\(\int \frac {e^{\frac {5}{3}+\frac {x-\log (x^2)}{x}} x+x \log (x)+(x \log (\frac {5}{x})+e^{\frac {x-\log (x^2)}{x}} (-2 e^{5/3} \log (\frac {5}{x})+e^{5/3} \log (\frac {5}{x}) \log (x^2))) \log (\log (\frac {5}{x}))}{x^2 \log (\frac {5}{x}) \log ^2(\log (\frac {5}{x}))} \, dx\) [604]

Optimal result

Integrand size = 108, antiderivative size = 31 \[ \int \frac {e^{\frac {5}{3}+\frac {x-\log \left (x^2\right )}{x}} x+x \log (x)+\left (x \log \left (\frac {5}{x}\right )+e^{\frac {x-\log \left (x^2\right )}{x}} \left (-2 e^{5/3} \log \left (\frac {5}{x}\right )+e^{5/3} \log \left (\frac {5}{x}\right ) \log \left (x^2\right )\right )\right ) \log \left (\log \left (\frac {5}{x}\right )\right )}{x^2 \log \left (\frac {5}{x}\right ) \log ^2\left (\log \left (\frac {5}{x}\right )\right )} \, dx=\frac {e^{\frac {5}{3}+\frac {x-\log \left (x^2\right )}{x}}+\log (x)}{\log \left (\log \left (\frac {5}{x}\right )\right )} \] Output:

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

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.90 \[ \int \frac {e^{\frac {5}{3}+\frac {x-\log \left (x^2\right )}{x}} x+x \log (x)+\left (x \log \left (\frac {5}{x}\right )+e^{\frac {x-\log \left (x^2\right )}{x}} \left (-2 e^{5/3} \log \left (\frac {5}{x}\right )+e^{5/3} \log \left (\frac {5}{x}\right ) \log \left (x^2\right )\right )\right ) \log \left (\log \left (\frac {5}{x}\right )\right )}{x^2 \log \left (\frac {5}{x}\right ) \log ^2\left (\log \left (\frac {5}{x}\right )\right )} \, dx=\frac {e^{8/3} \left (x^2\right )^{-1/x}+\log (x)}{\log \left (\log \left (\frac {5}{x}\right )\right )} \] Input:

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

Output:

(E^(8/3)/(x^2)^x^(-1) + Log[x])/Log[Log[5/x]]
 

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.

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

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (\frac {e^{8/3} \left (x^2\right )^{-\frac {1}{x}-1} \left (\log \left (\frac {5}{x}\right ) \log \left (x^2\right ) \log \left (\log \left (\frac {5}{x}\right )\right )+x-2 \log \left (\frac {5}{x}\right ) \log \left (\log \left (\frac {5}{x}\right )\right )\right )}{\log \left (\frac {5}{x}\right ) \log ^2\left (\log \left (\frac {5}{x}\right )\right )}+\frac {\log (x)+\log \left (\frac {5}{x}\right ) \log \left (\log \left (\frac {5}{x}\right )\right )}{x \log \left (\frac {5}{x}\right ) \log ^2\left (\log \left (\frac {5}{x}\right )\right )}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle e^{8/3} \int \frac {x \left (x^2\right )^{-1-\frac {1}{x}}}{\log \left (\frac {5}{x}\right ) \log ^2\left (\log \left (\frac {5}{x}\right )\right )}dx-2 e^{8/3} \int \frac {\left (x^2\right )^{-1-\frac {1}{x}}}{\log \left (\log \left (\frac {5}{x}\right )\right )}dx+e^{8/3} \int \frac {\left (x^2\right )^{-1-\frac {1}{x}} \log \left (x^2\right )}{\log \left (\log \left (\frac {5}{x}\right )\right )}dx+\int \frac {\log (x)}{x \log \left (\frac {5}{x}\right ) \log ^2\left (\log \left (\frac {5}{x}\right )\right )}dx-\operatorname {LogIntegral}\left (\log \left (\frac {5}{x}\right )\right )\)

Input:

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

Output:

$Aborted
 

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 1.14 (sec) , antiderivative size = 82, normalized size of antiderivative = 2.65

\[\frac {x^{-\frac {2}{x}} {\mathrm e}^{\frac {3 i \pi \operatorname {csgn}\left (i x^{2}\right )^{3}-6 i \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2}+3 i \pi \,\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x \right )^{2}+16 x}{6 x}}+\ln \left (x \right )}{\ln \left (\ln \left (5\right )-\ln \left (x \right )\right )}\]

Input:

int((((exp(5/3)*ln(5/x)*ln(x^2)-2*exp(5/3)*ln(5/x))*exp((-ln(x^2)+x)/x)+x* 
ln(5/x))*ln(ln(5/x))+x*exp(5/3)*exp((-ln(x^2)+x)/x)+x*ln(x))/x^2/ln(5/x)/l 
n(ln(5/x))^2,x)
 

Output:

(x^(-2/x)*exp(1/6*(3*I*Pi*csgn(I*x^2)^3-6*I*Pi*csgn(I*x)*csgn(I*x^2)^2+3*I 
*Pi*csgn(I*x^2)*csgn(I*x)^2+16*x)/x)+ln(x))/ln(ln(5)-ln(x))
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.39 \[ \int \frac {e^{\frac {5}{3}+\frac {x-\log \left (x^2\right )}{x}} x+x \log (x)+\left (x \log \left (\frac {5}{x}\right )+e^{\frac {x-\log \left (x^2\right )}{x}} \left (-2 e^{5/3} \log \left (\frac {5}{x}\right )+e^{5/3} \log \left (\frac {5}{x}\right ) \log \left (x^2\right )\right )\right ) \log \left (\log \left (\frac {5}{x}\right )\right )}{x^2 \log \left (\frac {5}{x}\right ) \log ^2\left (\log \left (\frac {5}{x}\right )\right )} \, dx=\frac {e^{\left (\frac {2 \, {\left (4 \, x - 3 \, \log \left (5\right ) + 3 \, \log \left (\frac {5}{x}\right )\right )}}{3 \, x}\right )} + \log \left (5\right ) - \log \left (\frac {5}{x}\right )}{\log \left (\log \left (\frac {5}{x}\right )\right )} \] Input:

integrate((((exp(5/3)*log(5/x)*log(x^2)-2*exp(5/3)*log(5/x))*exp((-log(x^2 
)+x)/x)+x*log(5/x))*log(log(5/x))+x*exp(5/3)*exp((-log(x^2)+x)/x)+x*log(x) 
)/x^2/log(5/x)/log(log(5/x))^2,x, algorithm="fricas")
 

Output:

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

Sympy [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.10 \[ \int \frac {e^{\frac {5}{3}+\frac {x-\log \left (x^2\right )}{x}} x+x \log (x)+\left (x \log \left (\frac {5}{x}\right )+e^{\frac {x-\log \left (x^2\right )}{x}} \left (-2 e^{5/3} \log \left (\frac {5}{x}\right )+e^{5/3} \log \left (\frac {5}{x}\right ) \log \left (x^2\right )\right )\right ) \log \left (\log \left (\frac {5}{x}\right )\right )}{x^2 \log \left (\frac {5}{x}\right ) \log ^2\left (\log \left (\frac {5}{x}\right )\right )} \, dx=\frac {e^{\frac {5}{3}} e^{\frac {x - 2 \log {\left (x \right )}}{x}}}{\log {\left (- \log {\left (x \right )} + \log {\left (5 \right )} \right )}} + \frac {\log {\left (x \right )}}{\log {\left (- \log {\left (x \right )} + \log {\left (5 \right )} \right )}} \] Input:

integrate((((exp(5/3)*ln(5/x)*ln(x**2)-2*exp(5/3)*ln(5/x))*exp((-ln(x**2)+ 
x)/x)+x*ln(5/x))*ln(ln(5/x))+x*exp(5/3)*exp((-ln(x**2)+x)/x)+x*ln(x))/x**2 
/ln(5/x)/ln(ln(5/x))**2,x)
 

Output:

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

Maxima [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.06 \[ \int \frac {e^{\frac {5}{3}+\frac {x-\log \left (x^2\right )}{x}} x+x \log (x)+\left (x \log \left (\frac {5}{x}\right )+e^{\frac {x-\log \left (x^2\right )}{x}} \left (-2 e^{5/3} \log \left (\frac {5}{x}\right )+e^{5/3} \log \left (\frac {5}{x}\right ) \log \left (x^2\right )\right )\right ) \log \left (\log \left (\frac {5}{x}\right )\right )}{x^2 \log \left (\frac {5}{x}\right ) \log ^2\left (\log \left (\frac {5}{x}\right )\right )} \, dx=\frac {x^{\frac {2}{x}} \log \left (x\right ) + e^{\frac {8}{3}}}{x^{\frac {2}{x}} \log \left (\log \left (5\right ) - \log \left (x\right )\right )} \] Input:

integrate((((exp(5/3)*log(5/x)*log(x^2)-2*exp(5/3)*log(5/x))*exp((-log(x^2 
)+x)/x)+x*log(5/x))*log(log(5/x))+x*exp(5/3)*exp((-log(x^2)+x)/x)+x*log(x) 
)/x^2/log(5/x)/log(log(5/x))^2,x, algorithm="maxima")
 

Output:

(x^(2/x)*log(x) + e^(8/3))/(x^(2/x)*log(log(5) - log(x)))
 

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.16 \[ \int \frac {e^{\frac {5}{3}+\frac {x-\log \left (x^2\right )}{x}} x+x \log (x)+\left (x \log \left (\frac {5}{x}\right )+e^{\frac {x-\log \left (x^2\right )}{x}} \left (-2 e^{5/3} \log \left (\frac {5}{x}\right )+e^{5/3} \log \left (\frac {5}{x}\right ) \log \left (x^2\right )\right )\right ) \log \left (\log \left (\frac {5}{x}\right )\right )}{x^2 \log \left (\frac {5}{x}\right ) \log ^2\left (\log \left (\frac {5}{x}\right )\right )} \, dx=\frac {\log \left (x\right )}{\log \left (\log \left (5\right ) - \log \left (x\right )\right )} + \frac {e^{\frac {8}{3}}}{x^{\frac {2}{x}} \log \left (\log \left (5\right ) - \log \left (x\right )\right )} \] Input:

integrate((((exp(5/3)*log(5/x)*log(x^2)-2*exp(5/3)*log(5/x))*exp((-log(x^2 
)+x)/x)+x*log(5/x))*log(log(5/x))+x*exp(5/3)*exp((-log(x^2)+x)/x)+x*log(x) 
)/x^2/log(5/x)/log(log(5/x))^2,x, algorithm="giac")
 

Output:

log(x)/log(log(5) - log(x)) + e^(8/3)/(x^(2/x)*log(log(5) - log(x)))
 

Mupad [B] (verification not implemented)

Time = 0.70 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.35 \[ \int \frac {e^{\frac {5}{3}+\frac {x-\log \left (x^2\right )}{x}} x+x \log (x)+\left (x \log \left (\frac {5}{x}\right )+e^{\frac {x-\log \left (x^2\right )}{x}} \left (-2 e^{5/3} \log \left (\frac {5}{x}\right )+e^{5/3} \log \left (\frac {5}{x}\right ) \log \left (x^2\right )\right )\right ) \log \left (\log \left (\frac {5}{x}\right )\right )}{x^2 \log \left (\frac {5}{x}\right ) \log ^2\left (\log \left (\frac {5}{x}\right )\right )} \, dx=\ln \left (\frac {1}{x}\right )+\ln \left (x\right )+\frac {\ln \left (x\right )}{\ln \left (\ln \left (\frac {1}{x}\right )+\ln \left (5\right )\right )}+\frac {{\mathrm {e}}^{8/3}}{\ln \left (\ln \left (\frac {1}{x}\right )+\ln \left (5\right )\right )\,{\left (x^2\right )}^{1/x}} \] Input:

int((log(log(5/x))*(x*log(5/x) - exp((x - log(x^2))/x)*(2*exp(5/3)*log(5/x 
) - log(x^2)*exp(5/3)*log(5/x))) + x*log(x) + x*exp((x - log(x^2))/x)*exp( 
5/3))/(x^2*log(log(5/x))^2*log(5/x)),x)
 

Output:

log(1/x) + log(x) + log(x)/log(log(1/x) + log(5)) + exp(8/3)/(log(log(1/x) 
 + log(5))*(x^2)^(1/x))
 

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.26 \[ \int \frac {e^{\frac {5}{3}+\frac {x-\log \left (x^2\right )}{x}} x+x \log (x)+\left (x \log \left (\frac {5}{x}\right )+e^{\frac {x-\log \left (x^2\right )}{x}} \left (-2 e^{5/3} \log \left (\frac {5}{x}\right )+e^{5/3} \log \left (\frac {5}{x}\right ) \log \left (x^2\right )\right )\right ) \log \left (\log \left (\frac {5}{x}\right )\right )}{x^2 \log \left (\frac {5}{x}\right ) \log ^2\left (\log \left (\frac {5}{x}\right )\right )} \, dx=\frac {e^{\frac {\mathrm {log}\left (x^{2}\right )}{x}} \mathrm {log}\left (x \right )+e^{\frac {8}{3}}}{e^{\frac {\mathrm {log}\left (x^{2}\right )}{x}} \mathrm {log}\left (\mathrm {log}\left (\frac {5}{x}\right )\right )} \] Input:

int((((exp(5/3)*log(5/x)*log(x^2)-2*exp(5/3)*log(5/x))*exp((-log(x^2)+x)/x 
)+x*log(5/x))*log(log(5/x))+x*exp(5/3)*exp((-log(x^2)+x)/x)+x*log(x))/x^2/ 
log(5/x)/log(log(5/x))^2,x)
 

Output:

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