+(-4-4 x) \log (3)-2 \log ^2(3)\right )\right ) \log \left (x+\log \left (-2-3 x-2 x^2+(-4-4 x) \log (3)-2 \log ^2(3)\right )\right ) \log \left (\log ^2\left (x+\log \left (-2-3 x-2 x^2+(-4-4 x) \log (3)-2 \log ^2(3)\right )\right )\right )}{\left (2 x+3 x^2+2 x^3+\left (4 x+4 x^2\right ) \log (3)+2 x \log ^2(3)+\left (2+3 x+2 x^2+(4+4 x) \log (3)+2 \log ^2(3)\right ) \log \left (-2-3 x-2 x^2+(-4-4 x) \log (3)-2 \log ^2(3)\right )\right ) \log \left (x+\log \left (-2-3 x-2 x^2+(-4-4 x) \log (3)-2 \log ^2(3)\right )\right )} \, dx=x e^{x} \log \left (\log \left (x + \log \left (-2 \, x^{2} - 4 \, {\left (x + 1\right )} \log \left (3\right ) - 2 \, \log \left (3\right )^{2} - 3 \, x - 2\right )\right )^{2}\right ) \]
[In]
[Out]
Timed out. \[ \int \frac {e^x \left (10 x+14 x^2+4 x^3+\left (16 x+8 x^2\right ) \log (3)+4 x \log ^2(3)\right )+\left (e^x \left (2 x+5 x^2+5 x^3+2 x^4+\left (4 x+8 x^2+4 x^3\right ) \log (3)+\left (2 x+2 x^2\right ) \log ^2(3)\right )+e^x \left (2+5 x+5 x^2+2 x^3+\left (4+8 x+4 x^2\right ) \log (3)+(2+2 x) \log ^2(3)\right ) \log \left (-2-3 x-2 x^2+(-4-4 x) \log (3)-2 \log ^2(3)\right )\right ) \log \left (x+\log \left (-2-3 x-2 x^2+(-4-4 x) \log (3)-2 \log ^2(3)\right )\right ) \log \left (\log ^2\left (x+\log \left (-2-3 x-2 x^2+(-4-4 x) \log (3)-2 \log ^2(3)\right )\right )\right )}{\left (2 x+3 x^2+2 x^3+\left (4 x+4 x^2\right ) \log (3)+2 x \log ^2(3)+\left (2+3 x+2 x^2+(4+4 x) \log (3)+2 \log ^2(3)\right ) \log \left (-2-3 x-2 x^2+(-4-4 x) \log (3)-2 \log ^2(3)\right )\right ) \log \left (x+\log \left (-2-3 x-2 x^2+(-4-4 x) \log (3)-2 \log ^2(3)\right )\right )} \, dx=\text {Timed out} \]
[In]
[Out]
none
Time = 0.42 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.44 \[ \int \frac {e^x \left (10 x+14 x^2+4 x^3+\left (16 x+8 x^2\right ) \log (3)+4 x \log ^2(3)\right )+\left (e^x \left (2 x+5 x^2+5 x^3+2 x^4+\left (4 x+8 x^2+4 x^3\right ) \log (3)+\left (2 x+2 x^2\right ) \log ^2(3)\right )+e^x \left (2+5 x+5 x^2+2 x^3+\left (4+8 x+4 x^2\right ) \log (3)+(2+2 x) \log ^2(3)\right ) \log \left (-2-3 x-2 x^2+(-4-4 x) \log (3)-2 \log ^2(3)\right )\right ) \log \left (x+\log \left (-2-3 x-2 x^2+(-4-4 x) \log (3)-2 \log ^2(3)\right )\right ) \log \left (\log ^2\left (x+\log \left (-2-3 x-2 x^2+(-4-4 x) \log (3)-2 \log ^2(3)\right )\right )\right )}{\left (2 x+3 x^2+2 x^3+\left (4 x+4 x^2\right ) \log (3)+2 x \log ^2(3)+\left (2+3 x+2 x^2+(4+4 x) \log (3)+2 \log ^2(3)\right ) \log \left (-2-3 x-2 x^2+(-4-4 x) \log (3)-2 \log ^2(3)\right )\right ) \log \left (x+\log \left (-2-3 x-2 x^2+(-4-4 x) \log (3)-2 \log ^2(3)\right )\right )} \, dx=2 \, x e^{x} \log \left (\log \left (x + \log \left (-2 \, x^{2} - x {\left (4 \, \log \left (3\right ) + 3\right )} - 2 \, \log \left (3\right )^{2} - 4 \, \log \left (3\right ) - 2\right )\right )\right ) \]
[In]
[Out]
none
Time = 20.29 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.44 \[ \int \frac {e^x \left (10 x+14 x^2+4 x^3+\left (16 x+8 x^2\right ) \log (3)+4 x \log ^2(3)\right )+\left (e^x \left (2 x+5 x^2+5 x^3+2 x^4+\left (4 x+8 x^2+4 x^3\right ) \log (3)+\left (2 x+2 x^2\right ) \log ^2(3)\right )+e^x \left (2+5 x+5 x^2+2 x^3+\left (4+8 x+4 x^2\right ) \log (3)+(2+2 x) \log ^2(3)\right ) \log \left (-2-3 x-2 x^2+(-4-4 x) \log (3)-2 \log ^2(3)\right )\right ) \log \left (x+\log \left (-2-3 x-2 x^2+(-4-4 x) \log (3)-2 \log ^2(3)\right )\right ) \log \left (\log ^2\left (x+\log \left (-2-3 x-2 x^2+(-4-4 x) \log (3)-2 \log ^2(3)\right )\right )\right )}{\left (2 x+3 x^2+2 x^3+\left (4 x+4 x^2\right ) \log (3)+2 x \log ^2(3)+\left (2+3 x+2 x^2+(4+4 x) \log (3)+2 \log ^2(3)\right ) \log \left (-2-3 x-2 x^2+(-4-4 x) \log (3)-2 \log ^2(3)\right )\right ) \log \left (x+\log \left (-2-3 x-2 x^2+(-4-4 x) \log (3)-2 \log ^2(3)\right )\right )} \, dx=x e^{x} \log \left (\log \left (x + \log \left (-2 \, x^{2} - 4 \, x \log \left (3\right ) - 2 \, \log \left (3\right )^{2} - 3 \, x - 4 \, \log \left (3\right ) - 2\right )\right )^{2}\right ) \]
[In]
[Out]
Timed out. \[ \int \frac {e^x \left (10 x+14 x^2+4 x^3+\left (16 x+8 x^2\right ) \log (3)+4 x \log ^2(3)\right )+\left (e^x \left (2 x+5 x^2+5 x^3+2 x^4+\left (4 x+8 x^2+4 x^3\right ) \log (3)+\left (2 x+2 x^2\right ) \log ^2(3)\right )+e^x \left (2+5 x+5 x^2+2 x^3+\left (4+8 x+4 x^2\right ) \log (3)+(2+2 x) \log ^2(3)\right ) \log \left (-2-3 x-2 x^2+(-4-4 x) \log (3)-2 \log ^2(3)\right )\right ) \log \left (x+\log \left (-2-3 x-2 x^2+(-4-4 x) \log (3)-2 \log ^2(3)\right )\right ) \log \left (\log ^2\left (x+\log \left (-2-3 x-2 x^2+(-4-4 x) \log (3)-2 \log ^2(3)\right )\right )\right )}{\left (2 x+3 x^2+2 x^3+\left (4 x+4 x^2\right ) \log (3)+2 x \log ^2(3)+\left (2+3 x+2 x^2+(4+4 x) \log (3)+2 \log ^2(3)\right ) \log \left (-2-3 x-2 x^2+(-4-4 x) \log (3)-2 \log ^2(3)\right )\right ) \log \left (x+\log \left (-2-3 x-2 x^2+(-4-4 x) \log (3)-2 \log ^2(3)\right )\right )} \, dx=\int \frac {{\mathrm {e}}^x\,\left (10\,x+\ln \left (3\right )\,\left (8\,x^2+16\,x\right )+4\,x\,{\ln \left (3\right )}^2+14\,x^2+4\,x^3\right )+\ln \left ({\ln \left (x+\ln \left (-3\,x-\ln \left (3\right )\,\left (4\,x+4\right )-2\,{\ln \left (3\right )}^2-2\,x^2-2\right )\right )}^2\right )\,\ln \left (x+\ln \left (-3\,x-\ln \left (3\right )\,\left (4\,x+4\right )-2\,{\ln \left (3\right )}^2-2\,x^2-2\right )\right )\,\left ({\mathrm {e}}^x\,\left (2\,x+\ln \left (3\right )\,\left (4\,x^3+8\,x^2+4\,x\right )+{\ln \left (3\right )}^2\,\left (2\,x^2+2\,x\right )+5\,x^2+5\,x^3+2\,x^4\right )+\ln \left (-3\,x-\ln \left (3\right )\,\left (4\,x+4\right )-2\,{\ln \left (3\right )}^2-2\,x^2-2\right )\,{\mathrm {e}}^x\,\left (5\,x+\ln \left (3\right )\,\left (4\,x^2+8\,x+4\right )+{\ln \left (3\right )}^2\,\left (2\,x+2\right )+5\,x^2+2\,x^3+2\right )\right )}{\ln \left (x+\ln \left (-3\,x-\ln \left (3\right )\,\left (4\,x+4\right )-2\,{\ln \left (3\right )}^2-2\,x^2-2\right )\right )\,\left (2\,x+\ln \left (-3\,x-\ln \left (3\right )\,\left (4\,x+4\right )-2\,{\ln \left (3\right )}^2-2\,x^2-2\right )\,\left (3\,x+\ln \left (3\right )\,\left (4\,x+4\right )+2\,{\ln \left (3\right )}^2+2\,x^2+2\right )+\ln \left (3\right )\,\left (4\,x^2+4\,x\right )+2\,x\,{\ln \left (3\right )}^2+3\,x^2+2\,x^3\right )} \,d x \]
[In]
[Out]