\(\int \frac {e^{\frac {e^{16-x} \log ^2(x)}{-x \log (5)+x \log (\frac {16}{x^2})}} ((-2 e^{16-x} \log (5)+2 e^{16-x} \log (\frac {16}{x^2})) \log (x)+(e^{16-x} (2+(1+x) \log (5))+e^{16-x} (-1-x) \log (\frac {16}{x^2})) \log ^2(x))}{x^2 \log ^2(5)-2 x^2 \log (5) \log (\frac {16}{x^2})+x^2 \log ^2(\frac {16}{x^2})} \, dx\) [830]

Optimal result

Integrand size = 139, antiderivative size = 30 \[ \int \frac {e^{\frac {e^{16-x} \log ^2(x)}{-x \log (5)+x \log \left (\frac {16}{x^2}\right )}} \left (\left (-2 e^{16-x} \log (5)+2 e^{16-x} \log \left (\frac {16}{x^2}\right )\right ) \log (x)+\left (e^{16-x} (2+(1+x) \log (5))+e^{16-x} (-1-x) \log \left (\frac {16}{x^2}\right )\right ) \log ^2(x)\right )}{x^2 \log ^2(5)-2 x^2 \log (5) \log \left (\frac {16}{x^2}\right )+x^2 \log ^2\left (\frac {16}{x^2}\right )} \, dx=e^{\frac {e^{16-x} \log ^2(x)}{x \left (-\log (5)+\log \left (\frac {16}{x^2}\right )\right )}} \] Output:

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

Mathematica [F]

\[ \int \frac {e^{\frac {e^{16-x} \log ^2(x)}{-x \log (5)+x \log \left (\frac {16}{x^2}\right )}} \left (\left (-2 e^{16-x} \log (5)+2 e^{16-x} \log \left (\frac {16}{x^2}\right )\right ) \log (x)+\left (e^{16-x} (2+(1+x) \log (5))+e^{16-x} (-1-x) \log \left (\frac {16}{x^2}\right )\right ) \log ^2(x)\right )}{x^2 \log ^2(5)-2 x^2 \log (5) \log \left (\frac {16}{x^2}\right )+x^2 \log ^2\left (\frac {16}{x^2}\right )} \, dx=\int \frac {e^{\frac {e^{16-x} \log ^2(x)}{-x \log (5)+x \log \left (\frac {16}{x^2}\right )}} \left (\left (-2 e^{16-x} \log (5)+2 e^{16-x} \log \left (\frac {16}{x^2}\right )\right ) \log (x)+\left (e^{16-x} (2+(1+x) \log (5))+e^{16-x} (-1-x) \log \left (\frac {16}{x^2}\right )\right ) \log ^2(x)\right )}{x^2 \log ^2(5)-2 x^2 \log (5) \log \left (\frac {16}{x^2}\right )+x^2 \log ^2\left (\frac {16}{x^2}\right )} \, dx \] Input:

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

Output:

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

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

\(\Big \downarrow \) 7239

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

\(\Big \downarrow \) 7293

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

$Aborted
 

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.49 (sec) , antiderivative size = 81, normalized size of antiderivative = 2.70

\[{\mathrm e}^{\frac {2 \,{\mathrm e}^{16-x} \ln \left (x \right )^{2}}{x \left (i \pi \operatorname {csgn}\left (i x^{2}\right )^{3}-2 i \pi \operatorname {csgn}\left (i x^{2}\right )^{2} \operatorname {csgn}\left (i x \right )+i \pi \,\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x \right )^{2}-4 \ln \left (x \right )+8 \ln \left (2\right )-2 \ln \left (5\right )\right )}}\]

Input:

int((((-1-x)*exp(16-x)*ln(16/x^2)+((1+x)*ln(5)+2)*exp(16-x))*ln(x)^2+(2*ex 
p(16-x)*ln(16/x^2)-2*ln(5)*exp(16-x))*ln(x))*exp(exp(16-x)*ln(x)^2/(x*ln(1 
6/x^2)-x*ln(5)))/(x^2*ln(16/x^2)^2-2*x^2*ln(5)*ln(16/x^2)+x^2*ln(5)^2),x)
 

Output:

exp(2*exp(16-x)*ln(x)^2/x/(I*Pi*csgn(I*x^2)^3-2*I*Pi*csgn(I*x^2)^2*csgn(I* 
x)+I*Pi*csgn(I*x^2)*csgn(I*x)^2-4*ln(x)+8*ln(2)-2*ln(5)))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 63 vs. \(2 (29) = 58\).

Time = 0.09 (sec) , antiderivative size = 63, normalized size of antiderivative = 2.10 \[ \int \frac {e^{\frac {e^{16-x} \log ^2(x)}{-x \log (5)+x \log \left (\frac {16}{x^2}\right )}} \left (\left (-2 e^{16-x} \log (5)+2 e^{16-x} \log \left (\frac {16}{x^2}\right )\right ) \log (x)+\left (e^{16-x} (2+(1+x) \log (5))+e^{16-x} (-1-x) \log \left (\frac {16}{x^2}\right )\right ) \log ^2(x)\right )}{x^2 \log ^2(5)-2 x^2 \log (5) \log \left (\frac {16}{x^2}\right )+x^2 \log ^2\left (\frac {16}{x^2}\right )} \, dx=e^{\left (-\frac {16 \, e^{\left (-x + 16\right )} \log \left (2\right )^{2} - 8 \, e^{\left (-x + 16\right )} \log \left (2\right ) \log \left (\frac {16}{x^{2}}\right ) + e^{\left (-x + 16\right )} \log \left (\frac {16}{x^{2}}\right )^{2}}{4 \, {\left (x \log \left (5\right ) - x \log \left (\frac {16}{x^{2}}\right )\right )}}\right )} \] Input:

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

Output:

e^(-1/4*(16*e^(-x + 16)*log(2)^2 - 8*e^(-x + 16)*log(2)*log(16/x^2) + e^(- 
x + 16)*log(16/x^2)^2)/(x*log(5) - x*log(16/x^2)))
 

Sympy [A] (verification not implemented)

Time = 0.91 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.87 \[ \int \frac {e^{\frac {e^{16-x} \log ^2(x)}{-x \log (5)+x \log \left (\frac {16}{x^2}\right )}} \left (\left (-2 e^{16-x} \log (5)+2 e^{16-x} \log \left (\frac {16}{x^2}\right )\right ) \log (x)+\left (e^{16-x} (2+(1+x) \log (5))+e^{16-x} (-1-x) \log \left (\frac {16}{x^2}\right )\right ) \log ^2(x)\right )}{x^2 \log ^2(5)-2 x^2 \log (5) \log \left (\frac {16}{x^2}\right )+x^2 \log ^2\left (\frac {16}{x^2}\right )} \, dx=e^{\frac {e^{16 - x} \log {\left (x \right )}^{2}}{x \left (- 2 \log {\left (x \right )} + \log {\left (16 \right )}\right ) - x \log {\left (5 \right )}}} \] Input:

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.58 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {e^{16-x} \log ^2(x)}{-x \log (5)+x \log \left (\frac {16}{x^2}\right )}} \left (\left (-2 e^{16-x} \log (5)+2 e^{16-x} \log \left (\frac {16}{x^2}\right )\right ) \log (x)+\left (e^{16-x} (2+(1+x) \log (5))+e^{16-x} (-1-x) \log \left (\frac {16}{x^2}\right )\right ) \log ^2(x)\right )}{x^2 \log ^2(5)-2 x^2 \log (5) \log \left (\frac {16}{x^2}\right )+x^2 \log ^2\left (\frac {16}{x^2}\right )} \, dx=e^{\left (-\frac {e^{16} \log \left (x\right )^{2}}{x {\left (\log \left (5\right ) - 4 \, \log \left (2\right )\right )} e^{x} + 2 \, x e^{x} \log \left (x\right )}\right )} \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.97 \[ \int \frac {e^{\frac {e^{16-x} \log ^2(x)}{-x \log (5)+x \log \left (\frac {16}{x^2}\right )}} \left (\left (-2 e^{16-x} \log (5)+2 e^{16-x} \log \left (\frac {16}{x^2}\right )\right ) \log (x)+\left (e^{16-x} (2+(1+x) \log (5))+e^{16-x} (-1-x) \log \left (\frac {16}{x^2}\right )\right ) \log ^2(x)\right )}{x^2 \log ^2(5)-2 x^2 \log (5) \log \left (\frac {16}{x^2}\right )+x^2 \log ^2\left (\frac {16}{x^2}\right )} \, dx=e^{\left (-\frac {e^{\left (-x + 16\right )} \log \left (x\right )^{2}}{x \log \left (5\right ) - x \log \left (\frac {16}{x^{2}}\right )}\right )} \] Input:

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

Output:

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

Mupad [F(-1)]

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

int((exp(-(exp(16 - x)*log(x)^2)/(x*log(5) - x*log(16/x^2)))*(log(x)^2*(ex 
p(16 - x)*(log(5)*(x + 1) + 2) - exp(16 - x)*log(16/x^2)*(x + 1)) - log(x) 
*(2*exp(16 - x)*log(5) - 2*exp(16 - x)*log(16/x^2))))/(x^2*log(5)^2 + x^2* 
log(16/x^2)^2 - 2*x^2*log(5)*log(16/x^2)),x)
 

Output:

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

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.07 \[ \int \frac {e^{\frac {e^{16-x} \log ^2(x)}{-x \log (5)+x \log \left (\frac {16}{x^2}\right )}} \left (\left (-2 e^{16-x} \log (5)+2 e^{16-x} \log \left (\frac {16}{x^2}\right )\right ) \log (x)+\left (e^{16-x} (2+(1+x) \log (5))+e^{16-x} (-1-x) \log \left (\frac {16}{x^2}\right )\right ) \log ^2(x)\right )}{x^2 \log ^2(5)-2 x^2 \log (5) \log \left (\frac {16}{x^2}\right )+x^2 \log ^2\left (\frac {16}{x^2}\right )} \, dx=e^{\frac {\mathrm {log}\left (x \right )^{2} e^{16}}{e^{x} \mathrm {log}\left (\frac {16}{x^{2}}\right ) x -e^{x} \mathrm {log}\left (5\right ) x}} \] Input:

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

Output:

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