\(\int \frac {(6 x-e^2 x+6 e^x x+2 x^2) \log (x^2)+(4 x-2 e^2 x+4 e^x x+4 x^2+(4-2 e^2+4 e^x+4 x) \log (\frac {1}{4} (4+e^4+4 e^{2 x}+e^2 (-4-4 x)+8 x+4 x^2+e^x (8-4 e^2+8 x)))) \log (x+\log (\frac {1}{4} (4+e^4+4 e^{2 x}+e^2 (-4-4 x)+8 x+4 x^2+e^x (8-4 e^2+8 x))))}{4 x^2-2 e^2 x^2+4 e^x x^2+4 x^3+(4 x-2 e^2 x+4 e^x x+4 x^2) \log (\frac {1}{4} (4+e^4+4 e^{2 x}+e^2 (-4-4 x)+8 x+4 x^2+e^x (8-4 e^2+8 x)))} \, dx\) [963]

Optimal result

Integrand size = 264, antiderivative size = 27 \[ \int \frac {\left (6 x-e^2 x+6 e^x x+2 x^2\right ) \log \left (x^2\right )+\left (4 x-2 e^2 x+4 e^x x+4 x^2+\left (4-2 e^2+4 e^x+4 x\right ) \log \left (\frac {1}{4} \left (4+e^4+4 e^{2 x}+e^2 (-4-4 x)+8 x+4 x^2+e^x \left (8-4 e^2+8 x\right )\right )\right )\right ) \log \left (x+\log \left (\frac {1}{4} \left (4+e^4+4 e^{2 x}+e^2 (-4-4 x)+8 x+4 x^2+e^x \left (8-4 e^2+8 x\right )\right )\right )\right )}{4 x^2-2 e^2 x^2+4 e^x x^2+4 x^3+\left (4 x-2 e^2 x+4 e^x x+4 x^2\right ) \log \left (\frac {1}{4} \left (4+e^4+4 e^{2 x}+e^2 (-4-4 x)+8 x+4 x^2+e^x \left (8-4 e^2+8 x\right )\right )\right )} \, dx=\frac {1}{2} \log \left (x^2\right ) \log \left (x+\log \left (\left (1-\frac {e^2}{2}+e^x+x\right )^2\right )\right ) \] Output:

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

Mathematica [A] (verified)

Time = 0.61 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.22 \[ \int \frac {\left (6 x-e^2 x+6 e^x x+2 x^2\right ) \log \left (x^2\right )+\left (4 x-2 e^2 x+4 e^x x+4 x^2+\left (4-2 e^2+4 e^x+4 x\right ) \log \left (\frac {1}{4} \left (4+e^4+4 e^{2 x}+e^2 (-4-4 x)+8 x+4 x^2+e^x \left (8-4 e^2+8 x\right )\right )\right )\right ) \log \left (x+\log \left (\frac {1}{4} \left (4+e^4+4 e^{2 x}+e^2 (-4-4 x)+8 x+4 x^2+e^x \left (8-4 e^2+8 x\right )\right )\right )\right )}{4 x^2-2 e^2 x^2+4 e^x x^2+4 x^3+\left (4 x-2 e^2 x+4 e^x x+4 x^2\right ) \log \left (\frac {1}{4} \left (4+e^4+4 e^{2 x}+e^2 (-4-4 x)+8 x+4 x^2+e^x \left (8-4 e^2+8 x\right )\right )\right )} \, dx=\frac {1}{2} \log \left (x^2\right ) \log \left (x+\log \left (\frac {1}{4} \left (2-e^2+2 e^x+2 x\right )^2\right )\right ) \] Input:

Integrate[((6*x - E^2*x + 6*E^x*x + 2*x^2)*Log[x^2] + (4*x - 2*E^2*x + 4*E 
^x*x + 4*x^2 + (4 - 2*E^2 + 4*E^x + 4*x)*Log[(4 + E^4 + 4*E^(2*x) + E^2*(- 
4 - 4*x) + 8*x + 4*x^2 + E^x*(8 - 4*E^2 + 8*x))/4])*Log[x + Log[(4 + E^4 + 
 4*E^(2*x) + E^2*(-4 - 4*x) + 8*x + 4*x^2 + E^x*(8 - 4*E^2 + 8*x))/4]])/(4 
*x^2 - 2*E^2*x^2 + 4*E^x*x^2 + 4*x^3 + (4*x - 2*E^2*x + 4*E^x*x + 4*x^2)*L 
og[(4 + E^4 + 4*E^(2*x) + E^2*(-4 - 4*x) + 8*x + 4*x^2 + E^x*(8 - 4*E^2 + 
8*x))/4]),x]
 

Output:

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

\(\displaystyle \int \frac {\left (2 x^2+6 e^x x-e^2 x+6 x\right ) \log \left (x^2\right )+\left (4 x^2+\left (4 x+4 e^x-2 e^2+4\right ) \log \left (\frac {1}{4} \left (4 x^2+8 x+4 e^{2 x}+e^2 (-4 x-4)+e^x \left (8 x-4 e^2+8\right )+e^4+4\right )\right )+4 e^x x-2 e^2 x+4 x\right ) \log \left (\log \left (\frac {1}{4} \left (4 x^2+8 x+4 e^{2 x}+e^2 (-4 x-4)+e^x \left (8 x-4 e^2+8\right )+e^4+4\right )\right )+x\right )}{4 x^3+4 e^x x^2-2 e^2 x^2+4 x^2+\left (4 x^2+4 e^x x-2 e^2 x+4 x\right ) \log \left (\frac {1}{4} \left (4 x^2+8 x+4 e^{2 x}+e^2 (-4 x-4)+e^x \left (8 x-4 e^2+8\right )+e^4+4\right )\right )} \, dx\)

\(\Big \downarrow \) 6

\(\displaystyle \int \frac {\left (2 x^2+6 e^x x-e^2 x+6 x\right ) \log \left (x^2\right )+\left (4 x^2+\left (4 x+4 e^x-2 e^2+4\right ) \log \left (\frac {1}{4} \left (4 x^2+8 x+4 e^{2 x}+e^2 (-4 x-4)+e^x \left (8 x-4 e^2+8\right )+e^4+4\right )\right )+4 e^x x-2 e^2 x+4 x\right ) \log \left (\log \left (\frac {1}{4} \left (4 x^2+8 x+4 e^{2 x}+e^2 (-4 x-4)+e^x \left (8 x-4 e^2+8\right )+e^4+4\right )\right )+x\right )}{4 x^3+4 e^x x^2+\left (4-2 e^2\right ) x^2+\left (4 x^2+4 e^x x-2 e^2 x+4 x\right ) \log \left (\frac {1}{4} \left (4 x^2+8 x+4 e^{2 x}+e^2 (-4 x-4)+e^x \left (8 x-4 e^2+8\right )+e^4+4\right )\right )}dx\)

\(\Big \downarrow \) 7239

\(\displaystyle \int \left (\frac {\left (2 x+6 e^x+6 \left (1-\frac {e^2}{6}\right )\right ) \log \left (x^2\right )}{2 \left (2 x+2 e^x+2 \left (1-\frac {e^2}{2}\right )\right ) \left (x+\log \left (\frac {1}{4} \left (2 x+2 e^x-e^2+2\right )^2\right )\right )}+\frac {\log \left (x+\log \left (\frac {1}{4} \left (2 x+2 e^x-e^2+2\right )^2\right )\right )}{x}\right )dx\)

\(\Big \downarrow \) 2009

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

Input:

Int[((6*x - E^2*x + 6*E^x*x + 2*x^2)*Log[x^2] + (4*x - 2*E^2*x + 4*E^x*x + 
 4*x^2 + (4 - 2*E^2 + 4*E^x + 4*x)*Log[(4 + E^4 + 4*E^(2*x) + E^2*(-4 - 4* 
x) + 8*x + 4*x^2 + E^x*(8 - 4*E^2 + 8*x))/4])*Log[x + Log[(4 + E^4 + 4*E^( 
2*x) + E^2*(-4 - 4*x) + 8*x + 4*x^2 + E^x*(8 - 4*E^2 + 8*x))/4]])/(4*x^2 - 
 2*E^2*x^2 + 4*E^x*x^2 + 4*x^3 + (4*x - 2*E^2*x + 4*E^x*x + 4*x^2)*Log[(4 
+ E^4 + 4*E^(2*x) + E^2*(-4 - 4*x) + 8*x + 4*x^2 + E^x*(8 - 4*E^2 + 8*x))/ 
4]),x]
 

Output:

$Aborted
 

Maple [C] (warning: unable to verify)

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

Time = 37.40 (sec) , antiderivative size = 247, normalized size of antiderivative = 9.15

method	result	size
risch	\(\ln \left (x \right ) \ln \left (-2 \ln \left (2\right )+2 \ln \left ({\mathrm e}^{2}-2 \,{\mathrm e}^{x}-2 x -2\right )-\frac {i \pi \,\operatorname {csgn}\left (i \left ({\mathrm e}^{2}-2 \,{\mathrm e}^{x}-2 x -2\right )^{2}\right ) {\left (-\operatorname {csgn}\left (i \left ({\mathrm e}^{2}-2 \,{\mathrm e}^{x}-2 x -2\right )^{2}\right )+\operatorname {csgn}\left (i \left ({\mathrm e}^{2}-2 \,{\mathrm e}^{x}-2 x -2\right )\right )\right )}^{2}}{2}+x \right )-\frac {i \pi \,\operatorname {csgn}\left (i x^{2}\right ) \left (\operatorname {csgn}\left (i x^{2}\right )^{2}-2 \,\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x \right )+\operatorname {csgn}\left (i x \right )^{2}\right ) \ln \left (\ln \left ({\mathrm e}^{2}-2 \,{\mathrm e}^{x}-2 x -2\right )-\frac {i \left (\pi {\operatorname {csgn}\left (i \left ({\mathrm e}^{2}-2 \,{\mathrm e}^{x}-2 x -2\right )\right )}^{2} \operatorname {csgn}\left (i \left ({\mathrm e}^{2}-2 \,{\mathrm e}^{x}-2 x -2\right )^{2}\right )-2 \pi \,\operatorname {csgn}\left (i \left ({\mathrm e}^{2}-2 \,{\mathrm e}^{x}-2 x -2\right )\right ) {\operatorname {csgn}\left (i \left ({\mathrm e}^{2}-2 \,{\mathrm e}^{x}-2 x -2\right )^{2}\right )}^{2}+\pi {\operatorname {csgn}\left (i \left ({\mathrm e}^{2}-2 \,{\mathrm e}^{x}-2 x -2\right )^{2}\right )}^{3}-4 i \ln \left (2\right )+2 i x \right )}{4}\right )}{4}\)	\(247\)

Input:

int((((4*exp(x)-2*exp(2)+4*x+4)*ln(exp(x)^2+1/4*(-4*exp(2)+8*x+8)*exp(x)+1 
/4*exp(2)^2+1/4*(-4-4*x)*exp(2)+x^2+2*x+1)+4*exp(x)*x-2*exp(2)*x+4*x^2+4*x 
)*ln(ln(exp(x)^2+1/4*(-4*exp(2)+8*x+8)*exp(x)+1/4*exp(2)^2+1/4*(-4-4*x)*ex 
p(2)+x^2+2*x+1)+x)+(6*exp(x)*x-exp(2)*x+2*x^2+6*x)*ln(x^2))/((4*exp(x)*x-2 
*exp(2)*x+4*x^2+4*x)*ln(exp(x)^2+1/4*(-4*exp(2)+8*x+8)*exp(x)+1/4*exp(2)^2 
+1/4*(-4-4*x)*exp(2)+x^2+2*x+1)+4*exp(x)*x^2-2*x^2*exp(2)+4*x^3+4*x^2),x,m 
ethod=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.67 \[ \int \frac {\left (6 x-e^2 x+6 e^x x+2 x^2\right ) \log \left (x^2\right )+\left (4 x-2 e^2 x+4 e^x x+4 x^2+\left (4-2 e^2+4 e^x+4 x\right ) \log \left (\frac {1}{4} \left (4+e^4+4 e^{2 x}+e^2 (-4-4 x)+8 x+4 x^2+e^x \left (8-4 e^2+8 x\right )\right )\right )\right ) \log \left (x+\log \left (\frac {1}{4} \left (4+e^4+4 e^{2 x}+e^2 (-4-4 x)+8 x+4 x^2+e^x \left (8-4 e^2+8 x\right )\right )\right )\right )}{4 x^2-2 e^2 x^2+4 e^x x^2+4 x^3+\left (4 x-2 e^2 x+4 e^x x+4 x^2\right ) \log \left (\frac {1}{4} \left (4+e^4+4 e^{2 x}+e^2 (-4-4 x)+8 x+4 x^2+e^x \left (8-4 e^2+8 x\right )\right )\right )} \, dx=\frac {1}{2} \, \log \left (x^{2}\right ) \log \left (x + \log \left (x^{2} - {\left (x + 1\right )} e^{2} + {\left (2 \, x - e^{2} + 2\right )} e^{x} + 2 \, x + \frac {1}{4} \, e^{4} + e^{\left (2 \, x\right )} + 1\right )\right ) \] Input:

integrate((((4*exp(x)-2*exp(2)+4*x+4)*log(exp(x)^2+1/4*(-4*exp(2)+8*x+8)*e 
xp(x)+1/4*exp(2)^2+1/4*(-4-4*x)*exp(2)+x^2+2*x+1)+4*exp(x)*x-2*exp(2)*x+4* 
x^2+4*x)*log(log(exp(x)^2+1/4*(-4*exp(2)+8*x+8)*exp(x)+1/4*exp(2)^2+1/4*(- 
4-4*x)*exp(2)+x^2+2*x+1)+x)+(6*exp(x)*x-exp(2)*x+2*x^2+6*x)*log(x^2))/((4* 
exp(x)*x-2*exp(2)*x+4*x^2+4*x)*log(exp(x)^2+1/4*(-4*exp(2)+8*x+8)*exp(x)+1 
/4*exp(2)^2+1/4*(-4-4*x)*exp(2)+x^2+2*x+1)+4*exp(x)*x^2-2*x^2*exp(2)+4*x^3 
+4*x^2),x, algorithm="fricas")
 

Output:

1/2*log(x^2)*log(x + log(x^2 - (x + 1)*e^2 + (2*x - e^2 + 2)*e^x + 2*x + 1 
/4*e^4 + e^(2*x) + 1))
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (6 x-e^2 x+6 e^x x+2 x^2\right ) \log \left (x^2\right )+\left (4 x-2 e^2 x+4 e^x x+4 x^2+\left (4-2 e^2+4 e^x+4 x\right ) \log \left (\frac {1}{4} \left (4+e^4+4 e^{2 x}+e^2 (-4-4 x)+8 x+4 x^2+e^x \left (8-4 e^2+8 x\right )\right )\right )\right ) \log \left (x+\log \left (\frac {1}{4} \left (4+e^4+4 e^{2 x}+e^2 (-4-4 x)+8 x+4 x^2+e^x \left (8-4 e^2+8 x\right )\right )\right )\right )}{4 x^2-2 e^2 x^2+4 e^x x^2+4 x^3+\left (4 x-2 e^2 x+4 e^x x+4 x^2\right ) \log \left (\frac {1}{4} \left (4+e^4+4 e^{2 x}+e^2 (-4-4 x)+8 x+4 x^2+e^x \left (8-4 e^2+8 x\right )\right )\right )} \, dx=\text {Timed out} \] Input:

integrate((((4*exp(x)-2*exp(2)+4*x+4)*ln(exp(x)**2+1/4*(-4*exp(2)+8*x+8)*e 
xp(x)+1/4*exp(2)**2+1/4*(-4-4*x)*exp(2)+x**2+2*x+1)+4*exp(x)*x-2*exp(2)*x+ 
4*x**2+4*x)*ln(ln(exp(x)**2+1/4*(-4*exp(2)+8*x+8)*exp(x)+1/4*exp(2)**2+1/4 
*(-4-4*x)*exp(2)+x**2+2*x+1)+x)+(6*exp(x)*x-exp(2)*x+2*x**2+6*x)*ln(x**2)) 
/((4*exp(x)*x-2*exp(2)*x+4*x**2+4*x)*ln(exp(x)**2+1/4*(-4*exp(2)+8*x+8)*ex 
p(x)+1/4*exp(2)**2+1/4*(-4-4*x)*exp(2)+x**2+2*x+1)+4*exp(x)*x**2-2*x**2*ex 
p(2)+4*x**3+4*x**2),x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.33 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int \frac {\left (6 x-e^2 x+6 e^x x+2 x^2\right ) \log \left (x^2\right )+\left (4 x-2 e^2 x+4 e^x x+4 x^2+\left (4-2 e^2+4 e^x+4 x\right ) \log \left (\frac {1}{4} \left (4+e^4+4 e^{2 x}+e^2 (-4-4 x)+8 x+4 x^2+e^x \left (8-4 e^2+8 x\right )\right )\right )\right ) \log \left (x+\log \left (\frac {1}{4} \left (4+e^4+4 e^{2 x}+e^2 (-4-4 x)+8 x+4 x^2+e^x \left (8-4 e^2+8 x\right )\right )\right )\right )}{4 x^2-2 e^2 x^2+4 e^x x^2+4 x^3+\left (4 x-2 e^2 x+4 e^x x+4 x^2\right ) \log \left (\frac {1}{4} \left (4+e^4+4 e^{2 x}+e^2 (-4-4 x)+8 x+4 x^2+e^x \left (8-4 e^2+8 x\right )\right )\right )} \, dx=\log \left (x - 2 \, \log \left (2\right ) + 2 \, \log \left (2 \, x - e^{2} + 2 \, e^{x} + 2\right )\right ) \log \left (x\right ) \] Input:

integrate((((4*exp(x)-2*exp(2)+4*x+4)*log(exp(x)^2+1/4*(-4*exp(2)+8*x+8)*e 
xp(x)+1/4*exp(2)^2+1/4*(-4-4*x)*exp(2)+x^2+2*x+1)+4*exp(x)*x-2*exp(2)*x+4* 
x^2+4*x)*log(log(exp(x)^2+1/4*(-4*exp(2)+8*x+8)*exp(x)+1/4*exp(2)^2+1/4*(- 
4-4*x)*exp(2)+x^2+2*x+1)+x)+(6*exp(x)*x-exp(2)*x+2*x^2+6*x)*log(x^2))/((4* 
exp(x)*x-2*exp(2)*x+4*x^2+4*x)*log(exp(x)^2+1/4*(-4*exp(2)+8*x+8)*exp(x)+1 
/4*exp(2)^2+1/4*(-4-4*x)*exp(2)+x^2+2*x+1)+4*exp(x)*x^2-2*x^2*exp(2)+4*x^3 
+4*x^2),x, algorithm="maxima")
 

Output:

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

Giac [F]

\[ \int \frac {\left (6 x-e^2 x+6 e^x x+2 x^2\right ) \log \left (x^2\right )+\left (4 x-2 e^2 x+4 e^x x+4 x^2+\left (4-2 e^2+4 e^x+4 x\right ) \log \left (\frac {1}{4} \left (4+e^4+4 e^{2 x}+e^2 (-4-4 x)+8 x+4 x^2+e^x \left (8-4 e^2+8 x\right )\right )\right )\right ) \log \left (x+\log \left (\frac {1}{4} \left (4+e^4+4 e^{2 x}+e^2 (-4-4 x)+8 x+4 x^2+e^x \left (8-4 e^2+8 x\right )\right )\right )\right )}{4 x^2-2 e^2 x^2+4 e^x x^2+4 x^3+\left (4 x-2 e^2 x+4 e^x x+4 x^2\right ) \log \left (\frac {1}{4} \left (4+e^4+4 e^{2 x}+e^2 (-4-4 x)+8 x+4 x^2+e^x \left (8-4 e^2+8 x\right )\right )\right )} \, dx=\int { \frac {{\left (2 \, x^{2} - x e^{2} + 6 \, x e^{x} + 6 \, x\right )} \log \left (x^{2}\right ) + 2 \, {\left (2 \, x^{2} - x e^{2} + 2 \, x e^{x} + {\left (2 \, x - e^{2} + 2 \, e^{x} + 2\right )} \log \left (x^{2} - {\left (x + 1\right )} e^{2} + {\left (2 \, x - e^{2} + 2\right )} e^{x} + 2 \, x + \frac {1}{4} \, e^{4} + e^{\left (2 \, x\right )} + 1\right ) + 2 \, x\right )} \log \left (x + \log \left (x^{2} - {\left (x + 1\right )} e^{2} + {\left (2 \, x - e^{2} + 2\right )} e^{x} + 2 \, x + \frac {1}{4} \, e^{4} + e^{\left (2 \, x\right )} + 1\right )\right )}{2 \, {\left (2 \, x^{3} - x^{2} e^{2} + 2 \, x^{2} e^{x} + 2 \, x^{2} + {\left (2 \, x^{2} - x e^{2} + 2 \, x e^{x} + 2 \, x\right )} \log \left (x^{2} - {\left (x + 1\right )} e^{2} + {\left (2 \, x - e^{2} + 2\right )} e^{x} + 2 \, x + \frac {1}{4} \, e^{4} + e^{\left (2 \, x\right )} + 1\right )\right )}} \,d x } \] Input:

integrate((((4*exp(x)-2*exp(2)+4*x+4)*log(exp(x)^2+1/4*(-4*exp(2)+8*x+8)*e 
xp(x)+1/4*exp(2)^2+1/4*(-4-4*x)*exp(2)+x^2+2*x+1)+4*exp(x)*x-2*exp(2)*x+4* 
x^2+4*x)*log(log(exp(x)^2+1/4*(-4*exp(2)+8*x+8)*exp(x)+1/4*exp(2)^2+1/4*(- 
4-4*x)*exp(2)+x^2+2*x+1)+x)+(6*exp(x)*x-exp(2)*x+2*x^2+6*x)*log(x^2))/((4* 
exp(x)*x-2*exp(2)*x+4*x^2+4*x)*log(exp(x)^2+1/4*(-4*exp(2)+8*x+8)*exp(x)+1 
/4*exp(2)^2+1/4*(-4-4*x)*exp(2)+x^2+2*x+1)+4*exp(x)*x^2-2*x^2*exp(2)+4*x^3 
+4*x^2),x, algorithm="giac")
 

Output:

integrate(1/2*((2*x^2 - x*e^2 + 6*x*e^x + 6*x)*log(x^2) + 2*(2*x^2 - x*e^2 
 + 2*x*e^x + (2*x - e^2 + 2*e^x + 2)*log(x^2 - (x + 1)*e^2 + (2*x - e^2 + 
2)*e^x + 2*x + 1/4*e^4 + e^(2*x) + 1) + 2*x)*log(x + log(x^2 - (x + 1)*e^2 
 + (2*x - e^2 + 2)*e^x + 2*x + 1/4*e^4 + e^(2*x) + 1)))/(2*x^3 - x^2*e^2 + 
 2*x^2*e^x + 2*x^2 + (2*x^2 - x*e^2 + 2*x*e^x + 2*x)*log(x^2 - (x + 1)*e^2 
 + (2*x - e^2 + 2)*e^x + 2*x + 1/4*e^4 + e^(2*x) + 1)), x)
 

Mupad [B] (verification not implemented)

Time = 3.43 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.78 \[ \int \frac {\left (6 x-e^2 x+6 e^x x+2 x^2\right ) \log \left (x^2\right )+\left (4 x-2 e^2 x+4 e^x x+4 x^2+\left (4-2 e^2+4 e^x+4 x\right ) \log \left (\frac {1}{4} \left (4+e^4+4 e^{2 x}+e^2 (-4-4 x)+8 x+4 x^2+e^x \left (8-4 e^2+8 x\right )\right )\right )\right ) \log \left (x+\log \left (\frac {1}{4} \left (4+e^4+4 e^{2 x}+e^2 (-4-4 x)+8 x+4 x^2+e^x \left (8-4 e^2+8 x\right )\right )\right )\right )}{4 x^2-2 e^2 x^2+4 e^x x^2+4 x^3+\left (4 x-2 e^2 x+4 e^x x+4 x^2\right ) \log \left (\frac {1}{4} \left (4+e^4+4 e^{2 x}+e^2 (-4-4 x)+8 x+4 x^2+e^x \left (8-4 e^2+8 x\right )\right )\right )} \, dx=\frac {\ln \left (x^2\right )\,\ln \left (x+\ln \left (2\,x+{\mathrm {e}}^{2\,x}+\frac {{\mathrm {e}}^4}{4}+\frac {{\mathrm {e}}^x\,\left (8\,x-4\,{\mathrm {e}}^2+8\right )}{4}+x^2-\frac {{\mathrm {e}}^2\,\left (4\,x+4\right )}{4}+1\right )\right )}{2} \] Input:

int((log(x^2)*(6*x - x*exp(2) + 6*x*exp(x) + 2*x^2) + log(x + log(2*x + ex 
p(2*x) + exp(4)/4 + (exp(x)*(8*x - 4*exp(2) + 8))/4 + x^2 - (exp(2)*(4*x + 
 4))/4 + 1))*(4*x - 2*x*exp(2) + log(2*x + exp(2*x) + exp(4)/4 + (exp(x)*( 
8*x - 4*exp(2) + 8))/4 + x^2 - (exp(2)*(4*x + 4))/4 + 1)*(4*x - 2*exp(2) + 
 4*exp(x) + 4) + 4*x*exp(x) + 4*x^2))/(4*x^2*exp(x) + log(2*x + exp(2*x) + 
 exp(4)/4 + (exp(x)*(8*x - 4*exp(2) + 8))/4 + x^2 - (exp(2)*(4*x + 4))/4 + 
 1)*(4*x - 2*x*exp(2) + 4*x*exp(x) + 4*x^2) - 2*x^2*exp(2) + 4*x^2 + 4*x^3 
),x)
 

Output:

(log(x^2)*log(x + log(2*x + exp(2*x) + exp(4)/4 + (exp(x)*(8*x - 4*exp(2) 
+ 8))/4 + x^2 - (exp(2)*(4*x + 4))/4 + 1)))/2
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 58, normalized size of antiderivative = 2.15 \[ \int \frac {\left (6 x-e^2 x+6 e^x x+2 x^2\right ) \log \left (x^2\right )+\left (4 x-2 e^2 x+4 e^x x+4 x^2+\left (4-2 e^2+4 e^x+4 x\right ) \log \left (\frac {1}{4} \left (4+e^4+4 e^{2 x}+e^2 (-4-4 x)+8 x+4 x^2+e^x \left (8-4 e^2+8 x\right )\right )\right )\right ) \log \left (x+\log \left (\frac {1}{4} \left (4+e^4+4 e^{2 x}+e^2 (-4-4 x)+8 x+4 x^2+e^x \left (8-4 e^2+8 x\right )\right )\right )\right )}{4 x^2-2 e^2 x^2+4 e^x x^2+4 x^3+\left (4 x-2 e^2 x+4 e^x x+4 x^2\right ) \log \left (\frac {1}{4} \left (4+e^4+4 e^{2 x}+e^2 (-4-4 x)+8 x+4 x^2+e^x \left (8-4 e^2+8 x\right )\right )\right )} \, dx=\frac {\mathrm {log}\left (x^{2}\right ) \mathrm {log}\left (\mathrm {log}\left (e^{2 x}-e^{x} e^{2}+2 e^{x} x +2 e^{x}+\frac {e^{4}}{4}-e^{2} x -e^{2}+x^{2}+2 x +1\right )+x \right )}{2} \] Input:

int((((4*exp(x)-2*exp(2)+4*x+4)*log(exp(x)^2+1/4*(-4*exp(2)+8*x+8)*exp(x)+ 
1/4*exp(2)^2+1/4*(-4-4*x)*exp(2)+x^2+2*x+1)+4*exp(x)*x-2*exp(2)*x+4*x^2+4* 
x)*log(log(exp(x)^2+1/4*(-4*exp(2)+8*x+8)*exp(x)+1/4*exp(2)^2+1/4*(-4-4*x) 
*exp(2)+x^2+2*x+1)+x)+(6*exp(x)*x-exp(2)*x+2*x^2+6*x)*log(x^2))/((4*exp(x) 
*x-2*exp(2)*x+4*x^2+4*x)*log(exp(x)^2+1/4*(-4*exp(2)+8*x+8)*exp(x)+1/4*exp 
(2)^2+1/4*(-4-4*x)*exp(2)+x^2+2*x+1)+4*exp(x)*x^2-2*x^2*exp(2)+4*x^3+4*x^2 
),x)
 

Output:

(log(x**2)*log(log((4*e**(2*x) - 4*e**x*e**2 + 8*e**x*x + 8*e**x + e**4 - 
4*e**2*x - 4*e**2 + 4*x**2 + 8*x + 4)/4) + x))/2