\(\int \frac {e^{\frac {x^2+3 x \log (5)+3 \log (5) \log (x)}{x \log (5)}} (-72 x^2-216 \log (5)+216 \log (5) \log (x))}{(4 x^2 \log (5)+4 e^{\frac {x^2+3 x \log (5)+3 \log (5) \log (x)}{x \log (5)}} x^2 \log (5)+(x^2 \log (5)+e^{\frac {x^2+3 x \log (5)+3 \log (5) \log (x)}{x \log (5)}} x^2 \log (5)) \log (1+e^{\frac {x^2+3 x \log (5)+3 \log (5) \log (x)}{x \log (5)}})) \log ^3(4+\log (1+e^{\frac {x^2+3 x \log (5)+3 \log (5) \log (x)}{x \log (5)}}))} \, dx\) [1214]

Optimal result

Integrand size = 184, antiderivative size = 28 \[ \int \frac {e^{\frac {x^2+3 x \log (5)+3 \log (5) \log (x)}{x \log (5)}} \left (-72 x^2-216 \log (5)+216 \log (5) \log (x)\right )}{\left (4 x^2 \log (5)+4 e^{\frac {x^2+3 x \log (5)+3 \log (5) \log (x)}{x \log (5)}} x^2 \log (5)+\left (x^2 \log (5)+e^{\frac {x^2+3 x \log (5)+3 \log (5) \log (x)}{x \log (5)}} x^2 \log (5)\right ) \log \left (1+e^{\frac {x^2+3 x \log (5)+3 \log (5) \log (x)}{x \log (5)}}\right )\right ) \log ^3\left (4+\log \left (1+e^{\frac {x^2+3 x \log (5)+3 \log (5) \log (x)}{x \log (5)}}\right )\right )} \, dx=\frac {36}{\log ^2\left (4+\log \left (1+e^{\frac {x}{\log (5)}+\frac {3 (x+\log (x))}{x}}\right )\right )} \] Output:

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

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {x^2+3 x \log (5)+3 \log (5) \log (x)}{x \log (5)}} \left (-72 x^2-216 \log (5)+216 \log (5) \log (x)\right )}{\left (4 x^2 \log (5)+4 e^{\frac {x^2+3 x \log (5)+3 \log (5) \log (x)}{x \log (5)}} x^2 \log (5)+\left (x^2 \log (5)+e^{\frac {x^2+3 x \log (5)+3 \log (5) \log (x)}{x \log (5)}} x^2 \log (5)\right ) \log \left (1+e^{\frac {x^2+3 x \log (5)+3 \log (5) \log (x)}{x \log (5)}}\right )\right ) \log ^3\left (4+\log \left (1+e^{\frac {x^2+3 x \log (5)+3 \log (5) \log (x)}{x \log (5)}}\right )\right )} \, dx=\frac {36}{\log ^2\left (4+\log \left (1+e^{3+\frac {x}{\log (5)}} x^{3/x}\right )\right )} \] Input:

Integrate[(E^((x^2 + 3*x*Log[5] + 3*Log[5]*Log[x])/(x*Log[5]))*(-72*x^2 - 
216*Log[5] + 216*Log[5]*Log[x]))/((4*x^2*Log[5] + 4*E^((x^2 + 3*x*Log[5] + 
 3*Log[5]*Log[x])/(x*Log[5]))*x^2*Log[5] + (x^2*Log[5] + E^((x^2 + 3*x*Log 
[5] + 3*Log[5]*Log[x])/(x*Log[5]))*x^2*Log[5])*Log[1 + E^((x^2 + 3*x*Log[5 
] + 3*Log[5]*Log[x])/(x*Log[5]))])*Log[4 + Log[1 + E^((x^2 + 3*x*Log[5] + 
3*Log[5]*Log[x])/(x*Log[5]))]]^3),x]
 

Output:

36/Log[4 + Log[1 + E^(3 + x/Log[5])*x^(3/x)]]^2
 

Rubi [A] (verified)

Time = 1.69 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.21, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.005, Rules used = {7237}

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.

\(\displaystyle \int \frac {e^{\frac {x^2+3 x \log (5)+3 \log (5) \log (x)}{x \log (5)}} \left (-72 x^2+216 \log (5) \log (x)-216 \log (5)\right )}{\left (4 x^2 \log (5) e^{\frac {x^2+3 x \log (5)+3 \log (5) \log (x)}{x \log (5)}}+4 x^2 \log (5)+\left (x^2 \log (5) e^{\frac {x^2+3 x \log (5)+3 \log (5) \log (x)}{x \log (5)}}+x^2 \log (5)\right ) \log \left (e^{\frac {x^2+3 x \log (5)+3 \log (5) \log (x)}{x \log (5)}}+1\right )\right ) \log ^3\left (\log \left (e^{\frac {x^2+3 x \log (5)+3 \log (5) \log (x)}{x \log (5)}}+1\right )+4\right )} \, dx\)

\(\Big \downarrow \) 7237

\(\displaystyle \frac {36}{\log ^2\left (\log \left (x^{3/x} 5^{\frac {3}{\log (5)}} e^{\frac {x}{\log (5)}}+1\right )+4\right )}\)

Input:

Int[(E^((x^2 + 3*x*Log[5] + 3*Log[5]*Log[x])/(x*Log[5]))*(-72*x^2 - 216*Lo 
g[5] + 216*Log[5]*Log[x]))/((4*x^2*Log[5] + 4*E^((x^2 + 3*x*Log[5] + 3*Log 
[5]*Log[x])/(x*Log[5]))*x^2*Log[5] + (x^2*Log[5] + E^((x^2 + 3*x*Log[5] + 
3*Log[5]*Log[x])/(x*Log[5]))*x^2*Log[5])*Log[1 + E^((x^2 + 3*x*Log[5] + 3* 
Log[5]*Log[x])/(x*Log[5]))])*Log[4 + Log[1 + E^((x^2 + 3*x*Log[5] + 3*Log[ 
5]*Log[x])/(x*Log[5]))]]^3),x]
 

Output:

36/Log[4 + Log[1 + 5^(3/Log[5])*E^(x/Log[5])*x^(3/x)]]^2
 

Defintions of rubi rules used

rule 7237

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Si 
mp[q*(y^(m + 1)/(m + 1)), x] /;  !FalseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]
 

Maple [A] (verified)

Time = 2405.16 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.11

method	result	size
risch	\(\frac {36}{{\ln \left (\ln \left (x^{\frac {3}{x}} {\mathrm e}^{\frac {3 \ln \left (5\right )+x}{\ln \left (5\right )}}+1\right )+4\right )}^{2}}\)	\(31\)
parallelrisch	\(\frac {36}{{\ln \left (\ln \left ({\mathrm e}^{\frac {3 \ln \left (5\right ) \ln \left (x \right )+3 x \ln \left (5\right )+x^{2}}{x \ln \left (5\right )}}+1\right )+4\right )}^{2}}\)	\(35\)

Input:

int((216*ln(5)*ln(x)-216*ln(5)-72*x^2)*exp((3*ln(5)*ln(x)+3*x*ln(5)+x^2)/x 
/ln(5))/((x^2*ln(5)*exp((3*ln(5)*ln(x)+3*x*ln(5)+x^2)/x/ln(5))+x^2*ln(5))* 
ln(exp((3*ln(5)*ln(x)+3*x*ln(5)+x^2)/x/ln(5))+1)+4*x^2*ln(5)*exp((3*ln(5)* 
ln(x)+3*x*ln(5)+x^2)/x/ln(5))+4*x^2*ln(5))/ln(ln(exp((3*ln(5)*ln(x)+3*x*ln 
(5)+x^2)/x/ln(5))+1)+4)^3,x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.21 \[ \int \frac {e^{\frac {x^2+3 x \log (5)+3 \log (5) \log (x)}{x \log (5)}} \left (-72 x^2-216 \log (5)+216 \log (5) \log (x)\right )}{\left (4 x^2 \log (5)+4 e^{\frac {x^2+3 x \log (5)+3 \log (5) \log (x)}{x \log (5)}} x^2 \log (5)+\left (x^2 \log (5)+e^{\frac {x^2+3 x \log (5)+3 \log (5) \log (x)}{x \log (5)}} x^2 \log (5)\right ) \log \left (1+e^{\frac {x^2+3 x \log (5)+3 \log (5) \log (x)}{x \log (5)}}\right )\right ) \log ^3\left (4+\log \left (1+e^{\frac {x^2+3 x \log (5)+3 \log (5) \log (x)}{x \log (5)}}\right )\right )} \, dx=\frac {36}{\log \left (\log \left (e^{\left (\frac {x^{2} + 3 \, x \log \left (5\right ) + 3 \, \log \left (5\right ) \log \left (x\right )}{x \log \left (5\right )}\right )} + 1\right ) + 4\right )^{2}} \] Input:

integrate((216*log(5)*log(x)-216*log(5)-72*x^2)*exp((3*log(5)*log(x)+3*x*l 
og(5)+x^2)/x/log(5))/((x^2*log(5)*exp((3*log(5)*log(x)+3*x*log(5)+x^2)/x/l 
og(5))+x^2*log(5))*log(exp((3*log(5)*log(x)+3*x*log(5)+x^2)/x/log(5))+1)+4 
*x^2*log(5)*exp((3*log(5)*log(x)+3*x*log(5)+x^2)/x/log(5))+4*x^2*log(5))/l 
og(log(exp((3*log(5)*log(x)+3*x*log(5)+x^2)/x/log(5))+1)+4)^3,x, algorithm 
="fricas")
 

Output:

36/log(log(e^((x^2 + 3*x*log(5) + 3*log(5)*log(x))/(x*log(5))) + 1) + 4)^2
 

Sympy [F(-1)]

Timed out. \[ \int \frac {e^{\frac {x^2+3 x \log (5)+3 \log (5) \log (x)}{x \log (5)}} \left (-72 x^2-216 \log (5)+216 \log (5) \log (x)\right )}{\left (4 x^2 \log (5)+4 e^{\frac {x^2+3 x \log (5)+3 \log (5) \log (x)}{x \log (5)}} x^2 \log (5)+\left (x^2 \log (5)+e^{\frac {x^2+3 x \log (5)+3 \log (5) \log (x)}{x \log (5)}} x^2 \log (5)\right ) \log \left (1+e^{\frac {x^2+3 x \log (5)+3 \log (5) \log (x)}{x \log (5)}}\right )\right ) \log ^3\left (4+\log \left (1+e^{\frac {x^2+3 x \log (5)+3 \log (5) \log (x)}{x \log (5)}}\right )\right )} \, dx=\text {Timed out} \] Input:

integrate((216*ln(5)*ln(x)-216*ln(5)-72*x**2)*exp((3*ln(5)*ln(x)+3*x*ln(5) 
+x**2)/x/ln(5))/((x**2*ln(5)*exp((3*ln(5)*ln(x)+3*x*ln(5)+x**2)/x/ln(5))+x 
**2*ln(5))*ln(exp((3*ln(5)*ln(x)+3*x*ln(5)+x**2)/x/ln(5))+1)+4*x**2*ln(5)* 
exp((3*ln(5)*ln(x)+3*x*ln(5)+x**2)/x/ln(5))+4*x**2*ln(5))/ln(ln(exp((3*ln( 
5)*ln(x)+3*x*ln(5)+x**2)/x/ln(5))+1)+4)**3,x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93 \[ \int \frac {e^{\frac {x^2+3 x \log (5)+3 \log (5) \log (x)}{x \log (5)}} \left (-72 x^2-216 \log (5)+216 \log (5) \log (x)\right )}{\left (4 x^2 \log (5)+4 e^{\frac {x^2+3 x \log (5)+3 \log (5) \log (x)}{x \log (5)}} x^2 \log (5)+\left (x^2 \log (5)+e^{\frac {x^2+3 x \log (5)+3 \log (5) \log (x)}{x \log (5)}} x^2 \log (5)\right ) \log \left (1+e^{\frac {x^2+3 x \log (5)+3 \log (5) \log (x)}{x \log (5)}}\right )\right ) \log ^3\left (4+\log \left (1+e^{\frac {x^2+3 x \log (5)+3 \log (5) \log (x)}{x \log (5)}}\right )\right )} \, dx=\frac {36}{\log \left (\log \left (e^{\left (\frac {x}{\log \left (5\right )} + \frac {3 \, \log \left (x\right )}{x} + 3\right )} + 1\right ) + 4\right )^{2}} \] Input:

integrate((216*log(5)*log(x)-216*log(5)-72*x^2)*exp((3*log(5)*log(x)+3*x*l 
og(5)+x^2)/x/log(5))/((x^2*log(5)*exp((3*log(5)*log(x)+3*x*log(5)+x^2)/x/l 
og(5))+x^2*log(5))*log(exp((3*log(5)*log(x)+3*x*log(5)+x^2)/x/log(5))+1)+4 
*x^2*log(5)*exp((3*log(5)*log(x)+3*x*log(5)+x^2)/x/log(5))+4*x^2*log(5))/l 
og(log(exp((3*log(5)*log(x)+3*x*log(5)+x^2)/x/log(5))+1)+4)^3,x, algorithm 
="maxima")
 

Output:

36/log(log(e^(x/log(5) + 3*log(x)/x + 3) + 1) + 4)^2
 

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.21 \[ \int \frac {e^{\frac {x^2+3 x \log (5)+3 \log (5) \log (x)}{x \log (5)}} \left (-72 x^2-216 \log (5)+216 \log (5) \log (x)\right )}{\left (4 x^2 \log (5)+4 e^{\frac {x^2+3 x \log (5)+3 \log (5) \log (x)}{x \log (5)}} x^2 \log (5)+\left (x^2 \log (5)+e^{\frac {x^2+3 x \log (5)+3 \log (5) \log (x)}{x \log (5)}} x^2 \log (5)\right ) \log \left (1+e^{\frac {x^2+3 x \log (5)+3 \log (5) \log (x)}{x \log (5)}}\right )\right ) \log ^3\left (4+\log \left (1+e^{\frac {x^2+3 x \log (5)+3 \log (5) \log (x)}{x \log (5)}}\right )\right )} \, dx=\frac {36}{\log \left (\log \left (e^{\left (\frac {x^{2} + 3 \, x \log \left (5\right ) + 3 \, \log \left (5\right ) \log \left (x\right )}{x \log \left (5\right )}\right )} + 1\right ) + 4\right )^{2}} \] Input:

integrate((216*log(5)*log(x)-216*log(5)-72*x^2)*exp((3*log(5)*log(x)+3*x*l 
og(5)+x^2)/x/log(5))/((x^2*log(5)*exp((3*log(5)*log(x)+3*x*log(5)+x^2)/x/l 
og(5))+x^2*log(5))*log(exp((3*log(5)*log(x)+3*x*log(5)+x^2)/x/log(5))+1)+4 
*x^2*log(5)*exp((3*log(5)*log(x)+3*x*log(5)+x^2)/x/log(5))+4*x^2*log(5))/l 
og(log(exp((3*log(5)*log(x)+3*x*log(5)+x^2)/x/log(5))+1)+4)^3,x, algorithm 
="giac")
 

Output:

36/log(log(e^((x^2 + 3*x*log(5) + 3*log(5)*log(x))/(x*log(5))) + 1) + 4)^2
 

Mupad [B] (verification not implemented)

Time = 3.36 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.96 \[ \int \frac {e^{\frac {x^2+3 x \log (5)+3 \log (5) \log (x)}{x \log (5)}} \left (-72 x^2-216 \log (5)+216 \log (5) \log (x)\right )}{\left (4 x^2 \log (5)+4 e^{\frac {x^2+3 x \log (5)+3 \log (5) \log (x)}{x \log (5)}} x^2 \log (5)+\left (x^2 \log (5)+e^{\frac {x^2+3 x \log (5)+3 \log (5) \log (x)}{x \log (5)}} x^2 \log (5)\right ) \log \left (1+e^{\frac {x^2+3 x \log (5)+3 \log (5) \log (x)}{x \log (5)}}\right )\right ) \log ^3\left (4+\log \left (1+e^{\frac {x^2+3 x \log (5)+3 \log (5) \log (x)}{x \log (5)}}\right )\right )} \, dx=\frac {36}{{\ln \left (\ln \left (x^{3/x}\,{\mathrm {e}}^{\frac {x}{\ln \left (5\right )}}\,{\mathrm {e}}^3+1\right )+4\right )}^2} \] Input:

int(-(exp((3*x*log(5) + 3*log(5)*log(x) + x^2)/(x*log(5)))*(216*log(5) - 2 
16*log(5)*log(x) + 72*x^2))/(log(log(exp((3*x*log(5) + 3*log(5)*log(x) + x 
^2)/(x*log(5))) + 1) + 4)^3*(4*x^2*log(5) + log(exp((3*x*log(5) + 3*log(5) 
*log(x) + x^2)/(x*log(5))) + 1)*(x^2*log(5) + x^2*exp((3*x*log(5) + 3*log( 
5)*log(x) + x^2)/(x*log(5)))*log(5)) + 4*x^2*exp((3*x*log(5) + 3*log(5)*lo 
g(x) + x^2)/(x*log(5)))*log(5))),x)
 

Output:

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

Reduce [F]

\[ \int \frac {e^{\frac {x^2+3 x \log (5)+3 \log (5) \log (x)}{x \log (5)}} \left (-72 x^2-216 \log (5)+216 \log (5) \log (x)\right )}{\left (4 x^2 \log (5)+4 e^{\frac {x^2+3 x \log (5)+3 \log (5) \log (x)}{x \log (5)}} x^2 \log (5)+\left (x^2 \log (5)+e^{\frac {x^2+3 x \log (5)+3 \log (5) \log (x)}{x \log (5)}} x^2 \log (5)\right ) \log \left (1+e^{\frac {x^2+3 x \log (5)+3 \log (5) \log (x)}{x \log (5)}}\right )\right ) \log ^3\left (4+\log \left (1+e^{\frac {x^2+3 x \log (5)+3 \log (5) \log (x)}{x \log (5)}}\right )\right )} \, dx =\text {Too large to display} \] Input:

int((216*log(5)*log(x)-216*log(5)-72*x^2)*exp((3*log(5)*log(x)+3*x*log(5)+ 
x^2)/x/log(5))/((x^2*log(5)*exp((3*log(5)*log(x)+3*x*log(5)+x^2)/x/log(5)) 
+x^2*log(5))*log(exp((3*log(5)*log(x)+3*x*log(5)+x^2)/x/log(5))+1)+4*x^2*l 
og(5)*exp((3*log(5)*log(x)+3*x*log(5)+x^2)/x/log(5))+4*x^2*log(5))/log(log 
(exp((3*log(5)*log(x)+3*x*log(5)+x^2)/x/log(5))+1)+4)^3,x)
 

Output:

(72*e**3*( - 3*int(e**((3*log(x)*log(5) + x**2)/(log(5)*x))/(e**((3*log(x) 
*log(5) + x**2)/(log(5)*x))*log(log(e**((3*log(x)*log(5) + x**2)/(log(5)*x 
))*e**3 + 1) + 4)**3*log(e**((3*log(x)*log(5) + x**2)/(log(5)*x))*e**3 + 1 
)*e**3*x**2 + 4*e**((3*log(x)*log(5) + x**2)/(log(5)*x))*log(log(e**((3*lo 
g(x)*log(5) + x**2)/(log(5)*x))*e**3 + 1) + 4)**3*e**3*x**2 + log(log(e**( 
(3*log(x)*log(5) + x**2)/(log(5)*x))*e**3 + 1) + 4)**3*log(e**((3*log(x)*l 
og(5) + x**2)/(log(5)*x))*e**3 + 1)*x**2 + 4*log(log(e**((3*log(x)*log(5) 
+ x**2)/(log(5)*x))*e**3 + 1) + 4)**3*x**2),x)*log(5) - int(e**((3*log(x)* 
log(5) + x**2)/(log(5)*x))/(e**((3*log(x)*log(5) + x**2)/(log(5)*x))*log(l 
og(e**((3*log(x)*log(5) + x**2)/(log(5)*x))*e**3 + 1) + 4)**3*log(e**((3*l 
og(x)*log(5) + x**2)/(log(5)*x))*e**3 + 1)*e**3 + 4*e**((3*log(x)*log(5) + 
 x**2)/(log(5)*x))*log(log(e**((3*log(x)*log(5) + x**2)/(log(5)*x))*e**3 + 
 1) + 4)**3*e**3 + log(log(e**((3*log(x)*log(5) + x**2)/(log(5)*x))*e**3 + 
 1) + 4)**3*log(e**((3*log(x)*log(5) + x**2)/(log(5)*x))*e**3 + 1) + 4*log 
(log(e**((3*log(x)*log(5) + x**2)/(log(5)*x))*e**3 + 1) + 4)**3),x) + 3*in 
t((e**((3*log(x)*log(5) + x**2)/(log(5)*x))*log(x))/(e**((3*log(x)*log(5) 
+ x**2)/(log(5)*x))*log(log(e**((3*log(x)*log(5) + x**2)/(log(5)*x))*e**3 
+ 1) + 4)**3*log(e**((3*log(x)*log(5) + x**2)/(log(5)*x))*e**3 + 1)*e**3*x 
**2 + 4*e**((3*log(x)*log(5) + x**2)/(log(5)*x))*log(log(e**((3*log(x)*log 
(5) + x**2)/(log(5)*x))*e**3 + 1) + 4)**3*e**3*x**2 + log(log(e**((3*lo...