\(\int \frac {(16+8 x+x^2+e^x (-32-16 x-2 x^2)+e^{2 x} (16+8 x+x^2)) \log ^2(x)+e^{\frac {-4 x-8 \log (x)}{(-4-x+e^x (4+x)) \log (x)}} (-16-4 x+e^x (16+4 x)+(16+e^x (-16+16 x+4 x^2)) \log (x)+(-8+e^x (40+8 x)) \log ^2(x))}{(16+8 x+x^2+e^x (-32-16 x-2 x^2)+e^{2 x} (16+8 x+x^2)) \log ^2(x)} \, dx\) [232]

Optimal result

Integrand size = 164, antiderivative size = 33 \[ \int \frac {\left (16+8 x+x^2+e^x \left (-32-16 x-2 x^2\right )+e^{2 x} \left (16+8 x+x^2\right )\right ) \log ^2(x)+e^{\frac {-4 x-8 \log (x)}{\left (-4-x+e^x (4+x)\right ) \log (x)}} \left (-16-4 x+e^x (16+4 x)+\left (16+e^x \left (-16+16 x+4 x^2\right )\right ) \log (x)+\left (-8+e^x (40+8 x)\right ) \log ^2(x)\right )}{\left (16+8 x+x^2+e^x \left (-32-16 x-2 x^2\right )+e^{2 x} \left (16+8 x+x^2\right )\right ) \log ^2(x)} \, dx=e^{\frac {2 x \left (\frac {4}{x}+\frac {2}{\log (x)}\right )}{\left (1-e^x\right ) (4+x)}}+x \] Output:

exp(2*(2/ln(x)+4/x)/(1-exp(x))*x/(4+x))+x
 

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.15 \[ \int \frac {\left (16+8 x+x^2+e^x \left (-32-16 x-2 x^2\right )+e^{2 x} \left (16+8 x+x^2\right )\right ) \log ^2(x)+e^{\frac {-4 x-8 \log (x)}{\left (-4-x+e^x (4+x)\right ) \log (x)}} \left (-16-4 x+e^x (16+4 x)+\left (16+e^x \left (-16+16 x+4 x^2\right )\right ) \log (x)+\left (-8+e^x (40+8 x)\right ) \log ^2(x)\right )}{\left (16+8 x+x^2+e^x \left (-32-16 x-2 x^2\right )+e^{2 x} \left (16+8 x+x^2\right )\right ) \log ^2(x)} \, dx=e^{-\frac {8}{\left (-1+e^x\right ) (4+x)}-\frac {4 x}{\left (-1+e^x\right ) (4+x) \log (x)}}+x \] Input:

Integrate[((16 + 8*x + x^2 + E^x*(-32 - 16*x - 2*x^2) + E^(2*x)*(16 + 8*x 
+ x^2))*Log[x]^2 + E^((-4*x - 8*Log[x])/((-4 - x + E^x*(4 + x))*Log[x]))*( 
-16 - 4*x + E^x*(16 + 4*x) + (16 + E^x*(-16 + 16*x + 4*x^2))*Log[x] + (-8 
+ E^x*(40 + 8*x))*Log[x]^2))/((16 + 8*x + x^2 + E^x*(-32 - 16*x - 2*x^2) + 
 E^(2*x)*(16 + 8*x + x^2))*Log[x]^2),x]
 

Output:

E^(-8/((-1 + E^x)*(4 + x)) - (4*x)/((-1 + E^x)*(4 + x)*Log[x])) + 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.

\(\displaystyle \int \frac {\left (\left (e^x \left (4 x^2+16 x-16\right )+16\right ) \log (x)-4 x+e^x (4 x+16)+\left (e^x (8 x+40)-8\right ) \log ^2(x)-16\right ) \exp \left (\frac {-4 x-8 \log (x)}{\left (-x+e^x (x+4)-4\right ) \log (x)}\right )+\left (x^2+e^x \left (-2 x^2-16 x-32\right )+e^{2 x} \left (x^2+8 x+16\right )+8 x+16\right ) \log ^2(x)}{\left (x^2+e^x \left (-2 x^2-16 x-32\right )+e^{2 x} \left (x^2+8 x+16\right )+8 x+16\right ) \log ^2(x)} \, dx\)

\(\Big \downarrow \) 7239

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

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (\frac {4 \left (e^x x^2 \log (x)+e^x x-x+4 e^x+2 e^x x \log ^2(x)+10 e^x \log ^2(x)-2 \log ^2(x)+4 e^x x \log (x)-4 e^x \log (x)+4 \log (x)-4\right ) \exp \left (-\frac {4 (x+2 \log (x))}{\left (e^x-1\right ) (x+4) \log (x)}\right )}{\left (1-e^x\right )^2 (x+4)^2 \log ^2(x)}+1\right )dx\)

\(\Big \downarrow \) 7299

\(\displaystyle \int \left (\frac {4 \left (e^x x^2 \log (x)+e^x x-x+4 e^x+2 e^x x \log ^2(x)+10 e^x \log ^2(x)-2 \log ^2(x)+4 e^x x \log (x)-4 e^x \log (x)+4 \log (x)-4\right ) \exp \left (-\frac {4 (x+2 \log (x))}{\left (e^x-1\right ) (x+4) \log (x)}\right )}{\left (1-e^x\right )^2 (x+4)^2 \log ^2(x)}+1\right )dx\)

Input:

Int[((16 + 8*x + x^2 + E^x*(-32 - 16*x - 2*x^2) + E^(2*x)*(16 + 8*x + x^2) 
)*Log[x]^2 + E^((-4*x - 8*Log[x])/((-4 - x + E^x*(4 + x))*Log[x]))*(-16 - 
4*x + E^x*(16 + 4*x) + (16 + E^x*(-16 + 16*x + 4*x^2))*Log[x] + (-8 + E^x* 
(40 + 8*x))*Log[x]^2))/((16 + 8*x + x^2 + E^x*(-32 - 16*x - 2*x^2) + E^(2* 
x)*(16 + 8*x + x^2))*Log[x]^2),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 240.12 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.82

method	result	size
risch	\(x +{\mathrm e}^{-\frac {4 \left (2 \ln \left (x \right )+x \right )}{\left (4+x \right ) \left ({\mathrm e}^{x}-1\right ) \ln \left (x \right )}}\)	\(27\)
parallelrisch	\(x +{\mathrm e}^{-\frac {4 \left (2 \ln \left (x \right )+x \right )}{\left ({\mathrm e}^{x} x +4 \,{\mathrm e}^{x}-x -4\right ) \ln \left (x \right )}}-6\)	\(32\)

Input:

int(((((8*x+40)*exp(x)-8)*ln(x)^2+((4*x^2+16*x-16)*exp(x)+16)*ln(x)+(4*x+1 
6)*exp(x)-16-4*x)*exp((-8*ln(x)-4*x)/((4+x)*exp(x)-x-4)/ln(x))+((x^2+8*x+1 
6)*exp(x)^2+(-2*x^2-16*x-32)*exp(x)+x^2+8*x+16)*ln(x)^2)/((x^2+8*x+16)*exp 
(x)^2+(-2*x^2-16*x-32)*exp(x)+x^2+8*x+16)/ln(x)^2,x,method=_RETURNVERBOSE)
 

Output:

x+exp(-4*(2*ln(x)+x)/(4+x)/(exp(x)-1)/ln(x))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.85 \[ \int \frac {\left (16+8 x+x^2+e^x \left (-32-16 x-2 x^2\right )+e^{2 x} \left (16+8 x+x^2\right )\right ) \log ^2(x)+e^{\frac {-4 x-8 \log (x)}{\left (-4-x+e^x (4+x)\right ) \log (x)}} \left (-16-4 x+e^x (16+4 x)+\left (16+e^x \left (-16+16 x+4 x^2\right )\right ) \log (x)+\left (-8+e^x (40+8 x)\right ) \log ^2(x)\right )}{\left (16+8 x+x^2+e^x \left (-32-16 x-2 x^2\right )+e^{2 x} \left (16+8 x+x^2\right )\right ) \log ^2(x)} \, dx=x + e^{\left (-\frac {4 \, {\left (x + 2 \, \log \left (x\right )\right )}}{{\left ({\left (x + 4\right )} e^{x} - x - 4\right )} \log \left (x\right )}\right )} \] Input:

integrate(((((8*x+40)*exp(x)-8)*log(x)^2+((4*x^2+16*x-16)*exp(x)+16)*log(x 
)+(4*x+16)*exp(x)-16-4*x)*exp((-8*log(x)-4*x)/((4+x)*exp(x)-x-4)/log(x))+( 
(x^2+8*x+16)*exp(x)^2+(-2*x^2-16*x-32)*exp(x)+x^2+8*x+16)*log(x)^2)/((x^2+ 
8*x+16)*exp(x)^2+(-2*x^2-16*x-32)*exp(x)+x^2+8*x+16)/log(x)^2,x, algorithm 
="fricas")
 

Output:

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

Sympy [A] (verification not implemented)

Time = 1.25 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.79 \[ \int \frac {\left (16+8 x+x^2+e^x \left (-32-16 x-2 x^2\right )+e^{2 x} \left (16+8 x+x^2\right )\right ) \log ^2(x)+e^{\frac {-4 x-8 \log (x)}{\left (-4-x+e^x (4+x)\right ) \log (x)}} \left (-16-4 x+e^x (16+4 x)+\left (16+e^x \left (-16+16 x+4 x^2\right )\right ) \log (x)+\left (-8+e^x (40+8 x)\right ) \log ^2(x)\right )}{\left (16+8 x+x^2+e^x \left (-32-16 x-2 x^2\right )+e^{2 x} \left (16+8 x+x^2\right )\right ) \log ^2(x)} \, dx=x + e^{\frac {- 4 x - 8 \log {\left (x \right )}}{\left (- x + \left (x + 4\right ) e^{x} - 4\right ) \log {\left (x \right )}}} \] Input:

integrate(((((8*x+40)*exp(x)-8)*ln(x)**2+((4*x**2+16*x-16)*exp(x)+16)*ln(x 
)+(4*x+16)*exp(x)-16-4*x)*exp((-8*ln(x)-4*x)/((4+x)*exp(x)-x-4)/ln(x))+((x 
**2+8*x+16)*exp(x)**2+(-2*x**2-16*x-32)*exp(x)+x**2+8*x+16)*ln(x)**2)/((x* 
*2+8*x+16)*exp(x)**2+(-2*x**2-16*x-32)*exp(x)+x**2+8*x+16)/ln(x)**2,x)
 

Output:

x + exp((-4*x - 8*log(x))/((-x + (x + 4)*exp(x) - 4)*log(x)))
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 82 vs. \(2 (27) = 54\).

Time = 0.20 (sec) , antiderivative size = 82, normalized size of antiderivative = 2.48 \[ \int \frac {\left (16+8 x+x^2+e^x \left (-32-16 x-2 x^2\right )+e^{2 x} \left (16+8 x+x^2\right )\right ) \log ^2(x)+e^{\frac {-4 x-8 \log (x)}{\left (-4-x+e^x (4+x)\right ) \log (x)}} \left (-16-4 x+e^x (16+4 x)+\left (16+e^x \left (-16+16 x+4 x^2\right )\right ) \log (x)+\left (-8+e^x (40+8 x)\right ) \log ^2(x)\right )}{\left (16+8 x+x^2+e^x \left (-32-16 x-2 x^2\right )+e^{2 x} \left (16+8 x+x^2\right )\right ) \log ^2(x)} \, dx={\left (x e^{\left (\frac {8}{{\left (x + 4\right )} e^{x} - x - 4} + \frac {4}{{\left (e^{x} - 1\right )} \log \left (x\right )}\right )} + e^{\left (\frac {16}{{\left ({\left (x + 4\right )} e^{x} - x - 4\right )} \log \left (x\right )}\right )}\right )} e^{\left (-\frac {8}{{\left (x + 4\right )} e^{x} - x - 4} - \frac {4}{{\left (e^{x} - 1\right )} \log \left (x\right )}\right )} \] Input:

integrate(((((8*x+40)*exp(x)-8)*log(x)^2+((4*x^2+16*x-16)*exp(x)+16)*log(x 
)+(4*x+16)*exp(x)-16-4*x)*exp((-8*log(x)-4*x)/((4+x)*exp(x)-x-4)/log(x))+( 
(x^2+8*x+16)*exp(x)^2+(-2*x^2-16*x-32)*exp(x)+x^2+8*x+16)*log(x)^2)/((x^2+ 
8*x+16)*exp(x)^2+(-2*x^2-16*x-32)*exp(x)+x^2+8*x+16)/log(x)^2,x, algorithm 
="maxima")
 

Output:

(x*e^(8/((x + 4)*e^x - x - 4) + 4/((e^x - 1)*log(x))) + e^(16/(((x + 4)*e^ 
x - x - 4)*log(x))))*e^(-8/((x + 4)*e^x - x - 4) - 4/((e^x - 1)*log(x)))
 

Giac [F(-2)]

Exception generated. \[ \int \frac {\left (16+8 x+x^2+e^x \left (-32-16 x-2 x^2\right )+e^{2 x} \left (16+8 x+x^2\right )\right ) \log ^2(x)+e^{\frac {-4 x-8 \log (x)}{\left (-4-x+e^x (4+x)\right ) \log (x)}} \left (-16-4 x+e^x (16+4 x)+\left (16+e^x \left (-16+16 x+4 x^2\right )\right ) \log (x)+\left (-8+e^x (40+8 x)\right ) \log ^2(x)\right )}{\left (16+8 x+x^2+e^x \left (-32-16 x-2 x^2\right )+e^{2 x} \left (16+8 x+x^2\right )\right ) \log ^2(x)} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(((((8*x+40)*exp(x)-8)*log(x)^2+((4*x^2+16*x-16)*exp(x)+16)*log(x 
)+(4*x+16)*exp(x)-16-4*x)*exp((-8*log(x)-4*x)/((4+x)*exp(x)-x-4)/log(x))+( 
(x^2+8*x+16)*exp(x)^2+(-2*x^2-16*x-32)*exp(x)+x^2+8*x+16)*log(x)^2)/((x^2+ 
8*x+16)*exp(x)^2+(-2*x^2-16*x-32)*exp(x)+x^2+8*x+16)/log(x)^2,x, algorithm 
="giac")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:exp(sageVARx)^2=exp(2*sageVARx)exp( 
sageVARx)^2=exp(2*sageVARx)exp(sageVARx)^2=exp(2*sageVARx)exp(sageVARx)^2= 
exp(2*sag
 

Mupad [B] (verification not implemented)

Time = 4.48 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.79 \[ \int \frac {\left (16+8 x+x^2+e^x \left (-32-16 x-2 x^2\right )+e^{2 x} \left (16+8 x+x^2\right )\right ) \log ^2(x)+e^{\frac {-4 x-8 \log (x)}{\left (-4-x+e^x (4+x)\right ) \log (x)}} \left (-16-4 x+e^x (16+4 x)+\left (16+e^x \left (-16+16 x+4 x^2\right )\right ) \log (x)+\left (-8+e^x (40+8 x)\right ) \log ^2(x)\right )}{\left (16+8 x+x^2+e^x \left (-32-16 x-2 x^2\right )+e^{2 x} \left (16+8 x+x^2\right )\right ) \log ^2(x)} \, dx=x+x^{\frac {8}{4\,\ln \left (x\right )-4\,{\mathrm {e}}^x\,\ln \left (x\right )+x\,\ln \left (x\right )-x\,{\mathrm {e}}^x\,\ln \left (x\right )}}\,{\mathrm {e}}^{\frac {4\,x}{4\,\ln \left (x\right )-4\,{\mathrm {e}}^x\,\ln \left (x\right )+x\,\ln \left (x\right )-x\,{\mathrm {e}}^x\,\ln \left (x\right )}} \] Input:

int((log(x)^2*(8*x + exp(2*x)*(8*x + x^2 + 16) - exp(x)*(16*x + 2*x^2 + 32 
) + x^2 + 16) + exp((4*x + 8*log(x))/(log(x)*(x - exp(x)*(x + 4) + 4)))*(l 
og(x)*(exp(x)*(16*x + 4*x^2 - 16) + 16) - 4*x + exp(x)*(4*x + 16) + log(x) 
^2*(exp(x)*(8*x + 40) - 8) - 16))/(log(x)^2*(8*x + exp(2*x)*(8*x + x^2 + 1 
6) - exp(x)*(16*x + 2*x^2 + 32) + x^2 + 16)),x)
 

Output:

x + x^(8/(4*log(x) - 4*exp(x)*log(x) + x*log(x) - x*exp(x)*log(x)))*exp((4 
*x)/(4*log(x) - 4*exp(x)*log(x) + x*log(x) - x*exp(x)*log(x)))
 

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 81, normalized size of antiderivative = 2.45 \[ \int \frac {\left (16+8 x+x^2+e^x \left (-32-16 x-2 x^2\right )+e^{2 x} \left (16+8 x+x^2\right )\right ) \log ^2(x)+e^{\frac {-4 x-8 \log (x)}{\left (-4-x+e^x (4+x)\right ) \log (x)}} \left (-16-4 x+e^x (16+4 x)+\left (16+e^x \left (-16+16 x+4 x^2\right )\right ) \log (x)+\left (-8+e^x (40+8 x)\right ) \log ^2(x)\right )}{\left (16+8 x+x^2+e^x \left (-32-16 x-2 x^2\right )+e^{2 x} \left (16+8 x+x^2\right )\right ) \log ^2(x)} \, dx=\frac {e^{\frac {8 \,\mathrm {log}\left (x \right )+4 x}{e^{x} \mathrm {log}\left (x \right ) x +4 e^{x} \mathrm {log}\left (x \right )-\mathrm {log}\left (x \right ) x -4 \,\mathrm {log}\left (x \right )}} x +1}{e^{\frac {8 \,\mathrm {log}\left (x \right )+4 x}{e^{x} \mathrm {log}\left (x \right ) x +4 e^{x} \mathrm {log}\left (x \right )-\mathrm {log}\left (x \right ) x -4 \,\mathrm {log}\left (x \right )}}} \] Input:

int(((((8*x+40)*exp(x)-8)*log(x)^2+((4*x^2+16*x-16)*exp(x)+16)*log(x)+(4*x 
+16)*exp(x)-16-4*x)*exp((-8*log(x)-4*x)/((4+x)*exp(x)-x-4)/log(x))+((x^2+8 
*x+16)*exp(x)^2+(-2*x^2-16*x-32)*exp(x)+x^2+8*x+16)*log(x)^2)/((x^2+8*x+16 
)*exp(x)^2+(-2*x^2-16*x-32)*exp(x)+x^2+8*x+16)/log(x)^2,x)
 

Output:

(e**((8*log(x) + 4*x)/(e**x*log(x)*x + 4*e**x*log(x) - log(x)*x - 4*log(x) 
))*x + 1)/e**((8*log(x) + 4*x)/(e**x*log(x)*x + 4*e**x*log(x) - log(x)*x - 
 4*log(x)))