\(\int \frac {e^{\log ^2(\frac {1}{2} (4+x^2+\log (2+e^x (-2+x))))} (8 x+e^x (-2-6 x+4 x^2)) \log (\frac {1}{2} (4+x^2+\log (2+e^x (-2+x))))}{8+2 x^2+e^x (-8+4 x-2 x^2+x^3)+(2+e^x (-2+x)) \log (2+e^x (-2+x))} \, dx\) [1579]

Optimal result

Integrand size = 109, antiderivative size = 28 \[ \int \frac {e^{\log ^2\left (\frac {1}{2} \left (4+x^2+\log \left (2+e^x (-2+x)\right )\right )\right )} \left (8 x+e^x \left (-2-6 x+4 x^2\right )\right ) \log \left (\frac {1}{2} \left (4+x^2+\log \left (2+e^x (-2+x)\right )\right )\right )}{8+2 x^2+e^x \left (-8+4 x-2 x^2+x^3\right )+\left (2+e^x (-2+x)\right ) \log \left (2+e^x (-2+x)\right )} \, dx=e^{\log ^2\left (2+\frac {1}{2} \left (x^2+\log \left (2-e^x (2-x)\right )\right )\right )} \] Output:

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

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86 \[ \int \frac {e^{\log ^2\left (\frac {1}{2} \left (4+x^2+\log \left (2+e^x (-2+x)\right )\right )\right )} \left (8 x+e^x \left (-2-6 x+4 x^2\right )\right ) \log \left (\frac {1}{2} \left (4+x^2+\log \left (2+e^x (-2+x)\right )\right )\right )}{8+2 x^2+e^x \left (-8+4 x-2 x^2+x^3\right )+\left (2+e^x (-2+x)\right ) \log \left (2+e^x (-2+x)\right )} \, dx=e^{\log ^2\left (\frac {1}{2} \left (4+x^2+\log \left (2+e^x (-2+x)\right )\right )\right )} \] Input:

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

Output:

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

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

\(\Big \downarrow \) 7292

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

\(\Big \downarrow \) 7293

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

$Aborted
 

Maple [A] (verified)

Time = 96.33 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.82

method	result	size
risch	\({\mathrm e}^{{\ln \left (\frac {\ln \left ({\mathrm e}^{x} \left (-2+x \right )+2\right )}{2}+\frac {x^{2}}{2}+2\right )}^{2}}\)	\(23\)
parallelrisch	\({\mathrm e}^{{\ln \left (\frac {\ln \left ({\mathrm e}^{x} \left (-2+x \right )+2\right )}{2}+\frac {x^{2}}{2}+2\right )}^{2}}\)	\(23\)

Input:

int(((4*x^2-6*x-2)*exp(x)+8*x)*ln(1/2*ln(exp(x)*(-2+x)+2)+1/2*x^2+2)*exp(l 
n(1/2*ln(exp(x)*(-2+x)+2)+1/2*x^2+2)^2)/((exp(x)*(-2+x)+2)*ln(exp(x)*(-2+x 
)+2)+(x^3-2*x^2+4*x-8)*exp(x)+2*x^2+8),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.79 \[ \int \frac {e^{\log ^2\left (\frac {1}{2} \left (4+x^2+\log \left (2+e^x (-2+x)\right )\right )\right )} \left (8 x+e^x \left (-2-6 x+4 x^2\right )\right ) \log \left (\frac {1}{2} \left (4+x^2+\log \left (2+e^x (-2+x)\right )\right )\right )}{8+2 x^2+e^x \left (-8+4 x-2 x^2+x^3\right )+\left (2+e^x (-2+x)\right ) \log \left (2+e^x (-2+x)\right )} \, dx=e^{\left (\log \left (\frac {1}{2} \, x^{2} + \frac {1}{2} \, \log \left ({\left (x - 2\right )} e^{x} + 2\right ) + 2\right )^{2}\right )} \] Input:

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.50 \[ \int \frac {e^{\log ^2\left (\frac {1}{2} \left (4+x^2+\log \left (2+e^x (-2+x)\right )\right )\right )} \left (8 x+e^x \left (-2-6 x+4 x^2\right )\right ) \log \left (\frac {1}{2} \left (4+x^2+\log \left (2+e^x (-2+x)\right )\right )\right )}{8+2 x^2+e^x \left (-8+4 x-2 x^2+x^3\right )+\left (2+e^x (-2+x)\right ) \log \left (2+e^x (-2+x)\right )} \, dx=e^{\left (\log \left (2\right )^{2} - 2 \, \log \left (2\right ) \log \left (x^{2} + \log \left ({\left (x - 2\right )} e^{x} + 2\right ) + 4\right ) + \log \left (x^{2} + \log \left ({\left (x - 2\right )} e^{x} + 2\right ) + 4\right )^{2}\right )} \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86 \[ \int \frac {e^{\log ^2\left (\frac {1}{2} \left (4+x^2+\log \left (2+e^x (-2+x)\right )\right )\right )} \left (8 x+e^x \left (-2-6 x+4 x^2\right )\right ) \log \left (\frac {1}{2} \left (4+x^2+\log \left (2+e^x (-2+x)\right )\right )\right )}{8+2 x^2+e^x \left (-8+4 x-2 x^2+x^3\right )+\left (2+e^x (-2+x)\right ) \log \left (2+e^x (-2+x)\right )} \, dx=e^{\left (\log \left (\frac {1}{2} \, x^{2} + \frac {1}{2} \, \log \left (x e^{x} - 2 \, e^{x} + 2\right ) + 2\right )^{2}\right )} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 4.02 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86 \[ \int \frac {e^{\log ^2\left (\frac {1}{2} \left (4+x^2+\log \left (2+e^x (-2+x)\right )\right )\right )} \left (8 x+e^x \left (-2-6 x+4 x^2\right )\right ) \log \left (\frac {1}{2} \left (4+x^2+\log \left (2+e^x (-2+x)\right )\right )\right )}{8+2 x^2+e^x \left (-8+4 x-2 x^2+x^3\right )+\left (2+e^x (-2+x)\right ) \log \left (2+e^x (-2+x)\right )} \, dx={\mathrm {e}}^{{\ln \left (\frac {\ln \left (x\,{\mathrm {e}}^x-2\,{\mathrm {e}}^x+2\right )}{2}+\frac {x^2}{2}+2\right )}^2} \] Input:

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

Output:

exp(log(log(x*exp(x) - 2*exp(x) + 2)/2 + x^2/2 + 2)^2)
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.96 \[ \int \frac {e^{\log ^2\left (\frac {1}{2} \left (4+x^2+\log \left (2+e^x (-2+x)\right )\right )\right )} \left (8 x+e^x \left (-2-6 x+4 x^2\right )\right ) \log \left (\frac {1}{2} \left (4+x^2+\log \left (2+e^x (-2+x)\right )\right )\right )}{8+2 x^2+e^x \left (-8+4 x-2 x^2+x^3\right )+\left (2+e^x (-2+x)\right ) \log \left (2+e^x (-2+x)\right )} \, dx=e^{{\mathrm {log}\left (\frac {\mathrm {log}\left (e^{x} x -2 e^{x}+2\right )}{2}+\frac {x^{2}}{2}+2\right )}^{2}} \] Input:

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

Output:

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