\(\int \frac {e^x (40+30 x-10 x^2)+(20-5 x+e^x (80 x+20 x^2-10 x^3)) \log (\log (2))+(-10 e^x x+(-5 x-10 e^x x^2) \log (\log (2))) \log (\frac {2 e^x x+(x+2 e^x x^2) \log (\log (2))}{\log (\log (2))})}{2 e^x x+(x+2 e^x x^2) \log (\log (2))} \, dx\) [1256]

Optimal result

Integrand size = 118, antiderivative size = 26 \[ \int \frac {e^x \left (40+30 x-10 x^2\right )+\left (20-5 x+e^x \left (80 x+20 x^2-10 x^3\right )\right ) \log (\log (2))+\left (-10 e^x x+\left (-5 x-10 e^x x^2\right ) \log (\log (2))\right ) \log \left (\frac {2 e^x x+\left (x+2 e^x x^2\right ) \log (\log (2))}{\log (\log (2))}\right )}{2 e^x x+\left (x+2 e^x x^2\right ) \log (\log (2))} \, dx=5 (4-x) \log \left (x+e^x x \left (2 x+\frac {2}{\log (\log (2))}\right )\right ) \] Output:

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

Mathematica [A] (verified)

Time = 3.86 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.73 \[ \int \frac {e^x \left (40+30 x-10 x^2\right )+\left (20-5 x+e^x \left (80 x+20 x^2-10 x^3\right )\right ) \log (\log (2))+\left (-10 e^x x+\left (-5 x-10 e^x x^2\right ) \log (\log (2))\right ) \log \left (\frac {2 e^x x+\left (x+2 e^x x^2\right ) \log (\log (2))}{\log (\log (2))}\right )}{2 e^x x+\left (x+2 e^x x^2\right ) \log (\log (2))} \, dx=20 \log (x)-5 x \log \left (x+2 e^x x \left (x+\frac {1}{\log (\log (2))}\right )\right )+20 \log \left (2 e^x+\log (\log (2))+2 e^x x \log (\log (2))\right ) \] Input:

Integrate[(E^x*(40 + 30*x - 10*x^2) + (20 - 5*x + E^x*(80*x + 20*x^2 - 10* 
x^3))*Log[Log[2]] + (-10*E^x*x + (-5*x - 10*E^x*x^2)*Log[Log[2]])*Log[(2*E 
^x*x + (x + 2*E^x*x^2)*Log[Log[2]])/Log[Log[2]]])/(2*E^x*x + (x + 2*E^x*x^ 
2)*Log[Log[2]]),x]
 

Output:

20*Log[x] - 5*x*Log[x + 2*E^x*x*(x + Log[Log[2]]^(-1))] + 20*Log[2*E^x + L 
og[Log[2]] + 2*E^x*x*Log[Log[2]]]
 

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.

Failed to integrate

Input:

Int[(E^x*(40 + 30*x - 10*x^2) + (20 - 5*x + E^x*(80*x + 20*x^2 - 10*x^3))* 
Log[Log[2]] + (-10*E^x*x + (-5*x - 10*E^x*x^2)*Log[Log[2]])*Log[(2*E^x*x + 
 (x + 2*E^x*x^2)*Log[Log[2]])/Log[Log[2]]])/(2*E^x*x + (x + 2*E^x*x^2)*Log 
[Log[2]]),x]
 

Output:

$Aborted
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(55\) vs. \(2(25)=50\).

Time = 1.19 (sec) , antiderivative size = 56, normalized size of antiderivative = 2.15

method	result	size
default	\(20 \ln \left (-x \left (2 \,{\mathrm e}^{x} \ln \left (\ln \left (2\right )\right ) x +2 \,{\mathrm e}^{x}+\ln \left (\ln \left (2\right )\right )\right )\right )+5 \ln \left (-\ln \left (\ln \left (2\right )\right )\right ) x -5 x \ln \left (-x \left (2 \,{\mathrm e}^{x} \ln \left (\ln \left (2\right )\right ) x +2 \,{\mathrm e}^{x}+\ln \left (\ln \left (2\right )\right )\right )\right )\)	\(56\)
norman	\(20 \ln \left (\frac {\left (2 \,{\mathrm e}^{x} x^{2}+x \right ) \ln \left (\ln \left (2\right )\right )+2 \,{\mathrm e}^{x} x}{\ln \left (\ln \left (2\right )\right )}\right )-5 x \ln \left (\frac {\left (2 \,{\mathrm e}^{x} x^{2}+x \right ) \ln \left (\ln \left (2\right )\right )+2 \,{\mathrm e}^{x} x}{\ln \left (\ln \left (2\right )\right )}\right )\)	\(59\)
parallelrisch	\(20 \ln \left (\frac {\left (2 \,{\mathrm e}^{x} x^{2}+x \right ) \ln \left (\ln \left (2\right )\right )+2 \,{\mathrm e}^{x} x}{\ln \left (\ln \left (2\right )\right )}\right )-5 x \ln \left (\frac {\left (2 \,{\mathrm e}^{x} x^{2}+x \right ) \ln \left (\ln \left (2\right )\right )+2 \,{\mathrm e}^{x} x}{\ln \left (\ln \left (2\right )\right )}\right )\)	\(59\)
risch	\(-5 x \ln \left (\left ({\mathrm e}^{x} x +\frac {1}{2}\right ) \ln \left (\ln \left (2\right )\right )+{\mathrm e}^{x}\right )-5 x \ln \left (x \right )+\frac {5 i \pi x \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i \left (\left ({\mathrm e}^{x} x +\frac {1}{2}\right ) \ln \left (\ln \left (2\right )\right )+{\mathrm e}^{x}\right )\right ) \operatorname {csgn}\left (i x \left (\left ({\mathrm e}^{x} x +\frac {1}{2}\right ) \ln \left (\ln \left (2\right )\right )+{\mathrm e}^{x}\right )\right )}{2}-\frac {5 i \pi x \,\operatorname {csgn}\left (i x \right ) {\operatorname {csgn}\left (i x \left (\left ({\mathrm e}^{x} x +\frac {1}{2}\right ) \ln \left (\ln \left (2\right )\right )+{\mathrm e}^{x}\right )\right )}^{2}}{2}-\frac {5 i \pi x \,\operatorname {csgn}\left (i \left (\left ({\mathrm e}^{x} x +\frac {1}{2}\right ) \ln \left (\ln \left (2\right )\right )+{\mathrm e}^{x}\right )\right ) {\operatorname {csgn}\left (i x \left (\left ({\mathrm e}^{x} x +\frac {1}{2}\right ) \ln \left (\ln \left (2\right )\right )+{\mathrm e}^{x}\right )\right )}^{2}}{2}+5 i \pi x {\operatorname {csgn}\left (i x \left (\left ({\mathrm e}^{x} x +\frac {1}{2}\right ) \ln \left (\ln \left (2\right )\right )+{\mathrm e}^{x}\right )\right )}^{2}-\frac {5 i \pi x {\operatorname {csgn}\left (i x \left (\left ({\mathrm e}^{x} x +\frac {1}{2}\right ) \ln \left (\ln \left (2\right )\right )+{\mathrm e}^{x}\right )\right )}^{3}}{2}-5 i \pi x -5 x \ln \left (2\right )+5 \ln \left (-\ln \left (\ln \left (2\right )\right )\right ) x +20 \ln \left (x^{2} \ln \left (\ln \left (2\right )\right )+x \right )+20 \ln \left ({\mathrm e}^{x}+\frac {\ln \left (\ln \left (2\right )\right )}{2 x \ln \left (\ln \left (2\right )\right )+2}\right )\)	\(242\)

Input:

int((((-10*exp(x)*x^2-5*x)*ln(ln(2))-10*exp(x)*x)*ln(((2*exp(x)*x^2+x)*ln( 
ln(2))+2*exp(x)*x)/ln(ln(2)))+((-10*x^3+20*x^2+80*x)*exp(x)-5*x+20)*ln(ln( 
2))+(-10*x^2+30*x+40)*exp(x))/((2*exp(x)*x^2+x)*ln(ln(2))+2*exp(x)*x),x,me 
thod=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.19 \[ \int \frac {e^x \left (40+30 x-10 x^2\right )+\left (20-5 x+e^x \left (80 x+20 x^2-10 x^3\right )\right ) \log (\log (2))+\left (-10 e^x x+\left (-5 x-10 e^x x^2\right ) \log (\log (2))\right ) \log \left (\frac {2 e^x x+\left (x+2 e^x x^2\right ) \log (\log (2))}{\log (\log (2))}\right )}{2 e^x x+\left (x+2 e^x x^2\right ) \log (\log (2))} \, dx=-5 \, {\left (x - 4\right )} \log \left (\frac {2 \, x e^{x} + {\left (2 \, x^{2} e^{x} + x\right )} \log \left (\log \left (2\right )\right )}{\log \left (\log \left (2\right )\right )}\right ) \] Input:

integrate((((-10*exp(x)*x^2-5*x)*log(log(2))-10*exp(x)*x)*log(((2*exp(x)*x 
^2+x)*log(log(2))+2*exp(x)*x)/log(log(2)))+((-10*x^3+20*x^2+80*x)*exp(x)-5 
*x+20)*log(log(2))+(-10*x^2+30*x+40)*exp(x))/((2*exp(x)*x^2+x)*log(log(2)) 
+2*exp(x)*x),x, algorithm="fricas")
 

Output:

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 66 vs. \(2 (22) = 44\).

Time = 0.47 (sec) , antiderivative size = 66, normalized size of antiderivative = 2.54 \[ \int \frac {e^x \left (40+30 x-10 x^2\right )+\left (20-5 x+e^x \left (80 x+20 x^2-10 x^3\right )\right ) \log (\log (2))+\left (-10 e^x x+\left (-5 x-10 e^x x^2\right ) \log (\log (2))\right ) \log \left (\frac {2 e^x x+\left (x+2 e^x x^2\right ) \log (\log (2))}{\log (\log (2))}\right )}{2 e^x x+\left (x+2 e^x x^2\right ) \log (\log (2))} \, dx=- 5 x \log {\left (\frac {2 x e^{x} + \left (2 x^{2} e^{x} + x\right ) \log {\left (\log {\left (2 \right )} \right )}}{\log {\left (\log {\left (2 \right )} \right )}} \right )} + 20 \log {\left (x^{2} \log {\left (\log {\left (2 \right )} \right )} + x \right )} + 20 \log {\left (e^{x} + \frac {\log {\left (\log {\left (2 \right )} \right )}}{2 x \log {\left (\log {\left (2 \right )} \right )} + 2} \right )} \] Input:

integrate((((-10*exp(x)*x**2-5*x)*ln(ln(2))-10*exp(x)*x)*ln(((2*exp(x)*x** 
2+x)*ln(ln(2))+2*exp(x)*x)/ln(ln(2)))+((-10*x**3+20*x**2+80*x)*exp(x)-5*x+ 
20)*ln(ln(2))+(-10*x**2+30*x+40)*exp(x))/((2*exp(x)*x**2+x)*ln(ln(2))+2*ex 
p(x)*x),x)
 

Output:

-5*x*log((2*x*exp(x) + (2*x**2*exp(x) + x)*log(log(2)))/log(log(2))) + 20* 
log(x**2*log(log(2)) + x) + 20*log(exp(x) + log(log(2))/(2*x*log(log(2)) + 
 2))
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 75 vs. \(2 (20) = 40\).

Time = 0.17 (sec) , antiderivative size = 75, normalized size of antiderivative = 2.88 \[ \int \frac {e^x \left (40+30 x-10 x^2\right )+\left (20-5 x+e^x \left (80 x+20 x^2-10 x^3\right )\right ) \log (\log (2))+\left (-10 e^x x+\left (-5 x-10 e^x x^2\right ) \log (\log (2))\right ) \log \left (\frac {2 e^x x+\left (x+2 e^x x^2\right ) \log (\log (2))}{\log (\log (2))}\right )}{2 e^x x+\left (x+2 e^x x^2\right ) \log (\log (2))} \, dx=-5 \, x \log \left (2 \, {\left (x \log \left (\log \left (2\right )\right ) + 1\right )} e^{x} + \log \left (\log \left (2\right )\right )\right ) - 5 \, x \log \left (x\right ) + 5 \, x \log \left (\log \left (\log \left (2\right )\right )\right ) + 20 \, \log \left (x \log \left (\log \left (2\right )\right ) + 1\right ) + 20 \, \log \left (x\right ) + 20 \, \log \left (\frac {2 \, {\left (x \log \left (\log \left (2\right )\right ) + 1\right )} e^{x} + \log \left (\log \left (2\right )\right )}{2 \, {\left (x \log \left (\log \left (2\right )\right ) + 1\right )}}\right ) \] Input:

integrate((((-10*exp(x)*x^2-5*x)*log(log(2))-10*exp(x)*x)*log(((2*exp(x)*x 
^2+x)*log(log(2))+2*exp(x)*x)/log(log(2)))+((-10*x^3+20*x^2+80*x)*exp(x)-5 
*x+20)*log(log(2))+(-10*x^2+30*x+40)*exp(x))/((2*exp(x)*x^2+x)*log(log(2)) 
+2*exp(x)*x),x, algorithm="maxima")
 

Output:

-5*x*log(2*(x*log(log(2)) + 1)*e^x + log(log(2))) - 5*x*log(x) + 5*x*log(l 
og(log(2))) + 20*log(x*log(log(2)) + 1) + 20*log(x) + 20*log(1/2*(2*(x*log 
(log(2)) + 1)*e^x + log(log(2)))/(x*log(log(2)) + 1))
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 56 vs. \(2 (20) = 40\).

Time = 0.17 (sec) , antiderivative size = 56, normalized size of antiderivative = 2.15 \[ \int \frac {e^x \left (40+30 x-10 x^2\right )+\left (20-5 x+e^x \left (80 x+20 x^2-10 x^3\right )\right ) \log (\log (2))+\left (-10 e^x x+\left (-5 x-10 e^x x^2\right ) \log (\log (2))\right ) \log \left (\frac {2 e^x x+\left (x+2 e^x x^2\right ) \log (\log (2))}{\log (\log (2))}\right )}{2 e^x x+\left (x+2 e^x x^2\right ) \log (\log (2))} \, dx=-5 \, x \log \left (2 \, x^{2} e^{x} \log \left (\log \left (2\right )\right ) + 2 \, x e^{x} + x \log \left (\log \left (2\right )\right )\right ) + 5 \, x \log \left (\log \left (\log \left (2\right )\right )\right ) + 20 \, \log \left (2 \, x e^{x} \log \left (\log \left (2\right )\right ) + 2 \, e^{x} + \log \left (\log \left (2\right )\right )\right ) + 20 \, \log \left (x\right ) \] Input:

integrate((((-10*exp(x)*x^2-5*x)*log(log(2))-10*exp(x)*x)*log(((2*exp(x)*x 
^2+x)*log(log(2))+2*exp(x)*x)/log(log(2)))+((-10*x^3+20*x^2+80*x)*exp(x)-5 
*x+20)*log(log(2))+(-10*x^2+30*x+40)*exp(x))/((2*exp(x)*x^2+x)*log(log(2)) 
+2*exp(x)*x),x, algorithm="giac")
 

Output:

-5*x*log(2*x^2*e^x*log(log(2)) + 2*x*e^x + x*log(log(2))) + 5*x*log(log(lo 
g(2))) + 20*log(2*x*e^x*log(log(2)) + 2*e^x + log(log(2))) + 20*log(x)
 

Mupad [B] (verification not implemented)

Time = 4.29 (sec) , antiderivative size = 71, normalized size of antiderivative = 2.73 \[ \int \frac {e^x \left (40+30 x-10 x^2\right )+\left (20-5 x+e^x \left (80 x+20 x^2-10 x^3\right )\right ) \log (\log (2))+\left (-10 e^x x+\left (-5 x-10 e^x x^2\right ) \log (\log (2))\right ) \log \left (\frac {2 e^x x+\left (x+2 e^x x^2\right ) \log (\log (2))}{\log (\log (2))}\right )}{2 e^x x+\left (x+2 e^x x^2\right ) \log (\log (2))} \, dx=20\,\ln \left (\frac {\ln \left (\ln \left (2\right )\right )+2\,{\mathrm {e}}^x+2\,x\,{\mathrm {e}}^x\,\ln \left (\ln \left (2\right )\right )}{x\,\ln \left (\ln \left (2\right )\right )+1}\right )+20\,\ln \left (\ln \left (\ln \left (2\right )\right )\,x^2+x\right )-5\,x\,\ln \left (\frac {\ln \left (\ln \left (2\right )\right )\,\left (x+2\,x^2\,{\mathrm {e}}^x\right )+2\,x\,{\mathrm {e}}^x}{\ln \left (\ln \left (2\right )\right )}\right ) \] Input:

int((log(log(2))*(exp(x)*(80*x + 20*x^2 - 10*x^3) - 5*x + 20) - log((log(l 
og(2))*(x + 2*x^2*exp(x)) + 2*x*exp(x))/log(log(2)))*(log(log(2))*(5*x + 1 
0*x^2*exp(x)) + 10*x*exp(x)) + exp(x)*(30*x - 10*x^2 + 40))/(log(log(2))*( 
x + 2*x^2*exp(x)) + 2*x*exp(x)),x)
 

Output:

20*log((log(log(2)) + 2*exp(x) + 2*x*exp(x)*log(log(2)))/(x*log(log(2)) + 
1)) + 20*log(x + x^2*log(log(2))) - 5*x*log((log(log(2))*(x + 2*x^2*exp(x) 
) + 2*x*exp(x))/log(log(2)))
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 59, normalized size of antiderivative = 2.27 \[ \int \frac {e^x \left (40+30 x-10 x^2\right )+\left (20-5 x+e^x \left (80 x+20 x^2-10 x^3\right )\right ) \log (\log (2))+\left (-10 e^x x+\left (-5 x-10 e^x x^2\right ) \log (\log (2))\right ) \log \left (\frac {2 e^x x+\left (x+2 e^x x^2\right ) \log (\log (2))}{\log (\log (2))}\right )}{2 e^x x+\left (x+2 e^x x^2\right ) \log (\log (2))} \, dx=20 \,\mathrm {log}\left (2 e^{x} \mathrm {log}\left (\mathrm {log}\left (2\right )\right ) x +2 e^{x}+\mathrm {log}\left (\mathrm {log}\left (2\right )\right )\right )-5 \,\mathrm {log}\left (\frac {2 e^{x} \mathrm {log}\left (\mathrm {log}\left (2\right )\right ) x^{2}+2 e^{x} x +\mathrm {log}\left (\mathrm {log}\left (2\right )\right ) x}{\mathrm {log}\left (\mathrm {log}\left (2\right )\right )}\right ) x +20 \,\mathrm {log}\left (x \right ) \] Input:

int((((-10*exp(x)*x^2-5*x)*log(log(2))-10*exp(x)*x)*log(((2*exp(x)*x^2+x)* 
log(log(2))+2*exp(x)*x)/log(log(2)))+((-10*x^3+20*x^2+80*x)*exp(x)-5*x+20) 
*log(log(2))+(-10*x^2+30*x+40)*exp(x))/((2*exp(x)*x^2+x)*log(log(2))+2*exp 
(x)*x),x)
 

Output:

5*(4*log(2*e**x*log(log(2))*x + 2*e**x + log(log(2))) - log((2*e**x*log(lo 
g(2))*x**2 + 2*e**x*x + log(log(2))*x)/log(log(2)))*x + 4*log(x))