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

3.15.95.1 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))
 

3.15.95.2 Mathematica [B] (verified)

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

Time = 0.71 (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])))
 

3.15.95.3 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
 

3.15.95.3.1 Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2720

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Simp[v/D[v, x] 
   Subst[Int[FunctionOfExponentialFunction[u, x]/x, x], x, v], x]] /; Funct 
ionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; FreeQ 
[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x)) 
*(F_)[v_] /; FreeQ[{a, b, c}, x] && InverseFunctionQ[F[x]]]
 

rule 7293

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v] 
]
 

3.15.95.4 Maple [A] (verified)

Time = 0.68 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.23

method	result	size
parallelrisch	\(5 x +{\mathrm e}^{\frac {{\mathrm e}^{-{\mathrm e}^{5}+x} \left (5+\ln \left (4 \ln \left (2\right )^{2}\right )\right )}{4 \,{\mathrm e}^{-{\mathrm e}^{5}+x}+\ln \left (5\right )}}\)	\(37\)
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

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

3.15.95.5 Fricas [A] (verification not implemented)

Time = 0.26 (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=\
 

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)))
 

3.15.95.6 Sympy [A] (verification not implemented)

Time = 0.22 (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)))
 

3.15.95.7 Maxima [B] (verification not implemented)

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

Time = 0.32 (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=\
 

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))
 

3.15.95.8 Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 110 vs. \(2 (29) = 58\).

Time = 0.44 (sec) , antiderivative size = 110, normalized size of antiderivative = 3.67 \[ \int \frac {80 e^{-2 e^5+2 x}+40 e^{-e^5+x} \log (5)+5 \log ^2(5)+e^{\frac {5 e^{-e^5+x}+2 e^{-e^5+x} \log (\log (4))}{4 e^{-e^5+x}+\log (5)}} \left (5 e^{-e^5+x} \log (5)+2 e^{-e^5+x} \log (5) \log (\log (4))\right )}{16 e^{-2 e^5+2 x}+8 e^{-e^5+x} \log (5)+\log ^2(5)} \, dx={\left (5 \, x e^{x} + 5 \, e^{x} \log \left (4 \, e^{\left (x - e^{5}\right )} + \log \left (5\right )\right ) - 5 \, e^{x} \log \left (-4 \, e^{\left (x - e^{5}\right )} - \log \left (5\right )\right ) + e^{\left (\frac {4 \, x e^{\left (x - e^{5}\right )} + x \log \left (5\right ) + 2 \, e^{\left (x - e^{5}\right )} \log \left (2\right ) + 2 \, e^{\left (x - e^{5}\right )} \log \left (\log \left (2\right )\right ) + 5 \, e^{\left (x - e^{5}\right )}}{4 \, e^{\left (x - e^{5}\right )} + \log \left (5\right )}\right )}\right )} e^{\left (-x\right )} \]

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=\
 

output

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

3.15.95.9 Mupad [B] (verification not implemented)

Time = 0.53 (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)))