\(\int \frac {e^{e^x} (5-6 x+x^2) \log (16)+(-5+6 x-x^2) \log (25-10 x+x^2)+(2 x+e^{e^x+x} (5 x-x^2) \log (16)) \log (e^{-x} x) \log (\log (e^{-x} x))}{(-5 x+x^2) \log (e^{-x} x)} \, dx\) [1635]

Optimal result

Integrand size = 98, antiderivative size = 26 \[ \int \frac {e^{e^x} \left (5-6 x+x^2\right ) \log (16)+\left (-5+6 x-x^2\right ) \log \left (25-10 x+x^2\right )+\left (2 x+e^{e^x+x} \left (5 x-x^2\right ) \log (16)\right ) \log \left (e^{-x} x\right ) \log \left (\log \left (e^{-x} x\right )\right )}{\left (-5 x+x^2\right ) \log \left (e^{-x} x\right )} \, dx=\left (-e^{e^x} \log (16)+\log \left ((-5+x)^2\right )\right ) \log \left (\log \left (e^{-x} x\right )\right ) \] Output:

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

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {e^{e^x} \left (5-6 x+x^2\right ) \log (16)+\left (-5+6 x-x^2\right ) \log \left (25-10 x+x^2\right )+\left (2 x+e^{e^x+x} \left (5 x-x^2\right ) \log (16)\right ) \log \left (e^{-x} x\right ) \log \left (\log \left (e^{-x} x\right )\right )}{\left (-5 x+x^2\right ) \log \left (e^{-x} x\right )} \, dx=\left (-e^{e^x} \log (16)+\log \left ((-5+x)^2\right )\right ) \log \left (\log \left (e^{-x} x\right )\right ) \] Input:

Integrate[(E^E^x*(5 - 6*x + x^2)*Log[16] + (-5 + 6*x - x^2)*Log[25 - 10*x 
+ x^2] + (2*x + E^(E^x + x)*(5*x - x^2)*Log[16])*Log[x/E^x]*Log[Log[x/E^x] 
])/((-5*x + x^2)*Log[x/E^x]),x]
 

Output:

(-(E^E^x*Log[16]) + Log[(-5 + x)^2])*Log[Log[x/E^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.

\(\displaystyle \int \frac {e^{e^x} \left (x^2-6 x+5\right ) \log (16)+\left (-x^2+6 x-5\right ) \log \left (x^2-10 x+25\right )+\left (e^{x+e^x} \left (5 x-x^2\right ) \log (16)+2 x\right ) \log \left (e^{-x} x\right ) \log \left (\log \left (e^{-x} x\right )\right )}{\left (x^2-5 x\right ) \log \left (e^{-x} x\right )} \, dx\)

\(\Big \downarrow \) 2026

\(\displaystyle \int \frac {e^{e^x} \left (x^2-6 x+5\right ) \log (16)+\left (-x^2+6 x-5\right ) \log \left (x^2-10 x+25\right )+\left (e^{x+e^x} \left (5 x-x^2\right ) \log (16)+2 x\right ) \log \left (e^{-x} x\right ) \log \left (\log \left (e^{-x} x\right )\right )}{(x-5) x \log \left (e^{-x} x\right )}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (\frac {x^2 \left (-\log \left ((x-5)^2\right )\right )+e^{e^x} x^2 \log (16)+6 x \log \left ((x-5)^2\right )+2 x \log \left (e^{-x} x\right ) \log \left (\log \left (e^{-x} x\right )\right )-6 e^{e^x} x \log (16)-5 \log \left ((x-5)^2\right )+5 e^{e^x} \log (16)}{(x-5) x \log \left (e^{-x} x\right )}-e^{x+e^x} \log (16) \log \left (\log \left (e^{-x} x\right )\right )\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \log (16) \int \frac {e^{e^x}}{\log \left (e^{-x} x\right )}dx-\log (16) \int \frac {e^{e^x}}{x \log \left (e^{-x} x\right )}dx-\int \frac {\log \left ((x-5)^2\right )}{\log \left (e^{-x} x\right )}dx+\int \frac {\log \left ((x-5)^2\right )}{x \log \left (e^{-x} x\right )}dx-\log (16) \int e^{x+e^x} \log \left (\log \left (e^{-x} x\right )\right )dx+2 \int \frac {\log \left (\log \left (e^{-x} x\right )\right )}{x-5}dx\)

Input:

Int[(E^E^x*(5 - 6*x + x^2)*Log[16] + (-5 + 6*x - x^2)*Log[25 - 10*x + x^2] 
 + (2*x + E^(E^x + x)*(5*x - x^2)*Log[16])*Log[x/E^x]*Log[Log[x/E^x]])/((- 
5*x + x^2)*Log[x/E^x]),x]
 

Output:

$Aborted
 

Maple [C] (warning: unable to verify)

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

Time = 0.17 (sec) , antiderivative size = 220, normalized size of antiderivative = 8.46

\[\left (-4 \ln \left (2\right ) {\mathrm e}^{{\mathrm e}^{x}}+2 \ln \left (-5+x \right )\right ) \ln \left (\ln \left (x \right )-\ln \left ({\mathrm e}^{x}\right )-\frac {i \pi \,\operatorname {csgn}\left (i x \,{\mathrm e}^{-x}\right ) \left (-\operatorname {csgn}\left (i x \,{\mathrm e}^{-x}\right )+\operatorname {csgn}\left (i x \right )\right ) \left (-\operatorname {csgn}\left (i x \,{\mathrm e}^{-x}\right )+\operatorname {csgn}\left (i {\mathrm e}^{-x}\right )\right )}{2}\right )-\frac {i \pi \,\operatorname {csgn}\left (i \left (-5+x \right )^{2}\right ) \left (\operatorname {csgn}\left (i \left (-5+x \right )\right )^{2}-2 \,\operatorname {csgn}\left (i \left (-5+x \right )^{2}\right ) \operatorname {csgn}\left (i \left (-5+x \right )\right )+\operatorname {csgn}\left (i \left (-5+x \right )^{2}\right )^{2}\right ) \ln \left (\ln \left ({\mathrm e}^{x}\right )-\frac {i \left (\pi \,\operatorname {csgn}\left (i {\mathrm e}^{-x}\right ) \operatorname {csgn}\left (i x \,{\mathrm e}^{-x}\right )^{2}-\pi \,\operatorname {csgn}\left (i {\mathrm e}^{-x}\right ) \operatorname {csgn}\left (i x \,{\mathrm e}^{-x}\right ) \operatorname {csgn}\left (i x \right )-\pi \operatorname {csgn}\left (i x \,{\mathrm e}^{-x}\right )^{3}+\pi \operatorname {csgn}\left (i x \,{\mathrm e}^{-x}\right )^{2} \operatorname {csgn}\left (i x \right )-2 i \ln \left (x \right )\right )}{2}\right )}{2}\]

Input:

int(((4*(-x^2+5*x)*ln(2)*exp(x)*exp(exp(x))+2*x)*ln(x/exp(x))*ln(ln(x/exp( 
x)))+4*(x^2-6*x+5)*ln(2)*exp(exp(x))+(-x^2+6*x-5)*ln(x^2-10*x+25))/(x^2-5* 
x)/ln(x/exp(x)),x)
 

Output:

(-4*ln(2)*exp(exp(x))+2*ln(-5+x))*ln(ln(x)-ln(exp(x))-1/2*I*Pi*csgn(I*x*ex 
p(-x))*(-csgn(I*x*exp(-x))+csgn(I*x))*(-csgn(I*x*exp(-x))+csgn(I*exp(-x))) 
)-1/2*I*Pi*csgn(I*(-5+x)^2)*(csgn(I*(-5+x))^2-2*csgn(I*(-5+x)^2)*csgn(I*(- 
5+x))+csgn(I*(-5+x)^2)^2)*ln(ln(exp(x))-1/2*I*(Pi*csgn(I*exp(-x))*csgn(I*x 
*exp(-x))^2-Pi*csgn(I*exp(-x))*csgn(I*x*exp(-x))*csgn(I*x)-Pi*csgn(I*x*exp 
(-x))^3+Pi*csgn(I*x*exp(-x))^2*csgn(I*x)-2*I*ln(x)))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.42 \[ \int \frac {e^{e^x} \left (5-6 x+x^2\right ) \log (16)+\left (-5+6 x-x^2\right ) \log \left (25-10 x+x^2\right )+\left (2 x+e^{e^x+x} \left (5 x-x^2\right ) \log (16)\right ) \log \left (e^{-x} x\right ) \log \left (\log \left (e^{-x} x\right )\right )}{\left (-5 x+x^2\right ) \log \left (e^{-x} x\right )} \, dx=-{\left (4 \, e^{\left (x + e^{x}\right )} \log \left (2\right ) - e^{x} \log \left (x^{2} - 10 \, x + 25\right )\right )} e^{\left (-x\right )} \log \left (\log \left (x e^{\left (-x\right )}\right )\right ) \] Input:

integrate(((4*(-x^2+5*x)*log(2)*exp(x)*exp(exp(x))+2*x)*log(x/exp(x))*log( 
log(x/exp(x)))+4*(x^2-6*x+5)*log(2)*exp(exp(x))+(-x^2+6*x-5)*log(x^2-10*x+ 
25))/(x^2-5*x)/log(x/exp(x)),x, algorithm="fricas")
 

Output:

-(4*e^(x + e^x)*log(2) - e^x*log(x^2 - 10*x + 25))*e^(-x)*log(log(x*e^(-x) 
))
 

Sympy [F(-1)]

Timed out. \[ \int \frac {e^{e^x} \left (5-6 x+x^2\right ) \log (16)+\left (-5+6 x-x^2\right ) \log \left (25-10 x+x^2\right )+\left (2 x+e^{e^x+x} \left (5 x-x^2\right ) \log (16)\right ) \log \left (e^{-x} x\right ) \log \left (\log \left (e^{-x} x\right )\right )}{\left (-5 x+x^2\right ) \log \left (e^{-x} x\right )} \, dx=\text {Timed out} \] Input:

integrate(((4*(-x**2+5*x)*ln(2)*exp(x)*exp(exp(x))+2*x)*ln(x/exp(x))*ln(ln 
(x/exp(x)))+4*(x**2-6*x+5)*ln(2)*exp(exp(x))+(-x**2+6*x-5)*ln(x**2-10*x+25 
))/(x**2-5*x)/ln(x/exp(x)),x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.88 \[ \int \frac {e^{e^x} \left (5-6 x+x^2\right ) \log (16)+\left (-5+6 x-x^2\right ) \log \left (25-10 x+x^2\right )+\left (2 x+e^{e^x+x} \left (5 x-x^2\right ) \log (16)\right ) \log \left (e^{-x} x\right ) \log \left (\log \left (e^{-x} x\right )\right )}{\left (-5 x+x^2\right ) \log \left (e^{-x} x\right )} \, dx=-2 \, {\left (2 \, e^{\left (e^{x}\right )} \log \left (2\right ) - \log \left (x - 5\right )\right )} \log \left (-x + \log \left (x\right )\right ) \] Input:

integrate(((4*(-x^2+5*x)*log(2)*exp(x)*exp(exp(x))+2*x)*log(x/exp(x))*log( 
log(x/exp(x)))+4*(x^2-6*x+5)*log(2)*exp(exp(x))+(-x^2+6*x-5)*log(x^2-10*x+ 
25))/(x^2-5*x)/log(x/exp(x)),x, algorithm="maxima")
 

Output:

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

Giac [F]

\[ \int \frac {e^{e^x} \left (5-6 x+x^2\right ) \log (16)+\left (-5+6 x-x^2\right ) \log \left (25-10 x+x^2\right )+\left (2 x+e^{e^x+x} \left (5 x-x^2\right ) \log (16)\right ) \log \left (e^{-x} x\right ) \log \left (\log \left (e^{-x} x\right )\right )}{\left (-5 x+x^2\right ) \log \left (e^{-x} x\right )} \, dx=\int { \frac {4 \, {\left (x^{2} - 6 \, x + 5\right )} e^{\left (e^{x}\right )} \log \left (2\right ) - 2 \, {\left (2 \, {\left (x^{2} - 5 \, x\right )} e^{\left (x + e^{x}\right )} \log \left (2\right ) - x\right )} \log \left (x e^{\left (-x\right )}\right ) \log \left (\log \left (x e^{\left (-x\right )}\right )\right ) - {\left (x^{2} - 6 \, x + 5\right )} \log \left (x^{2} - 10 \, x + 25\right )}{{\left (x^{2} - 5 \, x\right )} \log \left (x e^{\left (-x\right )}\right )} \,d x } \] Input:

integrate(((4*(-x^2+5*x)*log(2)*exp(x)*exp(exp(x))+2*x)*log(x/exp(x))*log( 
log(x/exp(x)))+4*(x^2-6*x+5)*log(2)*exp(exp(x))+(-x^2+6*x-5)*log(x^2-10*x+ 
25))/(x^2-5*x)/log(x/exp(x)),x, algorithm="giac")
 

Output:

integrate((4*(x^2 - 6*x + 5)*e^(e^x)*log(2) - 2*(2*(x^2 - 5*x)*e^(x + e^x) 
*log(2) - x)*log(x*e^(-x))*log(log(x*e^(-x))) - (x^2 - 6*x + 5)*log(x^2 - 
10*x + 25))/((x^2 - 5*x)*log(x*e^(-x))), x)
 

Mupad [B] (verification not implemented)

Time = 2.16 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.23 \[ \int \frac {e^{e^x} \left (5-6 x+x^2\right ) \log (16)+\left (-5+6 x-x^2\right ) \log \left (25-10 x+x^2\right )+\left (2 x+e^{e^x+x} \left (5 x-x^2\right ) \log (16)\right ) \log \left (e^{-x} x\right ) \log \left (\log \left (e^{-x} x\right )\right )}{\left (-5 x+x^2\right ) \log \left (e^{-x} x\right )} \, dx=\ln \left (x^2-10\,x+25\right )\,\ln \left (\ln \left (x\right )-x\right )-4\,{\mathrm {e}}^{{\mathrm {e}}^x}\,\ln \left (2\right )\,\ln \left (\ln \left (x\right )-x\right ) \] Input:

int(-(4*exp(exp(x))*log(2)*(x^2 - 6*x + 5) - log(x^2 - 10*x + 25)*(x^2 - 6 
*x + 5) + log(x*exp(-x))*log(log(x*exp(-x)))*(2*x + 4*exp(exp(x))*exp(x)*l 
og(2)*(5*x - x^2)))/(log(x*exp(-x))*(5*x - x^2)),x)
 

Output:

log(x^2 - 10*x + 25)*log(log(x) - x) - 4*exp(exp(x))*log(2)*log(log(x) - x 
)
 

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.12 \[ \int \frac {e^{e^x} \left (5-6 x+x^2\right ) \log (16)+\left (-5+6 x-x^2\right ) \log \left (25-10 x+x^2\right )+\left (2 x+e^{e^x+x} \left (5 x-x^2\right ) \log (16)\right ) \log \left (e^{-x} x\right ) \log \left (\log \left (e^{-x} x\right )\right )}{\left (-5 x+x^2\right ) \log \left (e^{-x} x\right )} \, dx=\mathrm {log}\left (\mathrm {log}\left (\frac {x}{e^{x}}\right )\right ) \left (-4 e^{e^{x}} \mathrm {log}\left (2\right )+\mathrm {log}\left (x^{2}-10 x +25\right )\right ) \] Input:

int(((4*(-x^2+5*x)*log(2)*exp(x)*exp(exp(x))+2*x)*log(x/exp(x))*log(log(x/ 
exp(x)))+4*(x^2-6*x+5)*log(2)*exp(exp(x))+(-x^2+6*x-5)*log(x^2-10*x+25))/( 
x^2-5*x)/log(x/exp(x)),x)
 

Output:

log(log(x/e**x))*( - 4*e**(e**x)*log(2) + log(x**2 - 10*x + 25))