Integrand size = 309, antiderivative size = 34 \begin {dmath*} \int \frac {-20 x+12 e^{x/5} x+\left (-60 e^{x/5}+20 x\right ) \log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right ) \log \left (\log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right )\right ) \log \left (\log \left (\log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right )\right )\right )+\left (15 e^{x/5}-5 x\right ) \log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right ) \log \left (\log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right )\right ) \log ^2\left (\log \left (\log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right )\right )\right )}{\left (60 e^{x/5}-20 x\right ) \log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right ) \log \left (\log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right )\right ) \log ^2\left (\log \left (\log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right )\right )\right )} \, dx=\frac {x}{4}-\frac {x}{\log \left (\log \left (\log \left (\frac {1}{4} e^2 \left (-3 e^{x/5}+x\right ) \log (2)\right )\right )\right )} \end {dmath*}
Time = 0.30 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00 \begin {dmath*} \int \frac {-20 x+12 e^{x/5} x+\left (-60 e^{x/5}+20 x\right ) \log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right ) \log \left (\log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right )\right ) \log \left (\log \left (\log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right )\right )\right )+\left (15 e^{x/5}-5 x\right ) \log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right ) \log \left (\log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right )\right ) \log ^2\left (\log \left (\log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right )\right )\right )}{\left (60 e^{x/5}-20 x\right ) \log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right ) \log \left (\log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right )\right ) \log ^2\left (\log \left (\log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right )\right )\right )} \, dx=\frac {x}{4}-\frac {x}{\log \left (\log \left (2+\log \left (-3 e^{x/5}+x\right )+\log \left (\frac {\log (2)}{4}\right )\right )\right )} \end {dmath*}
Integrate[(-20*x + 12*E^(x/5)*x + (-60*E^(x/5) + 20*x)*Log[(-3*E^(2 + x/5) *Log[2] + E^2*x*Log[2])/4]*Log[Log[(-3*E^(2 + x/5)*Log[2] + E^2*x*Log[2])/ 4]]*Log[Log[Log[(-3*E^(2 + x/5)*Log[2] + E^2*x*Log[2])/4]]] + (15*E^(x/5) - 5*x)*Log[(-3*E^(2 + x/5)*Log[2] + E^2*x*Log[2])/4]*Log[Log[(-3*E^(2 + x/ 5)*Log[2] + E^2*x*Log[2])/4]]*Log[Log[Log[(-3*E^(2 + x/5)*Log[2] + E^2*x*L og[2])/4]]]^2)/((60*E^(x/5) - 20*x)*Log[(-3*E^(2 + x/5)*Log[2] + E^2*x*Log [2])/4]*Log[Log[(-3*E^(2 + x/5)*Log[2] + E^2*x*Log[2])/4]]*Log[Log[Log[(-3 *E^(2 + x/5)*Log[2] + E^2*x*Log[2])/4]]]^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.
| \(\displaystyle \int \frac {12 e^{x/5} x-20 x+\left (15 e^{x/5}-5 x\right ) \log \left (\frac {1}{4} \left (e^2 x \log (2)-3 e^{\frac {x}{5}+2} \log (2)\right )\right ) \log \left (\log \left (\frac {1}{4} \left (e^2 x \log (2)-3 e^{\frac {x}{5}+2} \log (2)\right )\right )\right ) \log ^2\left (\log \left (\log \left (\frac {1}{4} \left (e^2 x \log (2)-3 e^{\frac {x}{5}+2} \log (2)\right )\right )\right )\right )+\left (20 x-60 e^{x/5}\right ) \log \left (\frac {1}{4} \left (e^2 x \log (2)-3 e^{\frac {x}{5}+2} \log (2)\right )\right ) \log \left (\log \left (\frac {1}{4} \left (e^2 x \log (2)-3 e^{\frac {x}{5}+2} \log (2)\right )\right )\right ) \log \left (\log \left (\log \left (\frac {1}{4} \left (e^2 x \log (2)-3 e^{\frac {x}{5}+2} \log (2)\right )\right )\right )\right )}{\left (60 e^{x/5}-20 x\right ) \log \left (\frac {1}{4} \left (e^2 x \log (2)-3 e^{\frac {x}{5}+2} \log (2)\right )\right ) \log \left (\log \left (\frac {1}{4} \left (e^2 x \log (2)-3 e^{\frac {x}{5}+2} \log (2)\right )\right )\right ) \log ^2\left (\log \left (\log \left (\frac {1}{4} \left (e^2 x \log (2)-3 e^{\frac {x}{5}+2} \log (2)\right )\right )\right )\right )} \, dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle \int \left (\frac {\left (3 e^{x/5}-5\right ) x}{5 \left (3 e^{x/5}-x\right ) \left (\log \left (x-3 e^{x/5}\right )+2 \left (1+\frac {1}{2} \log \left (\frac {\log (2)}{4}\right )\right )\right ) \log \left (\log \left (x-3 e^{x/5}\right )+2 \left (1+\frac {1}{2} \log \left (\frac {\log (2)}{4}\right )\right )\right ) \log ^2\left (\log \left (\log \left (x-3 e^{x/5}\right )+2 \left (1+\frac {1}{2} \log \left (\frac {\log (2)}{4}\right )\right )\right )\right )}-\frac {1}{\log \left (\log \left (\log \left (x-3 e^{x/5}\right )+2 \left (1+\frac {1}{2} \log \left (\frac {\log (2)}{4}\right )\right )\right )\right )}+\frac {1}{4}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -5 \text {Subst}\left (\int \frac {1}{\log \left (\log \left (\log \left (5 x-3 e^x\right )+2 \left (1+\frac {1}{2} \log \left (\frac {\log (2)}{4}\right )\right )\right )\right )}dx,x,\frac {x}{5}\right )+\frac {1}{5} \int \frac {x^2}{\left (3 e^{x/5}-x\right ) \left (\log \left (x-3 e^{x/5}\right )+2 \left (1+\frac {1}{2} \log \left (\frac {\log (2)}{4}\right )\right )\right ) \log \left (\log \left (x-3 e^{x/5}\right )+2 \left (1+\frac {1}{2} \log \left (\frac {\log (2)}{4}\right )\right )\right ) \log ^2\left (\log \left (\log \left (x-3 e^{x/5}\right )+2 \left (1+\frac {1}{2} \log \left (\frac {\log (2)}{4}\right )\right )\right )\right )}dx+\frac {1}{5} \int \frac {x}{\left (\log \left (x-3 e^{x/5}\right )+2 \left (1+\frac {1}{2} \log \left (\frac {\log (2)}{4}\right )\right )\right ) \log \left (\log \left (x-3 e^{x/5}\right )+2 \left (1+\frac {1}{2} \log \left (\frac {\log (2)}{4}\right )\right )\right ) \log ^2\left (\log \left (\log \left (x-3 e^{x/5}\right )+2 \left (1+\frac {1}{2} \log \left (\frac {\log (2)}{4}\right )\right )\right )\right )}dx+\int \frac {x}{\left (x-3 e^{x/5}\right ) \left (\log \left (x-3 e^{x/5}\right )+2 \left (1+\frac {1}{2} \log \left (\frac {\log (2)}{4}\right )\right )\right ) \log \left (\log \left (x-3 e^{x/5}\right )+2 \left (1+\frac {1}{2} \log \left (\frac {\log (2)}{4}\right )\right )\right ) \log ^2\left (\log \left (\log \left (x-3 e^{x/5}\right )+2 \left (1+\frac {1}{2} \log \left (\frac {\log (2)}{4}\right )\right )\right )\right )}dx+\frac {x}{4}\) |
Int[(-20*x + 12*E^(x/5)*x + (-60*E^(x/5) + 20*x)*Log[(-3*E^(2 + x/5)*Log[2 ] + E^2*x*Log[2])/4]*Log[Log[(-3*E^(2 + x/5)*Log[2] + E^2*x*Log[2])/4]]*Lo g[Log[Log[(-3*E^(2 + x/5)*Log[2] + E^2*x*Log[2])/4]]] + (15*E^(x/5) - 5*x) *Log[(-3*E^(2 + x/5)*Log[2] + E^2*x*Log[2])/4]*Log[Log[(-3*E^(2 + x/5)*Log [2] + E^2*x*Log[2])/4]]*Log[Log[Log[(-3*E^(2 + x/5)*Log[2] + E^2*x*Log[2]) /4]]]^2)/((60*E^(x/5) - 20*x)*Log[(-3*E^(2 + x/5)*Log[2] + E^2*x*Log[2])/4 ]*Log[Log[(-3*E^(2 + x/5)*Log[2] + E^2*x*Log[2])/4]]*Log[Log[Log[(-3*E^(2 + x/5)*Log[2] + E^2*x*Log[2])/4]]]^2),x]
3.12.32.3.1 Defintions of rubi rules used
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 10.09 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.91
| method | result | size |
| risch | \(\frac {x}{4}-\frac {x}{\ln \left (\ln \left (\ln \left (-\frac {3 \ln \left (2\right ) {\mathrm e}^{2+\frac {x}{5}}}{4}+\frac {x \,{\mathrm e}^{2} \ln \left (2\right )}{4}\right )\right )\right )}\) | \(31\) |
| parallelrisch | \(\frac {30 \ln \left (\ln \left (\ln \left (-\frac {{\mathrm e}^{2} \ln \left (2\right ) \left (3 \,{\mathrm e}^{\frac {x}{5}}-x \right )}{4}\right )\right )\right ) x -120 x}{120 \ln \left (\ln \left (\ln \left (-\frac {{\mathrm e}^{2} \ln \left (2\right ) \left (3 \,{\mathrm e}^{\frac {x}{5}}-x \right )}{4}\right )\right )\right )}\) | \(50\) |
int(((15*exp(1/5*x)-5*x)*ln(-3/4*exp(2)*ln(2)*exp(1/5*x)+1/4*x*exp(2)*ln(2 ))*ln(ln(-3/4*exp(2)*ln(2)*exp(1/5*x)+1/4*x*exp(2)*ln(2)))*ln(ln(ln(-3/4*e xp(2)*ln(2)*exp(1/5*x)+1/4*x*exp(2)*ln(2))))^2+(-60*exp(1/5*x)+20*x)*ln(-3 /4*exp(2)*ln(2)*exp(1/5*x)+1/4*x*exp(2)*ln(2))*ln(ln(-3/4*exp(2)*ln(2)*exp (1/5*x)+1/4*x*exp(2)*ln(2)))*ln(ln(ln(-3/4*exp(2)*ln(2)*exp(1/5*x)+1/4*x*e xp(2)*ln(2))))+12*x*exp(1/5*x)-20*x)/(60*exp(1/5*x)-20*x)/ln(-3/4*exp(2)*l n(2)*exp(1/5*x)+1/4*x*exp(2)*ln(2))/ln(ln(-3/4*exp(2)*ln(2)*exp(1/5*x)+1/4 *x*exp(2)*ln(2)))/ln(ln(ln(-3/4*exp(2)*ln(2)*exp(1/5*x)+1/4*x*exp(2)*ln(2) )))^2,x,method=_RETURNVERBOSE)
Time = 0.28 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.53 \begin {dmath*} \int \frac {-20 x+12 e^{x/5} x+\left (-60 e^{x/5}+20 x\right ) \log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right ) \log \left (\log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right )\right ) \log \left (\log \left (\log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right )\right )\right )+\left (15 e^{x/5}-5 x\right ) \log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right ) \log \left (\log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right )\right ) \log ^2\left (\log \left (\log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right )\right )\right )}{\left (60 e^{x/5}-20 x\right ) \log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right ) \log \left (\log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right )\right ) \log ^2\left (\log \left (\log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right )\right )\right )} \, dx=\frac {x \log \left (\log \left (\log \left (\frac {1}{4} \, x e^{2} \log \left (2\right ) - \frac {3}{4} \, e^{\left (\frac {1}{5} \, x + 2\right )} \log \left (2\right )\right )\right )\right ) - 4 \, x}{4 \, \log \left (\log \left (\log \left (\frac {1}{4} \, x e^{2} \log \left (2\right ) - \frac {3}{4} \, e^{\left (\frac {1}{5} \, x + 2\right )} \log \left (2\right )\right )\right )\right )} \end {dmath*}
integrate(((15*exp(1/5*x)-5*x)*log(-3/4*exp(2)*log(2)*exp(1/5*x)+1/4*x*exp (2)*log(2))*log(log(-3/4*exp(2)*log(2)*exp(1/5*x)+1/4*x*exp(2)*log(2)))*lo g(log(log(-3/4*exp(2)*log(2)*exp(1/5*x)+1/4*x*exp(2)*log(2))))^2+(-60*exp( 1/5*x)+20*x)*log(-3/4*exp(2)*log(2)*exp(1/5*x)+1/4*x*exp(2)*log(2))*log(lo g(-3/4*exp(2)*log(2)*exp(1/5*x)+1/4*x*exp(2)*log(2)))*log(log(log(-3/4*exp (2)*log(2)*exp(1/5*x)+1/4*x*exp(2)*log(2))))+12*x*exp(1/5*x)-20*x)/(60*exp (1/5*x)-20*x)/log(-3/4*exp(2)*log(2)*exp(1/5*x)+1/4*x*exp(2)*log(2))/log(l og(-3/4*exp(2)*log(2)*exp(1/5*x)+1/4*x*exp(2)*log(2)))/log(log(log(-3/4*ex p(2)*log(2)*exp(1/5*x)+1/4*x*exp(2)*log(2))))^2,x, algorithm=\
1/4*(x*log(log(log(1/4*x*e^2*log(2) - 3/4*e^(1/5*x + 2)*log(2)))) - 4*x)/l og(log(log(1/4*x*e^2*log(2) - 3/4*e^(1/5*x + 2)*log(2))))
Time = 7.64 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00 \begin {dmath*} \int \frac {-20 x+12 e^{x/5} x+\left (-60 e^{x/5}+20 x\right ) \log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right ) \log \left (\log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right )\right ) \log \left (\log \left (\log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right )\right )\right )+\left (15 e^{x/5}-5 x\right ) \log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right ) \log \left (\log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right )\right ) \log ^2\left (\log \left (\log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right )\right )\right )}{\left (60 e^{x/5}-20 x\right ) \log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right ) \log \left (\log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right )\right ) \log ^2\left (\log \left (\log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right )\right )\right )} \, dx=\frac {x}{4} - \frac {x}{\log {\left (\log {\left (\log {\left (\frac {x e^{2} \log {\left (2 \right )}}{4} - \frac {3 e^{2} e^{\frac {x}{5}} \log {\left (2 \right )}}{4} \right )} \right )} \right )}} \end {dmath*}
integrate(((15*exp(1/5*x)-5*x)*ln(-3/4*exp(2)*ln(2)*exp(1/5*x)+1/4*x*exp(2 )*ln(2))*ln(ln(-3/4*exp(2)*ln(2)*exp(1/5*x)+1/4*x*exp(2)*ln(2)))*ln(ln(ln( -3/4*exp(2)*ln(2)*exp(1/5*x)+1/4*x*exp(2)*ln(2))))**2+(-60*exp(1/5*x)+20*x )*ln(-3/4*exp(2)*ln(2)*exp(1/5*x)+1/4*x*exp(2)*ln(2))*ln(ln(-3/4*exp(2)*ln (2)*exp(1/5*x)+1/4*x*exp(2)*ln(2)))*ln(ln(ln(-3/4*exp(2)*ln(2)*exp(1/5*x)+ 1/4*x*exp(2)*ln(2))))+12*x*exp(1/5*x)-20*x)/(60*exp(1/5*x)-20*x)/ln(-3/4*e xp(2)*ln(2)*exp(1/5*x)+1/4*x*exp(2)*ln(2))/ln(ln(-3/4*exp(2)*ln(2)*exp(1/5 *x)+1/4*x*exp(2)*ln(2)))/ln(ln(ln(-3/4*exp(2)*ln(2)*exp(1/5*x)+1/4*x*exp(2 )*ln(2))))**2,x)
Time = 0.47 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.47 \begin {dmath*} \int \frac {-20 x+12 e^{x/5} x+\left (-60 e^{x/5}+20 x\right ) \log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right ) \log \left (\log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right )\right ) \log \left (\log \left (\log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right )\right )\right )+\left (15 e^{x/5}-5 x\right ) \log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right ) \log \left (\log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right )\right ) \log ^2\left (\log \left (\log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right )\right )\right )}{\left (60 e^{x/5}-20 x\right ) \log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right ) \log \left (\log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right )\right ) \log ^2\left (\log \left (\log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right )\right )\right )} \, dx=\frac {x \log \left (\log \left (-2 \, \log \left (2\right ) + \log \left (x - 3 \, e^{\left (\frac {1}{5} \, x\right )}\right ) + \log \left (\log \left (2\right )\right ) + 2\right )\right ) - 4 \, x}{4 \, \log \left (\log \left (-2 \, \log \left (2\right ) + \log \left (x - 3 \, e^{\left (\frac {1}{5} \, x\right )}\right ) + \log \left (\log \left (2\right )\right ) + 2\right )\right )} \end {dmath*}
integrate(((15*exp(1/5*x)-5*x)*log(-3/4*exp(2)*log(2)*exp(1/5*x)+1/4*x*exp (2)*log(2))*log(log(-3/4*exp(2)*log(2)*exp(1/5*x)+1/4*x*exp(2)*log(2)))*lo g(log(log(-3/4*exp(2)*log(2)*exp(1/5*x)+1/4*x*exp(2)*log(2))))^2+(-60*exp( 1/5*x)+20*x)*log(-3/4*exp(2)*log(2)*exp(1/5*x)+1/4*x*exp(2)*log(2))*log(lo g(-3/4*exp(2)*log(2)*exp(1/5*x)+1/4*x*exp(2)*log(2)))*log(log(log(-3/4*exp (2)*log(2)*exp(1/5*x)+1/4*x*exp(2)*log(2))))+12*x*exp(1/5*x)-20*x)/(60*exp (1/5*x)-20*x)/log(-3/4*exp(2)*log(2)*exp(1/5*x)+1/4*x*exp(2)*log(2))/log(l og(-3/4*exp(2)*log(2)*exp(1/5*x)+1/4*x*exp(2)*log(2)))/log(log(log(-3/4*ex p(2)*log(2)*exp(1/5*x)+1/4*x*exp(2)*log(2))))^2,x, algorithm=\
1/4*(x*log(log(-2*log(2) + log(x - 3*e^(1/5*x)) + log(log(2)) + 2)) - 4*x) /log(log(-2*log(2) + log(x - 3*e^(1/5*x)) + log(log(2)) + 2))
Time = 0.67 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.47 \begin {dmath*} \int \frac {-20 x+12 e^{x/5} x+\left (-60 e^{x/5}+20 x\right ) \log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right ) \log \left (\log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right )\right ) \log \left (\log \left (\log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right )\right )\right )+\left (15 e^{x/5}-5 x\right ) \log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right ) \log \left (\log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right )\right ) \log ^2\left (\log \left (\log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right )\right )\right )}{\left (60 e^{x/5}-20 x\right ) \log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right ) \log \left (\log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right )\right ) \log ^2\left (\log \left (\log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right )\right )\right )} \, dx=\frac {x \log \left (\log \left (-2 \, \log \left (2\right ) + \log \left (x - 3 \, e^{\left (\frac {1}{5} \, x\right )}\right ) + \log \left (\log \left (2\right )\right ) + 2\right )\right ) - 4 \, x}{4 \, \log \left (\log \left (-2 \, \log \left (2\right ) + \log \left (x - 3 \, e^{\left (\frac {1}{5} \, x\right )}\right ) + \log \left (\log \left (2\right )\right ) + 2\right )\right )} \end {dmath*}
integrate(((15*exp(1/5*x)-5*x)*log(-3/4*exp(2)*log(2)*exp(1/5*x)+1/4*x*exp (2)*log(2))*log(log(-3/4*exp(2)*log(2)*exp(1/5*x)+1/4*x*exp(2)*log(2)))*lo g(log(log(-3/4*exp(2)*log(2)*exp(1/5*x)+1/4*x*exp(2)*log(2))))^2+(-60*exp( 1/5*x)+20*x)*log(-3/4*exp(2)*log(2)*exp(1/5*x)+1/4*x*exp(2)*log(2))*log(lo g(-3/4*exp(2)*log(2)*exp(1/5*x)+1/4*x*exp(2)*log(2)))*log(log(log(-3/4*exp (2)*log(2)*exp(1/5*x)+1/4*x*exp(2)*log(2))))+12*x*exp(1/5*x)-20*x)/(60*exp (1/5*x)-20*x)/log(-3/4*exp(2)*log(2)*exp(1/5*x)+1/4*x*exp(2)*log(2))/log(l og(-3/4*exp(2)*log(2)*exp(1/5*x)+1/4*x*exp(2)*log(2)))/log(log(log(-3/4*ex p(2)*log(2)*exp(1/5*x)+1/4*x*exp(2)*log(2))))^2,x, algorithm=\
1/4*(x*log(log(-2*log(2) + log(x - 3*e^(1/5*x)) + log(log(2)) + 2)) - 4*x) /log(log(-2*log(2) + log(x - 3*e^(1/5*x)) + log(log(2)) + 2))
Time = 20.77 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.88 \begin {dmath*} \int \frac {-20 x+12 e^{x/5} x+\left (-60 e^{x/5}+20 x\right ) \log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right ) \log \left (\log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right )\right ) \log \left (\log \left (\log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right )\right )\right )+\left (15 e^{x/5}-5 x\right ) \log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right ) \log \left (\log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right )\right ) \log ^2\left (\log \left (\log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right )\right )\right )}{\left (60 e^{x/5}-20 x\right ) \log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right ) \log \left (\log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right )\right ) \log ^2\left (\log \left (\log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right )\right )\right )} \, dx=\frac {x}{4}-\frac {x}{\ln \left (\ln \left (\ln \left (\frac {x\,{\mathrm {e}}^2\,\ln \left (2\right )}{4}-\frac {3\,{\mathrm {e}}^2\,\ln \left (2\right )\,{\left ({\mathrm {e}}^x\right )}^{1/5}}{4}\right )\right )\right )} \end {dmath*}
int((20*x - 12*x*exp(x/5) - log(log((x*exp(2)*log(2))/4 - (3*exp(x/5)*exp( 2)*log(2))/4))*log(log(log((x*exp(2)*log(2))/4 - (3*exp(x/5)*exp(2)*log(2) )/4)))*log((x*exp(2)*log(2))/4 - (3*exp(x/5)*exp(2)*log(2))/4)*(20*x - 60* exp(x/5)) + log(log((x*exp(2)*log(2))/4 - (3*exp(x/5)*exp(2)*log(2))/4))*l og(log(log((x*exp(2)*log(2))/4 - (3*exp(x/5)*exp(2)*log(2))/4)))^2*log((x* exp(2)*log(2))/4 - (3*exp(x/5)*exp(2)*log(2))/4)*(5*x - 15*exp(x/5)))/(log (log((x*exp(2)*log(2))/4 - (3*exp(x/5)*exp(2)*log(2))/4))*log(log(log((x*e xp(2)*log(2))/4 - (3*exp(x/5)*exp(2)*log(2))/4)))^2*log((x*exp(2)*log(2))/ 4 - (3*exp(x/5)*exp(2)*log(2))/4)*(20*x - 60*exp(x/5))),x)