Optimal. Leaf size=30 \[ \left (-e^{e^x}+x+\left (e^4+x\right )^2\right )^2 \log \left (\frac {1}{4} \log \left (16 x^2\right )\right ) \]
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Rubi [F] time = 7.93, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {2 e^{16}+2 e^{2 e^x}+8 e^{12} x+2 x^2+4 x^3+2 x^4+e^{e^x} \left (-4 e^8-4 x-8 e^4 x-4 x^2\right )+e^8 \left (4 x+12 x^2\right )+e^4 \left (8 x^2+8 x^3\right )+\left (2 e^{2 e^x+x} x \log \left (16 x^2\right )+e^{e^x} \left (-2 x-4 e^4 x-4 x^2+e^x \left (-2 e^8 x-2 x^2-4 e^4 x^2-2 x^3\right )\right ) \log \left (16 x^2\right )+\left (4 e^{12} x+2 x^2+6 x^3+4 x^4+e^8 \left (2 x+12 x^2\right )+e^4 \left (8 x^2+12 x^3\right )\right ) \log \left (16 x^2\right )\right ) \log \left (\frac {1}{4} \log \left (16 x^2\right )\right )}{x \log \left (16 x^2\right )} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {2 \left (e^8-e^{e^x}+\left (1+2 e^4\right ) x+x^2\right ) \left (e^8-e^{e^x}+\left (1+2 e^4\right ) x+x^2+x \left (1+2 e^4-e^{e^x+x}+2 x\right ) \log \left (16 x^2\right ) \log \left (\log (2)+\frac {\log \left (x^2\right )}{4}\right )\right )}{x \log \left (16 x^2\right )} \, dx\\ &=2 \int \frac {\left (e^8-e^{e^x}+\left (1+2 e^4\right ) x+x^2\right ) \left (e^8-e^{e^x}+\left (1+2 e^4\right ) x+x^2+x \left (1+2 e^4-e^{e^x+x}+2 x\right ) \log \left (16 x^2\right ) \log \left (\log (2)+\frac {\log \left (x^2\right )}{4}\right )\right )}{x \log \left (16 x^2\right )} \, dx\\ &=2 \int \left (\frac {\left (e^8-e^{e^x}+\left (1+2 e^4\right ) x+x^2\right ) \left (e^8-e^{e^x}+\left (1+2 e^4\right ) x+x^2+\left (1+2 e^4\right ) x \log \left (16 x^2\right ) \log \left (\log (2)+\frac {\log \left (x^2\right )}{4}\right )+2 x^2 \log \left (16 x^2\right ) \log \left (\log (2)+\frac {\log \left (x^2\right )}{4}\right )\right )}{x \log \left (16 x^2\right )}+e^{e^x+x} \left (-e^8+e^{e^x}-\left (1+2 e^4\right ) x-x^2\right ) \log \left (\frac {1}{4} \log \left (16 x^2\right )\right )\right ) \, dx\\ &=2 \int \frac {\left (e^8-e^{e^x}+\left (1+2 e^4\right ) x+x^2\right ) \left (e^8-e^{e^x}+\left (1+2 e^4\right ) x+x^2+\left (1+2 e^4\right ) x \log \left (16 x^2\right ) \log \left (\log (2)+\frac {\log \left (x^2\right )}{4}\right )+2 x^2 \log \left (16 x^2\right ) \log \left (\log (2)+\frac {\log \left (x^2\right )}{4}\right )\right )}{x \log \left (16 x^2\right )} \, dx+2 \int e^{e^x+x} \left (-e^8+e^{e^x}-\left (1+2 e^4\right ) x-x^2\right ) \log \left (\frac {1}{4} \log \left (16 x^2\right )\right ) \, dx\\ &=e^{2 e^x} \log \left (\frac {1}{4} \log \left (16 x^2\right )\right )-2 e^{8+e^x} \log \left (\frac {1}{4} \log \left (16 x^2\right )\right )+2 \int \frac {\left (e^8-e^{e^x}+\left (1+2 e^4\right ) x+x^2\right ) \left (e^8-e^{e^x}+\left (1+2 e^4\right ) x+x^2+x \left (1+2 e^4+2 x\right ) \log \left (16 x^2\right ) \log \left (\log (2)+\frac {\log \left (x^2\right )}{4}\right )\right )}{x \log \left (16 x^2\right )} \, dx-2 \int \frac {e^{2 e^x}-2 e^{8+e^x}-2 \left (1+2 e^4\right ) \int e^{e^x+x} x \, dx-2 \int e^{e^x+x} x^2 \, dx}{x \log \left (16 x^2\right )} \, dx-\left (2 \log \left (\frac {1}{4} \log \left (16 x^2\right )\right )\right ) \int e^{e^x+x} x^2 \, dx-\left (2 \left (1+2 e^4\right ) \log \left (\frac {1}{4} \log \left (16 x^2\right )\right )\right ) \int e^{e^x+x} x \, dx\\ &=e^{2 e^x} \log \left (\frac {1}{4} \log \left (16 x^2\right )\right )-2 e^{8+e^x} \log \left (\frac {1}{4} \log \left (16 x^2\right )\right )+2 \int \left (\frac {e^{2 e^x}}{x \log \left (16 x^2\right )}+\frac {e^{e^x} \left (-2 e^8-2 \left (1+2 e^4\right ) x-2 x^2-\left (1+2 e^4\right ) x \log \left (16 x^2\right ) \log \left (\log (2)+\frac {\log \left (x^2\right )}{4}\right )-2 x^2 \log \left (16 x^2\right ) \log \left (\log (2)+\frac {\log \left (x^2\right )}{4}\right )\right )}{x \log \left (16 x^2\right )}+\frac {\left (e^8+\left (1+2 e^4\right ) x+x^2\right ) \left (e^8+\left (1+2 e^4\right ) x+x^2+\left (1+2 e^4\right ) x \log \left (16 x^2\right ) \log \left (\log (2)+\frac {\log \left (x^2\right )}{4}\right )+2 x^2 \log \left (16 x^2\right ) \log \left (\log (2)+\frac {\log \left (x^2\right )}{4}\right )\right )}{x \log \left (16 x^2\right )}\right ) \, dx-2 \int \left (\frac {e^{2 e^x}}{x \log \left (16 x^2\right )}-\frac {2 e^{8+e^x}}{x \log \left (16 x^2\right )}+\frac {2 \left (-\left (\left (1+2 e^4\right ) \int e^{e^x+x} x \, dx\right )-\int e^{e^x+x} x^2 \, dx\right )}{x \log \left (16 x^2\right )}\right ) \, dx-\left (2 \log \left (\frac {1}{4} \log \left (16 x^2\right )\right )\right ) \int e^{e^x+x} x^2 \, dx-\left (2 \left (1+2 e^4\right ) \log \left (\frac {1}{4} \log \left (16 x^2\right )\right )\right ) \int e^{e^x+x} x \, dx\\ &=e^{2 e^x} \log \left (\frac {1}{4} \log \left (16 x^2\right )\right )-2 e^{8+e^x} \log \left (\frac {1}{4} \log \left (16 x^2\right )\right )+2 \int \frac {e^{e^x} \left (-2 e^8-2 \left (1+2 e^4\right ) x-2 x^2-\left (1+2 e^4\right ) x \log \left (16 x^2\right ) \log \left (\log (2)+\frac {\log \left (x^2\right )}{4}\right )-2 x^2 \log \left (16 x^2\right ) \log \left (\log (2)+\frac {\log \left (x^2\right )}{4}\right )\right )}{x \log \left (16 x^2\right )} \, dx+2 \int \frac {\left (e^8+\left (1+2 e^4\right ) x+x^2\right ) \left (e^8+\left (1+2 e^4\right ) x+x^2+\left (1+2 e^4\right ) x \log \left (16 x^2\right ) \log \left (\log (2)+\frac {\log \left (x^2\right )}{4}\right )+2 x^2 \log \left (16 x^2\right ) \log \left (\log (2)+\frac {\log \left (x^2\right )}{4}\right )\right )}{x \log \left (16 x^2\right )} \, dx+4 \int \frac {e^{8+e^x}}{x \log \left (16 x^2\right )} \, dx-4 \int \frac {-\left (\left (1+2 e^4\right ) \int e^{e^x+x} x \, dx\right )-\int e^{e^x+x} x^2 \, dx}{x \log \left (16 x^2\right )} \, dx-\left (2 \log \left (\frac {1}{4} \log \left (16 x^2\right )\right )\right ) \int e^{e^x+x} x^2 \, dx-\left (2 \left (1+2 e^4\right ) \log \left (\frac {1}{4} \log \left (16 x^2\right )\right )\right ) \int e^{e^x+x} x \, dx\\ &=e^{2 e^x} \log \left (\frac {1}{4} \log \left (16 x^2\right )\right )-2 e^{8+e^x} \log \left (\frac {1}{4} \log \left (16 x^2\right )\right )+2 \int e^{e^x} \left (-\frac {2 \left (e^8+x+2 e^4 x+x^2\right )}{x \log \left (16 x^2\right )}-\left (1+2 e^4+2 x\right ) \log \left (\log (2)+\frac {\log \left (x^2\right )}{4}\right )\right ) \, dx+2 \int \frac {\left (e^8+\left (1+2 e^4\right ) x+x^2\right ) \left (e^8+\left (1+2 e^4\right ) x+x^2+x \left (1+2 e^4+2 x\right ) \log \left (16 x^2\right ) \log \left (\log (2)+\frac {\log \left (x^2\right )}{4}\right )\right )}{x \log \left (16 x^2\right )} \, dx+4 \int \frac {e^{8+e^x}}{x \log \left (16 x^2\right )} \, dx-4 \int \left (-\frac {\left (1+2 e^4\right ) \int e^{e^x+x} x \, dx}{x \log \left (16 x^2\right )}-\frac {\int e^{e^x+x} x^2 \, dx}{x \log \left (16 x^2\right )}\right ) \, dx-\left (2 \log \left (\frac {1}{4} \log \left (16 x^2\right )\right )\right ) \int e^{e^x+x} x^2 \, dx-\left (2 \left (1+2 e^4\right ) \log \left (\frac {1}{4} \log \left (16 x^2\right )\right )\right ) \int e^{e^x+x} x \, dx\\ &=e^{2 e^x} \log \left (\frac {1}{4} \log \left (16 x^2\right )\right )-2 e^{8+e^x} \log \left (\frac {1}{4} \log \left (16 x^2\right )\right )+2 \int \left (\frac {2 e^{e^x} \left (-e^8-\left (1+2 e^4\right ) x-x^2\right )}{x \log \left (16 x^2\right )}+e^{e^x} \left (-1-2 e^4-2 x\right ) \log \left (\frac {1}{4} \log \left (16 x^2\right )\right )\right ) \, dx+2 \int \left (\frac {\left (e^8+\left (1+2 e^4\right ) x+x^2\right )^2}{x \log \left (16 x^2\right )}+\left (1+2 e^4+2 x\right ) \left (e^8+\left (1+2 e^4\right ) x+x^2\right ) \log \left (\frac {1}{4} \log \left (16 x^2\right )\right )\right ) \, dx+4 \int \frac {e^{8+e^x}}{x \log \left (16 x^2\right )} \, dx+4 \int \frac {\int e^{e^x+x} x^2 \, dx}{x \log \left (16 x^2\right )} \, dx+\left (4 \left (1+2 e^4\right )\right ) \int \frac {\int e^{e^x+x} x \, dx}{x \log \left (16 x^2\right )} \, dx-\left (2 \log \left (\frac {1}{4} \log \left (16 x^2\right )\right )\right ) \int e^{e^x+x} x^2 \, dx-\left (2 \left (1+2 e^4\right ) \log \left (\frac {1}{4} \log \left (16 x^2\right )\right )\right ) \int e^{e^x+x} x \, dx\\ &=e^{2 e^x} \log \left (\frac {1}{4} \log \left (16 x^2\right )\right )-2 e^{8+e^x} \log \left (\frac {1}{4} \log \left (16 x^2\right )\right )+2 \int \frac {\left (e^8+\left (1+2 e^4\right ) x+x^2\right )^2}{x \log \left (16 x^2\right )} \, dx+2 \int e^{e^x} \left (-1-2 e^4-2 x\right ) \log \left (\frac {1}{4} \log \left (16 x^2\right )\right ) \, dx+2 \int \left (1+2 e^4+2 x\right ) \left (e^8+\left (1+2 e^4\right ) x+x^2\right ) \log \left (\frac {1}{4} \log \left (16 x^2\right )\right ) \, dx+4 \int \frac {e^{8+e^x}}{x \log \left (16 x^2\right )} \, dx+4 \int \frac {e^{e^x} \left (-e^8-\left (1+2 e^4\right ) x-x^2\right )}{x \log \left (16 x^2\right )} \, dx+4 \int \frac {\int e^{e^x+x} x^2 \, dx}{x \log \left (16 x^2\right )} \, dx+\left (4 \left (1+2 e^4\right )\right ) \int \frac {\int e^{e^x+x} x \, dx}{x \log \left (16 x^2\right )} \, dx-\left (2 \log \left (\frac {1}{4} \log \left (16 x^2\right )\right )\right ) \int e^{e^x+x} x^2 \, dx-\left (2 \left (1+2 e^4\right ) \log \left (\frac {1}{4} \log \left (16 x^2\right )\right )\right ) \int e^{e^x+x} x \, dx\\ &=e^{2 e^x} \log \left (\frac {1}{4} \log \left (16 x^2\right )\right )-2 e^{8+e^x} \log \left (\frac {1}{4} \log \left (16 x^2\right )\right )+\left (e^8+\left (1+2 e^4\right ) x+x^2\right )^2 \log \left (\frac {1}{4} \log \left (16 x^2\right )\right )-2 \left (1+2 e^4\right ) \text {Ei}\left (e^x\right ) \log \left (\frac {1}{4} \log \left (16 x^2\right )\right )-2 \int \frac {2 \left (-\left (\left (1+2 e^4\right ) \text {Ei}\left (e^x\right )\right )-2 \int e^{e^x} x \, dx\right )}{x \log \left (16 x^2\right )} \, dx+4 \int \left (-\frac {e^{e^x} \left (1+2 e^4\right )}{\log \left (16 x^2\right )}-\frac {e^{8+e^x}}{x \log \left (16 x^2\right )}-\frac {e^{e^x} x}{\log \left (16 x^2\right )}\right ) \, dx+4 \int \frac {e^{8+e^x}}{x \log \left (16 x^2\right )} \, dx+4 \int \frac {\int e^{e^x+x} x^2 \, dx}{x \log \left (16 x^2\right )} \, dx+\left (4 \left (1+2 e^4\right )\right ) \int \frac {\int e^{e^x+x} x \, dx}{x \log \left (16 x^2\right )} \, dx-\left (2 \log \left (\frac {1}{4} \log \left (16 x^2\right )\right )\right ) \int e^{e^x+x} x^2 \, dx-\left (4 \log \left (\frac {1}{4} \log \left (16 x^2\right )\right )\right ) \int e^{e^x} x \, dx-\left (2 \left (1+2 e^4\right ) \log \left (\frac {1}{4} \log \left (16 x^2\right )\right )\right ) \int e^{e^x+x} x \, dx\\ &=e^{2 e^x} \log \left (\frac {1}{4} \log \left (16 x^2\right )\right )-2 e^{8+e^x} \log \left (\frac {1}{4} \log \left (16 x^2\right )\right )+\left (e^8+\left (1+2 e^4\right ) x+x^2\right )^2 \log \left (\frac {1}{4} \log \left (16 x^2\right )\right )-2 \left (1+2 e^4\right ) \text {Ei}\left (e^x\right ) \log \left (\frac {1}{4} \log \left (16 x^2\right )\right )-4 \int \frac {e^{e^x} x}{\log \left (16 x^2\right )} \, dx-4 \int \frac {-\left (\left (1+2 e^4\right ) \text {Ei}\left (e^x\right )\right )-2 \int e^{e^x} x \, dx}{x \log \left (16 x^2\right )} \, dx+4 \int \frac {\int e^{e^x+x} x^2 \, dx}{x \log \left (16 x^2\right )} \, dx-\left (4 \left (1+2 e^4\right )\right ) \int \frac {e^{e^x}}{\log \left (16 x^2\right )} \, dx+\left (4 \left (1+2 e^4\right )\right ) \int \frac {\int e^{e^x+x} x \, dx}{x \log \left (16 x^2\right )} \, dx-\left (2 \log \left (\frac {1}{4} \log \left (16 x^2\right )\right )\right ) \int e^{e^x+x} x^2 \, dx-\left (4 \log \left (\frac {1}{4} \log \left (16 x^2\right )\right )\right ) \int e^{e^x} x \, dx-\left (2 \left (1+2 e^4\right ) \log \left (\frac {1}{4} \log \left (16 x^2\right )\right )\right ) \int e^{e^x+x} x \, dx\\ &=e^{2 e^x} \log \left (\frac {1}{4} \log \left (16 x^2\right )\right )-2 e^{8+e^x} \log \left (\frac {1}{4} \log \left (16 x^2\right )\right )+\left (e^8+\left (1+2 e^4\right ) x+x^2\right )^2 \log \left (\frac {1}{4} \log \left (16 x^2\right )\right )-2 \left (1+2 e^4\right ) \text {Ei}\left (e^x\right ) \log \left (\frac {1}{4} \log \left (16 x^2\right )\right )-4 \int \frac {e^{e^x} x}{\log \left (16 x^2\right )} \, dx-4 \int \left (-\frac {\left (1+2 e^4\right ) \text {Ei}\left (e^x\right )}{x \log \left (16 x^2\right )}-\frac {2 \int e^{e^x} x \, dx}{x \log \left (16 x^2\right )}\right ) \, dx+4 \int \frac {\int e^{e^x+x} x^2 \, dx}{x \log \left (16 x^2\right )} \, dx-\left (4 \left (1+2 e^4\right )\right ) \int \frac {e^{e^x}}{\log \left (16 x^2\right )} \, dx+\left (4 \left (1+2 e^4\right )\right ) \int \frac {\int e^{e^x+x} x \, dx}{x \log \left (16 x^2\right )} \, dx-\left (2 \log \left (\frac {1}{4} \log \left (16 x^2\right )\right )\right ) \int e^{e^x+x} x^2 \, dx-\left (4 \log \left (\frac {1}{4} \log \left (16 x^2\right )\right )\right ) \int e^{e^x} x \, dx-\left (2 \left (1+2 e^4\right ) \log \left (\frac {1}{4} \log \left (16 x^2\right )\right )\right ) \int e^{e^x+x} x \, dx\\ &=e^{2 e^x} \log \left (\frac {1}{4} \log \left (16 x^2\right )\right )-2 e^{8+e^x} \log \left (\frac {1}{4} \log \left (16 x^2\right )\right )+\left (e^8+\left (1+2 e^4\right ) x+x^2\right )^2 \log \left (\frac {1}{4} \log \left (16 x^2\right )\right )-2 \left (1+2 e^4\right ) \text {Ei}\left (e^x\right ) \log \left (\frac {1}{4} \log \left (16 x^2\right )\right )-4 \int \frac {e^{e^x} x}{\log \left (16 x^2\right )} \, dx+4 \int \frac {\int e^{e^x+x} x^2 \, dx}{x \log \left (16 x^2\right )} \, dx+8 \int \frac {\int e^{e^x} x \, dx}{x \log \left (16 x^2\right )} \, dx-\left (4 \left (1+2 e^4\right )\right ) \int \frac {e^{e^x}}{\log \left (16 x^2\right )} \, dx+\left (4 \left (1+2 e^4\right )\right ) \int \frac {\text {Ei}\left (e^x\right )}{x \log \left (16 x^2\right )} \, dx+\left (4 \left (1+2 e^4\right )\right ) \int \frac {\int e^{e^x+x} x \, dx}{x \log \left (16 x^2\right )} \, dx-\left (2 \log \left (\frac {1}{4} \log \left (16 x^2\right )\right )\right ) \int e^{e^x+x} x^2 \, dx-\left (4 \log \left (\frac {1}{4} \log \left (16 x^2\right )\right )\right ) \int e^{e^x} x \, dx-\left (2 \left (1+2 e^4\right ) \log \left (\frac {1}{4} \log \left (16 x^2\right )\right )\right ) \int e^{e^x+x} x \, dx\\ \end {aligned} \end {gather*}
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Mathematica [F] time = 15.35, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {2 e^{16}+2 e^{2 e^x}+8 e^{12} x+2 x^2+4 x^3+2 x^4+e^{e^x} \left (-4 e^8-4 x-8 e^4 x-4 x^2\right )+e^8 \left (4 x+12 x^2\right )+e^4 \left (8 x^2+8 x^3\right )+\left (2 e^{2 e^x+x} x \log \left (16 x^2\right )+e^{e^x} \left (-2 x-4 e^4 x-4 x^2+e^x \left (-2 e^8 x-2 x^2-4 e^4 x^2-2 x^3\right )\right ) \log \left (16 x^2\right )+\left (4 e^{12} x+2 x^2+6 x^3+4 x^4+e^8 \left (2 x+12 x^2\right )+e^4 \left (8 x^2+12 x^3\right )\right ) \log \left (16 x^2\right )\right ) \log \left (\frac {1}{4} \log \left (16 x^2\right )\right )}{x \log \left (16 x^2\right )} \, dx \end {gather*}
Verification is not applicable to the result.
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fricas [B] time = 0.62, size = 73, normalized size = 2.43 \begin {gather*} {\left (x^{4} + 2 \, x^{3} + x^{2} + 4 \, x e^{12} + 2 \, {\left (3 \, x^{2} + x\right )} e^{8} + 4 \, {\left (x^{3} + x^{2}\right )} e^{4} - 2 \, {\left (x^{2} + 2 \, x e^{4} + x + e^{8}\right )} e^{\left (e^{x}\right )} + e^{16} + e^{\left (2 \, e^{x}\right )}\right )} \log \left (\frac {1}{4} \, \log \left (16 \, x^{2}\right )\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 0.85, size = 176, normalized size = 5.87
method | result | size |
risch | \(\left (4 x \,{\mathrm e}^{12}+6 x^{2} {\mathrm e}^{8}+4 x^{3} {\mathrm e}^{4}+x^{4}+2 x \,{\mathrm e}^{8}-2 \,{\mathrm e}^{{\mathrm e}^{x}+8}+4 x^{2} {\mathrm e}^{4}-4 x \,{\mathrm e}^{{\mathrm e}^{x}+4}+2 x^{3}-2 \,{\mathrm e}^{{\mathrm e}^{x}} x^{2}+x^{2}-2 x \,{\mathrm e}^{{\mathrm e}^{x}}+{\mathrm e}^{2 \,{\mathrm e}^{x}}\right ) \ln \left (\ln \relax (2)+\frac {\ln \relax (x )}{2}-\frac {i \pi \,\mathrm {csgn}\left (i x^{2}\right ) \left (-\mathrm {csgn}\left (i x^{2}\right )+\mathrm {csgn}\left (i x \right )\right )^{2}}{8}\right )+{\mathrm e}^{16} \ln \left (\ln \relax (x )-\frac {i \left (\pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )-2 \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}+\pi \mathrm {csgn}\left (i x^{2}\right )^{3}+8 i \ln \relax (2)\right )}{4}\right )\) | \(176\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.57, size = 168, normalized size = 5.60 \begin {gather*} -x^{4} \log \relax (2) - 2 \, {\left (2 \, e^{4} \log \relax (2) + \log \relax (2)\right )} x^{3} - {\left (6 \, e^{8} \log \relax (2) + 4 \, e^{4} \log \relax (2) + \log \relax (2)\right )} x^{2} - 2 \, {\left (2 \, e^{12} \log \relax (2) + e^{8} \log \relax (2)\right )} x + 2 \, {\left (x^{2} \log \relax (2) + {\left (2 \, e^{4} \log \relax (2) + \log \relax (2)\right )} x + e^{8} \log \relax (2)\right )} e^{\left (e^{x}\right )} - e^{\left (2 \, e^{x}\right )} \log \relax (2) + {\left (x^{4} + 2 \, x^{3} {\left (2 \, e^{4} + 1\right )} + x^{2} {\left (6 \, e^{8} + 4 \, e^{4} + 1\right )} + 2 \, x {\left (2 \, e^{12} + e^{8}\right )} - 2 \, {\left (x^{2} + x {\left (2 \, e^{4} + 1\right )} + e^{8}\right )} e^{\left (e^{x}\right )} + e^{16} + e^{\left (2 \, e^{x}\right )}\right )} \log \left (2 \, \log \relax (2) + \log \relax (x)\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.94, size = 95, normalized size = 3.17 \begin {gather*} \ln \left (\frac {\ln \left (16\,x^2\right )}{4}\right )\,\left ({\mathrm {e}}^{2\,{\mathrm {e}}^x}-{\mathrm {e}}^{{\mathrm {e}}^x}\,\left (2\,x+2\,{\mathrm {e}}^8+4\,x\,{\mathrm {e}}^4+2\,x^2\right )+\frac {x^5+\left (4\,{\mathrm {e}}^4+2\right )\,x^4+\left (4\,{\mathrm {e}}^4+6\,{\mathrm {e}}^8+1\right )\,x^3+2\,{\mathrm {e}}^8\,\left (2\,{\mathrm {e}}^4+1\right )\,x^2}{x}\right )+\ln \left (\ln \left (16\,x^2\right )\right )\,{\mathrm {e}}^{16} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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