\(\int \frac {2 \log (2)-4 e^{3+x} \log (2)-2 e^{3+x} x \log (2) \log (x^2)+(\log (2)-2 e^{3+x} \log (2)) \log (x^2) \log ((1-2 e^{3+x}) \log (x^2))}{(-x^2+2 e^{3+x} x^2) \log (x^2) \log ^2((1-2 e^{3+x}) \log (x^2))} \, dx\) [1048]

Optimal result

Integrand size = 102, antiderivative size = 24 \[ \int \frac {2 \log (2)-4 e^{3+x} \log (2)-2 e^{3+x} x \log (2) \log \left (x^2\right )+\left (\log (2)-2 e^{3+x} \log (2)\right ) \log \left (x^2\right ) \log \left (\left (1-2 e^{3+x}\right ) \log \left (x^2\right )\right )}{\left (-x^2+2 e^{3+x} x^2\right ) \log \left (x^2\right ) \log ^2\left (\left (1-2 e^{3+x}\right ) \log \left (x^2\right )\right )} \, dx=\frac {\log (2)}{x \log \left (-\left (\left (-1+2 e^{3+x}\right ) \log \left (x^2\right )\right )\right )} \] Output:

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

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96 \[ \int \frac {2 \log (2)-4 e^{3+x} \log (2)-2 e^{3+x} x \log (2) \log \left (x^2\right )+\left (\log (2)-2 e^{3+x} \log (2)\right ) \log \left (x^2\right ) \log \left (\left (1-2 e^{3+x}\right ) \log \left (x^2\right )\right )}{\left (-x^2+2 e^{3+x} x^2\right ) \log \left (x^2\right ) \log ^2\left (\left (1-2 e^{3+x}\right ) \log \left (x^2\right )\right )} \, dx=\frac {\log (2)}{x \log \left (\left (1-2 e^{3+x}\right ) \log \left (x^2\right )\right )} \] Input:

Integrate[(2*Log[2] - 4*E^(3 + x)*Log[2] - 2*E^(3 + x)*x*Log[2]*Log[x^2] + 
 (Log[2] - 2*E^(3 + x)*Log[2])*Log[x^2]*Log[(1 - 2*E^(3 + x))*Log[x^2]])/( 
(-x^2 + 2*E^(3 + x)*x^2)*Log[x^2]*Log[(1 - 2*E^(3 + x))*Log[x^2]]^2),x]
 

Output:

Log[2]/(x*Log[(1 - 2*E^(3 + x))*Log[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.

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

\(\Big \downarrow \) 7293

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

\(\Big \downarrow \) 2009

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

Input:

Int[(2*Log[2] - 4*E^(3 + x)*Log[2] - 2*E^(3 + x)*x*Log[2]*Log[x^2] + (Log[ 
2] - 2*E^(3 + x)*Log[2])*Log[x^2]*Log[(1 - 2*E^(3 + x))*Log[x^2]])/((-x^2 
+ 2*E^(3 + x)*x^2)*Log[x^2]*Log[(1 - 2*E^(3 + x))*Log[x^2]]^2),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 18.88 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96

method	result	size
parallelrisch	\(\frac {\ln \left (2\right )}{x \ln \left (\left (-2 \,{\mathrm e}^{3+x}+1\right ) \ln \left (x^{2}\right )\right )}\)	\(23\)
risch	\(\text {Expression too large to display}\)	\(833\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96 \[ \int \frac {2 \log (2)-4 e^{3+x} \log (2)-2 e^{3+x} x \log (2) \log \left (x^2\right )+\left (\log (2)-2 e^{3+x} \log (2)\right ) \log \left (x^2\right ) \log \left (\left (1-2 e^{3+x}\right ) \log \left (x^2\right )\right )}{\left (-x^2+2 e^{3+x} x^2\right ) \log \left (x^2\right ) \log ^2\left (\left (1-2 e^{3+x}\right ) \log \left (x^2\right )\right )} \, dx=\frac {\log \left (2\right )}{x \log \left (-{\left (2 \, e^{\left (x + 3\right )} - 1\right )} \log \left (x^{2}\right )\right )} \] Input:

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

Output:

log(2)/(x*log(-(2*e^(x + 3) - 1)*log(x^2)))
 

Sympy [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.79 \[ \int \frac {2 \log (2)-4 e^{3+x} \log (2)-2 e^{3+x} x \log (2) \log \left (x^2\right )+\left (\log (2)-2 e^{3+x} \log (2)\right ) \log \left (x^2\right ) \log \left (\left (1-2 e^{3+x}\right ) \log \left (x^2\right )\right )}{\left (-x^2+2 e^{3+x} x^2\right ) \log \left (x^2\right ) \log ^2\left (\left (1-2 e^{3+x}\right ) \log \left (x^2\right )\right )} \, dx=\frac {\log {\left (2 \right )}}{x \log {\left (\left (1 - 2 e^{x + 3}\right ) \log {\left (x^{2} \right )} \right )}} \] Input:

integrate(((-2*ln(2)*exp(3+x)+ln(2))*ln(x**2)*ln((-2*exp(3+x)+1)*ln(x**2)) 
-2*x*ln(2)*exp(3+x)*ln(x**2)-4*ln(2)*exp(3+x)+2*ln(2))/(2*x**2*exp(3+x)-x* 
*2)/ln(x**2)/ln((-2*exp(3+x)+1)*ln(x**2))**2,x)
 

Output:

log(2)/(x*log((1 - 2*exp(x + 3))*log(x**2)))
 

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.18 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.25 \[ \int \frac {2 \log (2)-4 e^{3+x} \log (2)-2 e^{3+x} x \log (2) \log \left (x^2\right )+\left (\log (2)-2 e^{3+x} \log (2)\right ) \log \left (x^2\right ) \log \left (\left (1-2 e^{3+x}\right ) \log \left (x^2\right )\right )}{\left (-x^2+2 e^{3+x} x^2\right ) \log \left (x^2\right ) \log ^2\left (\left (1-2 e^{3+x}\right ) \log \left (x^2\right )\right )} \, dx=\frac {\log \left (2\right )}{{\left (i \, \pi + \log \left (2\right )\right )} x + x \log \left (2 \, e^{\left (x + 3\right )} - 1\right ) + x \log \left (\log \left (x\right )\right )} \] Input:

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

Output:

log(2)/((I*pi + log(2))*x + x*log(2*e^(x + 3) - 1) + x*log(log(x)))
 

Giac [A] (verification not implemented)

Time = 0.32 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {2 \log (2)-4 e^{3+x} \log (2)-2 e^{3+x} x \log (2) \log \left (x^2\right )+\left (\log (2)-2 e^{3+x} \log (2)\right ) \log \left (x^2\right ) \log \left (\left (1-2 e^{3+x}\right ) \log \left (x^2\right )\right )}{\left (-x^2+2 e^{3+x} x^2\right ) \log \left (x^2\right ) \log ^2\left (\left (1-2 e^{3+x}\right ) \log \left (x^2\right )\right )} \, dx=\frac {\log \left (2\right )}{x \log \left (-2 \, e^{\left (x + 3\right )} \log \left (x^{2}\right ) + \log \left (x^{2}\right )\right )} \] Input:

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

Output:

log(2)/(x*log(-2*e^(x + 3)*log(x^2) + log(x^2)))
 

Mupad [F(-1)]

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

int((2*log(2) - 4*exp(x + 3)*log(2) + log(-log(x^2)*(2*exp(x + 3) - 1))*lo 
g(x^2)*(log(2) - 2*exp(x + 3)*log(2)) - 2*x*log(x^2)*exp(x + 3)*log(2))/(l 
og(-log(x^2)*(2*exp(x + 3) - 1))^2*log(x^2)*(2*x^2*exp(x + 3) - x^2)),x)
 

Output:

int((2*log(2) - 4*exp(x + 3)*log(2) + log(-log(x^2)*(2*exp(x + 3) - 1))*lo 
g(x^2)*(log(2) - 2*exp(x + 3)*log(2)) - 2*x*log(x^2)*exp(x + 3)*log(2))/(l 
og(-log(x^2)*(2*exp(x + 3) - 1))^2*log(x^2)*(2*x^2*exp(x + 3) - x^2)), x)
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08 \[ \int \frac {2 \log (2)-4 e^{3+x} \log (2)-2 e^{3+x} x \log (2) \log \left (x^2\right )+\left (\log (2)-2 e^{3+x} \log (2)\right ) \log \left (x^2\right ) \log \left (\left (1-2 e^{3+x}\right ) \log \left (x^2\right )\right )}{\left (-x^2+2 e^{3+x} x^2\right ) \log \left (x^2\right ) \log ^2\left (\left (1-2 e^{3+x}\right ) \log \left (x^2\right )\right )} \, dx=\frac {\mathrm {log}\left (2\right )}{\mathrm {log}\left (-2 e^{x} \mathrm {log}\left (x^{2}\right ) e^{3}+\mathrm {log}\left (x^{2}\right )\right ) x} \] Input:

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

Output:

log(2)/(log( - 2*e**x*log(x**2)*e**3 + log(x**2))*x)