\(\int \frac {e^{\frac {32 x+(-4+8 x+12 x^2) \log (\log (5))}{(1-2 x+x^2) \log (\log (5))}} (-32-32 x-32 x \log (\log (5)))}{(-1+3 x-3 x^2+x^3) \log (\log (5))} \, dx\) [2809]

Optimal result

Integrand size = 68, antiderivative size = 32 \[ \int \frac {e^{\frac {32 x+\left (-4+8 x+12 x^2\right ) \log (\log (5))}{\left (1-2 x+x^2\right ) \log (\log (5))}} (-32-32 x-32 x \log (\log (5)))}{\left (-1+3 x-3 x^2+x^3\right ) \log (\log (5))} \, dx=-3+e^{4 \left (-1+\frac {4 \left (x+\frac {2}{\log (\log (5))}\right )}{-4+\frac {(1+x)^2}{x}}\right )} \] Output:

exp(16*(x+2/ln(ln(5)))/((1+x)^2/x-4)-4)-3
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.82 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.09 \[ \int \frac {e^{\frac {32 x+\left (-4+8 x+12 x^2\right ) \log (\log (5))}{\left (1-2 x+x^2\right ) \log (\log (5))}} (-32-32 x-32 x \log (\log (5)))}{\left (-1+3 x-3 x^2+x^3\right ) \log (\log (5))} \, dx=e^{\frac {-4 \log (\log (5))+12 x^2 \log (\log (5))+8 x (4+\log (\log (5)))}{(-1+x)^2 \log (\log (5))}} \] Input:

Integrate[(E^((32*x + (-4 + 8*x + 12*x^2)*Log[Log[5]])/((1 - 2*x + x^2)*Lo 
g[Log[5]]))*(-32 - 32*x - 32*x*Log[Log[5]]))/((-1 + 3*x - 3*x^2 + x^3)*Log 
[Log[5]]),x]
 

Output:

E^((-4*Log[Log[5]] + 12*x^2*Log[Log[5]] + 8*x*(4 + Log[Log[5]]))/((-1 + x) 
^2*Log[Log[5]]))
 

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.

\(\displaystyle \int \frac {(-32 x-32 x \log (\log (5))-32) \exp \left (\frac {\left (12 x^2+8 x-4\right ) \log (\log (5))+32 x}{\left (x^2-2 x+1\right ) \log (\log (5))}\right )}{\left (x^3-3 x^2+3 x-1\right ) \log (\log (5))} \, dx\)

\(\Big \downarrow \) 6

\(\displaystyle \int \frac {(x (-32-32 \log (\log (5)))-32) \exp \left (\frac {\left (12 x^2+8 x-4\right ) \log (\log (5))+32 x}{\left (x^2-2 x+1\right ) \log (\log (5))}\right )}{\left (x^3-3 x^2+3 x-1\right ) \log (\log (5))}dx\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {32 e^{\frac {32 x}{\left (x^2-2 x+1\right ) \log (\log (5))}} \log ^{-\frac {-12 x^2-8 x+4}{\left (x^2-2 x+1\right ) \log (\log (5))}}(5) ((1+\log (\log (5))) x+1)}{-x^3+3 x^2-3 x+1}dx}{\log (\log (5))}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {32 \int \frac {e^{\frac {32 x}{\left (x^2-2 x+1\right ) \log (\log (5))}} \log ^{-\frac {4 \left (-3 x^2-2 x+1\right )}{\left (x^2-2 x+1\right ) \log (\log (5))}}(5) ((1+\log (\log (5))) x+1)}{-x^3+3 x^2-3 x+1}dx}{\log (\log (5))}\)

\(\Big \downarrow \) 2007

\(\displaystyle \frac {32 \int \frac {e^{\frac {32 x}{\left (x^2-2 x+1\right ) \log (\log (5))}} \log ^{-\frac {4 \left (-3 x^2-2 x+1\right )}{\left (x^2-2 x+1\right ) \log (\log (5))}}(5) ((1+\log (\log (5))) x+1)}{(1-x)^3}dx}{\log (\log (5))}\)

\(\Big \downarrow \) 7239

\(\displaystyle \frac {32 \int \frac {\exp \left (\frac {4 \left (3 x^2+2 \left (1+\frac {4}{\log (\log (5))}\right ) x-1\right )}{(x-1)^2}\right ) ((1+\log (\log (5))) x+1)}{(1-x)^3}dx}{\log (\log (5))}\)

\(\Big \downarrow \) 7293

\(\displaystyle \frac {32 \int \left (\frac {\exp \left (\frac {4 \left (3 x^2+2 \left (1+\frac {4}{\log (\log (5))}\right ) x-1\right )}{(x-1)^2}\right ) (-1-\log (\log (5)))}{(x-1)^2}+\frac {\exp \left (\frac {4 \left (3 x^2+2 \left (1+\frac {4}{\log (\log (5))}\right ) x-1\right )}{(x-1)^2}\right ) (-2-\log (\log (5)))}{(x-1)^3}\right )dx}{\log (\log (5))}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {32 \left (-\left ((2+\log (\log (5))) \int \frac {\exp \left (\frac {4 \left (3 x^2+2 \left (1+\frac {4}{\log (\log (5))}\right ) x-1\right )}{(x-1)^2}\right )}{(x-1)^3}dx\right )-(1+\log (\log (5))) \int \frac {\exp \left (\frac {4 \left (3 x^2+2 \left (1+\frac {4}{\log (\log (5))}\right ) x-1\right )}{(x-1)^2}\right )}{(x-1)^2}dx\right )}{\log (\log (5))}\)

Input:

Int[(E^((32*x + (-4 + 8*x + 12*x^2)*Log[Log[5]])/((1 - 2*x + x^2)*Log[Log[ 
5]]))*(-32 - 32*x - 32*x*Log[Log[5]]))/((-1 + 3*x - 3*x^2 + x^3)*Log[Log[5 
]]),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 0.44 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.12

method	result	size
parallelrisch	\({\mathrm e}^{\frac {\left (12 x^{2}+8 x -4\right ) \ln \left (\ln \left (5\right )\right )+32 x}{\left (x^{2}-2 x +1\right ) \ln \left (\ln \left (5\right )\right )}}\)	\(36\)
risch	\({\mathrm e}^{\frac {12 \ln \left (\ln \left (5\right )\right ) x^{2}+8 x \ln \left (\ln \left (5\right )\right )-4 \ln \left (\ln \left (5\right )\right )+32 x}{\ln \left (\ln \left (5\right )\right ) \left (-1+x \right )^{2}}}\)	\(37\)
gosper	\({\mathrm e}^{\frac {12 \ln \left (\ln \left (5\right )\right ) x^{2}+8 x \ln \left (\ln \left (5\right )\right )-4 \ln \left (\ln \left (5\right )\right )+32 x}{\left (x^{2}-2 x +1\right ) \ln \left (\ln \left (5\right )\right )}}\)	\(42\)
norman	\(\frac {x^{2} {\mathrm e}^{\frac {\left (12 x^{2}+8 x -4\right ) \ln \left (\ln \left (5\right )\right )+32 x}{\left (x^{2}-2 x +1\right ) \ln \left (\ln \left (5\right )\right )}}-2 x \,{\mathrm e}^{\frac {\left (12 x^{2}+8 x -4\right ) \ln \left (\ln \left (5\right )\right )+32 x}{\left (x^{2}-2 x +1\right ) \ln \left (\ln \left (5\right )\right )}}+{\mathrm e}^{\frac {\left (12 x^{2}+8 x -4\right ) \ln \left (\ln \left (5\right )\right )+32 x}{\left (x^{2}-2 x +1\right ) \ln \left (\ln \left (5\right )\right )}}}{\left (-1+x \right )^{2}}\)	\(120\)
default	\(\frac {-\frac {4 i {\mathrm e}^{12} \sqrt {\pi }\, {\mathrm e}^{-\frac {\left (32+\frac {32}{\ln \left (\ln \left (5\right )\right )}\right )^{2}}{4 \left (16+\frac {32}{\ln \left (\ln \left (5\right )\right )}\right )}} \operatorname {erf}\left (\frac {4 i \sqrt {1+\frac {2}{\ln \left (\ln \left (5\right )\right )}}}{-1+x}+\frac {i \left (32+\frac {32}{\ln \left (\ln \left (5\right )\right )}\right )}{8 \sqrt {1+\frac {2}{\ln \left (\ln \left (5\right )\right )}}}\right )}{\sqrt {1+\frac {2}{\ln \left (\ln \left (5\right )\right )}}}+64 \,{\mathrm e}^{12} \left (\frac {{\mathrm e}^{\frac {16+\frac {32}{\ln \left (\ln \left (5\right )\right )}}{\left (-1+x \right )^{2}}+\frac {32+\frac {32}{\ln \left (\ln \left (5\right )\right )}}{-1+x}}}{32+\frac {64}{\ln \left (\ln \left (5\right )\right )}}+\frac {i \left (32+\frac {32}{\ln \left (\ln \left (5\right )\right )}\right ) \sqrt {\pi }\, {\mathrm e}^{-\frac {\left (32+\frac {32}{\ln \left (\ln \left (5\right )\right )}\right )^{2}}{4 \left (16+\frac {32}{\ln \left (\ln \left (5\right )\right )}\right )}} \operatorname {erf}\left (\frac {4 i \sqrt {1+\frac {2}{\ln \left (\ln \left (5\right )\right )}}}{-1+x}+\frac {i \left (32+\frac {32}{\ln \left (\ln \left (5\right )\right )}\right )}{8 \sqrt {1+\frac {2}{\ln \left (\ln \left (5\right )\right )}}}\right )}{16 \left (16+\frac {32}{\ln \left (\ln \left (5\right )\right )}\right ) \sqrt {1+\frac {2}{\ln \left (\ln \left (5\right )\right )}}}\right )-\frac {4 i {\mathrm e}^{12} \ln \left (\ln \left (5\right )\right ) \sqrt {\pi }\, {\mathrm e}^{-\frac {\left (32+\frac {32}{\ln \left (\ln \left (5\right )\right )}\right )^{2}}{4 \left (16+\frac {32}{\ln \left (\ln \left (5\right )\right )}\right )}} \operatorname {erf}\left (\frac {4 i \sqrt {1+\frac {2}{\ln \left (\ln \left (5\right )\right )}}}{-1+x}+\frac {i \left (32+\frac {32}{\ln \left (\ln \left (5\right )\right )}\right )}{8 \sqrt {1+\frac {2}{\ln \left (\ln \left (5\right )\right )}}}\right )}{\sqrt {1+\frac {2}{\ln \left (\ln \left (5\right )\right )}}}+32 \,{\mathrm e}^{12} \ln \left (\ln \left (5\right )\right ) \left (\frac {{\mathrm e}^{\frac {16+\frac {32}{\ln \left (\ln \left (5\right )\right )}}{\left (-1+x \right )^{2}}+\frac {32+\frac {32}{\ln \left (\ln \left (5\right )\right )}}{-1+x}}}{32+\frac {64}{\ln \left (\ln \left (5\right )\right )}}+\frac {i \left (32+\frac {32}{\ln \left (\ln \left (5\right )\right )}\right ) \sqrt {\pi }\, {\mathrm e}^{-\frac {\left (32+\frac {32}{\ln \left (\ln \left (5\right )\right )}\right )^{2}}{4 \left (16+\frac {32}{\ln \left (\ln \left (5\right )\right )}\right )}} \operatorname {erf}\left (\frac {4 i \sqrt {1+\frac {2}{\ln \left (\ln \left (5\right )\right )}}}{-1+x}+\frac {i \left (32+\frac {32}{\ln \left (\ln \left (5\right )\right )}\right )}{8 \sqrt {1+\frac {2}{\ln \left (\ln \left (5\right )\right )}}}\right )}{16 \left (16+\frac {32}{\ln \left (\ln \left (5\right )\right )}\right ) \sqrt {1+\frac {2}{\ln \left (\ln \left (5\right )\right )}}}\right )}{\ln \left (\ln \left (5\right )\right )}\)	\(502\)
derivativedivides	\(-\frac {32 \left (\frac {i {\mathrm e}^{12} \sqrt {\pi }\, {\mathrm e}^{-\frac {\left (32+\frac {32}{\ln \left (\ln \left (5\right )\right )}\right )^{2}}{4 \left (16+\frac {32}{\ln \left (\ln \left (5\right )\right )}\right )}} \operatorname {erf}\left (\frac {4 i \sqrt {1+\frac {2}{\ln \left (\ln \left (5\right )\right )}}}{-1+x}+\frac {i \left (32+\frac {32}{\ln \left (\ln \left (5\right )\right )}\right )}{8 \sqrt {1+\frac {2}{\ln \left (\ln \left (5\right )\right )}}}\right )}{8 \sqrt {1+\frac {2}{\ln \left (\ln \left (5\right )\right )}}}-2 \,{\mathrm e}^{12} \left (\frac {{\mathrm e}^{\frac {16+\frac {32}{\ln \left (\ln \left (5\right )\right )}}{\left (-1+x \right )^{2}}+\frac {32+\frac {32}{\ln \left (\ln \left (5\right )\right )}}{-1+x}}}{32+\frac {64}{\ln \left (\ln \left (5\right )\right )}}+\frac {i \left (32+\frac {32}{\ln \left (\ln \left (5\right )\right )}\right ) \sqrt {\pi }\, {\mathrm e}^{-\frac {\left (32+\frac {32}{\ln \left (\ln \left (5\right )\right )}\right )^{2}}{4 \left (16+\frac {32}{\ln \left (\ln \left (5\right )\right )}\right )}} \operatorname {erf}\left (\frac {4 i \sqrt {1+\frac {2}{\ln \left (\ln \left (5\right )\right )}}}{-1+x}+\frac {i \left (32+\frac {32}{\ln \left (\ln \left (5\right )\right )}\right )}{8 \sqrt {1+\frac {2}{\ln \left (\ln \left (5\right )\right )}}}\right )}{16 \left (16+\frac {32}{\ln \left (\ln \left (5\right )\right )}\right ) \sqrt {1+\frac {2}{\ln \left (\ln \left (5\right )\right )}}}\right )+\frac {i {\mathrm e}^{12} \ln \left (\ln \left (5\right )\right ) \sqrt {\pi }\, {\mathrm e}^{-\frac {\left (32+\frac {32}{\ln \left (\ln \left (5\right )\right )}\right )^{2}}{4 \left (16+\frac {32}{\ln \left (\ln \left (5\right )\right )}\right )}} \operatorname {erf}\left (\frac {4 i \sqrt {1+\frac {2}{\ln \left (\ln \left (5\right )\right )}}}{-1+x}+\frac {i \left (32+\frac {32}{\ln \left (\ln \left (5\right )\right )}\right )}{8 \sqrt {1+\frac {2}{\ln \left (\ln \left (5\right )\right )}}}\right )}{8 \sqrt {1+\frac {2}{\ln \left (\ln \left (5\right )\right )}}}-{\mathrm e}^{12} \ln \left (\ln \left (5\right )\right ) \left (\frac {{\mathrm e}^{\frac {16+\frac {32}{\ln \left (\ln \left (5\right )\right )}}{\left (-1+x \right )^{2}}+\frac {32+\frac {32}{\ln \left (\ln \left (5\right )\right )}}{-1+x}}}{32+\frac {64}{\ln \left (\ln \left (5\right )\right )}}+\frac {i \left (32+\frac {32}{\ln \left (\ln \left (5\right )\right )}\right ) \sqrt {\pi }\, {\mathrm e}^{-\frac {\left (32+\frac {32}{\ln \left (\ln \left (5\right )\right )}\right )^{2}}{4 \left (16+\frac {32}{\ln \left (\ln \left (5\right )\right )}\right )}} \operatorname {erf}\left (\frac {4 i \sqrt {1+\frac {2}{\ln \left (\ln \left (5\right )\right )}}}{-1+x}+\frac {i \left (32+\frac {32}{\ln \left (\ln \left (5\right )\right )}\right )}{8 \sqrt {1+\frac {2}{\ln \left (\ln \left (5\right )\right )}}}\right )}{16 \left (16+\frac {32}{\ln \left (\ln \left (5\right )\right )}\right ) \sqrt {1+\frac {2}{\ln \left (\ln \left (5\right )\right )}}}\right )\right )}{\ln \left (\ln \left (5\right )\right )}\)	\(503\)

Input:

int((-32*x*ln(ln(5))-32*x-32)*exp(((12*x^2+8*x-4)*ln(ln(5))+32*x)/(x^2-2*x 
+1)/ln(ln(5)))/(x^3-3*x^2+3*x-1)/ln(ln(5)),x,method=_RETURNVERBOSE)
 

Output:

exp(((12*x^2+8*x-4)*ln(ln(5))+32*x)/(x^2-2*x+1)/ln(ln(5)))
                                                                                    
                                                                                    
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.12 \[ \int \frac {e^{\frac {32 x+\left (-4+8 x+12 x^2\right ) \log (\log (5))}{\left (1-2 x+x^2\right ) \log (\log (5))}} (-32-32 x-32 x \log (\log (5)))}{\left (-1+3 x-3 x^2+x^3\right ) \log (\log (5))} \, dx=e^{\left (\frac {4 \, {\left ({\left (3 \, x^{2} + 2 \, x - 1\right )} \log \left (\log \left (5\right )\right ) + 8 \, x\right )}}{{\left (x^{2} - 2 \, x + 1\right )} \log \left (\log \left (5\right )\right )}\right )} \] Input:

integrate((-32*x*log(log(5))-32*x-32)*exp(((12*x^2+8*x-4)*log(log(5))+32*x 
)/(x^2-2*x+1)/log(log(5)))/(x^3-3*x^2+3*x-1)/log(log(5)),x, algorithm="fri 
cas")
 

Output:

e^(4*((3*x^2 + 2*x - 1)*log(log(5)) + 8*x)/((x^2 - 2*x + 1)*log(log(5))))
 

Sympy [A] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {32 x+\left (-4+8 x+12 x^2\right ) \log (\log (5))}{\left (1-2 x+x^2\right ) \log (\log (5))}} (-32-32 x-32 x \log (\log (5)))}{\left (-1+3 x-3 x^2+x^3\right ) \log (\log (5))} \, dx=e^{\frac {32 x + \left (12 x^{2} + 8 x - 4\right ) \log {\left (\log {\left (5 \right )} \right )}}{\left (x^{2} - 2 x + 1\right ) \log {\left (\log {\left (5 \right )} \right )}}} \] Input:

integrate((-32*x*ln(ln(5))-32*x-32)*exp(((12*x**2+8*x-4)*ln(ln(5))+32*x)/( 
x**2-2*x+1)/ln(ln(5)))/(x**3-3*x**2+3*x-1)/ln(ln(5)),x)
 

Output:

exp((32*x + (12*x**2 + 8*x - 4)*log(log(5)))/((x**2 - 2*x + 1)*log(log(5)) 
))
 

Maxima [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.81 \[ \int \frac {e^{\frac {32 x+\left (-4+8 x+12 x^2\right ) \log (\log (5))}{\left (1-2 x+x^2\right ) \log (\log (5))}} (-32-32 x-32 x \log (\log (5)))}{\left (-1+3 x-3 x^2+x^3\right ) \log (\log (5))} \, dx=e^{\left (\frac {32}{x^{2} \log \left (\log \left (5\right )\right ) - 2 \, x \log \left (\log \left (5\right )\right ) + \log \left (\log \left (5\right )\right )} + \frac {16}{x^{2} - 2 \, x + 1} + \frac {32}{x \log \left (\log \left (5\right )\right ) - \log \left (\log \left (5\right )\right )} + \frac {32}{x - 1} + 12\right )} \] Input:

integrate((-32*x*log(log(5))-32*x-32)*exp(((12*x^2+8*x-4)*log(log(5))+32*x 
)/(x^2-2*x+1)/log(log(5)))/(x^3-3*x^2+3*x-1)/log(log(5)),x, algorithm="max 
ima")
 

Output:

e^(32/(x^2*log(log(5)) - 2*x*log(log(5)) + log(log(5))) + 16/(x^2 - 2*x + 
1) + 32/(x*log(log(5)) - log(log(5))) + 32/(x - 1) + 12)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 100 vs. \(2 (29) = 58\).

Time = 0.12 (sec) , antiderivative size = 100, normalized size of antiderivative = 3.12 \[ \int \frac {e^{\frac {32 x+\left (-4+8 x+12 x^2\right ) \log (\log (5))}{\left (1-2 x+x^2\right ) \log (\log (5))}} (-32-32 x-32 x \log (\log (5)))}{\left (-1+3 x-3 x^2+x^3\right ) \log (\log (5))} \, dx=e^{\left (\frac {12 \, x^{2} \log \left (\log \left (5\right )\right )}{x^{2} \log \left (\log \left (5\right )\right ) - 2 \, x \log \left (\log \left (5\right )\right ) + \log \left (\log \left (5\right )\right )} + \frac {8 \, x \log \left (\log \left (5\right )\right )}{x^{2} \log \left (\log \left (5\right )\right ) - 2 \, x \log \left (\log \left (5\right )\right ) + \log \left (\log \left (5\right )\right )} + \frac {32 \, x}{x^{2} \log \left (\log \left (5\right )\right ) - 2 \, x \log \left (\log \left (5\right )\right ) + \log \left (\log \left (5\right )\right )} - \frac {4 \, \log \left (\log \left (5\right )\right )}{x^{2} \log \left (\log \left (5\right )\right ) - 2 \, x \log \left (\log \left (5\right )\right ) + \log \left (\log \left (5\right )\right )}\right )} \] Input:

integrate((-32*x*log(log(5))-32*x-32)*exp(((12*x^2+8*x-4)*log(log(5))+32*x 
)/(x^2-2*x+1)/log(log(5)))/(x^3-3*x^2+3*x-1)/log(log(5)),x, algorithm="gia 
c")
 

Output:

e^(12*x^2*log(log(5))/(x^2*log(log(5)) - 2*x*log(log(5)) + log(log(5))) + 
8*x*log(log(5))/(x^2*log(log(5)) - 2*x*log(log(5)) + log(log(5))) + 32*x/( 
x^2*log(log(5)) - 2*x*log(log(5)) + log(log(5))) - 4*log(log(5))/(x^2*log( 
log(5)) - 2*x*log(log(5)) + log(log(5))))
 

Mupad [B] (verification not implemented)

Time = 3.71 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.62 \[ \int \frac {e^{\frac {32 x+\left (-4+8 x+12 x^2\right ) \log (\log (5))}{\left (1-2 x+x^2\right ) \log (\log (5))}} (-32-32 x-32 x \log (\log (5)))}{\left (-1+3 x-3 x^2+x^3\right ) \log (\log (5))} \, dx={\mathrm {e}}^{\frac {32\,x}{\ln \left (\ln \left (5\right )\right )\,x^2-2\,\ln \left (\ln \left (5\right )\right )\,x+\ln \left (\ln \left (5\right )\right )}}\,{\ln \left (5\right )}^{\frac {12\,x^2+8\,x-4}{\ln \left ({\ln \left (5\right )}^{x^2-2\,x+1}\right )}} \] Input:

int(-(exp((32*x + log(log(5))*(8*x + 12*x^2 - 4))/(log(log(5))*(x^2 - 2*x 
+ 1)))*(32*x + 32*x*log(log(5)) + 32))/(log(log(5))*(3*x - 3*x^2 + x^3 - 1 
)),x)
 

Output:

exp((32*x)/(log(log(5)) + x^2*log(log(5)) - 2*x*log(log(5))))*log(5)^((8*x 
 + 12*x^2 - 4)/log(log(5)^(x^2 - 2*x + 1)))
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.62 \[ \int \frac {e^{\frac {32 x+\left (-4+8 x+12 x^2\right ) \log (\log (5))}{\left (1-2 x+x^2\right ) \log (\log (5))}} (-32-32 x-32 x \log (\log (5)))}{\left (-1+3 x-3 x^2+x^3\right ) \log (\log (5))} \, dx=\frac {e^{\frac {32 \,\mathrm {log}\left (\mathrm {log}\left (5\right )\right ) x +32 x}{\mathrm {log}\left (\mathrm {log}\left (5\right )\right ) x^{2}-2 \,\mathrm {log}\left (\mathrm {log}\left (5\right )\right ) x +\mathrm {log}\left (\mathrm {log}\left (5\right )\right )}} e^{12}}{e^{\frac {16}{x^{2}-2 x +1}}} \] Input:

int((-32*x*log(log(5))-32*x-32)*exp(((12*x^2+8*x-4)*log(log(5))+32*x)/(x^2 
-2*x+1)/log(log(5)))/(x^3-3*x^2+3*x-1)/log(log(5)),x)
 

Output:

(e**((32*log(log(5))*x + 32*x)/(log(log(5))*x**2 - 2*log(log(5))*x + log(l 
og(5))))*e**12)/e**(16/(x**2 - 2*x + 1))