Integrand size = 241, antiderivative size = 31 \[ \int \frac {-50 e^{e^{2 x}} x^2+10 e^{2 x} x \log (4)+(5+25 x) \log (4)+\left (-60 e^{e^{2 x}} x^3+30 x^2 \log (4)+\left (12 e^{e^{2 x}} x^2-6 x \log (4)\right ) \log \left (\frac {e^{-e^{2 x}} \left (2 e^{e^{2 x}} x-\log (4)\right )}{x}\right )\right ) \log \left (-5 x+\log \left (\frac {e^{-e^{2 x}} \left (2 e^{e^{2 x}} x-\log (4)\right )}{x}\right )\right )}{\left (-10 e^{e^{2 x}} x^3+5 x^2 \log (4)+\left (2 e^{e^{2 x}} x^2-x \log (4)\right ) \log \left (\frac {e^{-e^{2 x}} \left (2 e^{e^{2 x}} x-\log (4)\right )}{x}\right )\right ) \log \left (-5 x+\log \left (\frac {e^{-e^{2 x}} \left (2 e^{e^{2 x}} x-\log (4)\right )}{x}\right )\right )} \, dx=x+5 \left (x+\log \left (\log \left (-5 x+\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right )\right )\right ) \]
Time = 0.18 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00 \[ \int \frac {-50 e^{e^{2 x}} x^2+10 e^{2 x} x \log (4)+(5+25 x) \log (4)+\left (-60 e^{e^{2 x}} x^3+30 x^2 \log (4)+\left (12 e^{e^{2 x}} x^2-6 x \log (4)\right ) \log \left (\frac {e^{-e^{2 x}} \left (2 e^{e^{2 x}} x-\log (4)\right )}{x}\right )\right ) \log \left (-5 x+\log \left (\frac {e^{-e^{2 x}} \left (2 e^{e^{2 x}} x-\log (4)\right )}{x}\right )\right )}{\left (-10 e^{e^{2 x}} x^3+5 x^2 \log (4)+\left (2 e^{e^{2 x}} x^2-x \log (4)\right ) \log \left (\frac {e^{-e^{2 x}} \left (2 e^{e^{2 x}} x-\log (4)\right )}{x}\right )\right ) \log \left (-5 x+\log \left (\frac {e^{-e^{2 x}} \left (2 e^{e^{2 x}} x-\log (4)\right )}{x}\right )\right )} \, dx=6 x+5 \log \left (\log \left (-5 x+\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right )\right ) \]
Integrate[(-50*E^E^(2*x)*x^2 + 10*E^(2*x)*x*Log[4] + (5 + 25*x)*Log[4] + ( -60*E^E^(2*x)*x^3 + 30*x^2*Log[4] + (12*E^E^(2*x)*x^2 - 6*x*Log[4])*Log[(2 *E^E^(2*x)*x - Log[4])/(E^E^(2*x)*x)])*Log[-5*x + Log[(2*E^E^(2*x)*x - Log [4])/(E^E^(2*x)*x)]])/((-10*E^E^(2*x)*x^3 + 5*x^2*Log[4] + (2*E^E^(2*x)*x^ 2 - x*Log[4])*Log[(2*E^E^(2*x)*x - Log[4])/(E^E^(2*x)*x)])*Log[-5*x + Log[ (2*E^E^(2*x)*x - Log[4])/(E^E^(2*x)*x)]]),x]
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {-50 e^{e^{2 x}} x^2+\left (-60 e^{e^{2 x}} x^3+30 x^2 \log (4)+\left (12 e^{e^{2 x}} x^2-6 x \log (4)\right ) \log \left (\frac {e^{-e^{2 x}} \left (2 e^{e^{2 x}} x-\log (4)\right )}{x}\right )\right ) \log \left (\log \left (\frac {e^{-e^{2 x}} \left (2 e^{e^{2 x}} x-\log (4)\right )}{x}\right )-5 x\right )+10 e^{2 x} x \log (4)+(25 x+5) \log (4)}{\left (-10 e^{e^{2 x}} x^3+5 x^2 \log (4)+\left (2 e^{e^{2 x}} x^2-x \log (4)\right ) \log \left (\frac {e^{-e^{2 x}} \left (2 e^{e^{2 x}} x-\log (4)\right )}{x}\right )\right ) \log \left (\log \left (\frac {e^{-e^{2 x}} \left (2 e^{e^{2 x}} x-\log (4)\right )}{x}\right )-5 x\right )} \, dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int \frac {50 e^{e^{2 x}} x^2-\left (-60 e^{e^{2 x}} x^3+30 x^2 \log (4)+\left (12 e^{e^{2 x}} x^2-6 x \log (4)\right ) \log \left (\frac {e^{-e^{2 x}} \left (2 e^{e^{2 x}} x-\log (4)\right )}{x}\right )\right ) \log \left (\log \left (\frac {e^{-e^{2 x}} \left (2 e^{e^{2 x}} x-\log (4)\right )}{x}\right )-5 x\right )-10 e^{2 x} x \log (4)-(25 x+5) \log (4)}{x \left (2 e^{e^{2 x}} x-\log (4)\right ) \left (5 x-\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right ) \log \left (\log \left (\frac {e^{-e^{2 x}} \left (2 e^{e^{2 x}} x-\log (4)\right )}{x}\right )-5 x\right )}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (\frac {60 e^{e^{2 x}} x^3 \log \left (\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )-5 x\right )+50 e^{e^{2 x}} x^2-12 e^{e^{2 x}} x^2 \log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right ) \log \left (\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )-5 x\right )-30 x^2 \log (4) \log \left (\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )-5 x\right )+6 x \log (4) \log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right ) \log \left (\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )-5 x\right )-25 x \log (4)-5 \log (4)}{x \left (2 e^{e^{2 x}} x-\log (4)\right ) \left (5 x-\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right ) \log \left (\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )-5 x\right )}-\frac {10 e^{2 x} \log (4)}{\left (2 e^{e^{2 x}} x-\log (4)\right ) \left (5 x-\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right ) \log \left (\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )-5 x\right )}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle 25 \int \frac {1}{\left (5 x-\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right ) \log \left (\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )-5 x\right )}dx-10 \log (4) \int \frac {e^{2 x}}{\left (2 e^{e^{2 x}} x-\log (4)\right ) \left (5 x-\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right ) \log \left (\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )-5 x\right )}dx-5 \log (4) \int \frac {1}{x \left (2 e^{e^{2 x}} x-\log (4)\right ) \left (5 x-\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )\right ) \log \left (\log \left (2-\frac {e^{-e^{2 x}} \log (4)}{x}\right )-5 x\right )}dx+6 x\) |
Int[(-50*E^E^(2*x)*x^2 + 10*E^(2*x)*x*Log[4] + (5 + 25*x)*Log[4] + (-60*E^ E^(2*x)*x^3 + 30*x^2*Log[4] + (12*E^E^(2*x)*x^2 - 6*x*Log[4])*Log[(2*E^E^( 2*x)*x - Log[4])/(E^E^(2*x)*x)])*Log[-5*x + Log[(2*E^E^(2*x)*x - Log[4])/( E^E^(2*x)*x)]])/((-10*E^E^(2*x)*x^3 + 5*x^2*Log[4] + (2*E^E^(2*x)*x^2 - x* Log[4])*Log[(2*E^E^(2*x)*x - Log[4])/(E^E^(2*x)*x)])*Log[-5*x + Log[(2*E^E ^(2*x)*x - Log[4])/(E^E^(2*x)*x)]]),x]
3.4.98.3.1 Defintions of rubi rules used
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.12 (sec) , antiderivative size = 317, normalized size of antiderivative = 10.23
\[6 x +5 \ln \left (\ln \left (\ln \left (2\right )+i \pi -\ln \left (x \right )-\ln \left ({\mathrm e}^{{\mathrm e}^{2 x}}\right )+\ln \left (-x \,{\mathrm e}^{{\mathrm e}^{2 x}}+\ln \left (2\right )\right )-\frac {i \pi \,\operatorname {csgn}\left (i {\mathrm e}^{-{\mathrm e}^{2 x}} \left (-x \,{\mathrm e}^{{\mathrm e}^{2 x}}+\ln \left (2\right )\right )\right ) \left (-\operatorname {csgn}\left (i {\mathrm e}^{-{\mathrm e}^{2 x}} \left (-x \,{\mathrm e}^{{\mathrm e}^{2 x}}+\ln \left (2\right )\right )\right )+\operatorname {csgn}\left (i {\mathrm e}^{-{\mathrm e}^{2 x}}\right )\right ) \left (-\operatorname {csgn}\left (i {\mathrm e}^{-{\mathrm e}^{2 x}} \left (-x \,{\mathrm e}^{{\mathrm e}^{2 x}}+\ln \left (2\right )\right )\right )+\operatorname {csgn}\left (i \left (-x \,{\mathrm e}^{{\mathrm e}^{2 x}}+\ln \left (2\right )\right )\right )\right )}{2}-\frac {i \pi \,\operatorname {csgn}\left (\frac {i \left (-x \,{\mathrm e}^{{\mathrm e}^{2 x}}+\ln \left (2\right )\right ) {\mathrm e}^{-{\mathrm e}^{2 x}}}{x}\right ) \left (-\operatorname {csgn}\left (\frac {i \left (-x \,{\mathrm e}^{{\mathrm e}^{2 x}}+\ln \left (2\right )\right ) {\mathrm e}^{-{\mathrm e}^{2 x}}}{x}\right )+\operatorname {csgn}\left (\frac {i}{x}\right )\right ) \left (-\operatorname {csgn}\left (\frac {i \left (-x \,{\mathrm e}^{{\mathrm e}^{2 x}}+\ln \left (2\right )\right ) {\mathrm e}^{-{\mathrm e}^{2 x}}}{x}\right )+\operatorname {csgn}\left (i {\mathrm e}^{-{\mathrm e}^{2 x}} \left (-x \,{\mathrm e}^{{\mathrm e}^{2 x}}+\ln \left (2\right )\right )\right )\right )}{2}+i \pi {\operatorname {csgn}\left (\frac {i \left (-x \,{\mathrm e}^{{\mathrm e}^{2 x}}+\ln \left (2\right )\right ) {\mathrm e}^{-{\mathrm e}^{2 x}}}{x}\right )}^{2} \left (\operatorname {csgn}\left (\frac {i \left (-x \,{\mathrm e}^{{\mathrm e}^{2 x}}+\ln \left (2\right )\right ) {\mathrm e}^{-{\mathrm e}^{2 x}}}{x}\right )-1\right )-5 x \right )\right )\]
int((((12*x^2*exp(exp(x)^2)-12*x*ln(2))*ln((2*x*exp(exp(x)^2)-2*ln(2))/x/e xp(exp(x)^2))-60*x^3*exp(exp(x)^2)+60*x^2*ln(2))*ln(ln((2*x*exp(exp(x)^2)- 2*ln(2))/x/exp(exp(x)^2))-5*x)-50*x^2*exp(exp(x)^2)+20*x*ln(2)*exp(x)^2+2* (25*x+5)*ln(2))/((2*x^2*exp(exp(x)^2)-2*x*ln(2))*ln((2*x*exp(exp(x)^2)-2*l n(2))/x/exp(exp(x)^2))-10*x^3*exp(exp(x)^2)+10*x^2*ln(2))/ln(ln((2*x*exp(e xp(x)^2)-2*ln(2))/x/exp(exp(x)^2))-5*x),x)
6*x+5*ln(ln(ln(2)+I*Pi-ln(x)-ln(exp(exp(2*x)))+ln(-x*exp(exp(2*x))+ln(2))- 1/2*I*Pi*csgn(I*exp(-exp(2*x))*(-x*exp(exp(2*x))+ln(2)))*(-csgn(I*exp(-exp (2*x))*(-x*exp(exp(2*x))+ln(2)))+csgn(I*exp(-exp(2*x))))*(-csgn(I*exp(-exp (2*x))*(-x*exp(exp(2*x))+ln(2)))+csgn(I*(-x*exp(exp(2*x))+ln(2))))-1/2*I*P i*csgn(I/x*(-x*exp(exp(2*x))+ln(2))*exp(-exp(2*x)))*(-csgn(I/x*(-x*exp(exp (2*x))+ln(2))*exp(-exp(2*x)))+csgn(I/x))*(-csgn(I/x*(-x*exp(exp(2*x))+ln(2 ))*exp(-exp(2*x)))+csgn(I*exp(-exp(2*x))*(-x*exp(exp(2*x))+ln(2))))+I*Pi*c sgn(I/x*(-x*exp(exp(2*x))+ln(2))*exp(-exp(2*x)))^2*(csgn(I/x*(-x*exp(exp(2 *x))+ln(2))*exp(-exp(2*x)))-1)-5*x))
Time = 0.28 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.19 \[ \int \frac {-50 e^{e^{2 x}} x^2+10 e^{2 x} x \log (4)+(5+25 x) \log (4)+\left (-60 e^{e^{2 x}} x^3+30 x^2 \log (4)+\left (12 e^{e^{2 x}} x^2-6 x \log (4)\right ) \log \left (\frac {e^{-e^{2 x}} \left (2 e^{e^{2 x}} x-\log (4)\right )}{x}\right )\right ) \log \left (-5 x+\log \left (\frac {e^{-e^{2 x}} \left (2 e^{e^{2 x}} x-\log (4)\right )}{x}\right )\right )}{\left (-10 e^{e^{2 x}} x^3+5 x^2 \log (4)+\left (2 e^{e^{2 x}} x^2-x \log (4)\right ) \log \left (\frac {e^{-e^{2 x}} \left (2 e^{e^{2 x}} x-\log (4)\right )}{x}\right )\right ) \log \left (-5 x+\log \left (\frac {e^{-e^{2 x}} \left (2 e^{e^{2 x}} x-\log (4)\right )}{x}\right )\right )} \, dx=6 \, x + 5 \, \log \left (\log \left (-5 \, x + \log \left (\frac {2 \, {\left (x e^{\left (e^{\left (2 \, x\right )}\right )} - \log \left (2\right )\right )} e^{\left (-e^{\left (2 \, x\right )}\right )}}{x}\right )\right )\right ) \]
integrate((((12*x^2*exp(exp(x)^2)-12*x*log(2))*log((2*x*exp(exp(x)^2)-2*lo g(2))/x/exp(exp(x)^2))-60*x^3*exp(exp(x)^2)+60*x^2*log(2))*log(log((2*x*ex p(exp(x)^2)-2*log(2))/x/exp(exp(x)^2))-5*x)-50*x^2*exp(exp(x)^2)+20*x*log( 2)*exp(x)^2+2*(25*x+5)*log(2))/((2*x^2*exp(exp(x)^2)-2*x*log(2))*log((2*x* exp(exp(x)^2)-2*log(2))/x/exp(exp(x)^2))-10*x^3*exp(exp(x)^2)+10*x^2*log(2 ))/log(log((2*x*exp(exp(x)^2)-2*log(2))/x/exp(exp(x)^2))-5*x),x, algorithm =\
Timed out. \[ \int \frac {-50 e^{e^{2 x}} x^2+10 e^{2 x} x \log (4)+(5+25 x) \log (4)+\left (-60 e^{e^{2 x}} x^3+30 x^2 \log (4)+\left (12 e^{e^{2 x}} x^2-6 x \log (4)\right ) \log \left (\frac {e^{-e^{2 x}} \left (2 e^{e^{2 x}} x-\log (4)\right )}{x}\right )\right ) \log \left (-5 x+\log \left (\frac {e^{-e^{2 x}} \left (2 e^{e^{2 x}} x-\log (4)\right )}{x}\right )\right )}{\left (-10 e^{e^{2 x}} x^3+5 x^2 \log (4)+\left (2 e^{e^{2 x}} x^2-x \log (4)\right ) \log \left (\frac {e^{-e^{2 x}} \left (2 e^{e^{2 x}} x-\log (4)\right )}{x}\right )\right ) \log \left (-5 x+\log \left (\frac {e^{-e^{2 x}} \left (2 e^{e^{2 x}} x-\log (4)\right )}{x}\right )\right )} \, dx=\text {Timed out} \]
integrate((((12*x**2*exp(exp(x)**2)-12*x*ln(2))*ln((2*x*exp(exp(x)**2)-2*l n(2))/x/exp(exp(x)**2))-60*x**3*exp(exp(x)**2)+60*x**2*ln(2))*ln(ln((2*x*e xp(exp(x)**2)-2*ln(2))/x/exp(exp(x)**2))-5*x)-50*x**2*exp(exp(x)**2)+20*x* ln(2)*exp(x)**2+2*(25*x+5)*ln(2))/((2*x**2*exp(exp(x)**2)-2*x*ln(2))*ln((2 *x*exp(exp(x)**2)-2*ln(2))/x/exp(exp(x)**2))-10*x**3*exp(exp(x)**2)+10*x** 2*ln(2))/ln(ln((2*x*exp(exp(x)**2)-2*ln(2))/x/exp(exp(x)**2))-5*x),x)
Time = 0.46 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.19 \[ \int \frac {-50 e^{e^{2 x}} x^2+10 e^{2 x} x \log (4)+(5+25 x) \log (4)+\left (-60 e^{e^{2 x}} x^3+30 x^2 \log (4)+\left (12 e^{e^{2 x}} x^2-6 x \log (4)\right ) \log \left (\frac {e^{-e^{2 x}} \left (2 e^{e^{2 x}} x-\log (4)\right )}{x}\right )\right ) \log \left (-5 x+\log \left (\frac {e^{-e^{2 x}} \left (2 e^{e^{2 x}} x-\log (4)\right )}{x}\right )\right )}{\left (-10 e^{e^{2 x}} x^3+5 x^2 \log (4)+\left (2 e^{e^{2 x}} x^2-x \log (4)\right ) \log \left (\frac {e^{-e^{2 x}} \left (2 e^{e^{2 x}} x-\log (4)\right )}{x}\right )\right ) \log \left (-5 x+\log \left (\frac {e^{-e^{2 x}} \left (2 e^{e^{2 x}} x-\log (4)\right )}{x}\right )\right )} \, dx=6 \, x + 5 \, \log \left (\log \left (-5 \, x - e^{\left (2 \, x\right )} + \log \left (2\right ) + \log \left (x e^{\left (e^{\left (2 \, x\right )}\right )} - \log \left (2\right )\right ) - \log \left (x\right )\right )\right ) \]
integrate((((12*x^2*exp(exp(x)^2)-12*x*log(2))*log((2*x*exp(exp(x)^2)-2*lo g(2))/x/exp(exp(x)^2))-60*x^3*exp(exp(x)^2)+60*x^2*log(2))*log(log((2*x*ex p(exp(x)^2)-2*log(2))/x/exp(exp(x)^2))-5*x)-50*x^2*exp(exp(x)^2)+20*x*log( 2)*exp(x)^2+2*(25*x+5)*log(2))/((2*x^2*exp(exp(x)^2)-2*x*log(2))*log((2*x* exp(exp(x)^2)-2*log(2))/x/exp(exp(x)^2))-10*x^3*exp(exp(x)^2)+10*x^2*log(2 ))/log(log((2*x*exp(exp(x)^2)-2*log(2))/x/exp(exp(x)^2))-5*x),x, algorithm =\
Time = 0.43 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.26 \[ \int \frac {-50 e^{e^{2 x}} x^2+10 e^{2 x} x \log (4)+(5+25 x) \log (4)+\left (-60 e^{e^{2 x}} x^3+30 x^2 \log (4)+\left (12 e^{e^{2 x}} x^2-6 x \log (4)\right ) \log \left (\frac {e^{-e^{2 x}} \left (2 e^{e^{2 x}} x-\log (4)\right )}{x}\right )\right ) \log \left (-5 x+\log \left (\frac {e^{-e^{2 x}} \left (2 e^{e^{2 x}} x-\log (4)\right )}{x}\right )\right )}{\left (-10 e^{e^{2 x}} x^3+5 x^2 \log (4)+\left (2 e^{e^{2 x}} x^2-x \log (4)\right ) \log \left (\frac {e^{-e^{2 x}} \left (2 e^{e^{2 x}} x-\log (4)\right )}{x}\right )\right ) \log \left (-5 x+\log \left (\frac {e^{-e^{2 x}} \left (2 e^{e^{2 x}} x-\log (4)\right )}{x}\right )\right )} \, dx=6 \, x + 5 \, \log \left (\log \left (-5 \, x + \log \left (2\right ) + \log \left ({\left (x e^{\left (e^{\left (2 \, x\right )}\right )} - \log \left (2\right )\right )} e^{\left (-e^{\left (2 \, x\right )}\right )}\right ) - \log \left (x\right )\right )\right ) \]
integrate((((12*x^2*exp(exp(x)^2)-12*x*log(2))*log((2*x*exp(exp(x)^2)-2*lo g(2))/x/exp(exp(x)^2))-60*x^3*exp(exp(x)^2)+60*x^2*log(2))*log(log((2*x*ex p(exp(x)^2)-2*log(2))/x/exp(exp(x)^2))-5*x)-50*x^2*exp(exp(x)^2)+20*x*log( 2)*exp(x)^2+2*(25*x+5)*log(2))/((2*x^2*exp(exp(x)^2)-2*x*log(2))*log((2*x* exp(exp(x)^2)-2*log(2))/x/exp(exp(x)^2))-10*x^3*exp(exp(x)^2)+10*x^2*log(2 ))/log(log((2*x*exp(exp(x)^2)-2*log(2))/x/exp(exp(x)^2))-5*x),x, algorithm =\
Time = 8.95 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.03 \[ \int \frac {-50 e^{e^{2 x}} x^2+10 e^{2 x} x \log (4)+(5+25 x) \log (4)+\left (-60 e^{e^{2 x}} x^3+30 x^2 \log (4)+\left (12 e^{e^{2 x}} x^2-6 x \log (4)\right ) \log \left (\frac {e^{-e^{2 x}} \left (2 e^{e^{2 x}} x-\log (4)\right )}{x}\right )\right ) \log \left (-5 x+\log \left (\frac {e^{-e^{2 x}} \left (2 e^{e^{2 x}} x-\log (4)\right )}{x}\right )\right )}{\left (-10 e^{e^{2 x}} x^3+5 x^2 \log (4)+\left (2 e^{e^{2 x}} x^2-x \log (4)\right ) \log \left (\frac {e^{-e^{2 x}} \left (2 e^{e^{2 x}} x-\log (4)\right )}{x}\right )\right ) \log \left (-5 x+\log \left (\frac {e^{-e^{2 x}} \left (2 e^{e^{2 x}} x-\log (4)\right )}{x}\right )\right )} \, dx=6\,x+5\,\ln \left (\ln \left (\ln \left (\frac {2\,x-2\,{\mathrm {e}}^{-{\mathrm {e}}^{2\,x}}\,\ln \left (2\right )}{x}\right )-5\,x\right )\right ) \]
int(-(2*log(2)*(25*x + 5) - log(log(-(exp(-exp(2*x))*(2*log(2) - 2*x*exp(e xp(2*x))))/x) - 5*x)*(log(-(exp(-exp(2*x))*(2*log(2) - 2*x*exp(exp(2*x)))) /x)*(12*x*log(2) - 12*x^2*exp(exp(2*x))) - 60*x^2*log(2) + 60*x^3*exp(exp( 2*x))) - 50*x^2*exp(exp(2*x)) + 20*x*exp(2*x)*log(2))/(log(log(-(exp(-exp( 2*x))*(2*log(2) - 2*x*exp(exp(2*x))))/x) - 5*x)*(log(-(exp(-exp(2*x))*(2*l og(2) - 2*x*exp(exp(2*x))))/x)*(2*x*log(2) - 2*x^2*exp(exp(2*x))) - 10*x^2 *log(2) + 10*x^3*exp(exp(2*x)))),x)