\(\int \frac {e^{\frac {60-5 x \log (\frac {e}{4+4 \log (5)+\log ^2(5)})}{x \log (\frac {e}{4+4 \log (5)+\log ^2(5)})}} (-60 \log (x)+x \log (\frac {e}{4+4 \log (5)+\log ^2(5)}))}{x^2 \log (\frac {e}{4+4 \log (5)+\log ^2(5)})} \, dx\) [169]

Optimal result

Integrand size = 86, antiderivative size = 23 \[ \int \frac {e^{\frac {60-5 x \log \left (\frac {e}{4+4 \log (5)+\log ^2(5)}\right )}{x \log \left (\frac {e}{4+4 \log (5)+\log ^2(5)}\right )}} \left (-60 \log (x)+x \log \left (\frac {e}{4+4 \log (5)+\log ^2(5)}\right )\right )}{x^2 \log \left (\frac {e}{4+4 \log (5)+\log ^2(5)}\right )} \, dx=e^{-5+\frac {60}{x \log \left (\frac {e}{(2+\log (5))^2}\right )}} \log (x) \] Output:

ln(x)*exp(60/x/ln(exp(1)/(2+ln(5))^2)-5)
 

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {60-5 x \log \left (\frac {e}{4+4 \log (5)+\log ^2(5)}\right )}{x \log \left (\frac {e}{4+4 \log (5)+\log ^2(5)}\right )}} \left (-60 \log (x)+x \log \left (\frac {e}{4+4 \log (5)+\log ^2(5)}\right )\right )}{x^2 \log \left (\frac {e}{4+4 \log (5)+\log ^2(5)}\right )} \, dx=e^{-5+\frac {60}{x \log \left (\frac {e}{(2+\log (5))^2}\right )}} \log (x) \] Input:

Integrate[(E^((60 - 5*x*Log[E/(4 + 4*Log[5] + Log[5]^2)])/(x*Log[E/(4 + 4* 
Log[5] + Log[5]^2)]))*(-60*Log[x] + x*Log[E/(4 + 4*Log[5] + Log[5]^2)]))/( 
x^2*Log[E/(4 + 4*Log[5] + Log[5]^2)]),x]
 

Output:

E^(-5 + 60/(x*Log[E/(2 + Log[5])^2]))*Log[x]
 

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(53\) vs. \(2(23)=46\).

Time = 0.28 (sec) , antiderivative size = 53, normalized size of antiderivative = 2.30, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.047, Rules used = {27, 25, 27, 2726}

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.

\(\displaystyle \int \frac {\left (x \log \left (\frac {e}{4+\log ^2(5)+4 \log (5)}\right )-60 \log (x)\right ) \exp \left (\frac {60-5 x \log \left (\frac {e}{4+\log ^2(5)+4 \log (5)}\right )}{x \log \left (\frac {e}{4+\log ^2(5)+4 \log (5)}\right )}\right )}{x^2 \log \left (\frac {e}{4+\log ^2(5)+4 \log (5)}\right )} \, dx\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int -\frac {\exp \left (\frac {60}{x \log \left (\frac {e}{(2+\log (5))^2}\right )}-\frac {5}{\log \left (\frac {e}{4+4 \log (5)+\log ^2(5)}\right )}\right ) \left (4+4 \log (5)+\log ^2(5)\right )^{\frac {5}{\log \left (\frac {e}{4+4 \log (5)+\log ^2(5)}\right )}} \left (60 \log (x)-x \log \left (\frac {e}{(2+\log (5))^2}\right )\right )}{x^2}dx}{\log \left (\frac {e}{(2+\log (5))^2}\right )}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {\int \frac {\exp \left (\frac {60}{x \log \left (\frac {e}{(2+\log (5))^2}\right )}-\frac {5}{\log \left (\frac {e}{(2+\log (5))^2}\right )}\right ) (2+\log (5))^{\frac {10}{\log \left (\frac {e}{(2+\log (5))^2}\right )}} \left (60 \log (x)-x \log \left (\frac {e}{(2+\log (5))^2}\right )\right )}{x^2}dx}{\log \left (\frac {e}{(2+\log (5))^2}\right )}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {(2+\log (5))^{\frac {10}{\log \left (\frac {e}{(2+\log (5))^2}\right )}} \int \frac {\exp \left (\frac {60}{x \log \left (\frac {e}{(2+\log (5))^2}\right )}-\frac {5}{\log \left (\frac {e}{(2+\log (5))^2}\right )}\right ) \left (60 \log (x)-x \log \left (\frac {e}{(2+\log (5))^2}\right )\right )}{x^2}dx}{\log \left (\frac {e}{(2+\log (5))^2}\right )}\)

\(\Big \downarrow \) 2726

\(\displaystyle (2+\log (5))^{\frac {10}{\log \left (\frac {e}{(2+\log (5))^2}\right )}} \log (x) \exp \left (\frac {60}{x \log \left (\frac {e}{(2+\log (5))^2}\right )}-\frac {5}{\log \left (\frac {e}{(2+\log (5))^2}\right )}\right )\)

Input:

Int[(E^((60 - 5*x*Log[E/(4 + 4*Log[5] + Log[5]^2)])/(x*Log[E/(4 + 4*Log[5] 
 + Log[5]^2)]))*(-60*Log[x] + x*Log[E/(4 + 4*Log[5] + Log[5]^2)]))/(x^2*Lo 
g[E/(4 + 4*Log[5] + Log[5]^2)]),x]
 

Output:

E^(-5/Log[E/(2 + Log[5])^2] + 60/(x*Log[E/(2 + Log[5])^2]))*(2 + Log[5])^( 
10/Log[E/(2 + Log[5])^2])*Log[x]
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 2726

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = v*(y/(Log[F]*D[u, 
 x]))}, Simp[F^u*z, x] /; EqQ[D[z, x], w*y]] /; FreeQ[F, x]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(47\) vs. \(2(23)=46\).

Time = 0.43 (sec) , antiderivative size = 48, normalized size of antiderivative = 2.09

method	result	size
default	\(\ln \left (x \right ) {\mathrm e}^{\frac {-5 x \ln \left (\frac {{\mathrm e}}{\ln \left (5\right )^{2}+4 \ln \left (5\right )+4}\right )+60}{x \ln \left (\frac {{\mathrm e}}{\ln \left (5\right )^{2}+4 \ln \left (5\right )+4}\right )}}\)	\(48\)
norman	\(\ln \left (x \right ) {\mathrm e}^{\frac {-5 x \ln \left (\frac {{\mathrm e}}{\ln \left (5\right )^{2}+4 \ln \left (5\right )+4}\right )+60}{x \ln \left (\frac {{\mathrm e}}{\ln \left (5\right )^{2}+4 \ln \left (5\right )+4}\right )}}\)	\(48\)
risch	\(-\frac {\left (2 \ln \left (2+\ln \left (5\right )\right )-1\right ) {\mathrm e}^{-\frac {5 \left (2 x \ln \left (2+\ln \left (5\right )\right )-x +12\right )}{x \left (2 \ln \left (2+\ln \left (5\right )\right )-1\right )}} \ln \left (x \right )}{-2 \ln \left (2+\ln \left (5\right )\right )+1}\)	\(55\)
orering	\(-\frac {\left (2 \ln \left (2+\ln \left (5\right )\right )-1\right ) \left (2 x \ln \left (2+\ln \left (5\right )\right )-x -120\right ) \left (-60 \ln \left (x \right )+x \ln \left (\frac {{\mathrm e}}{\ln \left (5\right )^{2}+4 \ln \left (5\right )+4}\right )\right ) {\mathrm e}^{\frac {-5 x \ln \left (\frac {{\mathrm e}}{\ln \left (5\right )^{2}+4 \ln \left (5\right )+4}\right )+60}{x \ln \left (\frac {{\mathrm e}}{\ln \left (5\right )^{2}+4 \ln \left (5\right )+4}\right )}}}{60 \left (2 x \ln \left (2+\ln \left (5\right )\right )-x +60\right ) \ln \left (\frac {{\mathrm e}}{\ln \left (5\right )^{2}+4 \ln \left (5\right )+4}\right )}-\frac {x^{4} \left (2 \ln \left (2+\ln \left (5\right )\right )-1\right )^{2} \left (\frac {\left (-\frac {60}{x}+\ln \left (\frac {{\mathrm e}}{\ln \left (5\right )^{2}+4 \ln \left (5\right )+4}\right )\right ) {\mathrm e}^{\frac {-5 x \ln \left (\frac {{\mathrm e}}{\ln \left (5\right )^{2}+4 \ln \left (5\right )+4}\right )+60}{x \ln \left (\frac {{\mathrm e}}{\ln \left (5\right )^{2}+4 \ln \left (5\right )+4}\right )}}}{x^{2} \ln \left (\frac {{\mathrm e}}{\ln \left (5\right )^{2}+4 \ln \left (5\right )+4}\right )}+\frac {\left (-60 \ln \left (x \right )+x \ln \left (\frac {{\mathrm e}}{\ln \left (5\right )^{2}+4 \ln \left (5\right )+4}\right )\right ) \left (-\frac {5}{x}-\frac {-5 x \ln \left (\frac {{\mathrm e}}{\ln \left (5\right )^{2}+4 \ln \left (5\right )+4}\right )+60}{x^{2} \ln \left (\frac {{\mathrm e}}{\ln \left (5\right )^{2}+4 \ln \left (5\right )+4}\right )}\right ) {\mathrm e}^{\frac {-5 x \ln \left (\frac {{\mathrm e}}{\ln \left (5\right )^{2}+4 \ln \left (5\right )+4}\right )+60}{x \ln \left (\frac {{\mathrm e}}{\ln \left (5\right )^{2}+4 \ln \left (5\right )+4}\right )}}}{x^{2} \ln \left (\frac {{\mathrm e}}{\ln \left (5\right )^{2}+4 \ln \left (5\right )+4}\right )}-\frac {2 \left (-60 \ln \left (x \right )+x \ln \left (\frac {{\mathrm e}}{\ln \left (5\right )^{2}+4 \ln \left (5\right )+4}\right )\right ) {\mathrm e}^{\frac {-5 x \ln \left (\frac {{\mathrm e}}{\ln \left (5\right )^{2}+4 \ln \left (5\right )+4}\right )+60}{x \ln \left (\frac {{\mathrm e}}{\ln \left (5\right )^{2}+4 \ln \left (5\right )+4}\right )}}}{x^{3} \ln \left (\frac {{\mathrm e}}{\ln \left (5\right )^{2}+4 \ln \left (5\right )+4}\right )}\right )}{60 \left (2 x \ln \left (2+\ln \left (5\right )\right )-x +60\right )}\)	\(475\)

Input:

int((-60*ln(x)+x*ln(exp(1)/(ln(5)^2+4*ln(5)+4)))*exp((-5*x*ln(exp(1)/(ln(5 
)^2+4*ln(5)+4))+60)/x/ln(exp(1)/(ln(5)^2+4*ln(5)+4)))/x^2/ln(exp(1)/(ln(5) 
^2+4*ln(5)+4)),x,method=_RETURNVERBOSE)
 

Output:

ln(x)*exp((-5*x*ln(exp(1)/(ln(5)^2+4*ln(5)+4))+60)/x/ln(exp(1)/(ln(5)^2+4* 
ln(5)+4)))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 47 vs. \(2 (23) = 46\).

Time = 0.09 (sec) , antiderivative size = 47, normalized size of antiderivative = 2.04 \[ \int \frac {e^{\frac {60-5 x \log \left (\frac {e}{4+4 \log (5)+\log ^2(5)}\right )}{x \log \left (\frac {e}{4+4 \log (5)+\log ^2(5)}\right )}} \left (-60 \log (x)+x \log \left (\frac {e}{4+4 \log (5)+\log ^2(5)}\right )\right )}{x^2 \log \left (\frac {e}{4+4 \log (5)+\log ^2(5)}\right )} \, dx=e^{\left (-\frac {5 \, {\left (x \log \left (\frac {e}{\log \left (5\right )^{2} + 4 \, \log \left (5\right ) + 4}\right ) - 12\right )}}{x \log \left (\frac {e}{\log \left (5\right )^{2} + 4 \, \log \left (5\right ) + 4}\right )}\right )} \log \left (x\right ) \] Input:

integrate((-60*log(x)+x*log(exp(1)/(log(5)^2+4*log(5)+4)))*exp((-5*x*log(e 
xp(1)/(log(5)^2+4*log(5)+4))+60)/x/log(exp(1)/(log(5)^2+4*log(5)+4)))/x^2/ 
log(exp(1)/(log(5)^2+4*log(5)+4)),x, algorithm="fricas")
 

Output:

e^(-5*(x*log(e/(log(5)^2 + 4*log(5) + 4)) - 12)/(x*log(e/(log(5)^2 + 4*log 
(5) + 4))))*log(x)
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 44 vs. \(2 (20) = 40\).

Time = 0.26 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.91 \[ \int \frac {e^{\frac {60-5 x \log \left (\frac {e}{4+4 \log (5)+\log ^2(5)}\right )}{x \log \left (\frac {e}{4+4 \log (5)+\log ^2(5)}\right )}} \left (-60 \log (x)+x \log \left (\frac {e}{4+4 \log (5)+\log ^2(5)}\right )\right )}{x^2 \log \left (\frac {e}{4+4 \log (5)+\log ^2(5)}\right )} \, dx=e^{\frac {- 5 x \log {\left (\frac {e}{\log {\left (5 \right )}^{2} + 4 + 4 \log {\left (5 \right )}} \right )} + 60}{x \log {\left (\frac {e}{\log {\left (5 \right )}^{2} + 4 + 4 \log {\left (5 \right )}} \right )}}} \log {\left (x \right )} \] Input:

integrate((-60*ln(x)+x*ln(exp(1)/(ln(5)**2+4*ln(5)+4)))*exp((-5*x*ln(exp(1 
)/(ln(5)**2+4*ln(5)+4))+60)/x/ln(exp(1)/(ln(5)**2+4*ln(5)+4)))/x**2/ln(exp 
(1)/(ln(5)**2+4*ln(5)+4)),x)
 

Output:

exp((-5*x*log(E/(log(5)**2 + 4 + 4*log(5))) + 60)/(x*log(E/(log(5)**2 + 4 
+ 4*log(5)))))*log(x)
 

Maxima [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.26 \[ \int \frac {e^{\frac {60-5 x \log \left (\frac {e}{4+4 \log (5)+\log ^2(5)}\right )}{x \log \left (\frac {e}{4+4 \log (5)+\log ^2(5)}\right )}} \left (-60 \log (x)+x \log \left (\frac {e}{4+4 \log (5)+\log ^2(5)}\right )\right )}{x^2 \log \left (\frac {e}{4+4 \log (5)+\log ^2(5)}\right )} \, dx=e^{\left (\frac {60}{x \log \left (\frac {e}{\log \left (5\right )^{2} + 4 \, \log \left (5\right ) + 4}\right )} - 5\right )} \log \left (x\right ) \] Input:

integrate((-60*log(x)+x*log(exp(1)/(log(5)^2+4*log(5)+4)))*exp((-5*x*log(e 
xp(1)/(log(5)^2+4*log(5)+4))+60)/x/log(exp(1)/(log(5)^2+4*log(5)+4)))/x^2/ 
log(exp(1)/(log(5)^2+4*log(5)+4)),x, algorithm="maxima")
 

Output:

e^(60/(x*log(e/(log(5)^2 + 4*log(5) + 4))) - 5)*log(x)
 

Giac [F]

\[ \int \frac {e^{\frac {60-5 x \log \left (\frac {e}{4+4 \log (5)+\log ^2(5)}\right )}{x \log \left (\frac {e}{4+4 \log (5)+\log ^2(5)}\right )}} \left (-60 \log (x)+x \log \left (\frac {e}{4+4 \log (5)+\log ^2(5)}\right )\right )}{x^2 \log \left (\frac {e}{4+4 \log (5)+\log ^2(5)}\right )} \, dx=\int { \frac {{\left (x \log \left (\frac {e}{\log \left (5\right )^{2} + 4 \, \log \left (5\right ) + 4}\right ) - 60 \, \log \left (x\right )\right )} e^{\left (-\frac {5 \, {\left (x \log \left (\frac {e}{\log \left (5\right )^{2} + 4 \, \log \left (5\right ) + 4}\right ) - 12\right )}}{x \log \left (\frac {e}{\log \left (5\right )^{2} + 4 \, \log \left (5\right ) + 4}\right )}\right )}}{x^{2} \log \left (\frac {e}{\log \left (5\right )^{2} + 4 \, \log \left (5\right ) + 4}\right )} \,d x } \] Input:

integrate((-60*log(x)+x*log(exp(1)/(log(5)^2+4*log(5)+4)))*exp((-5*x*log(e 
xp(1)/(log(5)^2+4*log(5)+4))+60)/x/log(exp(1)/(log(5)^2+4*log(5)+4)))/x^2/ 
log(exp(1)/(log(5)^2+4*log(5)+4)),x, algorithm="giac")
 

Output:

integrate((x*log(e/(log(5)^2 + 4*log(5) + 4)) - 60*log(x))*e^(-5*(x*log(e/ 
(log(5)^2 + 4*log(5) + 4)) - 12)/(x*log(e/(log(5)^2 + 4*log(5) + 4))))/(x^ 
2*log(e/(log(5)^2 + 4*log(5) + 4))), x)
 

Mupad [B] (verification not implemented)

Time = 2.72 (sec) , antiderivative size = 72, normalized size of antiderivative = 3.13 \[ \int \frac {e^{\frac {60-5 x \log \left (\frac {e}{4+4 \log (5)+\log ^2(5)}\right )}{x \log \left (\frac {e}{4+4 \log (5)+\log ^2(5)}\right )}} \left (-60 \log (x)+x \log \left (\frac {e}{4+4 \log (5)+\log ^2(5)}\right )\right )}{x^2 \log \left (\frac {e}{4+4 \log (5)+\log ^2(5)}\right )} \, dx=\frac {{\mathrm {e}}^{\frac {60}{x-x\,\ln \left (4\,\ln \left (5\right )+{\ln \left (5\right )}^2+4\right )}}\,{\mathrm {e}}^{\frac {5}{\ln \left (4\,\ln \left (5\right )+{\ln \left (5\right )}^2+4\right )-1}}\,\ln \left (x\right )}{{\left (4\,\ln \left (5\right )+{\ln \left (5\right )}^2+4\right )}^{\frac {5}{\ln \left (4\,\ln \left (5\right )+{\ln \left (5\right )}^2+4\right )-1}}} \] Input:

int(-(exp(-(5*x*log(exp(1)/(4*log(5) + log(5)^2 + 4)) - 60)/(x*log(exp(1)/ 
(4*log(5) + log(5)^2 + 4))))*(60*log(x) - x*log(exp(1)/(4*log(5) + log(5)^ 
2 + 4))))/(x^2*log(exp(1)/(4*log(5) + log(5)^2 + 4))),x)
 

Output:

(exp(60/(x - x*log(4*log(5) + log(5)^2 + 4)))*exp(5/(log(4*log(5) + log(5) 
^2 + 4) - 1))*log(x))/(4*log(5) + log(5)^2 + 4)^(5/(log(4*log(5) + log(5)^ 
2 + 4) - 1))
 

Reduce [B] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.30 \[ \int \frac {e^{\frac {60-5 x \log \left (\frac {e}{4+4 \log (5)+\log ^2(5)}\right )}{x \log \left (\frac {e}{4+4 \log (5)+\log ^2(5)}\right )}} \left (-60 \log (x)+x \log \left (\frac {e}{4+4 \log (5)+\log ^2(5)}\right )\right )}{x^2 \log \left (\frac {e}{4+4 \log (5)+\log ^2(5)}\right )} \, dx=\frac {e^{\frac {60}{\mathrm {log}\left (\frac {e}{\mathrm {log}\left (5\right )^{2}+4 \,\mathrm {log}\left (5\right )+4}\right ) x}} \mathrm {log}\left (x \right )}{e^{5}} \] Input:

int((-60*log(x)+x*log(exp(1)/(log(5)^2+4*log(5)+4)))*exp((-5*x*log(exp(1)/ 
(log(5)^2+4*log(5)+4))+60)/x/log(exp(1)/(log(5)^2+4*log(5)+4)))/x^2/log(ex 
p(1)/(log(5)^2+4*log(5)+4)),x)
 

Output:

(e**(60/(log(e/(log(5)**2 + 4*log(5) + 4))*x))*log(x))/e**5