\(\int \frac {512 x^5+768 x^6+256 x^7+e^x (-512 x^3+512 x^4+1152 x^5+384 x^6)+(-1024 x^4-1280 x^5-384 x^6+e^x (-1024 x^3-1280 x^4-384 x^5)) \log (\frac {e^x+x}{x})}{-e^x x^3-x^4+(3 e^x x^2+3 x^3) \log (\frac {e^x+x}{x})+(-3 e^x x-3 x^2) \log ^2(\frac {e^x+x}{x})+(e^x+x) \log ^3(\frac {e^x+x}{x})} \, dx\) [1631]

Optimal result

Integrand size = 173, antiderivative size = 29 \[ \int \frac {512 x^5+768 x^6+256 x^7+e^x \left (-512 x^3+512 x^4+1152 x^5+384 x^6\right )+\left (-1024 x^4-1280 x^5-384 x^6+e^x \left (-1024 x^3-1280 x^4-384 x^5\right )\right ) \log \left (\frac {e^x+x}{x}\right )}{-e^x x^3-x^4+\left (3 e^x x^2+3 x^3\right ) \log \left (\frac {e^x+x}{x}\right )+\left (-3 e^x x-3 x^2\right ) \log ^2\left (\frac {e^x+x}{x}\right )+\left (e^x+x\right ) \log ^3\left (\frac {e^x+x}{x}\right )} \, dx=-3+e-\frac {64 x^4 (2+x)^2}{\left (x-\log \left (\frac {e^x+x}{x}\right )\right )^2} \] Output:

exp(1)-3-64*x^4/(x-ln(1/x*(exp(x)+x)))^2*(2+x)^2
 

Mathematica [A] (verified)

Time = 0.37 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.90 \[ \int \frac {512 x^5+768 x^6+256 x^7+e^x \left (-512 x^3+512 x^4+1152 x^5+384 x^6\right )+\left (-1024 x^4-1280 x^5-384 x^6+e^x \left (-1024 x^3-1280 x^4-384 x^5\right )\right ) \log \left (\frac {e^x+x}{x}\right )}{-e^x x^3-x^4+\left (3 e^x x^2+3 x^3\right ) \log \left (\frac {e^x+x}{x}\right )+\left (-3 e^x x-3 x^2\right ) \log ^2\left (\frac {e^x+x}{x}\right )+\left (e^x+x\right ) \log ^3\left (\frac {e^x+x}{x}\right )} \, dx=-\frac {64 x^4 (2+x)^2}{\left (-x+\log \left (\frac {e^x+x}{x}\right )\right )^2} \] Input:

Integrate[(512*x^5 + 768*x^6 + 256*x^7 + E^x*(-512*x^3 + 512*x^4 + 1152*x^ 
5 + 384*x^6) + (-1024*x^4 - 1280*x^5 - 384*x^6 + E^x*(-1024*x^3 - 1280*x^4 
 - 384*x^5))*Log[(E^x + x)/x])/(-(E^x*x^3) - x^4 + (3*E^x*x^2 + 3*x^3)*Log 
[(E^x + x)/x] + (-3*E^x*x - 3*x^2)*Log[(E^x + x)/x]^2 + (E^x + x)*Log[(E^x 
 + x)/x]^3),x]
 

Output:

(-64*x^4*(2 + x)^2)/(-x + Log[(E^x + x)/x])^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.

\(\displaystyle \int \frac {256 x^7+768 x^6+512 x^5+e^x \left (384 x^6+1152 x^5+512 x^4-512 x^3\right )+\left (-384 x^6-1280 x^5-1024 x^4+e^x \left (-384 x^5-1280 x^4-1024 x^3\right )\right ) \log \left (\frac {x+e^x}{x}\right )}{-x^4-e^x x^3+\left (-3 x^2-3 e^x x\right ) \log ^2\left (\frac {x+e^x}{x}\right )+\left (3 x^3+3 e^x x^2\right ) \log \left (\frac {x+e^x}{x}\right )+\left (x+e^x\right ) \log ^3\left (\frac {x+e^x}{x}\right )} \, dx\)

\(\Big \downarrow \) 7239

\(\displaystyle \int \frac {128 x^3 (x+2) \left (-2 (x+1) x^2-e^x \left (3 x^2+3 x-2\right )+\left (x+e^x\right ) (3 x+4) \log \left (\frac {x+e^x}{x}\right )\right )}{\left (x+e^x\right ) \left (x-\log \left (\frac {x+e^x}{x}\right )\right )^3}dx\)

\(\Big \downarrow \) 27

\(\displaystyle 128 \int -\frac {x^3 (x+2) \left (2 (x+1) x^2-e^x \left (-3 x^2-3 x+2\right )-\left (x+e^x\right ) (3 x+4) \log \left (\frac {x+e^x}{x}\right )\right )}{\left (x+e^x\right ) \left (x-\log \left (\frac {x+e^x}{x}\right )\right )^3}dx\)

\(\Big \downarrow \) 25

\(\displaystyle -128 \int \frac {x^3 (x+2) \left (2 (x+1) x^2-e^x \left (-3 x^2-3 x+2\right )-\left (x+e^x\right ) (3 x+4) \log \left (\frac {x+e^x}{x}\right )\right )}{\left (x+e^x\right ) \left (x-\log \left (\frac {x+e^x}{x}\right )\right )^3}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle -128 \int \left (\frac {x^3 (x+2) \left (3 x^2-3 \log \left (\frac {x+e^x}{x}\right ) x+3 x-4 \log \left (\frac {x+e^x}{x}\right )-2\right )}{\left (x-\log \left (\frac {x+e^x}{x}\right )\right )^3}-\frac {(x-1) x^4 (x+2)^2}{\left (x+e^x\right ) \left (x-\log \left (\frac {x+e^x}{x}\right )\right )^3}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -128 \left (-\int \frac {x^7}{\left (x+e^x\right ) \left (x-\log \left (\frac {x+e^x}{x}\right )\right )^3}dx-3 \int \frac {x^6}{\left (x+e^x\right ) \left (x-\log \left (\frac {x+e^x}{x}\right )\right )^3}dx-\int \frac {x^5}{\left (x-\log \left (\frac {x+e^x}{x}\right )\right )^3}dx+3 \int \frac {x^5}{\left (x-\log \left (\frac {x+e^x}{x}\right )\right )^2}dx-4 \int \frac {x^4}{\left (x-\log \left (\frac {x+e^x}{x}\right )\right )^3}dx+4 \int \frac {x^4}{\left (x+e^x\right ) \left (x-\log \left (\frac {x+e^x}{x}\right )\right )^3}dx+10 \int \frac {x^4}{\left (x-\log \left (\frac {x+e^x}{x}\right )\right )^2}dx-4 \int \frac {x^3}{\left (x-\log \left (\frac {x+e^x}{x}\right )\right )^3}dx+8 \int \frac {x^3}{\left (x-\log \left (\frac {x+e^x}{x}\right )\right )^2}dx\right )\)

Input:

Int[(512*x^5 + 768*x^6 + 256*x^7 + E^x*(-512*x^3 + 512*x^4 + 1152*x^5 + 38 
4*x^6) + (-1024*x^4 - 1280*x^5 - 384*x^6 + E^x*(-1024*x^3 - 1280*x^4 - 384 
*x^5))*Log[(E^x + x)/x])/(-(E^x*x^3) - x^4 + (3*E^x*x^2 + 3*x^3)*Log[(E^x 
+ x)/x] + (-3*E^x*x - 3*x^2)*Log[(E^x + x)/x]^2 + (E^x + x)*Log[(E^x + x)/ 
x]^3),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 1.64 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.66

method	result	size
parallelrisch	\(\frac {-128 x^{6}-512 x^{5}-512 x^{4}}{2 x^{2}-4 x \ln \left (\frac {{\mathrm e}^{x}+x}{x}\right )+2 \ln \left (\frac {{\mathrm e}^{x}+x}{x}\right )^{2}}\)	\(48\)
risch	\(-\frac {256 \left (x^{2}+4 x +4\right ) x^{4}}{{\left (i \pi \,\operatorname {csgn}\left (\frac {i}{x}\right ) \operatorname {csgn}\left (i \left ({\mathrm e}^{x}+x \right )\right ) \operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{x}+x \right )}{x}\right )-i \pi \,\operatorname {csgn}\left (\frac {i}{x}\right ) {\operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{x}+x \right )}{x}\right )}^{2}-i \pi \,\operatorname {csgn}\left (i \left ({\mathrm e}^{x}+x \right )\right ) {\operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{x}+x \right )}{x}\right )}^{2}+i \pi {\operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{x}+x \right )}{x}\right )}^{3}+2 x +2 \ln \left (x \right )-2 \ln \left ({\mathrm e}^{x}+x \right )\right )}^{2}}\)	\(127\)

Input:

int((((-384*x^5-1280*x^4-1024*x^3)*exp(x)-384*x^6-1280*x^5-1024*x^4)*ln(1/ 
x*(exp(x)+x))+(384*x^6+1152*x^5+512*x^4-512*x^3)*exp(x)+256*x^7+768*x^6+51 
2*x^5)/((exp(x)+x)*ln(1/x*(exp(x)+x))^3+(-3*exp(x)*x-3*x^2)*ln(1/x*(exp(x) 
+x))^2+(3*exp(x)*x^2+3*x^3)*ln(1/x*(exp(x)+x))-exp(x)*x^3-x^4),x,method=_R 
ETURNVERBOSE)
 

Output:

1/2*(-128*x^6-512*x^5-512*x^4)/(x^2-2*x*ln(1/x*(exp(x)+x))+ln(1/x*(exp(x)+ 
x))^2)
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.55 \[ \int \frac {512 x^5+768 x^6+256 x^7+e^x \left (-512 x^3+512 x^4+1152 x^5+384 x^6\right )+\left (-1024 x^4-1280 x^5-384 x^6+e^x \left (-1024 x^3-1280 x^4-384 x^5\right )\right ) \log \left (\frac {e^x+x}{x}\right )}{-e^x x^3-x^4+\left (3 e^x x^2+3 x^3\right ) \log \left (\frac {e^x+x}{x}\right )+\left (-3 e^x x-3 x^2\right ) \log ^2\left (\frac {e^x+x}{x}\right )+\left (e^x+x\right ) \log ^3\left (\frac {e^x+x}{x}\right )} \, dx=-\frac {64 \, {\left (x^{6} + 4 \, x^{5} + 4 \, x^{4}\right )}}{x^{2} - 2 \, x \log \left (\frac {x + e^{x}}{x}\right ) + \log \left (\frac {x + e^{x}}{x}\right )^{2}} \] Input:

integrate((((-384*x^5-1280*x^4-1024*x^3)*exp(x)-384*x^6-1280*x^5-1024*x^4) 
*log(1/x*(exp(x)+x))+(384*x^6+1152*x^5+512*x^4-512*x^3)*exp(x)+256*x^7+768 
*x^6+512*x^5)/((exp(x)+x)*log(1/x*(exp(x)+x))^3+(-3*exp(x)*x-3*x^2)*log(1/ 
x*(exp(x)+x))^2+(3*exp(x)*x^2+3*x^3)*log(1/x*(exp(x)+x))-exp(x)*x^3-x^4),x 
, algorithm="fricas")
 

Output:

-64*(x^6 + 4*x^5 + 4*x^4)/(x^2 - 2*x*log((x + e^x)/x) + log((x + e^x)/x)^2 
)
 

Sympy [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.41 \[ \int \frac {512 x^5+768 x^6+256 x^7+e^x \left (-512 x^3+512 x^4+1152 x^5+384 x^6\right )+\left (-1024 x^4-1280 x^5-384 x^6+e^x \left (-1024 x^3-1280 x^4-384 x^5\right )\right ) \log \left (\frac {e^x+x}{x}\right )}{-e^x x^3-x^4+\left (3 e^x x^2+3 x^3\right ) \log \left (\frac {e^x+x}{x}\right )+\left (-3 e^x x-3 x^2\right ) \log ^2\left (\frac {e^x+x}{x}\right )+\left (e^x+x\right ) \log ^3\left (\frac {e^x+x}{x}\right )} \, dx=\frac {- 64 x^{6} - 256 x^{5} - 256 x^{4}}{x^{2} - 2 x \log {\left (\frac {x + e^{x}}{x} \right )} + \log {\left (\frac {x + e^{x}}{x} \right )}^{2}} \] Input:

integrate((((-384*x**5-1280*x**4-1024*x**3)*exp(x)-384*x**6-1280*x**5-1024 
*x**4)*ln(1/x*(exp(x)+x))+(384*x**6+1152*x**5+512*x**4-512*x**3)*exp(x)+25 
6*x**7+768*x**6+512*x**5)/((exp(x)+x)*ln(1/x*(exp(x)+x))**3+(-3*exp(x)*x-3 
*x**2)*ln(1/x*(exp(x)+x))**2+(3*exp(x)*x**2+3*x**3)*ln(1/x*(exp(x)+x))-exp 
(x)*x**3-x**4),x)
 

Output:

(-64*x**6 - 256*x**5 - 256*x**4)/(x**2 - 2*x*log((x + exp(x))/x) + log((x 
+ exp(x))/x)**2)
 

Maxima [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.69 \[ \int \frac {512 x^5+768 x^6+256 x^7+e^x \left (-512 x^3+512 x^4+1152 x^5+384 x^6\right )+\left (-1024 x^4-1280 x^5-384 x^6+e^x \left (-1024 x^3-1280 x^4-384 x^5\right )\right ) \log \left (\frac {e^x+x}{x}\right )}{-e^x x^3-x^4+\left (3 e^x x^2+3 x^3\right ) \log \left (\frac {e^x+x}{x}\right )+\left (-3 e^x x-3 x^2\right ) \log ^2\left (\frac {e^x+x}{x}\right )+\left (e^x+x\right ) \log ^3\left (\frac {e^x+x}{x}\right )} \, dx=-\frac {64 \, {\left (x^{6} + 4 \, x^{5} + 4 \, x^{4}\right )}}{x^{2} - 2 \, {\left (x + \log \left (x\right )\right )} \log \left (x + e^{x}\right ) + \log \left (x + e^{x}\right )^{2} + 2 \, x \log \left (x\right ) + \log \left (x\right )^{2}} \] Input:

integrate((((-384*x^5-1280*x^4-1024*x^3)*exp(x)-384*x^6-1280*x^5-1024*x^4) 
*log(1/x*(exp(x)+x))+(384*x^6+1152*x^5+512*x^4-512*x^3)*exp(x)+256*x^7+768 
*x^6+512*x^5)/((exp(x)+x)*log(1/x*(exp(x)+x))^3+(-3*exp(x)*x-3*x^2)*log(1/ 
x*(exp(x)+x))^2+(3*exp(x)*x^2+3*x^3)*log(1/x*(exp(x)+x))-exp(x)*x^3-x^4),x 
, algorithm="maxima")
 

Output:

-64*(x^6 + 4*x^5 + 4*x^4)/(x^2 - 2*(x + log(x))*log(x + e^x) + log(x + e^x 
)^2 + 2*x*log(x) + log(x)^2)
 

Giac [A] (verification not implemented)

Time = 0.49 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.55 \[ \int \frac {512 x^5+768 x^6+256 x^7+e^x \left (-512 x^3+512 x^4+1152 x^5+384 x^6\right )+\left (-1024 x^4-1280 x^5-384 x^6+e^x \left (-1024 x^3-1280 x^4-384 x^5\right )\right ) \log \left (\frac {e^x+x}{x}\right )}{-e^x x^3-x^4+\left (3 e^x x^2+3 x^3\right ) \log \left (\frac {e^x+x}{x}\right )+\left (-3 e^x x-3 x^2\right ) \log ^2\left (\frac {e^x+x}{x}\right )+\left (e^x+x\right ) \log ^3\left (\frac {e^x+x}{x}\right )} \, dx=-\frac {64 \, {\left (x^{6} + 4 \, x^{5} + 4 \, x^{4}\right )}}{x^{2} - 2 \, x \log \left (\frac {x + e^{x}}{x}\right ) + \log \left (\frac {x + e^{x}}{x}\right )^{2}} \] Input:

integrate((((-384*x^5-1280*x^4-1024*x^3)*exp(x)-384*x^6-1280*x^5-1024*x^4) 
*log(1/x*(exp(x)+x))+(384*x^6+1152*x^5+512*x^4-512*x^3)*exp(x)+256*x^7+768 
*x^6+512*x^5)/((exp(x)+x)*log(1/x*(exp(x)+x))^3+(-3*exp(x)*x-3*x^2)*log(1/ 
x*(exp(x)+x))^2+(3*exp(x)*x^2+3*x^3)*log(1/x*(exp(x)+x))-exp(x)*x^3-x^4),x 
, algorithm="giac")
 

Output:

-64*(x^6 + 4*x^5 + 4*x^4)/(x^2 - 2*x*log((x + e^x)/x) + log((x + e^x)/x)^2 
)
 

Mupad [B] (verification not implemented)

Time = 2.10 (sec) , antiderivative size = 540, normalized size of antiderivative = 18.62 \[ \int \frac {512 x^5+768 x^6+256 x^7+e^x \left (-512 x^3+512 x^4+1152 x^5+384 x^6\right )+\left (-1024 x^4-1280 x^5-384 x^6+e^x \left (-1024 x^3-1280 x^4-384 x^5\right )\right ) \log \left (\frac {e^x+x}{x}\right )}{-e^x x^3-x^4+\left (3 e^x x^2+3 x^3\right ) \log \left (\frac {e^x+x}{x}\right )+\left (-3 e^x x-3 x^2\right ) \log ^2\left (\frac {e^x+x}{x}\right )+\left (e^x+x\right ) \log ^3\left (\frac {e^x+x}{x}\right )} \, dx=\frac {\frac {64\,x\,\left (x+{\mathrm {e}}^x\right )\,\left (24\,x^5\,{\mathrm {e}}^x+48\,x^6\,{\mathrm {e}}^x+43\,x^7\,{\mathrm {e}}^x+19\,x^8\,{\mathrm {e}}^x+3\,x^9\,{\mathrm {e}}^x-8\,x^3\,{\mathrm {e}}^{2\,x}+22\,x^4\,{\mathrm {e}}^{2\,x}+47\,x^5\,{\mathrm {e}}^{2\,x}+18\,x^6\,{\mathrm {e}}^{2\,x}+16\,x^7+30\,x^8+12\,x^9\right )}{{\left ({\mathrm {e}}^x+x^2\right )}^3}-\frac {64\,x\,\ln \left (\frac {x+{\mathrm {e}}^x}{x}\right )\,\left (x+{\mathrm {e}}^x\right )\,\left (40\,x^4\,{\mathrm {e}}^x+68\,x^5\,{\mathrm {e}}^x+49\,x^6\,{\mathrm {e}}^x+19\,x^7\,{\mathrm {e}}^x+3\,x^8\,{\mathrm {e}}^x+32\,x^3\,{\mathrm {e}}^{2\,x}+50\,x^4\,{\mathrm {e}}^{2\,x}+18\,x^5\,{\mathrm {e}}^{2\,x}+24\,x^6+40\,x^7+15\,x^8\right )}{{\left ({\mathrm {e}}^x+x^2\right )}^3}}{x-\ln \left (\frac {x+{\mathrm {e}}^x}{x}\right )}+\frac {\frac {64\,x^4\,\left (x+2\right )\,\left (3\,x^2\,{\mathrm {e}}^x-2\,{\mathrm {e}}^x+3\,x\,{\mathrm {e}}^x+2\,x^2+2\,x^3\right )}{{\mathrm {e}}^x+x^2}-\frac {64\,x^4\,\ln \left (\frac {x+{\mathrm {e}}^x}{x}\right )\,\left (x+{\mathrm {e}}^x\right )\,\left (3\,x^2+10\,x+8\right )}{{\mathrm {e}}^x+x^2}}{x^2-2\,x\,\ln \left (\frac {x+{\mathrm {e}}^x}{x}\right )+{\ln \left (\frac {x+{\mathrm {e}}^x}{x}\right )}^2}-2048\,x^4-3200\,x^5-1152\,x^6-\frac {64\,\left (-3\,x^{11}+41\,x^{10}+13\,x^9-188\,x^8-28\,x^7+144\,x^6\right )}{\left ({\mathrm {e}}^x+x^2\right )\,\left (2\,x-x^2\right )}-\frac {64\,\left (-3\,x^{15}+8\,x^{14}+13\,x^{13}-46\,x^{12}+4\,x^{11}+56\,x^{10}-32\,x^9\right )}{\left (2\,x-x^2\right )\,\left ({\mathrm {e}}^{3\,x}+3\,x^4\,{\mathrm {e}}^x+3\,x^2\,{\mathrm {e}}^{2\,x}+x^6\right )}+\frac {64\,\left (-6\,x^{13}+31\,x^{12}+12\,x^{11}-151\,x^{10}+50\,x^9+144\,x^8-80\,x^7\right )}{\left (2\,x-x^2\right )\,\left ({\mathrm {e}}^{2\,x}+2\,x^2\,{\mathrm {e}}^x+x^4\right )} \] Input:

int(-(exp(x)*(512*x^4 - 512*x^3 + 1152*x^5 + 384*x^6) + 512*x^5 + 768*x^6 
+ 256*x^7 - log((x + exp(x))/x)*(exp(x)*(1024*x^3 + 1280*x^4 + 384*x^5) + 
1024*x^4 + 1280*x^5 + 384*x^6))/(x^3*exp(x) + log((x + exp(x))/x)^2*(3*x*e 
xp(x) + 3*x^2) - log((x + exp(x))/x)*(3*x^2*exp(x) + 3*x^3) - log((x + exp 
(x))/x)^3*(x + exp(x)) + x^4),x)
 

Output:

((64*x*(x + exp(x))*(24*x^5*exp(x) + 48*x^6*exp(x) + 43*x^7*exp(x) + 19*x^ 
8*exp(x) + 3*x^9*exp(x) - 8*x^3*exp(2*x) + 22*x^4*exp(2*x) + 47*x^5*exp(2* 
x) + 18*x^6*exp(2*x) + 16*x^7 + 30*x^8 + 12*x^9))/(exp(x) + x^2)^3 - (64*x 
*log((x + exp(x))/x)*(x + exp(x))*(40*x^4*exp(x) + 68*x^5*exp(x) + 49*x^6* 
exp(x) + 19*x^7*exp(x) + 3*x^8*exp(x) + 32*x^3*exp(2*x) + 50*x^4*exp(2*x) 
+ 18*x^5*exp(2*x) + 24*x^6 + 40*x^7 + 15*x^8))/(exp(x) + x^2)^3)/(x - log( 
(x + exp(x))/x)) + ((64*x^4*(x + 2)*(3*x^2*exp(x) - 2*exp(x) + 3*x*exp(x) 
+ 2*x^2 + 2*x^3))/(exp(x) + x^2) - (64*x^4*log((x + exp(x))/x)*(x + exp(x) 
)*(10*x + 3*x^2 + 8))/(exp(x) + x^2))/(log((x + exp(x))/x)^2 - 2*x*log((x 
+ exp(x))/x) + x^2) - 2048*x^4 - 3200*x^5 - 1152*x^6 - (64*(144*x^6 - 28*x 
^7 - 188*x^8 + 13*x^9 + 41*x^10 - 3*x^11))/((exp(x) + x^2)*(2*x - x^2)) - 
(64*(56*x^10 - 32*x^9 + 4*x^11 - 46*x^12 + 13*x^13 + 8*x^14 - 3*x^15))/((2 
*x - x^2)*(exp(3*x) + 3*x^4*exp(x) + 3*x^2*exp(2*x) + x^6)) + (64*(144*x^8 
 - 80*x^7 + 50*x^9 - 151*x^10 + 12*x^11 + 31*x^12 - 6*x^13))/((2*x - x^2)* 
(exp(2*x) + 2*x^2*exp(x) + x^4))
 

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.59 \[ \int \frac {512 x^5+768 x^6+256 x^7+e^x \left (-512 x^3+512 x^4+1152 x^5+384 x^6\right )+\left (-1024 x^4-1280 x^5-384 x^6+e^x \left (-1024 x^3-1280 x^4-384 x^5\right )\right ) \log \left (\frac {e^x+x}{x}\right )}{-e^x x^3-x^4+\left (3 e^x x^2+3 x^3\right ) \log \left (\frac {e^x+x}{x}\right )+\left (-3 e^x x-3 x^2\right ) \log ^2\left (\frac {e^x+x}{x}\right )+\left (e^x+x\right ) \log ^3\left (\frac {e^x+x}{x}\right )} \, dx=\frac {64 x^{4} \left (-x^{2}-4 x -4\right )}{\mathrm {log}\left (\frac {e^{x}+x}{x}\right )^{2}-2 \,\mathrm {log}\left (\frac {e^{x}+x}{x}\right ) x +x^{2}} \] Input:

int((((-384*x^5-1280*x^4-1024*x^3)*exp(x)-384*x^6-1280*x^5-1024*x^4)*log(1 
/x*(exp(x)+x))+(384*x^6+1152*x^5+512*x^4-512*x^3)*exp(x)+256*x^7+768*x^6+5 
12*x^5)/((exp(x)+x)*log(1/x*(exp(x)+x))^3+(-3*exp(x)*x-3*x^2)*log(1/x*(exp 
(x)+x))^2+(3*exp(x)*x^2+3*x^3)*log(1/x*(exp(x)+x))-exp(x)*x^3-x^4),x)
 

Output:

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