\(\int \frac {e^{\frac {\log (5) \log (\frac {x}{3})}{-x+(2+x) \log (\frac {x}{3})}} (-\log (5)+\log (5) \log (\frac {x}{3})-\log (5) \log ^2(\frac {x}{3}))}{x^2+(-4 x-2 x^2) \log (\frac {x}{3})+(4+4 x+x^2) \log ^2(\frac {x}{3})} \, dx\) [833]

Optimal result

Integrand size = 93, antiderivative size = 24 \[ \int \frac {e^{\frac {\log (5) \log \left (\frac {x}{3}\right )}{-x+(2+x) \log \left (\frac {x}{3}\right )}} \left (-\log (5)+\log (5) \log \left (\frac {x}{3}\right )-\log (5) \log ^2\left (\frac {x}{3}\right )\right )}{x^2+\left (-4 x-2 x^2\right ) \log \left (\frac {x}{3}\right )+\left (4+4 x+x^2\right ) \log ^2\left (\frac {x}{3}\right )} \, dx=5^{\frac {1}{2+x-\frac {x}{\log \left (\frac {x}{3}\right )}}}-e^5 \] Output:

exp(ln(5)/(x+2-x/ln(1/3*x)))-exp(5)
                                                                                    
                                                                                    
 

Mathematica [F]

\[ \int \frac {e^{\frac {\log (5) \log \left (\frac {x}{3}\right )}{-x+(2+x) \log \left (\frac {x}{3}\right )}} \left (-\log (5)+\log (5) \log \left (\frac {x}{3}\right )-\log (5) \log ^2\left (\frac {x}{3}\right )\right )}{x^2+\left (-4 x-2 x^2\right ) \log \left (\frac {x}{3}\right )+\left (4+4 x+x^2\right ) \log ^2\left (\frac {x}{3}\right )} \, dx=\int \frac {e^{\frac {\log (5) \log \left (\frac {x}{3}\right )}{-x+(2+x) \log \left (\frac {x}{3}\right )}} \left (-\log (5)+\log (5) \log \left (\frac {x}{3}\right )-\log (5) \log ^2\left (\frac {x}{3}\right )\right )}{x^2+\left (-4 x-2 x^2\right ) \log \left (\frac {x}{3}\right )+\left (4+4 x+x^2\right ) \log ^2\left (\frac {x}{3}\right )} \, dx \] Input:

Integrate[(E^((Log[5]*Log[x/3])/(-x + (2 + x)*Log[x/3]))*(-Log[5] + Log[5] 
*Log[x/3] - Log[5]*Log[x/3]^2))/(x^2 + (-4*x - 2*x^2)*Log[x/3] + (4 + 4*x 
+ x^2)*Log[x/3]^2),x]
 

Output:

Integrate[(E^((Log[5]*Log[x/3])/(-x + (2 + x)*Log[x/3]))*(-Log[5] + Log[5] 
*Log[x/3] - Log[5]*Log[x/3]^2))/(x^2 + (-4*x - 2*x^2)*Log[x/3] + (4 + 4*x 
+ x^2)*Log[x/3]^2), x]
 

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.

\(\displaystyle \int \frac {e^{\frac {\log (5) \log \left (\frac {x}{3}\right )}{(x+2) \log \left (\frac {x}{3}\right )-x}} \left (-\log (5) \log ^2\left (\frac {x}{3}\right )+\log (5) \log \left (\frac {x}{3}\right )-\log (5)\right )}{x^2+\left (x^2+4 x+4\right ) \log ^2\left (\frac {x}{3}\right )+\left (-2 x^2-4 x\right ) \log \left (\frac {x}{3}\right )} \, dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {\log (5) e^{\frac {\log (5) \log \left (\frac {x}{3}\right )}{(x+2) \log \left (\frac {x}{3}\right )-x}} \left (-\log ^2\left (\frac {x}{3}\right )+\log \left (\frac {x}{3}\right )-1\right )}{\left (x+x \left (-\log \left (\frac {x}{3}\right )\right )-2 \log \left (\frac {x}{3}\right )\right )^2}dx\)

\(\Big \downarrow \) 27

\(\displaystyle \log (5) \int -\frac {e^{-\frac {\log (5) \log \left (\frac {x}{3}\right )}{x-(x+2) \log \left (\frac {x}{3}\right )}} \left (\log ^2\left (\frac {x}{3}\right )-\log \left (\frac {x}{3}\right )+1\right )}{\left (-\log \left (\frac {x}{3}\right ) x+x-2 \log \left (\frac {x}{3}\right )\right )^2}dx\)

\(\Big \downarrow \) 25

\(\displaystyle -\log (5) \int \frac {e^{-\frac {\log (5) \log \left (\frac {x}{3}\right )}{x-(x+2) \log \left (\frac {x}{3}\right )}} \left (\log ^2\left (\frac {x}{3}\right )-\log \left (\frac {x}{3}\right )+1\right )}{\left (-\log \left (\frac {x}{3}\right ) x+x-2 \log \left (\frac {x}{3}\right )\right )^2}dx\)

\(\Big \downarrow \) 7293

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

\(\Big \downarrow \) 2009

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

Input:

Int[(E^((Log[5]*Log[x/3])/(-x + (2 + x)*Log[x/3]))*(-Log[5] + Log[5]*Log[x 
/3] - Log[5]*Log[x/3]^2))/(x^2 + (-4*x - 2*x^2)*Log[x/3] + (4 + 4*x + x^2) 
*Log[x/3]^2),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 2.01 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.12

method	result	size
parallelrisch	\({\mathrm e}^{\frac {\ln \left (5\right ) \ln \left (\frac {x}{3}\right )}{\ln \left (\frac {x}{3}\right ) x +2 \ln \left (\frac {x}{3}\right )-x}}\)	\(27\)
risch	\(\left (\frac {x}{3}\right )^{\frac {\ln \left (5\right )}{\left (-\ln \left (3\right )+\ln \left (x \right )\right ) x -2 \ln \left (3\right )+2 \ln \left (x \right )-x}}\)	\(31\)
norman	\(\frac {\ln \left (\frac {x}{3}\right ) x \,{\mathrm e}^{\frac {\ln \left (5\right ) \ln \left (\frac {x}{3}\right )}{\left (2+x \right ) \ln \left (\frac {x}{3}\right )-x}}-x \,{\mathrm e}^{\frac {\ln \left (5\right ) \ln \left (\frac {x}{3}\right )}{\left (2+x \right ) \ln \left (\frac {x}{3}\right )-x}}+2 \ln \left (\frac {x}{3}\right ) {\mathrm e}^{\frac {\ln \left (5\right ) \ln \left (\frac {x}{3}\right )}{\left (2+x \right ) \ln \left (\frac {x}{3}\right )-x}}}{\ln \left (\frac {x}{3}\right ) x +2 \ln \left (\frac {x}{3}\right )-x}\)	\(102\)

Input:

int((-ln(5)*ln(1/3*x)^2+ln(5)*ln(1/3*x)-ln(5))*exp(ln(5)*ln(1/3*x)/((2+x)* 
ln(1/3*x)-x))/((x^2+4*x+4)*ln(1/3*x)^2+(-2*x^2-4*x)*ln(1/3*x)+x^2),x,metho 
d=_RETURNVERBOSE)
 

Output:

exp(ln(5)*ln(1/3*x)/(ln(1/3*x)*x+2*ln(1/3*x)-x))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \frac {e^{\frac {\log (5) \log \left (\frac {x}{3}\right )}{-x+(2+x) \log \left (\frac {x}{3}\right )}} \left (-\log (5)+\log (5) \log \left (\frac {x}{3}\right )-\log (5) \log ^2\left (\frac {x}{3}\right )\right )}{x^2+\left (-4 x-2 x^2\right ) \log \left (\frac {x}{3}\right )+\left (4+4 x+x^2\right ) \log ^2\left (\frac {x}{3}\right )} \, dx=e^{\left (\frac {\log \left (5\right ) \log \left (\frac {1}{3} \, x\right )}{{\left (x + 2\right )} \log \left (\frac {1}{3} \, x\right ) - x}\right )} \] Input:

integrate((-log(5)*log(1/3*x)^2+log(5)*log(1/3*x)-log(5))*exp(log(5)*log(1 
/3*x)/((2+x)*log(1/3*x)-x))/((x^2+4*x+4)*log(1/3*x)^2+(-2*x^2-4*x)*log(1/3 
*x)+x^2),x, algorithm="fricas")
 

Output:

e^(log(5)*log(1/3*x)/((x + 2)*log(1/3*x) - x))
 

Sympy [A] (verification not implemented)

Time = 0.34 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.79 \[ \int \frac {e^{\frac {\log (5) \log \left (\frac {x}{3}\right )}{-x+(2+x) \log \left (\frac {x}{3}\right )}} \left (-\log (5)+\log (5) \log \left (\frac {x}{3}\right )-\log (5) \log ^2\left (\frac {x}{3}\right )\right )}{x^2+\left (-4 x-2 x^2\right ) \log \left (\frac {x}{3}\right )+\left (4+4 x+x^2\right ) \log ^2\left (\frac {x}{3}\right )} \, dx=e^{\frac {\log {\left (5 \right )} \log {\left (\frac {x}{3} \right )}}{- x + \left (x + 2\right ) \log {\left (\frac {x}{3} \right )}}} \] Input:

integrate((-ln(5)*ln(1/3*x)**2+ln(5)*ln(1/3*x)-ln(5))*exp(ln(5)*ln(1/3*x)/ 
((2+x)*ln(1/3*x)-x))/((x**2+4*x+4)*ln(1/3*x)**2+(-2*x**2-4*x)*ln(1/3*x)+x* 
*2),x)
 

Output:

exp(log(5)*log(x/3)/(-x + (x + 2)*log(x/3)))
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 53 vs. \(2 (21) = 42\).

Time = 0.22 (sec) , antiderivative size = 53, normalized size of antiderivative = 2.21 \[ \int \frac {e^{\frac {\log (5) \log \left (\frac {x}{3}\right )}{-x+(2+x) \log \left (\frac {x}{3}\right )}} \left (-\log (5)+\log (5) \log \left (\frac {x}{3}\right )-\log (5) \log ^2\left (\frac {x}{3}\right )\right )}{x^2+\left (-4 x-2 x^2\right ) \log \left (\frac {x}{3}\right )+\left (4+4 x+x^2\right ) \log ^2\left (\frac {x}{3}\right )} \, dx=e^{\left (\frac {\log \left (5\right ) \log \left (3\right )}{x {\left (\log \left (3\right ) + 1\right )} - {\left (x + 2\right )} \log \left (x\right ) + 2 \, \log \left (3\right )} - \frac {\log \left (5\right ) \log \left (x\right )}{x {\left (\log \left (3\right ) + 1\right )} - {\left (x + 2\right )} \log \left (x\right ) + 2 \, \log \left (3\right )}\right )} \] Input:

integrate((-log(5)*log(1/3*x)^2+log(5)*log(1/3*x)-log(5))*exp(log(5)*log(1 
/3*x)/((2+x)*log(1/3*x)-x))/((x^2+4*x+4)*log(1/3*x)^2+(-2*x^2-4*x)*log(1/3 
*x)+x^2),x, algorithm="maxima")
 

Output:

e^(log(5)*log(3)/(x*(log(3) + 1) - (x + 2)*log(x) + 2*log(3)) - log(5)*log 
(x)/(x*(log(3) + 1) - (x + 2)*log(x) + 2*log(3)))
 

Giac [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08 \[ \int \frac {e^{\frac {\log (5) \log \left (\frac {x}{3}\right )}{-x+(2+x) \log \left (\frac {x}{3}\right )}} \left (-\log (5)+\log (5) \log \left (\frac {x}{3}\right )-\log (5) \log ^2\left (\frac {x}{3}\right )\right )}{x^2+\left (-4 x-2 x^2\right ) \log \left (\frac {x}{3}\right )+\left (4+4 x+x^2\right ) \log ^2\left (\frac {x}{3}\right )} \, dx=e^{\left (\frac {\log \left (5\right ) \log \left (\frac {1}{3} \, x\right )}{x \log \left (\frac {1}{3} \, x\right ) - x + 2 \, \log \left (\frac {1}{3} \, x\right )}\right )} \] Input:

integrate((-log(5)*log(1/3*x)^2+log(5)*log(1/3*x)-log(5))*exp(log(5)*log(1 
/3*x)/((2+x)*log(1/3*x)-x))/((x^2+4*x+4)*log(1/3*x)^2+(-2*x^2-4*x)*log(1/3 
*x)+x^2),x, algorithm="giac")
 

Output:

e^(log(5)*log(1/3*x)/(x*log(1/3*x) - x + 2*log(1/3*x)))
 

Mupad [B] (verification not implemented)

Time = 3.13 (sec) , antiderivative size = 56, normalized size of antiderivative = 2.33 \[ \int \frac {e^{\frac {\log (5) \log \left (\frac {x}{3}\right )}{-x+(2+x) \log \left (\frac {x}{3}\right )}} \left (-\log (5)+\log (5) \log \left (\frac {x}{3}\right )-\log (5) \log ^2\left (\frac {x}{3}\right )\right )}{x^2+\left (-4 x-2 x^2\right ) \log \left (\frac {x}{3}\right )+\left (4+4 x+x^2\right ) \log ^2\left (\frac {x}{3}\right )} \, dx={\mathrm {e}}^{-\frac {\ln \left (5\right )\,\ln \left (x\right )}{x+2\,\ln \left (3\right )-2\,\ln \left (x\right )+x\,\ln \left (3\right )-x\,\ln \left (x\right )}}\,{\mathrm {e}}^{\frac {\ln \left (3\right )\,\ln \left (5\right )}{x+2\,\ln \left (3\right )-2\,\ln \left (x\right )+x\,\ln \left (3\right )-x\,\ln \left (x\right )}} \] Input:

int(-(exp(-(log(x/3)*log(5))/(x - log(x/3)*(x + 2)))*(log(5) - log(x/3)*lo 
g(5) + log(x/3)^2*log(5)))/(log(x/3)^2*(4*x + x^2 + 4) - log(x/3)*(4*x + 2 
*x^2) + x^2),x)
 

Output:

exp(-(log(5)*log(x))/(x + 2*log(3) - 2*log(x) + x*log(3) - x*log(x)))*exp( 
(log(3)*log(5))/(x + 2*log(3) - 2*log(x) + x*log(3) - x*log(x)))
 

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.12 \[ \int \frac {e^{\frac {\log (5) \log \left (\frac {x}{3}\right )}{-x+(2+x) \log \left (\frac {x}{3}\right )}} \left (-\log (5)+\log (5) \log \left (\frac {x}{3}\right )-\log (5) \log ^2\left (\frac {x}{3}\right )\right )}{x^2+\left (-4 x-2 x^2\right ) \log \left (\frac {x}{3}\right )+\left (4+4 x+x^2\right ) \log ^2\left (\frac {x}{3}\right )} \, dx=e^{\frac {\mathrm {log}\left (\frac {x}{3}\right ) \mathrm {log}\left (5\right )}{\mathrm {log}\left (\frac {x}{3}\right ) x +2 \,\mathrm {log}\left (\frac {x}{3}\right )-x}} \] Input:

int((-log(5)*log(1/3*x)^2+log(5)*log(1/3*x)-log(5))*exp(log(5)*log(1/3*x)/ 
((2+x)*log(1/3*x)-x))/((x^2+4*x+4)*log(1/3*x)^2+(-2*x^2-4*x)*log(1/3*x)+x^ 
2),x)
 

Output:

e**((log(x/3)*log(5))/(log(x/3)*x + 2*log(x/3) - x))