\(\int \frac {80 e^{-2 e^5+2 x}+40 e^{-e^5+x} \log (5)+5 \log ^2(5)+e^{\frac {5 e^{-e^5+x}+2 e^{-e^5+x} \log (\log (4))}{4 e^{-e^5+x}+\log (5)}} (5 e^{-e^5+x} \log (5)+2 e^{-e^5+x} \log (5) \log (\log (4)))}{16 e^{-2 e^5+2 x}+8 e^{-e^5+x} \log (5)+\log ^2(5)} \, dx\) [1495]

Optimal result
Mathematica [B] (verified)
Rubi [F]
Maple [A] (verified)
Fricas [A] (verification not implemented)
Sympy [A] (verification not implemented)
Maxima [B] (verification not implemented)
Giac [F]
Mupad [B] (verification not implemented)
Reduce [B] (verification not implemented)

Optimal result

Integrand size = 143, antiderivative size = 30 \[ \int \frac {80 e^{-2 e^5+2 x}+40 e^{-e^5+x} \log (5)+5 \log ^2(5)+e^{\frac {5 e^{-e^5+x}+2 e^{-e^5+x} \log (\log (4))}{4 e^{-e^5+x}+\log (5)}} \left (5 e^{-e^5+x} \log (5)+2 e^{-e^5+x} \log (5) \log (\log (4))\right )}{16 e^{-2 e^5+2 x}+8 e^{-e^5+x} \log (5)+\log ^2(5)} \, dx=e^{\frac {5+2 \log (\log (4))}{4+e^{e^5-x} \log (5)}}+5 x \] Output:

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

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(61\) vs. \(2(30)=60\).

Time = 0.44 (sec) , antiderivative size = 61, normalized size of antiderivative = 2.03 \[ \int \frac {80 e^{-2 e^5+2 x}+40 e^{-e^5+x} \log (5)+5 \log ^2(5)+e^{\frac {5 e^{-e^5+x}+2 e^{-e^5+x} \log (\log (4))}{4 e^{-e^5+x}+\log (5)}} \left (5 e^{-e^5+x} \log (5)+2 e^{-e^5+x} \log (5) \log (\log (4))\right )}{16 e^{-2 e^5+2 x}+8 e^{-e^5+x} \log (5)+\log ^2(5)} \, dx=5 x+5^{-\frac {5 e^{e^5}}{4 \left (4 e^x+e^{e^5} \log (5)\right )}} e^{5/4} \log ^{\frac {2 e^x}{4 e^x+e^{e^5} \log (5)}}(4) \] Input:

Integrate[(80*E^(-2*E^5 + 2*x) + 40*E^(-E^5 + x)*Log[5] + 5*Log[5]^2 + E^( 
(5*E^(-E^5 + x) + 2*E^(-E^5 + x)*Log[Log[4]])/(4*E^(-E^5 + x) + Log[5]))*( 
5*E^(-E^5 + x)*Log[5] + 2*E^(-E^5 + x)*Log[5]*Log[Log[4]]))/(16*E^(-2*E^5 
+ 2*x) + 8*E^(-E^5 + x)*Log[5] + Log[5]^2),x]
 

Output:

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

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.

\(\displaystyle \int \frac {\left (5 e^{x-e^5} \log (5)+2 e^{x-e^5} \log (5) \log (\log (4))\right ) \exp \left (\frac {5 e^{x-e^5}+2 e^{x-e^5} \log (\log (4))}{4 e^{x-e^5}+\log (5)}\right )+80 e^{2 x-2 e^5}+40 e^{x-e^5} \log (5)+5 \log ^2(5)}{16 e^{2 x-2 e^5}+8 e^{x-e^5} \log (5)+\log ^2(5)} \, dx\)

\(\Big \downarrow \) 2720

\(\displaystyle \int \frac {e^{e^5-x} \left (80 e^{2 x-2 e^5}+e^{x-e^5} \log (5) \left ((5+2 \log (\log (4))) e^{\frac {5 e^{x-e^5}}{4 e^{x-e^5}+\log (5)}} \log ^{\frac {2 e^{x-e^5}}{4 e^{x-e^5}+\log (5)}}(4)+40\right )+5 \log ^2(5)\right )}{\left (4 e^{x-e^5}+\log (5)\right )^2}de^{x-e^5}\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (5 e^{e^5-x}+\frac {\log (5) (5+2 \log (\log (4))) e^{\frac {5 e^{x-e^5}}{4 e^{x-e^5}+\log (5)}} \log ^{\frac {2 e^{x-e^5}}{4 e^{x-e^5}+\log (5)}}(4)}{\left (4 e^{x-e^5}+\log (5)\right )^2}\right )de^{x-e^5}\)

\(\Big \downarrow \) 2009

\(\displaystyle \log (5) (5+2 \log (\log (4))) \int \frac {e^{\frac {5 e^{x-e^5}}{4 e^{x-e^5}+\log (5)}} \log ^{\frac {2 e^{x-e^5}}{4 e^{x-e^5}+\log (5)}}(4)}{\left (4 e^{x-e^5}+\log (5)\right )^2}de^{x-e^5}+5 \log \left (e^{x-e^5}\right )\)

Input:

Int[(80*E^(-2*E^5 + 2*x) + 40*E^(-E^5 + x)*Log[5] + 5*Log[5]^2 + E^((5*E^( 
-E^5 + x) + 2*E^(-E^5 + x)*Log[Log[4]])/(4*E^(-E^5 + x) + Log[5]))*(5*E^(- 
E^5 + x)*Log[5] + 2*E^(-E^5 + x)*Log[5]*Log[Log[4]]))/(16*E^(-2*E^5 + 2*x) 
 + 8*E^(-E^5 + x)*Log[5] + Log[5]^2),x]
 

Output:

$Aborted
 
Maple [A] (verified)

Time = 0.82 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.23

method result size
parallelrisch \({\mathrm e}^{\frac {{\mathrm e}^{-{\mathrm e}^{5}+x} \left (5+\ln \left (4 \ln \left (2\right )^{2}\right )\right )}{4 \,{\mathrm e}^{-{\mathrm e}^{5}+x}+\ln \left (5\right )}}+5 x\) \(37\)
risch \(5 x +{\mathrm e}^{\frac {{\mathrm e}^{-{\mathrm e}^{5}+x} \left (5+2 \ln \left (2\right )+2 \ln \left (\ln \left (2\right )\right )\right )}{4 \,{\mathrm e}^{-{\mathrm e}^{5}+x}+\ln \left (5\right )}}\) \(39\)
parts \(5 x -\frac {\left (\ln \left (5\right ) \ln \left (4 \ln \left (2\right )^{2}\right )+5 \ln \left (5\right )\right ) \left (-4 \ln \left (5\right ) \ln \left (4 \ln \left (2\right )^{2}\right )-20 \ln \left (5\right )\right ) {\mathrm e}^{\frac {-\frac {\ln \left (5\right ) \ln \left (4 \ln \left (2\right )^{2}\right )}{4}-\frac {5 \ln \left (5\right )}{4}}{4 \,{\mathrm e}^{-{\mathrm e}^{5}+x}+\ln \left (5\right )}+\frac {5}{4}+\frac {\ln \left (4 \ln \left (2\right )^{2}\right )}{4}}}{4 \ln \left (5\right )^{2} \left (4 \ln \left (2\right )^{2}+8 \ln \left (2\right ) \ln \left (\ln \left (2\right )\right )+4 \ln \left (\ln \left (2\right )\right )^{2}+20 \ln \left (2\right )+20 \ln \left (\ln \left (2\right )\right )+25\right )}\) \(118\)
norman \(\frac {\ln \left (5\right ) {\mathrm e}^{\frac {{\mathrm e}^{-{\mathrm e}^{5}+x} \ln \left (4 \ln \left (2\right )^{2}\right )+5 \,{\mathrm e}^{-{\mathrm e}^{5}+x}}{4 \,{\mathrm e}^{-{\mathrm e}^{5}+x}+\ln \left (5\right )}}+20 x \,{\mathrm e}^{-{\mathrm e}^{5}+x}+5 x \ln \left (5\right )+4 \,{\mathrm e}^{-{\mathrm e}^{5}+x} {\mathrm e}^{\frac {{\mathrm e}^{-{\mathrm e}^{5}+x} \ln \left (4 \ln \left (2\right )^{2}\right )+5 \,{\mathrm e}^{-{\mathrm e}^{5}+x}}{4 \,{\mathrm e}^{-{\mathrm e}^{5}+x}+\ln \left (5\right )}}}{4 \,{\mathrm e}^{-{\mathrm e}^{5}+x}+\ln \left (5\right )}\) \(126\)
derivativedivides \(\text {Expression too large to display}\) \(1818\)
default \(\text {Expression too large to display}\) \(1818\)

Input:

int(((ln(5)*exp(-exp(5)+x)*ln(4*ln(2)^2)+5*ln(5)*exp(-exp(5)+x))*exp((exp( 
-exp(5)+x)*ln(4*ln(2)^2)+5*exp(-exp(5)+x))/(4*exp(-exp(5)+x)+ln(5)))+80*ex 
p(-exp(5)+x)^2+40*ln(5)*exp(-exp(5)+x)+5*ln(5)^2)/(16*exp(-exp(5)+x)^2+8*l 
n(5)*exp(-exp(5)+x)+ln(5)^2),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.50 \[ \int \frac {80 e^{-2 e^5+2 x}+40 e^{-e^5+x} \log (5)+5 \log ^2(5)+e^{\frac {5 e^{-e^5+x}+2 e^{-e^5+x} \log (\log (4))}{4 e^{-e^5+x}+\log (5)}} \left (5 e^{-e^5+x} \log (5)+2 e^{-e^5+x} \log (5) \log (\log (4))\right )}{16 e^{-2 e^5+2 x}+8 e^{-e^5+x} \log (5)+\log ^2(5)} \, dx=5 \, x + e^{\left (\frac {e^{\left (x - e^{5}\right )} \log \left (4 \, \log \left (2\right )^{2}\right ) + 5 \, e^{\left (x - e^{5}\right )}}{4 \, e^{\left (x - e^{5}\right )} + \log \left (5\right )}\right )} \] Input:

integrate(((log(5)*exp(-exp(5)+x)*log(4*log(2)^2)+5*log(5)*exp(-exp(5)+x)) 
*exp((exp(-exp(5)+x)*log(4*log(2)^2)+5*exp(-exp(5)+x))/(4*exp(-exp(5)+x)+l 
og(5)))+80*exp(-exp(5)+x)^2+40*log(5)*exp(-exp(5)+x)+5*log(5)^2)/(16*exp(- 
exp(5)+x)^2+8*log(5)*exp(-exp(5)+x)+log(5)^2),x, algorithm="fricas")
 

Output:

5*x + e^((e^(x - e^5)*log(4*log(2)^2) + 5*e^(x - e^5))/(4*e^(x - e^5) + lo 
g(5)))
 

Sympy [A] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.30 \[ \int \frac {80 e^{-2 e^5+2 x}+40 e^{-e^5+x} \log (5)+5 \log ^2(5)+e^{\frac {5 e^{-e^5+x}+2 e^{-e^5+x} \log (\log (4))}{4 e^{-e^5+x}+\log (5)}} \left (5 e^{-e^5+x} \log (5)+2 e^{-e^5+x} \log (5) \log (\log (4))\right )}{16 e^{-2 e^5+2 x}+8 e^{-e^5+x} \log (5)+\log ^2(5)} \, dx=5 x + e^{\frac {e^{x - e^{5}} \log {\left (4 \log {\left (2 \right )}^{2} \right )} + 5 e^{x - e^{5}}}{4 e^{x - e^{5}} + \log {\left (5 \right )}}} \] Input:

integrate(((ln(5)*exp(-exp(5)+x)*ln(4*ln(2)**2)+5*ln(5)*exp(-exp(5)+x))*ex 
p((exp(-exp(5)+x)*ln(4*ln(2)**2)+5*exp(-exp(5)+x))/(4*exp(-exp(5)+x)+ln(5) 
))+80*exp(-exp(5)+x)**2+40*ln(5)*exp(-exp(5)+x)+5*ln(5)**2)/(16*exp(-exp(5 
)+x)**2+8*ln(5)*exp(-exp(5)+x)+ln(5)**2),x)
 

Output:

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 279 vs. \(2 (29) = 58\).

Time = 0.16 (sec) , antiderivative size = 279, normalized size of antiderivative = 9.30 \[ \int \frac {80 e^{-2 e^5+2 x}+40 e^{-e^5+x} \log (5)+5 \log ^2(5)+e^{\frac {5 e^{-e^5+x}+2 e^{-e^5+x} \log (\log (4))}{4 e^{-e^5+x}+\log (5)}} \left (5 e^{-e^5+x} \log (5)+2 e^{-e^5+x} \log (5) \log (\log (4))\right )}{16 e^{-2 e^5+2 x}+8 e^{-e^5+x} \log (5)+\log ^2(5)} \, dx=5 \, {\left (\frac {1}{4 \, e^{\left (x - e^{5}\right )} \log \left (5\right ) + \log \left (5\right )^{2}} + \frac {x - e^{5}}{\log \left (5\right )^{2}} - \frac {\log \left (4 \, e^{\left (x - e^{5}\right )} + \log \left (5\right )\right )}{\log \left (5\right )^{2}}\right )} \log \left (5\right )^{2} + \frac {\sqrt {2} e^{\left (-\frac {\log \left (5\right ) \log \left (2\right )}{2 \, {\left (4 \, e^{\left (x - e^{5}\right )} + \log \left (5\right )\right )}} - \frac {\log \left (5\right ) \log \left (\log \left (2\right )\right )}{2 \, {\left (4 \, e^{\left (x - e^{5}\right )} + \log \left (5\right )\right )}} - \frac {5 \, \log \left (5\right )}{4 \, {\left (4 \, e^{\left (x - e^{5}\right )} + \log \left (5\right )\right )}} + \frac {5}{4}\right )} \log \left (5\right ) \sqrt {\log \left (2\right )} \log \left (4 \, \log \left (2\right )^{2}\right )}{{\left (2 \, \log \left (\log \left (2\right )\right ) + 5\right )} \log \left (5\right ) + 2 \, \log \left (5\right ) \log \left (2\right )} + \frac {5 \, \sqrt {2} e^{\left (-\frac {\log \left (5\right ) \log \left (2\right )}{2 \, {\left (4 \, e^{\left (x - e^{5}\right )} + \log \left (5\right )\right )}} - \frac {\log \left (5\right ) \log \left (\log \left (2\right )\right )}{2 \, {\left (4 \, e^{\left (x - e^{5}\right )} + \log \left (5\right )\right )}} - \frac {5 \, \log \left (5\right )}{4 \, {\left (4 \, e^{\left (x - e^{5}\right )} + \log \left (5\right )\right )}} + \frac {5}{4}\right )} \log \left (5\right ) \sqrt {\log \left (2\right )}}{{\left (2 \, \log \left (\log \left (2\right )\right ) + 5\right )} \log \left (5\right ) + 2 \, \log \left (5\right ) \log \left (2\right )} - \frac {5 \, \log \left (5\right )}{4 \, e^{\left (x - e^{5}\right )} + \log \left (5\right )} + 5 \, \log \left (4 \, e^{\left (x - e^{5}\right )} + \log \left (5\right )\right ) \] Input:

integrate(((log(5)*exp(-exp(5)+x)*log(4*log(2)^2)+5*log(5)*exp(-exp(5)+x)) 
*exp((exp(-exp(5)+x)*log(4*log(2)^2)+5*exp(-exp(5)+x))/(4*exp(-exp(5)+x)+l 
og(5)))+80*exp(-exp(5)+x)^2+40*log(5)*exp(-exp(5)+x)+5*log(5)^2)/(16*exp(- 
exp(5)+x)^2+8*log(5)*exp(-exp(5)+x)+log(5)^2),x, algorithm="maxima")
 

Output:

5*(1/(4*e^(x - e^5)*log(5) + log(5)^2) + (x - e^5)/log(5)^2 - log(4*e^(x - 
 e^5) + log(5))/log(5)^2)*log(5)^2 + sqrt(2)*e^(-1/2*log(5)*log(2)/(4*e^(x 
 - e^5) + log(5)) - 1/2*log(5)*log(log(2))/(4*e^(x - e^5) + log(5)) - 5/4* 
log(5)/(4*e^(x - e^5) + log(5)) + 5/4)*log(5)*sqrt(log(2))*log(4*log(2)^2) 
/((2*log(log(2)) + 5)*log(5) + 2*log(5)*log(2)) + 5*sqrt(2)*e^(-1/2*log(5) 
*log(2)/(4*e^(x - e^5) + log(5)) - 1/2*log(5)*log(log(2))/(4*e^(x - e^5) + 
 log(5)) - 5/4*log(5)/(4*e^(x - e^5) + log(5)) + 5/4)*log(5)*sqrt(log(2))/ 
((2*log(log(2)) + 5)*log(5) + 2*log(5)*log(2)) - 5*log(5)/(4*e^(x - e^5) + 
 log(5)) + 5*log(4*e^(x - e^5) + log(5))
 

Giac [F]

\[ \int \frac {80 e^{-2 e^5+2 x}+40 e^{-e^5+x} \log (5)+5 \log ^2(5)+e^{\frac {5 e^{-e^5+x}+2 e^{-e^5+x} \log (\log (4))}{4 e^{-e^5+x}+\log (5)}} \left (5 e^{-e^5+x} \log (5)+2 e^{-e^5+x} \log (5) \log (\log (4))\right )}{16 e^{-2 e^5+2 x}+8 e^{-e^5+x} \log (5)+\log ^2(5)} \, dx=\int { \frac {{\left (e^{\left (x - e^{5}\right )} \log \left (5\right ) \log \left (4 \, \log \left (2\right )^{2}\right ) + 5 \, e^{\left (x - e^{5}\right )} \log \left (5\right )\right )} e^{\left (\frac {e^{\left (x - e^{5}\right )} \log \left (4 \, \log \left (2\right )^{2}\right ) + 5 \, e^{\left (x - e^{5}\right )}}{4 \, e^{\left (x - e^{5}\right )} + \log \left (5\right )}\right )} + 40 \, e^{\left (x - e^{5}\right )} \log \left (5\right ) + 5 \, \log \left (5\right )^{2} + 80 \, e^{\left (2 \, x - 2 \, e^{5}\right )}}{8 \, e^{\left (x - e^{5}\right )} \log \left (5\right ) + \log \left (5\right )^{2} + 16 \, e^{\left (2 \, x - 2 \, e^{5}\right )}} \,d x } \] Input:

integrate(((log(5)*exp(-exp(5)+x)*log(4*log(2)^2)+5*log(5)*exp(-exp(5)+x)) 
*exp((exp(-exp(5)+x)*log(4*log(2)^2)+5*exp(-exp(5)+x))/(4*exp(-exp(5)+x)+l 
og(5)))+80*exp(-exp(5)+x)^2+40*log(5)*exp(-exp(5)+x)+5*log(5)^2)/(16*exp(- 
exp(5)+x)^2+8*log(5)*exp(-exp(5)+x)+log(5)^2),x, algorithm="giac")
 

Output:

integrate(((e^(x - e^5)*log(5)*log(4*log(2)^2) + 5*e^(x - e^5)*log(5))*e^( 
(e^(x - e^5)*log(4*log(2)^2) + 5*e^(x - e^5))/(4*e^(x - e^5) + log(5))) + 
40*e^(x - e^5)*log(5) + 5*log(5)^2 + 80*e^(2*x - 2*e^5))/(8*e^(x - e^5)*lo 
g(5) + log(5)^2 + 16*e^(2*x - 2*e^5)), x)
 

Mupad [B] (verification not implemented)

Time = 3.31 (sec) , antiderivative size = 80, normalized size of antiderivative = 2.67 \[ \int \frac {80 e^{-2 e^5+2 x}+40 e^{-e^5+x} \log (5)+5 \log ^2(5)+e^{\frac {5 e^{-e^5+x}+2 e^{-e^5+x} \log (\log (4))}{4 e^{-e^5+x}+\log (5)}} \left (5 e^{-e^5+x} \log (5)+2 e^{-e^5+x} \log (5) \log (\log (4))\right )}{16 e^{-2 e^5+2 x}+8 e^{-e^5+x} \log (5)+\log ^2(5)} \, dx=5\,x+2^{\frac {2\,{\mathrm {e}}^{x-{\mathrm {e}}^5}}{4\,{\mathrm {e}}^{x-{\mathrm {e}}^5}+\ln \left (5\right )}}\,{\mathrm {e}}^{\frac {5\,{\mathrm {e}}^{-{\mathrm {e}}^5}\,{\mathrm {e}}^x}{\ln \left (5\right )+4\,{\mathrm {e}}^{-{\mathrm {e}}^5}\,{\mathrm {e}}^x}}\,{\ln \left (2\right )}^{\frac {2\,{\mathrm {e}}^{x-{\mathrm {e}}^5}}{4\,{\mathrm {e}}^{x-{\mathrm {e}}^5}+\ln \left (5\right )}} \] Input:

int((80*exp(2*x - 2*exp(5)) + 5*log(5)^2 + 40*exp(x - exp(5))*log(5) + exp 
((5*exp(x - exp(5)) + exp(x - exp(5))*log(4*log(2)^2))/(4*exp(x - exp(5)) 
+ log(5)))*(5*exp(x - exp(5))*log(5) + exp(x - exp(5))*log(5)*log(4*log(2) 
^2)))/(16*exp(2*x - 2*exp(5)) + log(5)^2 + 8*exp(x - exp(5))*log(5)),x)
 

Output:

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

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.33 \[ \int \frac {80 e^{-2 e^5+2 x}+40 e^{-e^5+x} \log (5)+5 \log ^2(5)+e^{\frac {5 e^{-e^5+x}+2 e^{-e^5+x} \log (\log (4))}{4 e^{-e^5+x}+\log (5)}} \left (5 e^{-e^5+x} \log (5)+2 e^{-e^5+x} \log (5) \log (\log (4))\right )}{16 e^{-2 e^5+2 x}+8 e^{-e^5+x} \log (5)+\log ^2(5)} \, dx=e^{\frac {e^{x} \mathrm {log}\left (4 \mathrm {log}\left (2\right )^{2}\right )+5 e^{x}}{e^{e^{5}} \mathrm {log}\left (5\right )+4 e^{x}}}+5 x \] Input:

int(((log(5)*exp(-exp(5)+x)*log(4*log(2)^2)+5*log(5)*exp(-exp(5)+x))*exp(( 
exp(-exp(5)+x)*log(4*log(2)^2)+5*exp(-exp(5)+x))/(4*exp(-exp(5)+x)+log(5)) 
)+80*exp(-exp(5)+x)^2+40*log(5)*exp(-exp(5)+x)+5*log(5)^2)/(16*exp(-exp(5) 
+x)^2+8*log(5)*exp(-exp(5)+x)+log(5)^2),x)
 

Output:

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