\(\int \frac {e^{-3+\frac {8 e x^2+e^2 (4 x^2+e (-2 x+4 x^2))-8 e^3 \log (\log (x))}{e^3 x}} (-8 e^3+(8 e x^2+e^2 (4 x^2+4 e x^2)) \log (x)+8 e^3 \log (x) \log (\log (x)))}{x^2 \log (x)} \, dx\) [254]

Optimal result

Integrand size = 97, antiderivative size = 27 \[ \int \frac {e^{-3+\frac {8 e x^2+e^2 \left (4 x^2+e \left (-2 x+4 x^2\right )\right )-8 e^3 \log (\log (x))}{e^3 x}} \left (-8 e^3+\left (8 e x^2+e^2 \left (4 x^2+4 e x^2\right )\right ) \log (x)+8 e^3 \log (x) \log (\log (x))\right )}{x^2 \log (x)} \, dx=e^{-2+4 \left (x+\frac {2 x}{e^2}+\frac {x}{e}-\frac {2 \log (\log (x))}{x}\right )} \] Output:

exp(8*x/exp(2)+4*x-8*ln(ln(x))/x+4*x/exp(1)-2)
 

Mathematica [A] (verified)

Time = 0.37 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93 \[ \int \frac {e^{-3+\frac {8 e x^2+e^2 \left (4 x^2+e \left (-2 x+4 x^2\right )\right )-8 e^3 \log (\log (x))}{e^3 x}} \left (-8 e^3+\left (8 e x^2+e^2 \left (4 x^2+4 e x^2\right )\right ) \log (x)+8 e^3 \log (x) \log (\log (x))\right )}{x^2 \log (x)} \, dx=e^{-2+\frac {4 \left (2+e+e^2\right ) x}{e^2}} \log ^{-\frac {8}{x}}(x) \] Input:

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

Output:

E^(-2 + (4*(2 + E + E^2)*x)/E^2)/Log[x]^(8/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.

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

\(\Big \downarrow \) 7239

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

\(\Big \downarrow \) 7293

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

$Aborted
 

Maple [A] (verified)

Time = 5.36 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96

method	result	size
risch	\(\ln \left (x \right )^{-\frac {8}{x}} {\mathrm e}^{4 x +8 x \,{\mathrm e}^{-2}+4 x \,{\mathrm e}^{-1}-2}\)	\(26\)
parallelrisch	\({\mathrm e}^{\frac {\left (-8 \,{\mathrm e} \,{\mathrm e}^{2} \ln \left (\ln \left (x \right )\right )+\left (\left (4 x^{2}-2 x \right ) {\mathrm e}+4 x^{2}\right ) {\mathrm e}^{2}+8 x^{2} {\mathrm e}\right ) {\mathrm e}^{-1} {\mathrm e}^{-2}}{x}}\)	\(52\)

Input:

int((8*exp(1)*exp(2)*ln(x)*ln(ln(x))+((4*x^2*exp(1)+4*x^2)*exp(2)+8*x^2*ex 
p(1))*ln(x)-8*exp(1)*exp(2))*exp((-8*exp(1)*exp(2)*ln(ln(x))+((4*x^2-2*x)* 
exp(1)+4*x^2)*exp(2)+8*x^2*exp(1))/x/exp(1)/exp(2))/x^2/exp(1)/exp(2)/ln(x 
),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.52 \[ \int \frac {e^{-3+\frac {8 e x^2+e^2 \left (4 x^2+e \left (-2 x+4 x^2\right )\right )-8 e^3 \log (\log (x))}{e^3 x}} \left (-8 e^3+\left (8 e x^2+e^2 \left (4 x^2+4 e x^2\right )\right ) \log (x)+8 e^3 \log (x) \log (\log (x))\right )}{x^2 \log (x)} \, dx=e^{\left (\frac {{\left (4 \, x^{2} e + 8 \, x^{2} + {\left (4 \, x^{2} - 5 \, x\right )} e^{2} - 8 \, e^{2} \log \left (\log \left (x\right )\right )\right )} e^{\left (-2\right )}}{x} + 3\right )} \] Input:

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

Output:

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

Sympy [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.63 \[ \int \frac {e^{-3+\frac {8 e x^2+e^2 \left (4 x^2+e \left (-2 x+4 x^2\right )\right )-8 e^3 \log (\log (x))}{e^3 x}} \left (-8 e^3+\left (8 e x^2+e^2 \left (4 x^2+4 e x^2\right )\right ) \log (x)+8 e^3 \log (x) \log (\log (x))\right )}{x^2 \log (x)} \, dx=e^{\frac {8 e x^{2} + \left (4 x^{2} + e \left (4 x^{2} - 2 x\right )\right ) e^{2} - 8 e^{3} \log {\left (\log {\left (x \right )} \right )}}{x e^{3}}} \] Input:

integrate((8*exp(1)*exp(2)*ln(x)*ln(ln(x))+((4*x**2*exp(1)+4*x**2)*exp(2)+ 
8*x**2*exp(1))*ln(x)-8*exp(1)*exp(2))*exp((-8*exp(1)*exp(2)*ln(ln(x))+((4* 
x**2-2*x)*exp(1)+4*x**2)*exp(2)+8*x**2*exp(1))/x/exp(1)/exp(2))/x**2/exp(1 
)/exp(2)/ln(x),x)
 

Output:

exp((8*E*x**2 + (4*x**2 + E*(4*x**2 - 2*x))*exp(2) - 8*exp(3)*log(log(x))) 
*exp(-3)/x)
 

Maxima [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89 \[ \int \frac {e^{-3+\frac {8 e x^2+e^2 \left (4 x^2+e \left (-2 x+4 x^2\right )\right )-8 e^3 \log (\log (x))}{e^3 x}} \left (-8 e^3+\left (8 e x^2+e^2 \left (4 x^2+4 e x^2\right )\right ) \log (x)+8 e^3 \log (x) \log (\log (x))\right )}{x^2 \log (x)} \, dx=e^{\left (4 \, x e^{\left (-1\right )} + 8 \, x e^{\left (-2\right )} + 4 \, x - \frac {8 \, \log \left (\log \left (x\right )\right )}{x} - 2\right )} \] Input:

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

Output:

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

Giac [F]

\[ \int \frac {e^{-3+\frac {8 e x^2+e^2 \left (4 x^2+e \left (-2 x+4 x^2\right )\right )-8 e^3 \log (\log (x))}{e^3 x}} \left (-8 e^3+\left (8 e x^2+e^2 \left (4 x^2+4 e x^2\right )\right ) \log (x)+8 e^3 \log (x) \log (\log (x))\right )}{x^2 \log (x)} \, dx=\int { \frac {4 \, {\left (2 \, e^{3} \log \left (x\right ) \log \left (\log \left (x\right )\right ) + {\left (2 \, x^{2} e + {\left (x^{2} e + x^{2}\right )} e^{2}\right )} \log \left (x\right ) - 2 \, e^{3}\right )} e^{\left (\frac {2 \, {\left (4 \, x^{2} e + {\left (2 \, x^{2} + {\left (2 \, x^{2} - x\right )} e\right )} e^{2} - 4 \, e^{3} \log \left (\log \left (x\right )\right )\right )} e^{\left (-3\right )}}{x} - 3\right )}}{x^{2} \log \left (x\right )} \,d x } \] Input:

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

Output:

integrate(4*(2*e^3*log(x)*log(log(x)) + (2*x^2*e + (x^2*e + x^2)*e^2)*log( 
x) - 2*e^3)*e^(2*(4*x^2*e + (2*x^2 + (2*x^2 - x)*e)*e^2 - 4*e^3*log(log(x) 
))*e^(-3)/x - 3)/(x^2*log(x)), x)
 

Mupad [B] (verification not implemented)

Time = 4.03 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.07 \[ \int \frac {e^{-3+\frac {8 e x^2+e^2 \left (4 x^2+e \left (-2 x+4 x^2\right )\right )-8 e^3 \log (\log (x))}{e^3 x}} \left (-8 e^3+\left (8 e x^2+e^2 \left (4 x^2+4 e x^2\right )\right ) \log (x)+8 e^3 \log (x) \log (\log (x))\right )}{x^2 \log (x)} \, dx=\frac {{\mathrm {e}}^{4\,x}\,{\mathrm {e}}^{-2}\,{\mathrm {e}}^{4\,x\,{\mathrm {e}}^{-1}}\,{\mathrm {e}}^{8\,x\,{\mathrm {e}}^{-2}}}{{\ln \left (x\right )}^{8/x}} \] Input:

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

Output:

(exp(4*x)*exp(-2)*exp(4*x*exp(-1))*exp(8*x*exp(-2)))/log(x)^(8/x)
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.33 \[ \int \frac {e^{-3+\frac {8 e x^2+e^2 \left (4 x^2+e \left (-2 x+4 x^2\right )\right )-8 e^3 \log (\log (x))}{e^3 x}} \left (-8 e^3+\left (8 e x^2+e^2 \left (4 x^2+4 e x^2\right )\right ) \log (x)+8 e^3 \log (x) \log (\log (x))\right )}{x^2 \log (x)} \, dx=\frac {e^{\frac {4 e^{2} x +4 e x +8 x}{e^{2}}}}{e^{\frac {8 \,\mathrm {log}\left (\mathrm {log}\left (x \right )\right )}{x}} e^{2}} \] Input:

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

Output:

e**((4*e**2*x + 4*e*x + 8*x)/e**2)/(e**((8*log(log(x)))/x)*e**2)