Integrand size = 347, antiderivative size = 31 \[ \int \frac {-e^2+\left (-5 e^2-20 x\right ) \log \left (\frac {-e^2-4 x}{x}\right ) \log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right )+\left (e^2+4 x\right ) \log \left (\frac {-e^2-4 x}{x}\right ) \log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right ) \log \left (\log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right )\right )}{\left (25 e^2+100 x\right ) \log \left (\frac {-e^2-4 x}{x}\right ) \log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right )+\left (-10 e^2-40 x\right ) \log \left (\frac {-e^2-4 x}{x}\right ) \log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right ) \log \left (\log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right )\right )+\left (e^2+4 x\right ) \log \left (\frac {-e^2-4 x}{x}\right ) \log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right ) \log ^2\left (\log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right )\right )} \, dx=\frac {x}{-5+\log \left (\log \left (\frac {e^{2-e^4}}{\log \left (-4-\frac {e^2}{x}\right )}\right )\right )} \]
Time = 0.19 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.90 \[ \int \frac {-e^2+\left (-5 e^2-20 x\right ) \log \left (\frac {-e^2-4 x}{x}\right ) \log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right )+\left (e^2+4 x\right ) \log \left (\frac {-e^2-4 x}{x}\right ) \log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right ) \log \left (\log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right )\right )}{\left (25 e^2+100 x\right ) \log \left (\frac {-e^2-4 x}{x}\right ) \log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right )+\left (-10 e^2-40 x\right ) \log \left (\frac {-e^2-4 x}{x}\right ) \log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right ) \log \left (\log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right )\right )+\left (e^2+4 x\right ) \log \left (\frac {-e^2-4 x}{x}\right ) \log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right ) \log ^2\left (\log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right )\right )} \, dx=\frac {x}{-5+\log \left (2-e^4+\log \left (\frac {1}{\log \left (-4-\frac {e^2}{x}\right )}\right )\right )} \]
Integrate[(-E^2 + (-5*E^2 - 20*x)*Log[(-E^2 - 4*x)/x]*Log[E^(2 - E^4)/Log[ (-E^2 - 4*x)/x]] + (E^2 + 4*x)*Log[(-E^2 - 4*x)/x]*Log[E^(2 - E^4)/Log[(-E ^2 - 4*x)/x]]*Log[Log[E^(2 - E^4)/Log[(-E^2 - 4*x)/x]]])/((25*E^2 + 100*x) *Log[(-E^2 - 4*x)/x]*Log[E^(2 - E^4)/Log[(-E^2 - 4*x)/x]] + (-10*E^2 - 40* x)*Log[(-E^2 - 4*x)/x]*Log[E^(2 - E^4)/Log[(-E^2 - 4*x)/x]]*Log[Log[E^(2 - E^4)/Log[(-E^2 - 4*x)/x]]] + (E^2 + 4*x)*Log[(-E^2 - 4*x)/x]*Log[E^(2 - E ^4)/Log[(-E^2 - 4*x)/x]]*Log[Log[E^(2 - E^4)/Log[(-E^2 - 4*x)/x]]]^2),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.
| \(\displaystyle \int \frac {\left (-20 x-5 e^2\right ) \log \left (\frac {-4 x-e^2}{x}\right ) \log \left (\frac {e^{2-e^4}}{\log \left (\frac {-4 x-e^2}{x}\right )}\right )+\left (4 x+e^2\right ) \log \left (\frac {-4 x-e^2}{x}\right ) \log \left (\log \left (\frac {e^{2-e^4}}{\log \left (\frac {-4 x-e^2}{x}\right )}\right )\right ) \log \left (\frac {e^{2-e^4}}{\log \left (\frac {-4 x-e^2}{x}\right )}\right )-e^2}{\left (4 x+e^2\right ) \log \left (\frac {-4 x-e^2}{x}\right ) \log \left (\frac {e^{2-e^4}}{\log \left (\frac {-4 x-e^2}{x}\right )}\right ) \log ^2\left (\log \left (\frac {e^{2-e^4}}{\log \left (\frac {-4 x-e^2}{x}\right )}\right )\right )+\left (-40 x-10 e^2\right ) \log \left (\frac {-4 x-e^2}{x}\right ) \log \left (\frac {e^{2-e^4}}{\log \left (\frac {-4 x-e^2}{x}\right )}\right ) \log \left (\log \left (\frac {e^{2-e^4}}{\log \left (\frac {-4 x-e^2}{x}\right )}\right )\right )+\left (100 x+25 e^2\right ) \log \left (\frac {-4 x-e^2}{x}\right ) \log \left (\frac {e^{2-e^4}}{\log \left (\frac {-4 x-e^2}{x}\right )}\right )} \, dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle \int \frac {-\left (\left (4 x+e^2\right ) \log \left (-\frac {e^2}{x}-4\right ) \left (-\log \left (\frac {1}{\log \left (-\frac {e^2}{x}-4\right )}\right )+e^4-2\right ) \left (\log \left (\log \left (\frac {1}{\log \left (-\frac {e^2}{x}-4\right )}\right )-e^4+2\right )-5\right )\right )-e^2}{\left (4 x+e^2\right ) \log \left (-\frac {e^2}{x}-4\right ) \left (\log \left (\frac {1}{\log \left (-\frac {e^2}{x}-4\right )}\right )+2 \left (1-\frac {e^4}{2}\right )\right ) \left (5-\log \left (\log \left (\frac {1}{\log \left (-\frac {e^2}{x}-4\right )}\right )-e^4+2\right )\right )^2}dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle \int \frac {\log \left (\log \left (\frac {1}{\log \left (-\frac {e^2}{x}-4\right )}\right )-e^4+2\right )+\frac {e^2}{\left (4 x+e^2\right ) \log \left (-\frac {e^2}{x}-4\right ) \left (-\log \left (\frac {1}{\log \left (-\frac {e^2}{x}-4\right )}\right )+e^4-2\right )}-5}{\left (5-\log \left (\log \left (\frac {1}{\log \left (-\frac {e^2}{x}-4\right )}\right )-e^4+2\right )\right )^2}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (\frac {1}{\log \left (\log \left (\frac {1}{\log \left (-\frac {e^2}{x}-4\right )}\right )-e^4+2\right )-5}+\frac {e^2}{\left (-4 x-e^2\right ) \log \left (-\frac {e^2}{x}-4\right ) \left (5-\log \left (\log \left (\frac {1}{\log \left (-\frac {e^2}{x}-4\right )}\right )-e^4+2\right )\right )^2 \left (\log \left (\frac {1}{\log \left (-\frac {e^2}{x}-4\right )}\right )+2 \left (1-\frac {e^4}{2}\right )\right )}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle e^2 \int \frac {1}{\left (-4 x-e^2\right ) \log \left (-4-\frac {e^2}{x}\right ) \left (\log \left (\frac {1}{\log \left (-4-\frac {e^2}{x}\right )}\right )+2 \left (1-\frac {e^4}{2}\right )\right ) \left (5-\log \left (\log \left (\frac {1}{\log \left (-4-\frac {e^2}{x}\right )}\right )-e^4+2\right )\right )^2}dx+\int \frac {1}{\log \left (\log \left (\frac {1}{\log \left (-4-\frac {e^2}{x}\right )}\right )-e^4+2\right )-5}dx\) |
Int[(-E^2 + (-5*E^2 - 20*x)*Log[(-E^2 - 4*x)/x]*Log[E^(2 - E^4)/Log[(-E^2 - 4*x)/x]] + (E^2 + 4*x)*Log[(-E^2 - 4*x)/x]*Log[E^(2 - E^4)/Log[(-E^2 - 4 *x)/x]]*Log[Log[E^(2 - E^4)/Log[(-E^2 - 4*x)/x]]])/((25*E^2 + 100*x)*Log[( -E^2 - 4*x)/x]*Log[E^(2 - E^4)/Log[(-E^2 - 4*x)/x]] + (-10*E^2 - 40*x)*Log [(-E^2 - 4*x)/x]*Log[E^(2 - E^4)/Log[(-E^2 - 4*x)/x]]*Log[Log[E^(2 - E^4)/ Log[(-E^2 - 4*x)/x]]] + (E^2 + 4*x)*Log[(-E^2 - 4*x)/x]*Log[E^(2 - E^4)/Lo g[(-E^2 - 4*x)/x]]*Log[Log[E^(2 - E^4)/Log[(-E^2 - 4*x)/x]]]^2),x]
3.3.71.3.1 Defintions of rubi rules used
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 20.55 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00
| method | result | size |
| parallelrisch | \(\frac {x}{\ln \left (\ln \left (\frac {{\mathrm e}^{2-{\mathrm e}^{4}}}{\ln \left (-\frac {{\mathrm e}^{2}+4 x}{x}\right )}\right )\right )-5}\) | \(31\) |
int(((exp(2)+4*x)*ln((-exp(2)-4*x)/x)*ln(1/exp(exp(4)-2)/ln((-exp(2)-4*x)/ x))*ln(ln(1/exp(exp(4)-2)/ln((-exp(2)-4*x)/x)))+(-5*exp(2)-20*x)*ln((-exp( 2)-4*x)/x)*ln(1/exp(exp(4)-2)/ln((-exp(2)-4*x)/x))-exp(2))/((exp(2)+4*x)*l n((-exp(2)-4*x)/x)*ln(1/exp(exp(4)-2)/ln((-exp(2)-4*x)/x))*ln(ln(1/exp(exp (4)-2)/ln((-exp(2)-4*x)/x)))^2+(-10*exp(2)-40*x)*ln((-exp(2)-4*x)/x)*ln(1/ exp(exp(4)-2)/ln((-exp(2)-4*x)/x))*ln(ln(1/exp(exp(4)-2)/ln((-exp(2)-4*x)/ x)))+(25*exp(2)+100*x)*ln((-exp(2)-4*x)/x)*ln(1/exp(exp(4)-2)/ln((-exp(2)- 4*x)/x))),x,method=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.97 \[ \int \frac {-e^2+\left (-5 e^2-20 x\right ) \log \left (\frac {-e^2-4 x}{x}\right ) \log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right )+\left (e^2+4 x\right ) \log \left (\frac {-e^2-4 x}{x}\right ) \log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right ) \log \left (\log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right )\right )}{\left (25 e^2+100 x\right ) \log \left (\frac {-e^2-4 x}{x}\right ) \log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right )+\left (-10 e^2-40 x\right ) \log \left (\frac {-e^2-4 x}{x}\right ) \log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right ) \log \left (\log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right )\right )+\left (e^2+4 x\right ) \log \left (\frac {-e^2-4 x}{x}\right ) \log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right ) \log ^2\left (\log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right )\right )} \, dx=\frac {x}{\log \left (\log \left (\frac {e^{\left (-e^{4} + 2\right )}}{\log \left (-\frac {4 \, x + e^{2}}{x}\right )}\right )\right ) - 5} \]
integrate(((exp(2)+4*x)*log((-exp(2)-4*x)/x)*log(1/exp(exp(4)-2)/log((-exp (2)-4*x)/x))*log(log(1/exp(exp(4)-2)/log((-exp(2)-4*x)/x)))+(-5*exp(2)-20* x)*log((-exp(2)-4*x)/x)*log(1/exp(exp(4)-2)/log((-exp(2)-4*x)/x))-exp(2))/ ((exp(2)+4*x)*log((-exp(2)-4*x)/x)*log(1/exp(exp(4)-2)/log((-exp(2)-4*x)/x ))*log(log(1/exp(exp(4)-2)/log((-exp(2)-4*x)/x)))^2+(-10*exp(2)-40*x)*log( (-exp(2)-4*x)/x)*log(1/exp(exp(4)-2)/log((-exp(2)-4*x)/x))*log(log(1/exp(e xp(4)-2)/log((-exp(2)-4*x)/x)))+(25*exp(2)+100*x)*log((-exp(2)-4*x)/x)*log (1/exp(exp(4)-2)/log((-exp(2)-4*x)/x))),x, algorithm=\
Time = 0.20 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.84 \[ \int \frac {-e^2+\left (-5 e^2-20 x\right ) \log \left (\frac {-e^2-4 x}{x}\right ) \log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right )+\left (e^2+4 x\right ) \log \left (\frac {-e^2-4 x}{x}\right ) \log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right ) \log \left (\log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right )\right )}{\left (25 e^2+100 x\right ) \log \left (\frac {-e^2-4 x}{x}\right ) \log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right )+\left (-10 e^2-40 x\right ) \log \left (\frac {-e^2-4 x}{x}\right ) \log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right ) \log \left (\log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right )\right )+\left (e^2+4 x\right ) \log \left (\frac {-e^2-4 x}{x}\right ) \log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right ) \log ^2\left (\log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right )\right )} \, dx=\frac {x}{\log {\left (\log {\left (\frac {1}{e^{-2 + e^{4}} \log {\left (\frac {- 4 x - e^{2}}{x} \right )}} \right )} \right )} - 5} \]
integrate(((exp(2)+4*x)*ln((-exp(2)-4*x)/x)*ln(1/exp(exp(4)-2)/ln((-exp(2) -4*x)/x))*ln(ln(1/exp(exp(4)-2)/ln((-exp(2)-4*x)/x)))+(-5*exp(2)-20*x)*ln( (-exp(2)-4*x)/x)*ln(1/exp(exp(4)-2)/ln((-exp(2)-4*x)/x))-exp(2))/((exp(2)+ 4*x)*ln((-exp(2)-4*x)/x)*ln(1/exp(exp(4)-2)/ln((-exp(2)-4*x)/x))*ln(ln(1/e xp(exp(4)-2)/ln((-exp(2)-4*x)/x)))**2+(-10*exp(2)-40*x)*ln((-exp(2)-4*x)/x )*ln(1/exp(exp(4)-2)/ln((-exp(2)-4*x)/x))*ln(ln(1/exp(exp(4)-2)/ln((-exp(2 )-4*x)/x)))+(25*exp(2)+100*x)*ln((-exp(2)-4*x)/x)*ln(1/exp(exp(4)-2)/ln((- exp(2)-4*x)/x))),x)
Time = 0.32 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.97 \[ \int \frac {-e^2+\left (-5 e^2-20 x\right ) \log \left (\frac {-e^2-4 x}{x}\right ) \log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right )+\left (e^2+4 x\right ) \log \left (\frac {-e^2-4 x}{x}\right ) \log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right ) \log \left (\log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right )\right )}{\left (25 e^2+100 x\right ) \log \left (\frac {-e^2-4 x}{x}\right ) \log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right )+\left (-10 e^2-40 x\right ) \log \left (\frac {-e^2-4 x}{x}\right ) \log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right ) \log \left (\log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right )\right )+\left (e^2+4 x\right ) \log \left (\frac {-e^2-4 x}{x}\right ) \log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right ) \log ^2\left (\log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right )\right )} \, dx=\frac {x}{\log \left (-e^{4} - \log \left (-\log \left (x\right ) + \log \left (-4 \, x - e^{2}\right )\right ) + 2\right ) - 5} \]
integrate(((exp(2)+4*x)*log((-exp(2)-4*x)/x)*log(1/exp(exp(4)-2)/log((-exp (2)-4*x)/x))*log(log(1/exp(exp(4)-2)/log((-exp(2)-4*x)/x)))+(-5*exp(2)-20* x)*log((-exp(2)-4*x)/x)*log(1/exp(exp(4)-2)/log((-exp(2)-4*x)/x))-exp(2))/ ((exp(2)+4*x)*log((-exp(2)-4*x)/x)*log(1/exp(exp(4)-2)/log((-exp(2)-4*x)/x ))*log(log(1/exp(exp(4)-2)/log((-exp(2)-4*x)/x)))^2+(-10*exp(2)-40*x)*log( (-exp(2)-4*x)/x)*log(1/exp(exp(4)-2)/log((-exp(2)-4*x)/x))*log(log(1/exp(e xp(4)-2)/log((-exp(2)-4*x)/x)))+(25*exp(2)+100*x)*log((-exp(2)-4*x)/x)*log (1/exp(exp(4)-2)/log((-exp(2)-4*x)/x))),x, algorithm=\
\[ \int \frac {-e^2+\left (-5 e^2-20 x\right ) \log \left (\frac {-e^2-4 x}{x}\right ) \log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right )+\left (e^2+4 x\right ) \log \left (\frac {-e^2-4 x}{x}\right ) \log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right ) \log \left (\log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right )\right )}{\left (25 e^2+100 x\right ) \log \left (\frac {-e^2-4 x}{x}\right ) \log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right )+\left (-10 e^2-40 x\right ) \log \left (\frac {-e^2-4 x}{x}\right ) \log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right ) \log \left (\log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right )\right )+\left (e^2+4 x\right ) \log \left (\frac {-e^2-4 x}{x}\right ) \log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right ) \log ^2\left (\log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right )\right )} \, dx=\int { \frac {{\left (4 \, x + e^{2}\right )} \log \left (-\frac {4 \, x + e^{2}}{x}\right ) \log \left (\frac {e^{\left (-e^{4} + 2\right )}}{\log \left (-\frac {4 \, x + e^{2}}{x}\right )}\right ) \log \left (\log \left (\frac {e^{\left (-e^{4} + 2\right )}}{\log \left (-\frac {4 \, x + e^{2}}{x}\right )}\right )\right ) - 5 \, {\left (4 \, x + e^{2}\right )} \log \left (-\frac {4 \, x + e^{2}}{x}\right ) \log \left (\frac {e^{\left (-e^{4} + 2\right )}}{\log \left (-\frac {4 \, x + e^{2}}{x}\right )}\right ) - e^{2}}{{\left (4 \, x + e^{2}\right )} \log \left (-\frac {4 \, x + e^{2}}{x}\right ) \log \left (\frac {e^{\left (-e^{4} + 2\right )}}{\log \left (-\frac {4 \, x + e^{2}}{x}\right )}\right ) \log \left (\log \left (\frac {e^{\left (-e^{4} + 2\right )}}{\log \left (-\frac {4 \, x + e^{2}}{x}\right )}\right )\right )^{2} - 10 \, {\left (4 \, x + e^{2}\right )} \log \left (-\frac {4 \, x + e^{2}}{x}\right ) \log \left (\frac {e^{\left (-e^{4} + 2\right )}}{\log \left (-\frac {4 \, x + e^{2}}{x}\right )}\right ) \log \left (\log \left (\frac {e^{\left (-e^{4} + 2\right )}}{\log \left (-\frac {4 \, x + e^{2}}{x}\right )}\right )\right ) + 25 \, {\left (4 \, x + e^{2}\right )} \log \left (-\frac {4 \, x + e^{2}}{x}\right ) \log \left (\frac {e^{\left (-e^{4} + 2\right )}}{\log \left (-\frac {4 \, x + e^{2}}{x}\right )}\right )} \,d x } \]
integrate(((exp(2)+4*x)*log((-exp(2)-4*x)/x)*log(1/exp(exp(4)-2)/log((-exp (2)-4*x)/x))*log(log(1/exp(exp(4)-2)/log((-exp(2)-4*x)/x)))+(-5*exp(2)-20* x)*log((-exp(2)-4*x)/x)*log(1/exp(exp(4)-2)/log((-exp(2)-4*x)/x))-exp(2))/ ((exp(2)+4*x)*log((-exp(2)-4*x)/x)*log(1/exp(exp(4)-2)/log((-exp(2)-4*x)/x ))*log(log(1/exp(exp(4)-2)/log((-exp(2)-4*x)/x)))^2+(-10*exp(2)-40*x)*log( (-exp(2)-4*x)/x)*log(1/exp(exp(4)-2)/log((-exp(2)-4*x)/x))*log(log(1/exp(e xp(4)-2)/log((-exp(2)-4*x)/x)))+(25*exp(2)+100*x)*log((-exp(2)-4*x)/x)*log (1/exp(exp(4)-2)/log((-exp(2)-4*x)/x))),x, algorithm=\
integrate(((4*x + e^2)*log(-(4*x + e^2)/x)*log(e^(-e^4 + 2)/log(-(4*x + e^ 2)/x))*log(log(e^(-e^4 + 2)/log(-(4*x + e^2)/x))) - 5*(4*x + e^2)*log(-(4* x + e^2)/x)*log(e^(-e^4 + 2)/log(-(4*x + e^2)/x)) - e^2)/((4*x + e^2)*log( -(4*x + e^2)/x)*log(e^(-e^4 + 2)/log(-(4*x + e^2)/x))*log(log(e^(-e^4 + 2) /log(-(4*x + e^2)/x)))^2 - 10*(4*x + e^2)*log(-(4*x + e^2)/x)*log(e^(-e^4 + 2)/log(-(4*x + e^2)/x))*log(log(e^(-e^4 + 2)/log(-(4*x + e^2)/x))) + 25* (4*x + e^2)*log(-(4*x + e^2)/x)*log(e^(-e^4 + 2)/log(-(4*x + e^2)/x))), x)
Time = 11.09 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.97 \[ \int \frac {-e^2+\left (-5 e^2-20 x\right ) \log \left (\frac {-e^2-4 x}{x}\right ) \log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right )+\left (e^2+4 x\right ) \log \left (\frac {-e^2-4 x}{x}\right ) \log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right ) \log \left (\log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right )\right )}{\left (25 e^2+100 x\right ) \log \left (\frac {-e^2-4 x}{x}\right ) \log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right )+\left (-10 e^2-40 x\right ) \log \left (\frac {-e^2-4 x}{x}\right ) \log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right ) \log \left (\log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right )\right )+\left (e^2+4 x\right ) \log \left (\frac {-e^2-4 x}{x}\right ) \log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right ) \log ^2\left (\log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right )\right )} \, dx=\frac {x}{\ln \left (\ln \left (\frac {{\mathrm {e}}^{-{\mathrm {e}}^4}\,{\mathrm {e}}^2}{\ln \left (-\frac {4\,x+{\mathrm {e}}^2}{x}\right )}\right )\right )-5} \]
int(-(exp(2) + log(-(4*x + exp(2))/x)*log(exp(2 - exp(4))/log(-(4*x + exp( 2))/x))*(20*x + 5*exp(2)) - log(-(4*x + exp(2))/x)*log(log(exp(2 - exp(4)) /log(-(4*x + exp(2))/x)))*log(exp(2 - exp(4))/log(-(4*x + exp(2))/x))*(4*x + exp(2)))/(log(-(4*x + exp(2))/x)*log(exp(2 - exp(4))/log(-(4*x + exp(2) )/x))*(100*x + 25*exp(2)) - log(-(4*x + exp(2))/x)*log(log(exp(2 - exp(4)) /log(-(4*x + exp(2))/x)))*log(exp(2 - exp(4))/log(-(4*x + exp(2))/x))*(40* x + 10*exp(2)) + log(-(4*x + exp(2))/x)*log(log(exp(2 - exp(4))/log(-(4*x + exp(2))/x)))^2*log(exp(2 - exp(4))/log(-(4*x + exp(2))/x))*(4*x + exp(2) )),x)