\(\int \frac {e^{-\frac {5-e^x+x+\log (4)}{x}} \log ^{-1+e^{-\frac {5-e^x+x+\log (4)}{x}}}(x) (x+(5+e^x (-1+x)+\log (4)) \log (x) \log (\log (x)))}{x^2} \, dx\) [1847]

Optimal result

Integrand size = 62, antiderivative size = 20 \[ \int \frac {e^{-\frac {5-e^x+x+\log (4)}{x}} \log ^{-1+e^{-\frac {5-e^x+x+\log (4)}{x}}}(x) \left (x+\left (5+e^x (-1+x)+\log (4)\right ) \log (x) \log (\log (x))\right )}{x^2} \, dx=\log ^{e^{-\frac {5-e^x+x+\log (4)}{x}}}(x) \] Output:

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

Mathematica [A] (verified)

Time = 6.13 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.25 \[ \int \frac {e^{-\frac {5-e^x+x+\log (4)}{x}} \log ^{-1+e^{-\frac {5-e^x+x+\log (4)}{x}}}(x) \left (x+\left (5+e^x (-1+x)+\log (4)\right ) \log (x) \log (\log (x))\right )}{x^2} \, dx=\log ^{4^{-1/x} e^{\frac {-5+e^x-x}{x}}}(x) \] Input:

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

Output:

Log[x]^(E^((-5 + E^x - x)/x)/4^x^(-1))
 

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.

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

\(\Big \downarrow \) 7292

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

\(\Big \downarrow \) 7293

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

\(\Big \downarrow \) 2009

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

Input:

Int[(Log[x]^(-1 + E^(-((5 - E^x + x + Log[4])/x)))*(x + (5 + E^x*(-1 + x) 
+ Log[4])*Log[x]*Log[Log[x]]))/(E^((5 - E^x + x + Log[4])/x)*x^2),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 0.15 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.20

\[\ln \left (x \right )^{4^{-\frac {1}{x}} {\mathrm e}^{\frac {{\mathrm e}^{x}-5-x}{x}}}\]

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-\frac {5-e^x+x+\log (4)}{x}} \log ^{-1+e^{-\frac {5-e^x+x+\log (4)}{x}}}(x) \left (x+\left (5+e^x (-1+x)+\log (4)\right ) \log (x) \log (\log (x))\right )}{x^2} \, dx=\log \left (x\right )^{e^{\left (-\frac {x - e^{x} + 2 \, \log \left (2\right ) + 5}{x}\right )}} \] Input:

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.20 \[ \int \frac {e^{-\frac {5-e^x+x+\log (4)}{x}} \log ^{-1+e^{-\frac {5-e^x+x+\log (4)}{x}}}(x) \left (x+\left (5+e^x (-1+x)+\log (4)\right ) \log (x) \log (\log (x))\right )}{x^2} \, dx=\log \left (x\right )^{e^{\left (\frac {e^{x}}{x} - \frac {2 \, \log \left (2\right )}{x} - \frac {5}{x} - 1\right )}} \] Input:

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

Output:

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

Giac [F]

\[ \int \frac {e^{-\frac {5-e^x+x+\log (4)}{x}} \log ^{-1+e^{-\frac {5-e^x+x+\log (4)}{x}}}(x) \left (x+\left (5+e^x (-1+x)+\log (4)\right ) \log (x) \log (\log (x))\right )}{x^2} \, dx=\int { \frac {{\left ({\left ({\left (x - 1\right )} e^{x} + 2 \, \log \left (2\right ) + 5\right )} \log \left (x\right ) \log \left (\log \left (x\right )\right ) + x\right )} \log \left (x\right )^{e^{\left (-\frac {x - e^{x} + 2 \, \log \left (2\right ) + 5}{x}\right )}} e^{\left (-\frac {x - e^{x} + 2 \, \log \left (2\right ) + 5}{x}\right )}}{x^{2} \log \left (x\right )} \,d x } \] Input:

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

Output:

integrate((((x - 1)*e^x + 2*log(2) + 5)*log(x)*log(log(x)) + x)*log(x)^e^( 
-(x - e^x + 2*log(2) + 5)/x)*e^(-(x - e^x + 2*log(2) + 5)/x)/(x^2*log(x)), 
 x)
 

Mupad [B] (verification not implemented)

Time = 1.86 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.40 \[ \int \frac {e^{-\frac {5-e^x+x+\log (4)}{x}} \log ^{-1+e^{-\frac {5-e^x+x+\log (4)}{x}}}(x) \left (x+\left (5+e^x (-1+x)+\log (4)\right ) \log (x) \log (\log (x))\right )}{x^2} \, dx={\ln \left (x\right )}^{\frac {{\mathrm {e}}^{-1}\,{\mathrm {e}}^{\frac {{\mathrm {e}}^x}{x}}\,{\mathrm {e}}^{-\frac {5}{x}}}{2^{2/x}}} \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {e^{-\frac {5-e^x+x+\log (4)}{x}} \log ^{-1+e^{-\frac {5-e^x+x+\log (4)}{x}}}(x) \left (x+\left (5+e^x (-1+x)+\log (4)\right ) \log (x) \log (\log (x))\right )}{x^2} \, dx=\int \frac {\left (\left (\left (x -1\right ) {\mathrm e}^{x}+2 \,\mathrm {log}\left (2\right )+5\right ) \mathrm {log}\left (x \right ) \mathrm {log}\left (\mathrm {log}\left (x \right )\right )+x \right ) {\mathrm e}^{\frac {\mathrm {log}\left (\mathrm {log}\left (x \right )\right )}{{\mathrm e}^{\frac {-{\mathrm e}^{x}+2 \,\mathrm {log}\left (2\right )+5+x}{x}}}}}{x^{2} \mathrm {log}\left (x \right ) {\mathrm e}^{\frac {-{\mathrm e}^{x}+2 \,\mathrm {log}\left (2\right )+5+x}{x}}}d x \] Input:

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

Output:

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