\(\int \frac {-2 x+x^2-2 \log (\frac {e^x}{2}) \log (\log (\frac {e^x}{2})) \log (\log (\log (\frac {e^x}{2})))+(-1+x^2) \log (\frac {e^x}{2}) \log (\log (\frac {e^x}{2})) \log ^2(\log (\log (\frac {e^x}{2})))}{x^2 \log (\frac {e^x}{2}) \log (\log (\frac {e^x}{2})) \log ^2(\log (\log (\frac {e^x}{2})))} \, dx\) [2816]

Optimal result

Integrand size = 108, antiderivative size = 25 \[ \int \frac {-2 x+x^2-2 \log \left (\frac {e^x}{2}\right ) \log \left (\log \left (\frac {e^x}{2}\right )\right ) \log \left (\log \left (\log \left (\frac {e^x}{2}\right )\right )\right )+\left (-1+x^2\right ) \log \left (\frac {e^x}{2}\right ) \log \left (\log \left (\frac {e^x}{2}\right )\right ) \log ^2\left (\log \left (\log \left (\frac {e^x}{2}\right )\right )\right )}{x^2 \log \left (\frac {e^x}{2}\right ) \log \left (\log \left (\frac {e^x}{2}\right )\right ) \log ^2\left (\log \left (\log \left (\frac {e^x}{2}\right )\right )\right )} \, dx=x+\frac {1-\frac {-2+x}{\log \left (\log \left (\log \left (\frac {e^x}{2}\right )\right )\right )}}{x} \] Output:

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

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.04 \[ \int \frac {-2 x+x^2-2 \log \left (\frac {e^x}{2}\right ) \log \left (\log \left (\frac {e^x}{2}\right )\right ) \log \left (\log \left (\log \left (\frac {e^x}{2}\right )\right )\right )+\left (-1+x^2\right ) \log \left (\frac {e^x}{2}\right ) \log \left (\log \left (\frac {e^x}{2}\right )\right ) \log ^2\left (\log \left (\log \left (\frac {e^x}{2}\right )\right )\right )}{x^2 \log \left (\frac {e^x}{2}\right ) \log \left (\log \left (\frac {e^x}{2}\right )\right ) \log ^2\left (\log \left (\log \left (\frac {e^x}{2}\right )\right )\right )} \, dx=\frac {1}{x}+x+\frac {2-x}{x \log \left (\log \left (\log \left (\frac {e^x}{2}\right )\right )\right )} \] Input:

Integrate[(-2*x + x^2 - 2*Log[E^x/2]*Log[Log[E^x/2]]*Log[Log[Log[E^x/2]]] 
+ (-1 + x^2)*Log[E^x/2]*Log[Log[E^x/2]]*Log[Log[Log[E^x/2]]]^2)/(x^2*Log[E 
^x/2]*Log[Log[E^x/2]]*Log[Log[Log[E^x/2]]]^2),x]
 

Output:

x^(-1) + x + (2 - x)/(x*Log[Log[Log[E^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.

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

\(\Big \downarrow \) 7293

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

\(\Big \downarrow \) 2009

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

Input:

Int[(-2*x + x^2 - 2*Log[E^x/2]*Log[Log[E^x/2]]*Log[Log[Log[E^x/2]]] + (-1 
+ x^2)*Log[E^x/2]*Log[Log[E^x/2]]*Log[Log[Log[E^x/2]]]^2)/(x^2*Log[E^x/2]* 
Log[Log[E^x/2]]*Log[Log[Log[E^x/2]]]^2),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 2.97 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.24

method	result	size
risch	\(\frac {x^{2}+1}{x}-\frac {-2+x}{x \ln \left (\ln \left (-\ln \left (2\right )+\ln \left ({\mathrm e}^{x}\right )\right )\right )}\)	\(31\)
parallelrisch	\(-\frac {-12-6 \ln \left (\frac {{\mathrm e}^{x}}{2}\right ) \ln \left (\ln \left (\ln \left (\frac {{\mathrm e}^{x}}{2}\right )\right )\right ) x +6 x -6 \ln \left (\ln \left (\ln \left (\frac {{\mathrm e}^{x}}{2}\right )\right )\right )}{6 x \ln \left (\ln \left (\ln \left (\frac {{\mathrm e}^{x}}{2}\right )\right )\right )}\)	\(44\)

Input:

int(((x^2-1)*ln(1/2*exp(x))*ln(ln(1/2*exp(x)))*ln(ln(ln(1/2*exp(x))))^2-2* 
ln(1/2*exp(x))*ln(ln(1/2*exp(x)))*ln(ln(ln(1/2*exp(x))))+x^2-2*x)/x^2/ln(1 
/2*exp(x))/ln(ln(1/2*exp(x)))/ln(ln(ln(1/2*exp(x))))^2,x,method=_RETURNVER 
BOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.32 \[ \int \frac {-2 x+x^2-2 \log \left (\frac {e^x}{2}\right ) \log \left (\log \left (\frac {e^x}{2}\right )\right ) \log \left (\log \left (\log \left (\frac {e^x}{2}\right )\right )\right )+\left (-1+x^2\right ) \log \left (\frac {e^x}{2}\right ) \log \left (\log \left (\frac {e^x}{2}\right )\right ) \log ^2\left (\log \left (\log \left (\frac {e^x}{2}\right )\right )\right )}{x^2 \log \left (\frac {e^x}{2}\right ) \log \left (\log \left (\frac {e^x}{2}\right )\right ) \log ^2\left (\log \left (\log \left (\frac {e^x}{2}\right )\right )\right )} \, dx=\frac {{\left (x^{2} + 1\right )} \log \left (\log \left (x - \log \left (2\right )\right )\right ) - x + 2}{x \log \left (\log \left (x - \log \left (2\right )\right )\right )} \] Input:

integrate(((x^2-1)*log(1/2*exp(x))*log(log(1/2*exp(x)))*log(log(log(1/2*ex 
p(x))))^2-2*log(1/2*exp(x))*log(log(1/2*exp(x)))*log(log(log(1/2*exp(x)))) 
+x^2-2*x)/x^2/log(1/2*exp(x))/log(log(1/2*exp(x)))/log(log(log(1/2*exp(x)) 
))^2,x, algorithm="fricas")
 

Output:

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

Sympy [F(-1)]

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

integrate(((x**2-1)*ln(1/2*exp(x))*ln(ln(1/2*exp(x)))*ln(ln(ln(1/2*exp(x)) 
))**2-2*ln(1/2*exp(x))*ln(ln(1/2*exp(x)))*ln(ln(ln(1/2*exp(x))))+x**2-2*x) 
/x**2/ln(1/2*exp(x))/ln(ln(1/2*exp(x)))/ln(ln(ln(1/2*exp(x))))**2,x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.24 \[ \int \frac {-2 x+x^2-2 \log \left (\frac {e^x}{2}\right ) \log \left (\log \left (\frac {e^x}{2}\right )\right ) \log \left (\log \left (\log \left (\frac {e^x}{2}\right )\right )\right )+\left (-1+x^2\right ) \log \left (\frac {e^x}{2}\right ) \log \left (\log \left (\frac {e^x}{2}\right )\right ) \log ^2\left (\log \left (\log \left (\frac {e^x}{2}\right )\right )\right )}{x^2 \log \left (\frac {e^x}{2}\right ) \log \left (\log \left (\frac {e^x}{2}\right )\right ) \log ^2\left (\log \left (\log \left (\frac {e^x}{2}\right )\right )\right )} \, dx=x + \frac {1}{x} - \frac {1}{\log \left (\log \left (\log \left (\frac {1}{2} \, e^{x}\right )\right )\right )} + \frac {2}{x \log \left (\log \left (x - \log \left (2\right )\right )\right )} \] Input:

integrate(((x^2-1)*log(1/2*exp(x))*log(log(1/2*exp(x)))*log(log(log(1/2*ex 
p(x))))^2-2*log(1/2*exp(x))*log(log(1/2*exp(x)))*log(log(log(1/2*exp(x)))) 
+x^2-2*x)/x^2/log(1/2*exp(x))/log(log(1/2*exp(x)))/log(log(log(1/2*exp(x)) 
))^2,x, algorithm="maxima")
 

Output:

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

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92 \[ \int \frac {-2 x+x^2-2 \log \left (\frac {e^x}{2}\right ) \log \left (\log \left (\frac {e^x}{2}\right )\right ) \log \left (\log \left (\log \left (\frac {e^x}{2}\right )\right )\right )+\left (-1+x^2\right ) \log \left (\frac {e^x}{2}\right ) \log \left (\log \left (\frac {e^x}{2}\right )\right ) \log ^2\left (\log \left (\log \left (\frac {e^x}{2}\right )\right )\right )}{x^2 \log \left (\frac {e^x}{2}\right ) \log \left (\log \left (\frac {e^x}{2}\right )\right ) \log ^2\left (\log \left (\log \left (\frac {e^x}{2}\right )\right )\right )} \, dx=x + \frac {1}{x} - \frac {x - 2}{x \log \left (\log \left (x - \log \left (2\right )\right )\right )} \] Input:

integrate(((x^2-1)*log(1/2*exp(x))*log(log(1/2*exp(x)))*log(log(log(1/2*ex 
p(x))))^2-2*log(1/2*exp(x))*log(log(1/2*exp(x)))*log(log(log(1/2*exp(x)))) 
+x^2-2*x)/x^2/log(1/2*exp(x))/log(log(1/2*exp(x)))/log(log(log(1/2*exp(x)) 
))^2,x, algorithm="giac")
 

Output:

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

Mupad [B] (verification not implemented)

Time = 3.42 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.28 \[ \int \frac {-2 x+x^2-2 \log \left (\frac {e^x}{2}\right ) \log \left (\log \left (\frac {e^x}{2}\right )\right ) \log \left (\log \left (\log \left (\frac {e^x}{2}\right )\right )\right )+\left (-1+x^2\right ) \log \left (\frac {e^x}{2}\right ) \log \left (\log \left (\frac {e^x}{2}\right )\right ) \log ^2\left (\log \left (\log \left (\frac {e^x}{2}\right )\right )\right )}{x^2 \log \left (\frac {e^x}{2}\right ) \log \left (\log \left (\frac {e^x}{2}\right )\right ) \log ^2\left (\log \left (\log \left (\frac {e^x}{2}\right )\right )\right )} \, dx=x+\frac {2}{x\,\ln \left (\ln \left (x-\ln \left (2\right )\right )\right )}+\frac {1}{x}-\frac {1}{\ln \left (\ln \left (x-\ln \left (2\right )\right )\right )} \] Input:

int(-(2*x - x^2 + 2*log(log(exp(x)/2))*log(log(log(exp(x)/2)))*log(exp(x)/ 
2) - log(log(exp(x)/2))*log(log(log(exp(x)/2)))^2*log(exp(x)/2)*(x^2 - 1)) 
/(x^2*log(log(exp(x)/2))*log(log(log(exp(x)/2)))^2*log(exp(x)/2)),x)
 

Output:

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

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.56 \[ \int \frac {-2 x+x^2-2 \log \left (\frac {e^x}{2}\right ) \log \left (\log \left (\frac {e^x}{2}\right )\right ) \log \left (\log \left (\log \left (\frac {e^x}{2}\right )\right )\right )+\left (-1+x^2\right ) \log \left (\frac {e^x}{2}\right ) \log \left (\log \left (\frac {e^x}{2}\right )\right ) \log ^2\left (\log \left (\log \left (\frac {e^x}{2}\right )\right )\right )}{x^2 \log \left (\frac {e^x}{2}\right ) \log \left (\log \left (\frac {e^x}{2}\right )\right ) \log ^2\left (\log \left (\log \left (\frac {e^x}{2}\right )\right )\right )} \, dx=\frac {\mathrm {log}\left (\mathrm {log}\left (\mathrm {log}\left (\frac {e^{x}}{2}\right )\right )\right ) x^{2}+\mathrm {log}\left (\mathrm {log}\left (\mathrm {log}\left (\frac {e^{x}}{2}\right )\right )\right )-x +2}{\mathrm {log}\left (\mathrm {log}\left (\mathrm {log}\left (\frac {e^{x}}{2}\right )\right )\right ) x} \] Input:

int(((x^2-1)*log(1/2*exp(x))*log(log(1/2*exp(x)))*log(log(log(1/2*exp(x))) 
)^2-2*log(1/2*exp(x))*log(log(1/2*exp(x)))*log(log(log(1/2*exp(x))))+x^2-2 
*x)/x^2/log(1/2*exp(x))/log(log(1/2*exp(x)))/log(log(log(1/2*exp(x))))^2,x 
)
 

Output:

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