\(\int \frac {e^{\frac {-x+\log (x)}{(5+e^x) \log (\log (25 x^2))}} (10 x \log (2)+2 e^x x \log (2)+(-10 \log (2)-2 e^x \log (2)) \log (x)+((5-5 x) \log (2)+e^x (1-x+x^2) \log (2)-e^x x \log (2) \log (x)) \log (25 x^2) \log (\log (25 x^2)))}{(25 x+10 e^x x+e^{2 x} x) \log (25 x^2) \log ^2(\log (25 x^2))} \, dx\) [566]

Optimal result

Integrand size = 138, antiderivative size = 30 \[ \int \frac {e^{\frac {-x+\log (x)}{\left (5+e^x\right ) \log \left (\log \left (25 x^2\right )\right )}} \left (10 x \log (2)+2 e^x x \log (2)+\left (-10 \log (2)-2 e^x \log (2)\right ) \log (x)+\left ((5-5 x) \log (2)+e^x \left (1-x+x^2\right ) \log (2)-e^x x \log (2) \log (x)\right ) \log \left (25 x^2\right ) \log \left (\log \left (25 x^2\right )\right )\right )}{\left (25 x+10 e^x x+e^{2 x} x\right ) \log \left (25 x^2\right ) \log ^2\left (\log \left (25 x^2\right )\right )} \, dx=\left (-5+e^{\frac {-x+\log (x)}{\left (5+e^x\right ) \log \left (\log \left (25 x^2\right )\right )}}\right ) \log (2) \] Output:

(exp((ln(x)-x)/(exp(x)+5)/ln(ln(25*x^2)))-5)*ln(2)
 

Mathematica [F]

\[ \int \frac {e^{\frac {-x+\log (x)}{\left (5+e^x\right ) \log \left (\log \left (25 x^2\right )\right )}} \left (10 x \log (2)+2 e^x x \log (2)+\left (-10 \log (2)-2 e^x \log (2)\right ) \log (x)+\left ((5-5 x) \log (2)+e^x \left (1-x+x^2\right ) \log (2)-e^x x \log (2) \log (x)\right ) \log \left (25 x^2\right ) \log \left (\log \left (25 x^2\right )\right )\right )}{\left (25 x+10 e^x x+e^{2 x} x\right ) \log \left (25 x^2\right ) \log ^2\left (\log \left (25 x^2\right )\right )} \, dx=\int \frac {e^{\frac {-x+\log (x)}{\left (5+e^x\right ) \log \left (\log \left (25 x^2\right )\right )}} \left (10 x \log (2)+2 e^x x \log (2)+\left (-10 \log (2)-2 e^x \log (2)\right ) \log (x)+\left ((5-5 x) \log (2)+e^x \left (1-x+x^2\right ) \log (2)-e^x x \log (2) \log (x)\right ) \log \left (25 x^2\right ) \log \left (\log \left (25 x^2\right )\right )\right )}{\left (25 x+10 e^x x+e^{2 x} x\right ) \log \left (25 x^2\right ) \log ^2\left (\log \left (25 x^2\right )\right )} \, dx \] Input:

Integrate[(E^((-x + Log[x])/((5 + E^x)*Log[Log[25*x^2]]))*(10*x*Log[2] + 2 
*E^x*x*Log[2] + (-10*Log[2] - 2*E^x*Log[2])*Log[x] + ((5 - 5*x)*Log[2] + E 
^x*(1 - x + x^2)*Log[2] - E^x*x*Log[2]*Log[x])*Log[25*x^2]*Log[Log[25*x^2] 
]))/((25*x + 10*E^x*x + E^(2*x)*x)*Log[25*x^2]*Log[Log[25*x^2]]^2),x]
 

Output:

Integrate[(E^((-x + Log[x])/((5 + E^x)*Log[Log[25*x^2]]))*(10*x*Log[2] + 2 
*E^x*x*Log[2] + (-10*Log[2] - 2*E^x*Log[2])*Log[x] + ((5 - 5*x)*Log[2] + E 
^x*(1 - x + x^2)*Log[2] - E^x*x*Log[2]*Log[x])*Log[25*x^2]*Log[Log[25*x^2] 
]))/((25*x + 10*E^x*x + E^(2*x)*x)*Log[25*x^2]*Log[Log[25*x^2]]^2), 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.

\(\displaystyle \int \frac {e^{\frac {\log (x)-x}{\left (e^x+5\right ) \log \left (\log \left (25 x^2\right )\right )}} \left (\left (e^x \left (x^2-x+1\right ) \log (2)+(5-5 x) \log (2)-e^x x \log (2) \log (x)\right ) \log \left (25 x^2\right ) \log \left (\log \left (25 x^2\right )\right )+2 e^x x \log (2)+10 x \log (2)+\left (-2 e^x \log (2)-10 \log (2)\right ) \log (x)\right )}{\left (10 e^x x+e^{2 x} x+25 x\right ) \log \left (25 x^2\right ) \log ^2\left (\log \left (25 x^2\right )\right )} \, dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {e^{\frac {\log (x)-x}{\left (e^x+5\right ) \log \left (\log \left (25 x^2\right )\right )}} \left (\left (e^x \left (x^2-x+1\right ) \log (2)+(5-5 x) \log (2)-e^x x \log (2) \log (x)\right ) \log \left (25 x^2\right ) \log \left (\log \left (25 x^2\right )\right )+2 e^x x \log (2)+10 x \log (2)+\left (-2 e^x \log (2)-10 \log (2)\right ) \log (x)\right )}{\left (e^x+5\right )^2 x \log \left (25 x^2\right ) \log ^2\left (\log \left (25 x^2\right )\right )}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (\frac {\log (2) e^{\frac {\log (x)-x}{\left (e^x+5\right ) \log \left (\log \left (25 x^2\right )\right )}} \left (x^2 \log \left (25 x^2\right ) \log \left (\log \left (25 x^2\right )\right )-x \log (x) \log \left (25 x^2\right ) \log \left (\log \left (25 x^2\right )\right )-x \log \left (25 x^2\right ) \log \left (\log \left (25 x^2\right )\right )+\log \left (25 x^2\right ) \log \left (\log \left (25 x^2\right )\right )+2 x-2 \log (x)\right )}{\left (e^x+5\right ) x \log \left (25 x^2\right ) \log ^2\left (\log \left (25 x^2\right )\right )}-\frac {5 \log (2) e^{\frac {\log (x)-x}{\left (e^x+5\right ) \log \left (\log \left (25 x^2\right )\right )}} (x-\log (x))}{\left (e^x+5\right )^2 \log \left (\log \left (25 x^2\right )\right )}\right )dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (\frac {\log (2) e^{-\frac {x}{\left (e^x+5\right ) \log \left (\log \left (25 x^2\right )\right )}} x^{\frac {1}{\left (e^x+5\right ) \log \left (\log \left (25 x^2\right )\right )}-1} \left (x^2 \log \left (25 x^2\right ) \log \left (\log \left (25 x^2\right )\right )-x \log (x) \log \left (25 x^2\right ) \log \left (\log \left (25 x^2\right )\right )-x \log \left (25 x^2\right ) \log \left (\log \left (25 x^2\right )\right )+\log \left (25 x^2\right ) \log \left (\log \left (25 x^2\right )\right )+2 x-2 \log (x)\right )}{\left (e^x+5\right ) \log \left (25 x^2\right ) \log ^2\left (\log \left (25 x^2\right )\right )}-\frac {5 \log (2) e^{-\frac {x}{\left (e^x+5\right ) \log \left (\log \left (25 x^2\right )\right )}} x^{\frac {1}{\left (e^x+5\right ) \log \left (\log \left (25 x^2\right )\right )}} (x-\log (x))}{\left (e^x+5\right )^2 \log \left (\log \left (25 x^2\right )\right )}\right )dx\)

\(\Big \downarrow \) 7299

\(\displaystyle \int \left (\frac {\log (2) e^{-\frac {x}{\left (e^x+5\right ) \log \left (\log \left (25 x^2\right )\right )}} x^{\frac {1}{\left (e^x+5\right ) \log \left (\log \left (25 x^2\right )\right )}-1} \left (x^2 \log \left (25 x^2\right ) \log \left (\log \left (25 x^2\right )\right )-x \log (x) \log \left (25 x^2\right ) \log \left (\log \left (25 x^2\right )\right )-x \log \left (25 x^2\right ) \log \left (\log \left (25 x^2\right )\right )+\log \left (25 x^2\right ) \log \left (\log \left (25 x^2\right )\right )+2 x-2 \log (x)\right )}{\left (e^x+5\right ) \log \left (25 x^2\right ) \log ^2\left (\log \left (25 x^2\right )\right )}-\frac {5 \log (2) e^{-\frac {x}{\left (e^x+5\right ) \log \left (\log \left (25 x^2\right )\right )}} x^{\frac {1}{\left (e^x+5\right ) \log \left (\log \left (25 x^2\right )\right )}} (x-\log (x))}{\left (e^x+5\right )^2 \log \left (\log \left (25 x^2\right )\right )}\right )dx\)

Input:

Int[(E^((-x + Log[x])/((5 + E^x)*Log[Log[25*x^2]]))*(10*x*Log[2] + 2*E^x*x 
*Log[2] + (-10*Log[2] - 2*E^x*Log[2])*Log[x] + ((5 - 5*x)*Log[2] + E^x*(1 
- x + x^2)*Log[2] - E^x*x*Log[2]*Log[x])*Log[25*x^2]*Log[Log[25*x^2]]))/(( 
25*x + 10*E^x*x + E^(2*x)*x)*Log[25*x^2]*Log[Log[25*x^2]]^2),x]
 

Output:

$Aborted
 

Maple [C] (warning: unable to verify)

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

Time = 0.37 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.93

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

Input:

int(((-x*ln(2)*exp(x)*ln(x)+(x^2-x+1)*ln(2)*exp(x)+(-5*x+5)*ln(2))*ln(25*x 
^2)*ln(ln(25*x^2))+(-2*exp(x)*ln(2)-10*ln(2))*ln(x)+2*x*ln(2)*exp(x)+10*x* 
ln(2))*exp((ln(x)-x)/(exp(x)+5)/ln(ln(25*x^2)))/(x*exp(x)^2+10*exp(x)*x+25 
*x)/ln(25*x^2)/ln(ln(25*x^2))^2,x)
 

Output:

ln(2)*exp((ln(x)-x)/(exp(x)+5)/ln(2*ln(5)+2*ln(x)-1/2*I*Pi*csgn(I*x^2)*(-c 
sgn(I*x^2)+csgn(I*x))^2))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {-x+\log (x)}{\left (5+e^x\right ) \log \left (\log \left (25 x^2\right )\right )}} \left (10 x \log (2)+2 e^x x \log (2)+\left (-10 \log (2)-2 e^x \log (2)\right ) \log (x)+\left ((5-5 x) \log (2)+e^x \left (1-x+x^2\right ) \log (2)-e^x x \log (2) \log (x)\right ) \log \left (25 x^2\right ) \log \left (\log \left (25 x^2\right )\right )\right )}{\left (25 x+10 e^x x+e^{2 x} x\right ) \log \left (25 x^2\right ) \log ^2\left (\log \left (25 x^2\right )\right )} \, dx=e^{\left (-\frac {x - \log \left (x\right )}{{\left (e^{x} + 5\right )} \log \left (2 \, \log \left (5\right ) + 2 \, \log \left (x\right )\right )}\right )} \log \left (2\right ) \] Input:

integrate(((-x*log(2)*exp(x)*log(x)+(x^2-x+1)*log(2)*exp(x)+(-5*x+5)*log(2 
))*log(25*x^2)*log(log(25*x^2))+(-2*exp(x)*log(2)-10*log(2))*log(x)+2*x*lo 
g(2)*exp(x)+10*x*log(2))*exp((log(x)-x)/(exp(x)+5)/log(log(25*x^2)))/(x*ex 
p(x)^2+10*exp(x)*x+25*x)/log(25*x^2)/log(log(25*x^2))^2,x, algorithm="fric 
as")
 

Output:

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

Sympy [A] (verification not implemented)

Time = 27.84 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \[ \int \frac {e^{\frac {-x+\log (x)}{\left (5+e^x\right ) \log \left (\log \left (25 x^2\right )\right )}} \left (10 x \log (2)+2 e^x x \log (2)+\left (-10 \log (2)-2 e^x \log (2)\right ) \log (x)+\left ((5-5 x) \log (2)+e^x \left (1-x+x^2\right ) \log (2)-e^x x \log (2) \log (x)\right ) \log \left (25 x^2\right ) \log \left (\log \left (25 x^2\right )\right )\right )}{\left (25 x+10 e^x x+e^{2 x} x\right ) \log \left (25 x^2\right ) \log ^2\left (\log \left (25 x^2\right )\right )} \, dx=e^{\frac {- x + \log {\left (x \right )}}{\left (e^{x} + 5\right ) \log {\left (2 \log {\left (x \right )} + \log {\left (25 \right )} \right )}}} \log {\left (2 \right )} \] Input:

integrate(((-x*ln(2)*exp(x)*ln(x)+(x**2-x+1)*ln(2)*exp(x)+(-5*x+5)*ln(2))* 
ln(25*x**2)*ln(ln(25*x**2))+(-2*exp(x)*ln(2)-10*ln(2))*ln(x)+2*x*ln(2)*exp 
(x)+10*x*ln(2))*exp((ln(x)-x)/(exp(x)+5)/ln(ln(25*x**2)))/(x*exp(x)**2+10* 
exp(x)*x+25*x)/ln(25*x**2)/ln(ln(25*x**2))**2,x)
 

Output:

exp((-x + log(x))/((exp(x) + 5)*log(2*log(x) + log(25))))*log(2)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {e^{\frac {-x+\log (x)}{\left (5+e^x\right ) \log \left (\log \left (25 x^2\right )\right )}} \left (10 x \log (2)+2 e^x x \log (2)+\left (-10 \log (2)-2 e^x \log (2)\right ) \log (x)+\left ((5-5 x) \log (2)+e^x \left (1-x+x^2\right ) \log (2)-e^x x \log (2) \log (x)\right ) \log \left (25 x^2\right ) \log \left (\log \left (25 x^2\right )\right )\right )}{\left (25 x+10 e^x x+e^{2 x} x\right ) \log \left (25 x^2\right ) \log ^2\left (\log \left (25 x^2\right )\right )} \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate(((-x*log(2)*exp(x)*log(x)+(x^2-x+1)*log(2)*exp(x)+(-5*x+5)*log(2 
))*log(25*x^2)*log(log(25*x^2))+(-2*exp(x)*log(2)-10*log(2))*log(x)+2*x*lo 
g(2)*exp(x)+10*x*log(2))*exp((log(x)-x)/(exp(x)+5)/log(log(25*x^2)))/(x*ex 
p(x)^2+10*exp(x)*x+25*x)/log(25*x^2)/log(log(25*x^2))^2,x, algorithm="maxi 
ma")
 

Output:

Exception raised: RuntimeError >> ECL says: In function CAR, the value of 
the first argument is  0which is not of the expected type LIST
 

Giac [F]

\[ \int \frac {e^{\frac {-x+\log (x)}{\left (5+e^x\right ) \log \left (\log \left (25 x^2\right )\right )}} \left (10 x \log (2)+2 e^x x \log (2)+\left (-10 \log (2)-2 e^x \log (2)\right ) \log (x)+\left ((5-5 x) \log (2)+e^x \left (1-x+x^2\right ) \log (2)-e^x x \log (2) \log (x)\right ) \log \left (25 x^2\right ) \log \left (\log \left (25 x^2\right )\right )\right )}{\left (25 x+10 e^x x+e^{2 x} x\right ) \log \left (25 x^2\right ) \log ^2\left (\log \left (25 x^2\right )\right )} \, dx=\int { \frac {{\left (2 \, x e^{x} \log \left (2\right ) - {\left (x e^{x} \log \left (2\right ) \log \left (x\right ) - {\left (x^{2} - x + 1\right )} e^{x} \log \left (2\right ) + 5 \, {\left (x - 1\right )} \log \left (2\right )\right )} \log \left (25 \, x^{2}\right ) \log \left (\log \left (25 \, x^{2}\right )\right ) + 10 \, x \log \left (2\right ) - 2 \, {\left (e^{x} \log \left (2\right ) + 5 \, \log \left (2\right )\right )} \log \left (x\right )\right )} e^{\left (-\frac {x - \log \left (x\right )}{{\left (e^{x} + 5\right )} \log \left (\log \left (25 \, x^{2}\right )\right )}\right )}}{{\left (x e^{\left (2 \, x\right )} + 10 \, x e^{x} + 25 \, x\right )} \log \left (25 \, x^{2}\right ) \log \left (\log \left (25 \, x^{2}\right )\right )^{2}} \,d x } \] Input:

integrate(((-x*log(2)*exp(x)*log(x)+(x^2-x+1)*log(2)*exp(x)+(-5*x+5)*log(2 
))*log(25*x^2)*log(log(25*x^2))+(-2*exp(x)*log(2)-10*log(2))*log(x)+2*x*lo 
g(2)*exp(x)+10*x*log(2))*exp((log(x)-x)/(exp(x)+5)/log(log(25*x^2)))/(x*ex 
p(x)^2+10*exp(x)*x+25*x)/log(25*x^2)/log(log(25*x^2))^2,x, algorithm="giac 
")
 

Output:

undef
 

Mupad [B] (verification not implemented)

Time = 4.56 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.77 \[ \int \frac {e^{\frac {-x+\log (x)}{\left (5+e^x\right ) \log \left (\log \left (25 x^2\right )\right )}} \left (10 x \log (2)+2 e^x x \log (2)+\left (-10 \log (2)-2 e^x \log (2)\right ) \log (x)+\left ((5-5 x) \log (2)+e^x \left (1-x+x^2\right ) \log (2)-e^x x \log (2) \log (x)\right ) \log \left (25 x^2\right ) \log \left (\log \left (25 x^2\right )\right )\right )}{\left (25 x+10 e^x x+e^{2 x} x\right ) \log \left (25 x^2\right ) \log ^2\left (\log \left (25 x^2\right )\right )} \, dx=x^{\frac {1}{5\,\ln \left (\ln \left (25\,x^2\right )\right )+\ln \left (\ln \left (25\,x^2\right )\right )\,{\mathrm {e}}^x}}\,{\mathrm {e}}^{-\frac {x}{5\,\ln \left (\ln \left (25\,x^2\right )\right )+\ln \left (\ln \left (25\,x^2\right )\right )\,{\mathrm {e}}^x}}\,\ln \left (2\right ) \] Input:

int(-(exp(-(x - log(x))/(log(log(25*x^2))*(exp(x) + 5)))*(log(x)*(10*log(2 
) + 2*exp(x)*log(2)) - 10*x*log(2) - 2*x*exp(x)*log(2) + log(log(25*x^2))* 
log(25*x^2)*(log(2)*(5*x - 5) - exp(x)*log(2)*(x^2 - x + 1) + x*exp(x)*log 
(2)*log(x))))/(log(log(25*x^2))^2*log(25*x^2)*(25*x + x*exp(2*x) + 10*x*ex 
p(x))),x)
 

Output:

x^(1/(5*log(log(25*x^2)) + log(log(25*x^2))*exp(x)))*exp(-x/(5*log(log(25* 
x^2)) + log(log(25*x^2))*exp(x)))*log(2)
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 60, normalized size of antiderivative = 2.00 \[ \int \frac {e^{\frac {-x+\log (x)}{\left (5+e^x\right ) \log \left (\log \left (25 x^2\right )\right )}} \left (10 x \log (2)+2 e^x x \log (2)+\left (-10 \log (2)-2 e^x \log (2)\right ) \log (x)+\left ((5-5 x) \log (2)+e^x \left (1-x+x^2\right ) \log (2)-e^x x \log (2) \log (x)\right ) \log \left (25 x^2\right ) \log \left (\log \left (25 x^2\right )\right )\right )}{\left (25 x+10 e^x x+e^{2 x} x\right ) \log \left (25 x^2\right ) \log ^2\left (\log \left (25 x^2\right )\right )} \, dx=\frac {e^{\frac {\mathrm {log}\left (x \right )}{e^{x} \mathrm {log}\left (\mathrm {log}\left (25 x^{2}\right )\right )+5 \,\mathrm {log}\left (\mathrm {log}\left (25 x^{2}\right )\right )}} \mathrm {log}\left (2\right )}{e^{\frac {x}{e^{x} \mathrm {log}\left (\mathrm {log}\left (25 x^{2}\right )\right )+5 \,\mathrm {log}\left (\mathrm {log}\left (25 x^{2}\right )\right )}}} \] Input:

int(((-x*log(2)*exp(x)*log(x)+(x^2-x+1)*log(2)*exp(x)+(-5*x+5)*log(2))*log 
(25*x^2)*log(log(25*x^2))+(-2*exp(x)*log(2)-10*log(2))*log(x)+2*x*log(2)*e 
xp(x)+10*x*log(2))*exp((log(x)-x)/(exp(x)+5)/log(log(25*x^2)))/(x*exp(x)^2 
+10*exp(x)*x+25*x)/log(25*x^2)/log(log(25*x^2))^2,x)
 

Output:

(e**(log(x)/(e**x*log(log(25*x**2)) + 5*log(log(25*x**2))))*log(2))/e**(x/ 
(e**x*log(log(25*x**2)) + 5*log(log(25*x**2))))