\(\int \frac {-20 x+12 e^{x/5} x+(-60 e^{x/5}+20 x) \log (\frac {1}{4} (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2))) \log (\log (\frac {1}{4} (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)))) \log (\log (\log (\frac {1}{4} (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)))))+(15 e^{x/5}-5 x) \log (\frac {1}{4} (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2))) \log (\log (\frac {1}{4} (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)))) \log ^2(\log (\log (\frac {1}{4} (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)))))}{(60 e^{x/5}-20 x) \log (\frac {1}{4} (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2))) \log (\log (\frac {1}{4} (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)))) \log ^2(\log (\log (\frac {1}{4} (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)))))} \, dx\) [1132]

Optimal result

Integrand size = 309, antiderivative size = 34 \[ \int \frac {-20 x+12 e^{x/5} x+\left (-60 e^{x/5}+20 x\right ) \log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right ) \log \left (\log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right )\right ) \log \left (\log \left (\log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right )\right )\right )+\left (15 e^{x/5}-5 x\right ) \log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right ) \log \left (\log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right )\right ) \log ^2\left (\log \left (\log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right )\right )\right )}{\left (60 e^{x/5}-20 x\right ) \log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right ) \log \left (\log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right )\right ) \log ^2\left (\log \left (\log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right )\right )\right )} \, dx=\frac {x}{4}-\frac {x}{\log \left (\log \left (\log \left (\frac {1}{4} e^2 \left (-3 e^{x/5}+x\right ) \log (2)\right )\right )\right )} \] Output:

1/4*x-x/ln(ln(ln(1/4*(x-3*exp(1/5*x))*ln(2)*exp(2))))
 

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00 \[ \int \frac {-20 x+12 e^{x/5} x+\left (-60 e^{x/5}+20 x\right ) \log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right ) \log \left (\log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right )\right ) \log \left (\log \left (\log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right )\right )\right )+\left (15 e^{x/5}-5 x\right ) \log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right ) \log \left (\log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right )\right ) \log ^2\left (\log \left (\log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right )\right )\right )}{\left (60 e^{x/5}-20 x\right ) \log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right ) \log \left (\log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right )\right ) \log ^2\left (\log \left (\log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right )\right )\right )} \, dx=\frac {x}{4}-\frac {x}{\log \left (\log \left (2+\log \left (-3 e^{x/5}+x\right )+\log \left (\frac {\log (2)}{4}\right )\right )\right )} \] Input:

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]
 

Output:

x/4 - x/Log[Log[2 + Log[-3*E^(x/5) + x] + Log[Log[2]/4]]]
 

Rubi [F]

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

\(\displaystyle -5 \text {Subst}\left (\int \frac {1}{\log \left (\log \left (\log \left (5 x-3 e^x\right )+2 \left (1+\frac {1}{2} \log \left (\frac {\log (2)}{4}\right )\right )\right )\right )}dx,x,\frac {x}{5}\right )+\frac {1}{5} \int \frac {x^2}{\left (3 e^{x/5}-x\right ) \left (\log \left (x-3 e^{x/5}\right )+2 \left (1+\frac {1}{2} \log \left (\frac {\log (2)}{4}\right )\right )\right ) \log \left (\log \left (x-3 e^{x/5}\right )+2 \left (1+\frac {1}{2} \log \left (\frac {\log (2)}{4}\right )\right )\right ) \log ^2\left (\log \left (\log \left (x-3 e^{x/5}\right )+2 \left (1+\frac {1}{2} \log \left (\frac {\log (2)}{4}\right )\right )\right )\right )}dx+\frac {1}{5} \int \frac {x}{\left (\log \left (x-3 e^{x/5}\right )+2 \left (1+\frac {1}{2} \log \left (\frac {\log (2)}{4}\right )\right )\right ) \log \left (\log \left (x-3 e^{x/5}\right )+2 \left (1+\frac {1}{2} \log \left (\frac {\log (2)}{4}\right )\right )\right ) \log ^2\left (\log \left (\log \left (x-3 e^{x/5}\right )+2 \left (1+\frac {1}{2} \log \left (\frac {\log (2)}{4}\right )\right )\right )\right )}dx+\int \frac {x}{\left (x-3 e^{x/5}\right ) \left (\log \left (x-3 e^{x/5}\right )+2 \left (1+\frac {1}{2} \log \left (\frac {\log (2)}{4}\right )\right )\right ) \log \left (\log \left (x-3 e^{x/5}\right )+2 \left (1+\frac {1}{2} \log \left (\frac {\log (2)}{4}\right )\right )\right ) \log ^2\left (\log \left (\log \left (x-3 e^{x/5}\right )+2 \left (1+\frac {1}{2} \log \left (\frac {\log (2)}{4}\right )\right )\right )\right )}dx+\frac {x}{4}\)

Input:

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]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 7.29 (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	\(\frac {30 \ln \left (\ln \left (\ln \left (-\frac {{\mathrm e}^{2} \ln \left (2\right ) \left (3 \,{\mathrm e}^{\frac {x}{5}}-x \right )}{4}\right )\right )\right ) x -120 x}{120 \ln \left (\ln \left (\ln \left (-\frac {{\mathrm e}^{2} \ln \left (2\right ) \left (3 \,{\mathrm e}^{\frac {x}{5}}-x \right )}{4}\right )\right )\right )}\)	\(50\)

Input:

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)
 

Output:

1/4*x-x/ln(ln(ln(-3/4*ln(2)*exp(2+1/5*x)+1/4*x*exp(2)*ln(2))))
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.53 \[ \int \frac {-20 x+12 e^{x/5} x+\left (-60 e^{x/5}+20 x\right ) \log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right ) \log \left (\log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right )\right ) \log \left (\log \left (\log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right )\right )\right )+\left (15 e^{x/5}-5 x\right ) \log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right ) \log \left (\log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right )\right ) \log ^2\left (\log \left (\log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right )\right )\right )}{\left (60 e^{x/5}-20 x\right ) \log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right ) \log \left (\log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right )\right ) \log ^2\left (\log \left (\log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right )\right )\right )} \, dx=\frac {x \log \left (\log \left (\log \left (\frac {1}{4} \, x e^{2} \log \left (2\right ) - \frac {3}{4} \, e^{\left (\frac {1}{5} \, x + 2\right )} \log \left (2\right )\right )\right )\right ) - 4 \, x}{4 \, \log \left (\log \left (\log \left (\frac {1}{4} \, x e^{2} \log \left (2\right ) - \frac {3}{4} \, e^{\left (\frac {1}{5} \, x + 2\right )} \log \left (2\right )\right )\right )\right )} \] Input:

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="fricas")
 

Output:

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))))
 

Sympy [A] (verification not implemented)

Time = 6.23 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00 \[ \int \frac {-20 x+12 e^{x/5} x+\left (-60 e^{x/5}+20 x\right ) \log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right ) \log \left (\log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right )\right ) \log \left (\log \left (\log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right )\right )\right )+\left (15 e^{x/5}-5 x\right ) \log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right ) \log \left (\log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right )\right ) \log ^2\left (\log \left (\log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right )\right )\right )}{\left (60 e^{x/5}-20 x\right ) \log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right ) \log \left (\log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right )\right ) \log ^2\left (\log \left (\log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right )\right )\right )} \, dx=\frac {x}{4} - \frac {x}{\log {\left (\log {\left (\log {\left (\frac {x e^{2} \log {\left (2 \right )}}{4} - \frac {3 e^{2} e^{\frac {x}{5}} \log {\left (2 \right )}}{4} \right )} \right )} \right )}} \] Input:

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)
 

Output:

x/4 - x/log(log(log(x*exp(2)*log(2)/4 - 3*exp(2)*exp(x/5)*log(2)/4)))
 

Maxima [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.47 \[ \int \frac {-20 x+12 e^{x/5} x+\left (-60 e^{x/5}+20 x\right ) \log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right ) \log \left (\log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right )\right ) \log \left (\log \left (\log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right )\right )\right )+\left (15 e^{x/5}-5 x\right ) \log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right ) \log \left (\log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right )\right ) \log ^2\left (\log \left (\log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right )\right )\right )}{\left (60 e^{x/5}-20 x\right ) \log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right ) \log \left (\log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right )\right ) \log ^2\left (\log \left (\log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right )\right )\right )} \, dx=\frac {x \log \left (\log \left (-2 \, \log \left (2\right ) + \log \left (x - 3 \, e^{\left (\frac {1}{5} \, x\right )}\right ) + \log \left (\log \left (2\right )\right ) + 2\right )\right ) - 4 \, x}{4 \, \log \left (\log \left (-2 \, \log \left (2\right ) + \log \left (x - 3 \, e^{\left (\frac {1}{5} \, x\right )}\right ) + \log \left (\log \left (2\right )\right ) + 2\right )\right )} \] Input:

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="maxima")
 

Output:

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))
 

Giac [A] (verification not implemented)

Time = 0.45 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.47 \[ \int \frac {-20 x+12 e^{x/5} x+\left (-60 e^{x/5}+20 x\right ) \log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right ) \log \left (\log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right )\right ) \log \left (\log \left (\log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right )\right )\right )+\left (15 e^{x/5}-5 x\right ) \log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right ) \log \left (\log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right )\right ) \log ^2\left (\log \left (\log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right )\right )\right )}{\left (60 e^{x/5}-20 x\right ) \log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right ) \log \left (\log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right )\right ) \log ^2\left (\log \left (\log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right )\right )\right )} \, dx=\frac {x \log \left (\log \left (-2 \, \log \left (2\right ) + \log \left (x - 3 \, e^{\left (\frac {1}{5} \, x\right )}\right ) + \log \left (\log \left (2\right )\right ) + 2\right )\right ) - 4 \, x}{4 \, \log \left (\log \left (-2 \, \log \left (2\right ) + \log \left (x - 3 \, e^{\left (\frac {1}{5} \, x\right )}\right ) + \log \left (\log \left (2\right )\right ) + 2\right )\right )} \] Input:

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="giac")
 

Output:

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))
 

Mupad [B] (verification not implemented)

Time = 12.42 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.88 \[ \int \frac {-20 x+12 e^{x/5} x+\left (-60 e^{x/5}+20 x\right ) \log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right ) \log \left (\log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right )\right ) \log \left (\log \left (\log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right )\right )\right )+\left (15 e^{x/5}-5 x\right ) \log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right ) \log \left (\log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right )\right ) \log ^2\left (\log \left (\log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right )\right )\right )}{\left (60 e^{x/5}-20 x\right ) \log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right ) \log \left (\log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right )\right ) \log ^2\left (\log \left (\log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right )\right )\right )} \, dx=\frac {x}{4}-\frac {x}{\ln \left (\ln \left (\ln \left (\frac {x\,{\mathrm {e}}^2\,\ln \left (2\right )}{4}-\frac {3\,{\mathrm {e}}^2\,\ln \left (2\right )\,{\left ({\mathrm {e}}^x\right )}^{1/5}}{4}\right )\right )\right )} \] Input:

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)
 

Output:

x/4 - x/log(log(log((x*exp(2)*log(2))/4 - (3*exp(2)*log(2)*exp(x)^(1/5))/4 
)))
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.62 \[ \int \frac {-20 x+12 e^{x/5} x+\left (-60 e^{x/5}+20 x\right ) \log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right ) \log \left (\log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right )\right ) \log \left (\log \left (\log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right )\right )\right )+\left (15 e^{x/5}-5 x\right ) \log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right ) \log \left (\log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right )\right ) \log ^2\left (\log \left (\log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right )\right )\right )}{\left (60 e^{x/5}-20 x\right ) \log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right ) \log \left (\log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right )\right ) \log ^2\left (\log \left (\log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right )\right )\right )} \, dx=\frac {x \left (\mathrm {log}\left (\mathrm {log}\left (\mathrm {log}\left (-\frac {3 e^{\frac {x}{5}} \mathrm {log}\left (2\right ) e^{2}}{4}+\frac {\mathrm {log}\left (2\right ) e^{2} x}{4}\right )\right )\right )-4\right )}{4 \,\mathrm {log}\left (\mathrm {log}\left (\mathrm {log}\left (-\frac {3 e^{\frac {x}{5}} \mathrm {log}\left (2\right ) e^{2}}{4}+\frac {\mathrm {log}\left (2\right ) e^{2} x}{4}\right )\right )\right )} \] Input:

int(((15*exp(1/5*x)-5*x)*log(-3/4*exp(2)*log(2)*exp(1/5*x)+1/4*x*exp(2)*lo 
g(2))*log(log(-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))))^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(log(-3/4 
*exp(2)*log(2)*exp(1/5*x)+1/4*x*exp(2)*log(2)))*log(log(log(-3/4*exp(2)*lo 
g(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(log(-3/ 
4*exp(2)*log(2)*exp(1/5*x)+1/4*x*exp(2)*log(2)))/log(log(log(-3/4*exp(2)*l 
og(2)*exp(1/5*x)+1/4*x*exp(2)*log(2))))^2,x)
 

Output:

(x*(log(log(log(( - 3*e**(x/5)*log(2)*e**2 + log(2)*e**2*x)/4))) - 4))/(4* 
log(log(log(( - 3*e**(x/5)*log(2)*e**2 + log(2)*e**2*x)/4))))