3.24.83 \(\int \frac {2 e^{16}+2 e^{2 e^x}+8 e^{12} x+2 x^2+4 x^3+2 x^4+e^{e^x} (-4 e^8-4 x-8 e^4 x-4 x^2)+e^8 (4 x+12 x^2)+e^4 (8 x^2+8 x^3)+(2 e^{2 e^x+x} x \log (16 x^2)+e^{e^x} (-2 x-4 e^4 x-4 x^2+e^x (-2 e^8 x-2 x^2-4 e^4 x^2-2 x^3)) \log (16 x^2)+(4 e^{12} x+2 x^2+6 x^3+4 x^4+e^8 (2 x+12 x^2)+e^4 (8 x^2+12 x^3)) \log (16 x^2)) \log (\frac {1}{4} \log (16 x^2))}{x \log (16 x^2)} \, dx\)

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.

[In]

Int[(2*E^16 + 2*E^(2*E^x) + 8*E^12*x + 2*x^2 + 4*x^3 + 2*x^4 + E^E^x*(-4*E^8 - 4*x - 8*E^4*x - 4*x^2) + E^8*(4
*x + 12*x^2) + E^4*(8*x^2 + 8*x^3) + (2*E^(2*E^x + x)*x*Log[16*x^2] + E^E^x*(-2*x - 4*E^4*x - 4*x^2 + E^x*(-2*
E^8*x - 2*x^2 - 4*E^4*x^2 - 2*x^3))*Log[16*x^2] + (4*E^12*x + 2*x^2 + 6*x^3 + 4*x^4 + E^8*(2*x + 12*x^2) + E^4
*(8*x^2 + 12*x^3))*Log[16*x^2])*Log[Log[16*x^2]/4])/(x*Log[16*x^2]),x]

[Out]

E^(2*E^x)*Log[Log[16*x^2]/4] - 2*E^(8 + E^x)*Log[Log[16*x^2]/4] + (E^8 + (1 + 2*E^4)*x + x^2)^2*Log[Log[16*x^2
]/4] - 2*(1 + 2*E^4)*ExpIntegralEi[E^x]*Log[Log[16*x^2]/4] - 4*Log[Log[16*x^2]/4]*Defer[Int][E^E^x*x, x] - 2*(
1 + 2*E^4)*Log[Log[16*x^2]/4]*Defer[Int][E^(E^x + x)*x, x] - 2*Log[Log[16*x^2]/4]*Defer[Int][E^(E^x + x)*x^2,
x] - 4*(1 + 2*E^4)*Defer[Int][E^E^x/Log[16*x^2], x] - 4*Defer[Int][(E^E^x*x)/Log[16*x^2], x] + 4*(1 + 2*E^4)*D
efer[Int][ExpIntegralEi[E^x]/(x*Log[16*x^2]), x] + 8*Defer[Int][Defer[Int][E^E^x*x, x]/(x*Log[16*x^2]), x] + 4
*(1 + 2*E^4)*Defer[Int][Defer[Int][E^(E^x + x)*x, x]/(x*Log[16*x^2]), x] + 4*Defer[Int][Defer[Int][E^(E^x + x)
*x^2, x]/(x*Log[16*x^2]), x]

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.

[In]

Integrate[(2*E^16 + 2*E^(2*E^x) + 8*E^12*x + 2*x^2 + 4*x^3 + 2*x^4 + E^E^x*(-4*E^8 - 4*x - 8*E^4*x - 4*x^2) +
E^8*(4*x + 12*x^2) + E^4*(8*x^2 + 8*x^3) + (2*E^(2*E^x + x)*x*Log[16*x^2] + E^E^x*(-2*x - 4*E^4*x - 4*x^2 + E^
x*(-2*E^8*x - 2*x^2 - 4*E^4*x^2 - 2*x^3))*Log[16*x^2] + (4*E^12*x + 2*x^2 + 6*x^3 + 4*x^4 + E^8*(2*x + 12*x^2)
+ E^4*(8*x^2 + 12*x^3))*Log[16*x^2])*Log[Log[16*x^2]/4])/(x*Log[16*x^2]),x]

[Out]

Integrate[(2*E^16 + 2*E^(2*E^x) + 8*E^12*x + 2*x^2 + 4*x^3 + 2*x^4 + E^E^x*(-4*E^8 - 4*x - 8*E^4*x - 4*x^2) +
E^8*(4*x + 12*x^2) + E^4*(8*x^2 + 8*x^3) + (2*E^(2*E^x + x)*x*Log[16*x^2] + E^E^x*(-2*x - 4*E^4*x - 4*x^2 + E^
x*(-2*E^8*x - 2*x^2 - 4*E^4*x^2 - 2*x^3))*Log[16*x^2] + (4*E^12*x + 2*x^2 + 6*x^3 + 4*x^4 + E^8*(2*x + 12*x^2)
+ E^4*(8*x^2 + 12*x^3))*Log[16*x^2])*Log[Log[16*x^2]/4])/(x*Log[16*x^2]), x]

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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.

[In]

integrate(((2*x*exp(x)*log(16*x^2)*exp(exp(x))^2+((-2*x*exp(4)^2-4*x^2*exp(4)-2*x^3-2*x^2)*exp(x)-4*x*exp(4)-4
*x^2-2*x)*log(16*x^2)*exp(exp(x))+(4*x*exp(4)^3+(12*x^2+2*x)*exp(4)^2+(12*x^3+8*x^2)*exp(4)+4*x^4+6*x^3+2*x^2)
*log(16*x^2))*log(1/4*log(16*x^2))+2*exp(exp(x))^2+(-4*exp(4)^2-8*x*exp(4)-4*x^2-4*x)*exp(exp(x))+2*exp(4)^4+8
*x*exp(4)^3+(12*x^2+4*x)*exp(4)^2+(8*x^3+8*x^2)*exp(4)+2*x^4+4*x^3+2*x^2)/x/log(16*x^2),x, algorithm="fricas")

[Out]

(x^4 + 2*x^3 + x^2 + 4*x*e^12 + 2*(3*x^2 + x)*e^8 + 4*(x^3 + x^2)*e^4 - 2*(x^2 + 2*x*e^4 + x + e^8)*e^(e^x) +
e^16 + e^(2*e^x))*log(1/4*log(16*x^2))

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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.

[In]

integrate(((2*x*exp(x)*log(16*x^2)*exp(exp(x))^2+((-2*x*exp(4)^2-4*x^2*exp(4)-2*x^3-2*x^2)*exp(x)-4*x*exp(4)-4
*x^2-2*x)*log(16*x^2)*exp(exp(x))+(4*x*exp(4)^3+(12*x^2+2*x)*exp(4)^2+(12*x^3+8*x^2)*exp(4)+4*x^4+6*x^3+2*x^2)
*log(16*x^2))*log(1/4*log(16*x^2))+2*exp(exp(x))^2+(-4*exp(4)^2-8*x*exp(4)-4*x^2-4*x)*exp(exp(x))+2*exp(4)^4+8
*x*exp(4)^3+(12*x^2+4*x)*exp(4)^2+(8*x^3+8*x^2)*exp(4)+2*x^4+4*x^3+2*x^2)/x/log(16*x^2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Warning, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Ch
eck [abs(sa

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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.

[In]

int(((2*x*exp(x)*ln(16*x^2)*exp(exp(x))^2+((-2*x*exp(4)^2-4*x^2*exp(4)-2*x^3-2*x^2)*exp(x)-4*x*exp(4)-4*x^2-2*
x)*ln(16*x^2)*exp(exp(x))+(4*x*exp(4)^3+(12*x^2+2*x)*exp(4)^2+(12*x^3+8*x^2)*exp(4)+4*x^4+6*x^3+2*x^2)*ln(16*x
^2))*ln(1/4*ln(16*x^2))+2*exp(exp(x))^2+(-4*exp(4)^2-8*x*exp(4)-4*x^2-4*x)*exp(exp(x))+2*exp(4)^4+8*x*exp(4)^3
+(12*x^2+4*x)*exp(4)^2+(8*x^3+8*x^2)*exp(4)+2*x^4+4*x^3+2*x^2)/x/ln(16*x^2),x,method=_RETURNVERBOSE)

[Out]

(4*x*exp(12)+6*x^2*exp(8)+4*x^3*exp(4)+x^4+2*x*exp(8)-2*exp(exp(x)+8)+4*x^2*exp(4)-4*x*exp(exp(x)+4)+2*x^3-2*e
xp(exp(x))*x^2+x^2-2*x*exp(exp(x))+exp(2*exp(x)))*ln(ln(2)+1/2*ln(x)-1/8*I*Pi*csgn(I*x^2)*(-csgn(I*x^2)+csgn(I
*x))^2)+exp(16)*ln(ln(x)-1/4*I*(Pi*csgn(I*x)^2*csgn(I*x^2)-2*Pi*csgn(I*x)*csgn(I*x^2)^2+Pi*csgn(I*x^2)^3+8*I*l
n(2)))

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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.

[In]

integrate(((2*x*exp(x)*log(16*x^2)*exp(exp(x))^2+((-2*x*exp(4)^2-4*x^2*exp(4)-2*x^3-2*x^2)*exp(x)-4*x*exp(4)-4
*x^2-2*x)*log(16*x^2)*exp(exp(x))+(4*x*exp(4)^3+(12*x^2+2*x)*exp(4)^2+(12*x^3+8*x^2)*exp(4)+4*x^4+6*x^3+2*x^2)
*log(16*x^2))*log(1/4*log(16*x^2))+2*exp(exp(x))^2+(-4*exp(4)^2-8*x*exp(4)-4*x^2-4*x)*exp(exp(x))+2*exp(4)^4+8
*x*exp(4)^3+(12*x^2+4*x)*exp(4)^2+(8*x^3+8*x^2)*exp(4)+2*x^4+4*x^3+2*x^2)/x/log(16*x^2),x, algorithm="maxima")

[Out]

-x^4*log(2) - 2*(2*e^4*log(2) + log(2))*x^3 - (6*e^8*log(2) + 4*e^4*log(2) + log(2))*x^2 - 2*(2*e^12*log(2) +
e^8*log(2))*x + 2*(x^2*log(2) + (2*e^4*log(2) + log(2))*x + e^8*log(2))*e^(e^x) - e^(2*e^x)*log(2) + (x^4 + 2*
x^3*(2*e^4 + 1) + x^2*(6*e^8 + 4*e^4 + 1) + 2*x*(2*e^12 + e^8) - 2*(x^2 + x*(2*e^4 + 1) + e^8)*e^(e^x) + e^16
+ e^(2*e^x))*log(2*log(2) + log(x))

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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.

[In]

int((2*exp(16) + 2*exp(2*exp(x)) + log(log(16*x^2)/4)*(log(16*x^2)*(exp(8)*(2*x + 12*x^2) + 4*x*exp(12) + exp(
4)*(8*x^2 + 12*x^3) + 2*x^2 + 6*x^3 + 4*x^4) - exp(exp(x))*log(16*x^2)*(2*x + 4*x*exp(4) + 4*x^2 + exp(x)*(2*x
*exp(8) + 4*x^2*exp(4) + 2*x^2 + 2*x^3)) + 2*x*exp(2*exp(x))*exp(x)*log(16*x^2)) - exp(exp(x))*(4*x + 4*exp(8)
+ 8*x*exp(4) + 4*x^2) + exp(8)*(4*x + 12*x^2) + 8*x*exp(12) + exp(4)*(8*x^2 + 8*x^3) + 2*x^2 + 4*x^3 + 2*x^4)
/(x*log(16*x^2)),x)

[Out]

log(log(16*x^2)/4)*(exp(2*exp(x)) - exp(exp(x))*(2*x + 2*exp(8) + 4*x*exp(4) + 2*x^2) + (x^3*(4*exp(4) + 6*exp
(8) + 1) + x^4*(4*exp(4) + 2) + x^5 + 2*x^2*exp(8)*(2*exp(4) + 1))/x) + log(log(16*x^2))*exp(16)

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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.

[In]

integrate(((2*x*exp(x)*ln(16*x**2)*exp(exp(x))**2+((-2*x*exp(4)**2-4*x**2*exp(4)-2*x**3-2*x**2)*exp(x)-4*x*exp
(4)-4*x**2-2*x)*ln(16*x**2)*exp(exp(x))+(4*x*exp(4)**3+(12*x**2+2*x)*exp(4)**2+(12*x**3+8*x**2)*exp(4)+4*x**4+
6*x**3+2*x**2)*ln(16*x**2))*ln(1/4*ln(16*x**2))+2*exp(exp(x))**2+(-4*exp(4)**2-8*x*exp(4)-4*x**2-4*x)*exp(exp(
x))+2*exp(4)**4+8*x*exp(4)**3+(12*x**2+4*x)*exp(4)**2+(8*x**3+8*x**2)*exp(4)+2*x**4+4*x**3+2*x**2)/x/ln(16*x**
2),x)

[Out]

Timed out

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