\(\int \frac {-\log ^2(x)+e^{e^{\frac {x+(-8 x-4 x^2-4 \log (4)) \log (x)}{\log (x)}}+\frac {x+(-8 x-4 x^2-4 \log (4)) \log (x)}{\log (x)}} (4-4 \log (x)+(32+32 x) \log ^2(x))}{4 \log ^2(x)} \, dx\) [2251]

Optimal result

Integrand size = 83, antiderivative size = 37 \[ \int \frac {-\log ^2(x)+e^{e^{\frac {x+\left (-8 x-4 x^2-4 \log (4)\right ) \log (x)}{\log (x)}}+\frac {x+\left (-8 x-4 x^2-4 \log (4)\right ) \log (x)}{\log (x)}} \left (4-4 \log (x)+(32+32 x) \log ^2(x)\right )}{4 \log ^2(x)} \, dx=4-e^{e^{4 \left (-2 x-x^2-\log (4)+\frac {x}{4 \log (x)}\right )}}-\frac {x}{4} \] Output:

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

Mathematica [A] (verified)

Time = 2.40 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.84 \[ \int \frac {-\log ^2(x)+e^{e^{\frac {x+\left (-8 x-4 x^2-4 \log (4)\right ) \log (x)}{\log (x)}}+\frac {x+\left (-8 x-4 x^2-4 \log (4)\right ) \log (x)}{\log (x)}} \left (4-4 \log (x)+(32+32 x) \log ^2(x)\right )}{4 \log ^2(x)} \, dx=-e^{\frac {1}{256} e^{-8 x-4 x^2+\frac {x}{\log (x)}}}-\frac {x}{4} \] Input:

Integrate[(-Log[x]^2 + E^(E^((x + (-8*x - 4*x^2 - 4*Log[4])*Log[x])/Log[x] 
) + (x + (-8*x - 4*x^2 - 4*Log[4])*Log[x])/Log[x])*(4 - 4*Log[x] + (32 + 3 
2*x)*Log[x]^2))/(4*Log[x]^2),x]
 

Output:

-E^(E^(-8*x - 4*x^2 + x/Log[x])/256) - x/4
 

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.

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

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{4} \int -\frac {\log ^2(x)-4 \exp \left (\frac {x-4 \left (x^2+2 x+\log (4)\right ) \log (x)}{\log (x)}+\frac {1}{256} e^{-4 x^2+\frac {x}{\log (x)}-8 x}\right ) \left (8 (x+1) \log ^2(x)-\log (x)+1\right )}{\log ^2(x)}dx\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {1}{4} \int \frac {\log ^2(x)-4 \exp \left (\frac {x-4 \left (x^2+2 x+\log (4)\right ) \log (x)}{\log (x)}+\frac {1}{256} e^{-4 x^2+\frac {x}{\log (x)}-8 x}\right ) \left (8 (x+1) \log ^2(x)-\log (x)+1\right )}{\log ^2(x)}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle -\frac {1}{4} \int \left (1-\frac {\exp \left (-4 (x+2) x+\frac {x}{\log (x)}+\frac {1}{256} e^{x \left (\frac {1}{\log (x)}-4 (x+2)\right )}\right ) \left (8 x \log ^2(x)+8 \log ^2(x)-\log (x)+1\right )}{64 \log ^2(x)}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{4} \left (\frac {1}{64} \int \frac {\exp \left (-4 (x+2) x+\frac {x}{\log (x)}+\frac {1}{256} e^{x \left (\frac {1}{\log (x)}-4 (x+2)\right )}\right )}{\log ^2(x)}dx+\frac {1}{8} \int \exp \left (-4 (x+2) x+\frac {x}{\log (x)}+\frac {1}{256} e^{x \left (\frac {1}{\log (x)}-4 (x+2)\right )}\right )dx+\frac {1}{8} \int \exp \left (-4 (x+2) x+\frac {x}{\log (x)}+\frac {1}{256} e^{x \left (\frac {1}{\log (x)}-4 (x+2)\right )}\right ) xdx-\frac {1}{64} \int \frac {\exp \left (-4 (x+2) x+\frac {x}{\log (x)}+\frac {1}{256} e^{x \left (\frac {1}{\log (x)}-4 (x+2)\right )}\right )}{\log (x)}dx-x\right )\)

Input:

Int[(-Log[x]^2 + E^(E^((x + (-8*x - 4*x^2 - 4*Log[4])*Log[x])/Log[x]) + (x 
 + (-8*x - 4*x^2 - 4*Log[4])*Log[x])/Log[x])*(4 - 4*Log[x] + (32 + 32*x)*L 
og[x]^2))/(4*Log[x]^2),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 1.04 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.78

method	result	size
risch	\(-\frac {x}{4}-{\mathrm e}^{\frac {{\mathrm e}^{-\frac {x \left (4 x \ln \left (x \right )+8 \ln \left (x \right )-1\right )}{\ln \left (x \right )}}}{256}}\)	\(29\)
parallelrisch	\(-\frac {x}{4}-{\mathrm e}^{{\mathrm e}^{\frac {\left (-8 \ln \left (2\right )-4 x^{2}-8 x \right ) \ln \left (x \right )+x}{\ln \left (x \right )}}}\)	\(32\)

Input:

int(1/4*(((32*x+32)*ln(x)^2-4*ln(x)+4)*exp(((-8*ln(2)-4*x^2-8*x)*ln(x)+x)/ 
ln(x))*exp(exp(((-8*ln(2)-4*x^2-8*x)*ln(x)+x)/ln(x)))-ln(x)^2)/ln(x)^2,x,m 
ethod=_RETURNVERBOSE)
 

Output:

-1/4*x-exp(1/256*exp(-x*(4*x*ln(x)+8*ln(x)-1)/ln(x)))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 114 vs. \(2 (28) = 56\).

Time = 0.09 (sec) , antiderivative size = 114, normalized size of antiderivative = 3.08 \[ \int \frac {-\log ^2(x)+e^{e^{\frac {x+\left (-8 x-4 x^2-4 \log (4)\right ) \log (x)}{\log (x)}}+\frac {x+\left (-8 x-4 x^2-4 \log (4)\right ) \log (x)}{\log (x)}} \left (4-4 \log (x)+(32+32 x) \log ^2(x)\right )}{4 \log ^2(x)} \, dx=-\frac {1}{4} \, {\left (x e^{\left (-\frac {4 \, {\left (x^{2} + 2 \, x + 2 \, \log \left (2\right )\right )} \log \left (x\right ) - x}{\log \left (x\right )}\right )} + 4 \, e^{\left (-\frac {4 \, {\left (x^{2} + 2 \, x + 2 \, \log \left (2\right )\right )} \log \left (x\right ) - e^{\left (-\frac {4 \, {\left (x^{2} + 2 \, x + 2 \, \log \left (2\right )\right )} \log \left (x\right ) - x}{\log \left (x\right )}\right )} \log \left (x\right ) - x}{\log \left (x\right )}\right )}\right )} e^{\left (\frac {4 \, {\left (x^{2} + 2 \, x + 2 \, \log \left (2\right )\right )} \log \left (x\right ) - x}{\log \left (x\right )}\right )} \] Input:

integrate(1/4*(((32*x+32)*log(x)^2-4*log(x)+4)*exp(((-8*log(2)-4*x^2-8*x)* 
log(x)+x)/log(x))*exp(exp(((-8*log(2)-4*x^2-8*x)*log(x)+x)/log(x)))-log(x) 
^2)/log(x)^2,x, algorithm="fricas")
 

Output:

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

Sympy [A] (verification not implemented)

Time = 0.50 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.84 \[ \int \frac {-\log ^2(x)+e^{e^{\frac {x+\left (-8 x-4 x^2-4 \log (4)\right ) \log (x)}{\log (x)}}+\frac {x+\left (-8 x-4 x^2-4 \log (4)\right ) \log (x)}{\log (x)}} \left (4-4 \log (x)+(32+32 x) \log ^2(x)\right )}{4 \log ^2(x)} \, dx=- \frac {x}{4} - e^{e^{\frac {x + \left (- 4 x^{2} - 8 x - 8 \log {\left (2 \right )}\right ) \log {\left (x \right )}}{\log {\left (x \right )}}}} \] Input:

integrate(1/4*(((32*x+32)*ln(x)**2-4*ln(x)+4)*exp(((-8*ln(2)-4*x**2-8*x)*l 
n(x)+x)/ln(x))*exp(exp(((-8*ln(2)-4*x**2-8*x)*ln(x)+x)/ln(x)))-ln(x)**2)/l 
n(x)**2,x)
 

Output:

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

Maxima [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.68 \[ \int \frac {-\log ^2(x)+e^{e^{\frac {x+\left (-8 x-4 x^2-4 \log (4)\right ) \log (x)}{\log (x)}}+\frac {x+\left (-8 x-4 x^2-4 \log (4)\right ) \log (x)}{\log (x)}} \left (4-4 \log (x)+(32+32 x) \log ^2(x)\right )}{4 \log ^2(x)} \, dx=-\frac {1}{4} \, x - e^{\left (\frac {1}{256} \, e^{\left (-4 \, x^{2} - 8 \, x + \frac {x}{\log \left (x\right )}\right )}\right )} \] Input:

integrate(1/4*(((32*x+32)*log(x)^2-4*log(x)+4)*exp(((-8*log(2)-4*x^2-8*x)* 
log(x)+x)/log(x))*exp(exp(((-8*log(2)-4*x^2-8*x)*log(x)+x)/log(x)))-log(x) 
^2)/log(x)^2,x, algorithm="maxima")
 

Output:

-1/4*x - e^(1/256*e^(-4*x^2 - 8*x + x/log(x)))
 

Giac [F]

\[ \int \frac {-\log ^2(x)+e^{e^{\frac {x+\left (-8 x-4 x^2-4 \log (4)\right ) \log (x)}{\log (x)}}+\frac {x+\left (-8 x-4 x^2-4 \log (4)\right ) \log (x)}{\log (x)}} \left (4-4 \log (x)+(32+32 x) \log ^2(x)\right )}{4 \log ^2(x)} \, dx=\int { \frac {4 \, {\left (8 \, {\left (x + 1\right )} \log \left (x\right )^{2} - \log \left (x\right ) + 1\right )} e^{\left (-\frac {4 \, {\left (x^{2} + 2 \, x + 2 \, \log \left (2\right )\right )} \log \left (x\right ) - x}{\log \left (x\right )} + e^{\left (-\frac {4 \, {\left (x^{2} + 2 \, x + 2 \, \log \left (2\right )\right )} \log \left (x\right ) - x}{\log \left (x\right )}\right )}\right )} - \log \left (x\right )^{2}}{4 \, \log \left (x\right )^{2}} \,d x } \] Input:

integrate(1/4*(((32*x+32)*log(x)^2-4*log(x)+4)*exp(((-8*log(2)-4*x^2-8*x)* 
log(x)+x)/log(x))*exp(exp(((-8*log(2)-4*x^2-8*x)*log(x)+x)/log(x)))-log(x) 
^2)/log(x)^2,x, algorithm="giac")
 

Output:

undef
 

Mupad [B] (verification not implemented)

Time = 2.67 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.70 \[ \int \frac {-\log ^2(x)+e^{e^{\frac {x+\left (-8 x-4 x^2-4 \log (4)\right ) \log (x)}{\log (x)}}+\frac {x+\left (-8 x-4 x^2-4 \log (4)\right ) \log (x)}{\log (x)}} \left (4-4 \log (x)+(32+32 x) \log ^2(x)\right )}{4 \log ^2(x)} \, dx=-\frac {x}{4}-{\mathrm {e}}^{\frac {{\mathrm {e}}^{-8\,x}\,{\mathrm {e}}^{\frac {x}{\ln \left (x\right )}}\,{\mathrm {e}}^{-4\,x^2}}{256}} \] Input:

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

Output:

- x/4 - exp((exp(-8*x)*exp(x/log(x))*exp(-4*x^2))/256)
 

Reduce [F]

\[ \int \frac {-\log ^2(x)+e^{e^{\frac {x+\left (-8 x-4 x^2-4 \log (4)\right ) \log (x)}{\log (x)}}+\frac {x+\left (-8 x-4 x^2-4 \log (4)\right ) \log (x)}{\log (x)}} \left (4-4 \log (x)+(32+32 x) \log ^2(x)\right )}{4 \log ^2(x)} \, dx=\frac {\left (\int \frac {e^{\frac {256 e^{4 x^{2}+8 x} x +e^{\frac {x}{\mathrm {log}\left (x \right )}} \mathrm {log}\left (x \right )}{256 e^{4 x^{2}+8 x} \mathrm {log}\left (x \right )}}}{e^{4 x^{2}+8 x}}d x \right )}{32}+\frac {\left (\int \frac {e^{\frac {256 e^{4 x^{2}+8 x} x +e^{\frac {x}{\mathrm {log}\left (x \right )}} \mathrm {log}\left (x \right )}{256 e^{4 x^{2}+8 x} \mathrm {log}\left (x \right )}}}{e^{4 x^{2}+8 x} \mathrm {log}\left (x \right )^{2}}d x \right )}{256}-\frac {\left (\int \frac {e^{\frac {256 e^{4 x^{2}+8 x} x +e^{\frac {x}{\mathrm {log}\left (x \right )}} \mathrm {log}\left (x \right )}{256 e^{4 x^{2}+8 x} \mathrm {log}\left (x \right )}}}{e^{4 x^{2}+8 x} \mathrm {log}\left (x \right )}d x \right )}{256}+\frac {\left (\int \frac {e^{\frac {256 e^{4 x^{2}+8 x} x +e^{\frac {x}{\mathrm {log}\left (x \right )}} \mathrm {log}\left (x \right )}{256 e^{4 x^{2}+8 x} \mathrm {log}\left (x \right )}} x}{e^{4 x^{2}+8 x}}d x \right )}{32}-\frac {x}{4} \] Input:

int(1/4*(((32*x+32)*log(x)^2-4*log(x)+4)*exp(((-8*log(2)-4*x^2-8*x)*log(x) 
+x)/log(x))*exp(exp(((-8*log(2)-4*x^2-8*x)*log(x)+x)/log(x)))-log(x)^2)/lo 
g(x)^2,x)
 

Output:

(8*int(e**((256*e**(4*x**2 + 8*x)*x + e**(x/log(x))*log(x))/(256*e**(4*x** 
2 + 8*x)*log(x)))/e**(4*x**2 + 8*x),x) + int(e**((256*e**(4*x**2 + 8*x)*x 
+ e**(x/log(x))*log(x))/(256*e**(4*x**2 + 8*x)*log(x)))/(e**(4*x**2 + 8*x) 
*log(x)**2),x) - int(e**((256*e**(4*x**2 + 8*x)*x + e**(x/log(x))*log(x))/ 
(256*e**(4*x**2 + 8*x)*log(x)))/(e**(4*x**2 + 8*x)*log(x)),x) + 8*int((e** 
((256*e**(4*x**2 + 8*x)*x + e**(x/log(x))*log(x))/(256*e**(4*x**2 + 8*x)*l 
og(x)))*x)/e**(4*x**2 + 8*x),x) - 64*x)/256