\(\int \frac {e^x (-16+32 x-18 x^2)+e^x (32 x-32 x^2) \log (\frac {x}{5})+(e^x x^3+e^x (16 x-32 x^2+16 x^3) \log (\frac {x}{5})) \log (\frac {1}{16} (x^2+(16-32 x+16 x^2) \log (\frac {x}{5})))}{(x^3+(16 x-32 x^2+16 x^3) \log (\frac {x}{5})) \log ^2(\frac {1}{16} (x^2+(16-32 x+16 x^2) \log (\frac {x}{5})))} \, dx\) [998]

Optimal result

Integrand size = 149, antiderivative size = 27 \[ \int \frac {e^x \left (-16+32 x-18 x^2\right )+e^x \left (32 x-32 x^2\right ) \log \left (\frac {x}{5}\right )+\left (e^x x^3+e^x \left (16 x-32 x^2+16 x^3\right ) \log \left (\frac {x}{5}\right )\right ) \log \left (\frac {1}{16} \left (x^2+\left (16-32 x+16 x^2\right ) \log \left (\frac {x}{5}\right )\right )\right )}{\left (x^3+\left (16 x-32 x^2+16 x^3\right ) \log \left (\frac {x}{5}\right )\right ) \log ^2\left (\frac {1}{16} \left (x^2+\left (16-32 x+16 x^2\right ) \log \left (\frac {x}{5}\right )\right )\right )} \, dx=\frac {e^x}{\log \left (\frac {x^2}{16}+(-1+x)^2 \log \left (\frac {x}{5}\right )\right )} \] Output:

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

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.04 \[ \int \frac {e^x \left (-16+32 x-18 x^2\right )+e^x \left (32 x-32 x^2\right ) \log \left (\frac {x}{5}\right )+\left (e^x x^3+e^x \left (16 x-32 x^2+16 x^3\right ) \log \left (\frac {x}{5}\right )\right ) \log \left (\frac {1}{16} \left (x^2+\left (16-32 x+16 x^2\right ) \log \left (\frac {x}{5}\right )\right )\right )}{\left (x^3+\left (16 x-32 x^2+16 x^3\right ) \log \left (\frac {x}{5}\right )\right ) \log ^2\left (\frac {1}{16} \left (x^2+\left (16-32 x+16 x^2\right ) \log \left (\frac {x}{5}\right )\right )\right )} \, dx=\frac {e^x}{\log \left (\frac {1}{16} \left (x^2+16 (-1+x)^2 \log \left (\frac {x}{5}\right )\right )\right )} \] Input:

Integrate[(E^x*(-16 + 32*x - 18*x^2) + E^x*(32*x - 32*x^2)*Log[x/5] + (E^x 
*x^3 + E^x*(16*x - 32*x^2 + 16*x^3)*Log[x/5])*Log[(x^2 + (16 - 32*x + 16*x 
^2)*Log[x/5])/16])/((x^3 + (16*x - 32*x^2 + 16*x^3)*Log[x/5])*Log[(x^2 + ( 
16 - 32*x + 16*x^2)*Log[x/5])/16]^2),x]
 

Output:

E^x/Log[(x^2 + 16*(-1 + x)^2*Log[x/5])/16]
 

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.

\(\displaystyle \int \frac {e^x \left (-18 x^2+32 x-16\right )+e^x \left (32 x-32 x^2\right ) \log \left (\frac {x}{5}\right )+\left (e^x x^3+e^x \left (16 x^3-32 x^2+16 x\right ) \log \left (\frac {x}{5}\right )\right ) \log \left (\frac {1}{16} \left (x^2+\left (16 x^2-32 x+16\right ) \log \left (\frac {x}{5}\right )\right )\right )}{\left (x^3+\left (16 x^3-32 x^2+16 x\right ) \log \left (\frac {x}{5}\right )\right ) \log ^2\left (\frac {1}{16} \left (x^2+\left (16 x^2-32 x+16\right ) \log \left (\frac {x}{5}\right )\right )\right )} \, dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (-\frac {18 e^x x}{\left (x^2+16 x^2 \log \left (\frac {x}{5}\right )-32 x \log \left (\frac {x}{5}\right )+16 \log \left (\frac {x}{5}\right )\right ) \log ^2\left (\frac {1}{16} \left (x^2+16 (x-1)^2 \log \left (\frac {x}{5}\right )\right )\right )}-\frac {32 e^x x \log \left (\frac {x}{5}\right )}{\left (x^2+16 x^2 \log \left (\frac {x}{5}\right )-32 x \log \left (\frac {x}{5}\right )+16 \log \left (\frac {x}{5}\right )\right ) \log ^2\left (\frac {1}{16} \left (x^2+16 (x-1)^2 \log \left (\frac {x}{5}\right )\right )\right )}+\frac {32 e^x}{\left (x^2+16 x^2 \log \left (\frac {x}{5}\right )-32 x \log \left (\frac {x}{5}\right )+16 \log \left (\frac {x}{5}\right )\right ) \log ^2\left (\frac {1}{16} \left (x^2+16 (x-1)^2 \log \left (\frac {x}{5}\right )\right )\right )}+\frac {32 e^x \log \left (\frac {x}{5}\right )}{\left (x^2+16 x^2 \log \left (\frac {x}{5}\right )-32 x \log \left (\frac {x}{5}\right )+16 \log \left (\frac {x}{5}\right )\right ) \log ^2\left (\frac {1}{16} \left (x^2+16 (x-1)^2 \log \left (\frac {x}{5}\right )\right )\right )}-\frac {16 e^x}{x \left (x^2+16 x^2 \log \left (\frac {x}{5}\right )-32 x \log \left (\frac {x}{5}\right )+16 \log \left (\frac {x}{5}\right )\right ) \log ^2\left (\frac {1}{16} \left (x^2+16 (x-1)^2 \log \left (\frac {x}{5}\right )\right )\right )}+\frac {e^x x^2}{\left (x^2+16 x^2 \log \left (\frac {x}{5}\right )-32 x \log \left (\frac {x}{5}\right )+16 \log \left (\frac {x}{5}\right )\right ) \log \left (\frac {1}{16} \left (x^2+16 (x-1)^2 \log \left (\frac {x}{5}\right )\right )\right )}+\frac {16 e^x x^2 \log \left (\frac {x}{5}\right )}{\left (x^2+16 x^2 \log \left (\frac {x}{5}\right )-32 x \log \left (\frac {x}{5}\right )+16 \log \left (\frac {x}{5}\right )\right ) \log \left (\frac {1}{16} \left (x^2+16 (x-1)^2 \log \left (\frac {x}{5}\right )\right )\right )}-\frac {32 e^x x \log \left (\frac {x}{5}\right )}{\left (x^2+16 x^2 \log \left (\frac {x}{5}\right )-32 x \log \left (\frac {x}{5}\right )+16 \log \left (\frac {x}{5}\right )\right ) \log \left (\frac {1}{16} \left (x^2+16 (x-1)^2 \log \left (\frac {x}{5}\right )\right )\right )}+\frac {16 e^x \log \left (\frac {x}{5}\right )}{\left (x^2+16 x^2 \log \left (\frac {x}{5}\right )-32 x \log \left (\frac {x}{5}\right )+16 \log \left (\frac {x}{5}\right )\right ) \log \left (\frac {1}{16} \left (x^2+16 (x-1)^2 \log \left (\frac {x}{5}\right )\right )\right )}\right )dx\)

\(\Big \downarrow \) 7239

\(\displaystyle \int \frac {e^x \left (-18 x^2+16 (x-1) x \log \left (\frac {x}{5}\right ) \left ((x-1) \log \left (\frac {1}{16} \left (x^2+16 (x-1)^2 \log \left (\frac {x}{5}\right )\right )\right )-2\right )+x^3 \log \left (\frac {1}{16} \left (x^2+16 (x-1)^2 \log \left (\frac {x}{5}\right )\right )\right )+32 x-16\right )}{x \left (x^2+16 (x-1)^2 \log \left (\frac {x}{5}\right )\right ) \log ^2\left (\frac {1}{16} \left (x^2+16 (x-1)^2 \log \left (\frac {x}{5}\right )\right )\right )}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (\frac {e^x}{\log \left (\frac {1}{16} \left (x^2+16 (x-1)^2 \log \left (\frac {x}{5}\right )\right )\right )}-\frac {2 e^x \left (9 x^2+16 x^2 \log \left (\frac {x}{5}\right )-16 x-16 x \log \left (\frac {x}{5}\right )+8\right )}{x \left (x^2+16 x^2 \log \left (\frac {x}{5}\right )-32 x \log \left (\frac {x}{5}\right )+16 \log \left (\frac {x}{5}\right )\right ) \log ^2\left (\frac {1}{16} \left (x^2+16 (x-1)^2 \log \left (\frac {x}{5}\right )\right )\right )}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle 160 \text {Subst}\left (\int \frac {e^{5 x}}{\left (25 x^2+16 (5 x-1)^2 \log (x)\right ) \log ^2\left (\frac {1}{16} \left (25 x^2+16 (5 x-1)^2 \log (x)\right )\right )}dx,x,\frac {x}{5}\right )+160 \text {Subst}\left (\int \frac {e^{5 x} \log (x)}{\left (25 x^2+16 (5 x-1)^2 \log (x)\right ) \log ^2\left (\frac {1}{16} \left (25 x^2+16 (5 x-1)^2 \log (x)\right )\right )}dx,x,\frac {x}{5}\right )+5 \text {Subst}\left (\int \frac {e^{5 x}}{\log \left (\frac {1}{16} \left (25 x^2+16 (5 x-1)^2 \log (x)\right )\right )}dx,x,\frac {x}{5}\right )-16 \int \frac {e^x}{x \left (16 \log \left (\frac {x}{5}\right ) (x-1)^2+x^2\right ) \log ^2\left (\frac {1}{16} \left (16 \log \left (\frac {x}{5}\right ) (x-1)^2+x^2\right )\right )}dx-18 \int \frac {e^x x}{\left (16 \log \left (\frac {x}{5}\right ) (x-1)^2+x^2\right ) \log ^2\left (\frac {1}{16} \left (16 \log \left (\frac {x}{5}\right ) (x-1)^2+x^2\right )\right )}dx-32 \int \frac {e^x x \log \left (\frac {x}{5}\right )}{\left (16 \log \left (\frac {x}{5}\right ) (x-1)^2+x^2\right ) \log ^2\left (\frac {1}{16} \left (16 \log \left (\frac {x}{5}\right ) (x-1)^2+x^2\right )\right )}dx\)

Input:

Int[(E^x*(-16 + 32*x - 18*x^2) + E^x*(32*x - 32*x^2)*Log[x/5] + (E^x*x^3 + 
 E^x*(16*x - 32*x^2 + 16*x^3)*Log[x/5])*Log[(x^2 + (16 - 32*x + 16*x^2)*Lo 
g[x/5])/16])/((x^3 + (16*x - 32*x^2 + 16*x^3)*Log[x/5])*Log[(x^2 + (16 - 3 
2*x + 16*x^2)*Log[x/5])/16]^2),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 4.12 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.07

method	result	size
risch	\(\frac {{\mathrm e}^{x}}{\ln \left (\frac {\left (16 x^{2}-32 x +16\right ) \ln \left (\frac {x}{5}\right )}{16}+\frac {x^{2}}{16}\right )}\)	\(29\)
parallelrisch	\(\frac {{\mathrm e}^{x}}{\ln \left (\frac {\left (16 x^{2}-32 x +16\right ) \ln \left (\frac {x}{5}\right )}{16}+\frac {x^{2}}{16}\right )}\)	\(29\)

Input:

int((((16*x^3-32*x^2+16*x)*exp(x)*ln(1/5*x)+exp(x)*x^3)*ln(1/16*(16*x^2-32 
*x+16)*ln(1/5*x)+1/16*x^2)+(-32*x^2+32*x)*exp(x)*ln(1/5*x)+(-18*x^2+32*x-1 
6)*exp(x))/((16*x^3-32*x^2+16*x)*ln(1/5*x)+x^3)/ln(1/16*(16*x^2-32*x+16)*l 
n(1/5*x)+1/16*x^2)^2,x,method=_RETURNVERBOSE)
 

Output:

exp(x)/ln(1/16*(16*x^2-32*x+16)*ln(1/5*x)+1/16*x^2)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93 \[ \int \frac {e^x \left (-16+32 x-18 x^2\right )+e^x \left (32 x-32 x^2\right ) \log \left (\frac {x}{5}\right )+\left (e^x x^3+e^x \left (16 x-32 x^2+16 x^3\right ) \log \left (\frac {x}{5}\right )\right ) \log \left (\frac {1}{16} \left (x^2+\left (16-32 x+16 x^2\right ) \log \left (\frac {x}{5}\right )\right )\right )}{\left (x^3+\left (16 x-32 x^2+16 x^3\right ) \log \left (\frac {x}{5}\right )\right ) \log ^2\left (\frac {1}{16} \left (x^2+\left (16-32 x+16 x^2\right ) \log \left (\frac {x}{5}\right )\right )\right )} \, dx=\frac {e^{x}}{\log \left (\frac {1}{16} \, x^{2} + {\left (x^{2} - 2 \, x + 1\right )} \log \left (\frac {1}{5} \, x\right )\right )} \] Input:

integrate((((16*x^3-32*x^2+16*x)*exp(x)*log(1/5*x)+exp(x)*x^3)*log(1/16*(1 
6*x^2-32*x+16)*log(1/5*x)+1/16*x^2)+(-32*x^2+32*x)*exp(x)*log(1/5*x)+(-18* 
x^2+32*x-16)*exp(x))/((16*x^3-32*x^2+16*x)*log(1/5*x)+x^3)/log(1/16*(16*x^ 
2-32*x+16)*log(1/5*x)+1/16*x^2)^2,x, algorithm="fricas")
 

Output:

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

Sympy [A] (verification not implemented)

Time = 0.42 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81 \[ \int \frac {e^x \left (-16+32 x-18 x^2\right )+e^x \left (32 x-32 x^2\right ) \log \left (\frac {x}{5}\right )+\left (e^x x^3+e^x \left (16 x-32 x^2+16 x^3\right ) \log \left (\frac {x}{5}\right )\right ) \log \left (\frac {1}{16} \left (x^2+\left (16-32 x+16 x^2\right ) \log \left (\frac {x}{5}\right )\right )\right )}{\left (x^3+\left (16 x-32 x^2+16 x^3\right ) \log \left (\frac {x}{5}\right )\right ) \log ^2\left (\frac {1}{16} \left (x^2+\left (16-32 x+16 x^2\right ) \log \left (\frac {x}{5}\right )\right )\right )} \, dx=\frac {e^{x}}{\log {\left (\frac {x^{2}}{16} + \left (x^{2} - 2 x + 1\right ) \log {\left (\frac {x}{5} \right )} \right )}} \] Input:

integrate((((16*x**3-32*x**2+16*x)*exp(x)*ln(1/5*x)+exp(x)*x**3)*ln(1/16*( 
16*x**2-32*x+16)*ln(1/5*x)+1/16*x**2)+(-32*x**2+32*x)*exp(x)*ln(1/5*x)+(-1 
8*x**2+32*x-16)*exp(x))/((16*x**3-32*x**2+16*x)*ln(1/5*x)+x**3)/ln(1/16*(1 
6*x**2-32*x+16)*ln(1/5*x)+1/16*x**2)**2,x)
 

Output:

exp(x)/log(x**2/16 + (x**2 - 2*x + 1)*log(x/5))
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 47 vs. \(2 (22) = 44\).

Time = 0.18 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.74 \[ \int \frac {e^x \left (-16+32 x-18 x^2\right )+e^x \left (32 x-32 x^2\right ) \log \left (\frac {x}{5}\right )+\left (e^x x^3+e^x \left (16 x-32 x^2+16 x^3\right ) \log \left (\frac {x}{5}\right )\right ) \log \left (\frac {1}{16} \left (x^2+\left (16-32 x+16 x^2\right ) \log \left (\frac {x}{5}\right )\right )\right )}{\left (x^3+\left (16 x-32 x^2+16 x^3\right ) \log \left (\frac {x}{5}\right )\right ) \log ^2\left (\frac {1}{16} \left (x^2+\left (16-32 x+16 x^2\right ) \log \left (\frac {x}{5}\right )\right )\right )} \, dx=-\frac {e^{x}}{4 \, \log \left (2\right ) - \log \left (-x^{2} {\left (16 \, \log \left (5\right ) - 1\right )} + 32 \, x \log \left (5\right ) + 16 \, {\left (x^{2} - 2 \, x + 1\right )} \log \left (x\right ) - 16 \, \log \left (5\right )\right )} \] Input:

integrate((((16*x^3-32*x^2+16*x)*exp(x)*log(1/5*x)+exp(x)*x^3)*log(1/16*(1 
6*x^2-32*x+16)*log(1/5*x)+1/16*x^2)+(-32*x^2+32*x)*exp(x)*log(1/5*x)+(-18* 
x^2+32*x-16)*exp(x))/((16*x^3-32*x^2+16*x)*log(1/5*x)+x^3)/log(1/16*(16*x^ 
2-32*x+16)*log(1/5*x)+1/16*x^2)^2,x, algorithm="maxima")
 

Output:

-e^x/(4*log(2) - log(-x^2*(16*log(5) - 1) + 32*x*log(5) + 16*(x^2 - 2*x + 
1)*log(x) - 16*log(5)))
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1234 vs. \(2 (22) = 44\).

Time = 0.41 (sec) , antiderivative size = 1234, normalized size of antiderivative = 45.70 \[ \int \frac {e^x \left (-16+32 x-18 x^2\right )+e^x \left (32 x-32 x^2\right ) \log \left (\frac {x}{5}\right )+\left (e^x x^3+e^x \left (16 x-32 x^2+16 x^3\right ) \log \left (\frac {x}{5}\right )\right ) \log \left (\frac {1}{16} \left (x^2+\left (16-32 x+16 x^2\right ) \log \left (\frac {x}{5}\right )\right )\right )}{\left (x^3+\left (16 x-32 x^2+16 x^3\right ) \log \left (\frac {x}{5}\right )\right ) \log ^2\left (\frac {1}{16} \left (x^2+\left (16-32 x+16 x^2\right ) \log \left (\frac {x}{5}\right )\right )\right )} \, dx=\text {Too large to display} \] Input:

integrate((((16*x^3-32*x^2+16*x)*exp(x)*log(1/5*x)+exp(x)*x^3)*log(1/16*(1 
6*x^2-32*x+16)*log(1/5*x)+1/16*x^2)+(-32*x^2+32*x)*exp(x)*log(1/5*x)+(-18* 
x^2+32*x-16)*exp(x))/((16*x^3-32*x^2+16*x)*log(1/5*x)+x^3)/log(1/16*(16*x^ 
2-32*x+16)*log(1/5*x)+1/16*x^2)^2,x, algorithm="giac")
 

Output:

-(256*x^4*e^x*log(5)*log(1/5*x) - 256*x^4*e^x*log(1/5*x)*log(x) + 16*x^4*e 
^x*log(5) - 144*x^4*e^x*log(1/5*x) - 768*x^3*e^x*log(5)*log(1/5*x) - 16*x^ 
4*e^x*log(x) + 768*x^3*e^x*log(1/5*x)*log(x) - 9*x^4*e^x - 16*x^3*e^x*log( 
5) + 544*x^3*e^x*log(1/5*x) + 768*x^2*e^x*log(5)*log(1/5*x) + 16*x^3*e^x*l 
og(x) - 768*x^2*e^x*log(1/5*x)*log(x) + 16*x^3*e^x - 784*x^2*e^x*log(1/5*x 
) - 256*x*e^x*log(5)*log(1/5*x) + 256*x*e^x*log(1/5*x)*log(x) - 8*x^2*e^x 
+ 512*x*e^x*log(1/5*x) - 128*e^x*log(1/5*x))/(1024*x^4*log(5)*log(2)*log(1 
/5*x) - 256*x^4*log(5)*log(16*x^2*log(1/5*x) + x^2 - 32*x*log(1/5*x) + 16* 
log(1/5*x))*log(1/5*x) - 1024*x^4*log(2)*log(1/5*x)*log(x) + 256*x^4*log(1 
6*x^2*log(1/5*x) + x^2 - 32*x*log(1/5*x) + 16*log(1/5*x))*log(1/5*x)*log(x 
) + 576*x^4*log(5)*log(2) - 144*x^4*log(5)*log(16*x^2*log(1/5*x) + x^2 - 3 
2*x*log(1/5*x) + 16*log(1/5*x)) - 64*x^4*log(2)*log(1/5*x) - 3072*x^3*log( 
5)*log(2)*log(1/5*x) + 16*x^4*log(16*x^2*log(1/5*x) + x^2 - 32*x*log(1/5*x 
) + 16*log(1/5*x))*log(1/5*x) + 768*x^3*log(5)*log(16*x^2*log(1/5*x) + x^2 
 - 32*x*log(1/5*x) + 16*log(1/5*x))*log(1/5*x) - 576*x^4*log(2)*log(x) + 1 
44*x^4*log(16*x^2*log(1/5*x) + x^2 - 32*x*log(1/5*x) + 16*log(1/5*x))*log( 
x) + 3072*x^3*log(2)*log(1/5*x)*log(x) - 768*x^3*log(16*x^2*log(1/5*x) + x 
^2 - 32*x*log(1/5*x) + 16*log(1/5*x))*log(1/5*x)*log(x) - 36*x^4*log(2) - 
2176*x^3*log(5)*log(2) + 9*x^4*log(16*x^2*log(1/5*x) + x^2 - 32*x*log(1/5* 
x) + 16*log(1/5*x)) + 544*x^3*log(5)*log(16*x^2*log(1/5*x) + x^2 - 32*x...
 

Mupad [F(-1)]

Timed out. \[ \int \frac {e^x \left (-16+32 x-18 x^2\right )+e^x \left (32 x-32 x^2\right ) \log \left (\frac {x}{5}\right )+\left (e^x x^3+e^x \left (16 x-32 x^2+16 x^3\right ) \log \left (\frac {x}{5}\right )\right ) \log \left (\frac {1}{16} \left (x^2+\left (16-32 x+16 x^2\right ) \log \left (\frac {x}{5}\right )\right )\right )}{\left (x^3+\left (16 x-32 x^2+16 x^3\right ) \log \left (\frac {x}{5}\right )\right ) \log ^2\left (\frac {1}{16} \left (x^2+\left (16-32 x+16 x^2\right ) \log \left (\frac {x}{5}\right )\right )\right )} \, dx=\int \frac {\ln \left (\frac {\ln \left (\frac {x}{5}\right )\,\left (16\,x^2-32\,x+16\right )}{16}+\frac {x^2}{16}\right )\,\left (x^3\,{\mathrm {e}}^x+\ln \left (\frac {x}{5}\right )\,{\mathrm {e}}^x\,\left (16\,x^3-32\,x^2+16\,x\right )\right )-{\mathrm {e}}^x\,\left (18\,x^2-32\,x+16\right )+\ln \left (\frac {x}{5}\right )\,{\mathrm {e}}^x\,\left (32\,x-32\,x^2\right )}{{\ln \left (\frac {\ln \left (\frac {x}{5}\right )\,\left (16\,x^2-32\,x+16\right )}{16}+\frac {x^2}{16}\right )}^2\,\left (\ln \left (\frac {x}{5}\right )\,\left (16\,x^3-32\,x^2+16\,x\right )+x^3\right )} \,d x \] Input:

int((log((log(x/5)*(16*x^2 - 32*x + 16))/16 + x^2/16)*(x^3*exp(x) + log(x/ 
5)*exp(x)*(16*x - 32*x^2 + 16*x^3)) - exp(x)*(18*x^2 - 32*x + 16) + log(x/ 
5)*exp(x)*(32*x - 32*x^2))/(log((log(x/5)*(16*x^2 - 32*x + 16))/16 + x^2/1 
6)^2*(log(x/5)*(16*x - 32*x^2 + 16*x^3) + x^3)),x)
 

Output:

int((log((log(x/5)*(16*x^2 - 32*x + 16))/16 + x^2/16)*(x^3*exp(x) + log(x/ 
5)*exp(x)*(16*x - 32*x^2 + 16*x^3)) - exp(x)*(18*x^2 - 32*x + 16) + log(x/ 
5)*exp(x)*(32*x - 32*x^2))/(log((log(x/5)*(16*x^2 - 32*x + 16))/16 + x^2/1 
6)^2*(log(x/5)*(16*x - 32*x^2 + 16*x^3) + x^3)), x)
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.19 \[ \int \frac {e^x \left (-16+32 x-18 x^2\right )+e^x \left (32 x-32 x^2\right ) \log \left (\frac {x}{5}\right )+\left (e^x x^3+e^x \left (16 x-32 x^2+16 x^3\right ) \log \left (\frac {x}{5}\right )\right ) \log \left (\frac {1}{16} \left (x^2+\left (16-32 x+16 x^2\right ) \log \left (\frac {x}{5}\right )\right )\right )}{\left (x^3+\left (16 x-32 x^2+16 x^3\right ) \log \left (\frac {x}{5}\right )\right ) \log ^2\left (\frac {1}{16} \left (x^2+\left (16-32 x+16 x^2\right ) \log \left (\frac {x}{5}\right )\right )\right )} \, dx=\frac {e^{x}}{\mathrm {log}\left (\mathrm {log}\left (\frac {x}{5}\right ) x^{2}-2 \,\mathrm {log}\left (\frac {x}{5}\right ) x +\mathrm {log}\left (\frac {x}{5}\right )+\frac {x^{2}}{16}\right )} \] Input:

int((((16*x^3-32*x^2+16*x)*exp(x)*log(1/5*x)+exp(x)*x^3)*log(1/16*(16*x^2- 
32*x+16)*log(1/5*x)+1/16*x^2)+(-32*x^2+32*x)*exp(x)*log(1/5*x)+(-18*x^2+32 
*x-16)*exp(x))/((16*x^3-32*x^2+16*x)*log(1/5*x)+x^3)/log(1/16*(16*x^2-32*x 
+16)*log(1/5*x)+1/16*x^2)^2,x)
 

Output:

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