\(\int \frac {(\frac {16}{e^{25}}+x) \log (\frac {16}{e^{25}}+x)+(\frac {16 x}{e^{25}}+x^2) \log (x) \log ^2(\frac {16}{e^{25}}+x)+(-x \log (x)+(-\frac {16}{e^{25}}-x) \log (x) \log (\frac {16}{e^{25}}+x)) \log (\log (x))}{(\frac {16 x^2}{e^{25}}+x^3) \log (x) \log ^2(\frac {16}{e^{25}}+x)} \, dx\) [25]

Optimal result

Integrand size = 99, antiderivative size = 20 \[ \int \frac {\left (\frac {16}{e^{25}}+x\right ) \log \left (\frac {16}{e^{25}}+x\right )+\left (\frac {16 x}{e^{25}}+x^2\right ) \log (x) \log ^2\left (\frac {16}{e^{25}}+x\right )+\left (-x \log (x)+\left (-\frac {16}{e^{25}}-x\right ) \log (x) \log \left (\frac {16}{e^{25}}+x\right )\right ) \log (\log (x))}{\left (\frac {16 x^2}{e^{25}}+x^3\right ) \log (x) \log ^2\left (\frac {16}{e^{25}}+x\right )} \, dx=\log \left (x \sqrt [x \log \left (\frac {16}{e^{25}}+x\right )]{\log (x)}\right ) \] Output:

ln(x*exp(ln(ln(x))/x/ln(exp(4*ln(2)-25)+x)))
 

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {\left (\frac {16}{e^{25}}+x\right ) \log \left (\frac {16}{e^{25}}+x\right )+\left (\frac {16 x}{e^{25}}+x^2\right ) \log (x) \log ^2\left (\frac {16}{e^{25}}+x\right )+\left (-x \log (x)+\left (-\frac {16}{e^{25}}-x\right ) \log (x) \log \left (\frac {16}{e^{25}}+x\right )\right ) \log (\log (x))}{\left (\frac {16 x^2}{e^{25}}+x^3\right ) \log (x) \log ^2\left (\frac {16}{e^{25}}+x\right )} \, dx=\log (x)+\frac {\log (\log (x))}{x \log \left (\frac {16}{e^{25}}+x\right )} \] Input:

Integrate[((16/E^25 + x)*Log[16/E^25 + x] + ((16*x)/E^25 + x^2)*Log[x]*Log 
[16/E^25 + x]^2 + (-(x*Log[x]) + (-16/E^25 - x)*Log[x]*Log[16/E^25 + x])*L 
og[Log[x]])/(((16*x^2)/E^25 + x^3)*Log[x]*Log[16/E^25 + x]^2),x]
 

Output:

Log[x] + Log[Log[x]]/(x*Log[16/E^25 + 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.

\(\displaystyle \int \frac {\left (x^2+\frac {16 x}{e^{25}}\right ) \log (x) \log ^2\left (x+\frac {16}{e^{25}}\right )+\left (x+\frac {16}{e^{25}}\right ) \log \left (x+\frac {16}{e^{25}}\right )+\left (\left (-x-\frac {16}{e^{25}}\right ) \log (x) \log \left (x+\frac {16}{e^{25}}\right )-x \log (x)\right ) \log (\log (x))}{\left (x^3+\frac {16 x^2}{e^{25}}\right ) \log (x) \log ^2\left (x+\frac {16}{e^{25}}\right )} \, dx\)

\(\Big \downarrow \) 2026

\(\displaystyle \int \frac {\left (x^2+\frac {16 x}{e^{25}}\right ) \log (x) \log ^2\left (x+\frac {16}{e^{25}}\right )+\left (x+\frac {16}{e^{25}}\right ) \log \left (x+\frac {16}{e^{25}}\right )+\left (\left (-x-\frac {16}{e^{25}}\right ) \log (x) \log \left (x+\frac {16}{e^{25}}\right )-x \log (x)\right ) \log (\log (x))}{x^2 \left (x+\frac {16}{e^{25}}\right ) \log (x) \log ^2\left (x+\frac {16}{e^{25}}\right )}dx\)

\(\Big \downarrow \) 7239

\(\displaystyle \int \frac {x-\frac {e^{25} x \log (\log (x))}{\left (e^{25} x+16\right ) \log ^2\left (x+\frac {16}{e^{25}}\right )}+\frac {1-\log (x) \log (\log (x))}{\log (x) \log \left (x+\frac {16}{e^{25}}\right )}}{x^2}dx\)

\(\Big \downarrow \) 2010

\(\displaystyle \int \left (\frac {x \log (x) \log \left (x+\frac {16}{e^{25}}\right )+1}{x^2 \log (x) \log \left (x+\frac {16}{e^{25}}\right )}-\frac {\left (e^{25} x+e^{25} x \log \left (x+\frac {16}{e^{25}}\right )+16 \log \left (x+\frac {16}{e^{25}}\right )\right ) \log (\log (x))}{x^2 \left (e^{25} x+16\right ) \log ^2\left (x+\frac {16}{e^{25}}\right )}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \int \frac {1}{x^2 \log (x) \log \left (x+\frac {16}{e^{25}}\right )}dx-\int \frac {\log (\log (x))}{x^2 \log \left (x+\frac {16}{e^{25}}\right )}dx-\frac {1}{16} e^{25} \int \frac {\log (\log (x))}{x \log ^2\left (x+\frac {16}{e^{25}}\right )}dx+\frac {1}{16} e^{50} \int \frac {\log (\log (x))}{\left (e^{25} x+16\right ) \log ^2\left (x+\frac {16}{e^{25}}\right )}dx+\log (x)\)

Input:

Int[((16/E^25 + x)*Log[16/E^25 + x] + ((16*x)/E^25 + x^2)*Log[x]*Log[16/E^ 
25 + x]^2 + (-(x*Log[x]) + (-16/E^25 - x)*Log[x]*Log[16/E^25 + x])*Log[Log 
[x]])/(((16*x^2)/E^25 + x^3)*Log[x]*Log[16/E^25 + x]^2),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 8.36 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00

method	result	size
risch	\(\frac {\ln \left (\ln \left (x \right )\right )}{x \ln \left (16 \,{\mathrm e}^{-25}+x \right )}+\ln \left (x \right )\)	\(20\)
parallelrisch	\(\frac {\left (2 \ln \left (x \right ) {\mathrm e}^{4 \ln \left (2\right )-25} x \ln \left ({\mathrm e}^{4 \ln \left (2\right )-25}+x \right )+2 \,{\mathrm e}^{4 \ln \left (2\right )-25} \ln \left (\ln \left (x \right )\right )\right ) {\mathrm e}^{25}}{32 x \ln \left ({\mathrm e}^{4 \ln \left (2\right )-25}+x \right )}\)	\(62\)

Input:

int((((-exp(4*ln(2)-25)-x)*ln(x)*ln(exp(4*ln(2)-25)+x)-x*ln(x))*ln(ln(x))+ 
(x*exp(4*ln(2)-25)+x^2)*ln(x)*ln(exp(4*ln(2)-25)+x)^2+(exp(4*ln(2)-25)+x)* 
ln(exp(4*ln(2)-25)+x))/(x^2*exp(4*ln(2)-25)+x^3)/ln(x)/ln(exp(4*ln(2)-25)+ 
x)^2,x,method=_RETURNVERBOSE)
 

Output:

1/x/ln(16*exp(-25)+x)*ln(ln(x))+ln(x)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.70 \[ \int \frac {\left (\frac {16}{e^{25}}+x\right ) \log \left (\frac {16}{e^{25}}+x\right )+\left (\frac {16 x}{e^{25}}+x^2\right ) \log (x) \log ^2\left (\frac {16}{e^{25}}+x\right )+\left (-x \log (x)+\left (-\frac {16}{e^{25}}-x\right ) \log (x) \log \left (\frac {16}{e^{25}}+x\right )\right ) \log (\log (x))}{\left (\frac {16 x^2}{e^{25}}+x^3\right ) \log (x) \log ^2\left (\frac {16}{e^{25}}+x\right )} \, dx=\frac {x \log \left (x + e^{\left (4 \, \log \left (2\right ) - 25\right )}\right ) \log \left (x\right ) + \log \left (\log \left (x\right )\right )}{x \log \left (x + e^{\left (4 \, \log \left (2\right ) - 25\right )}\right )} \] Input:

integrate((((-exp(4*log(2)-25)-x)*log(x)*log(exp(4*log(2)-25)+x)-x*log(x)) 
*log(log(x))+(x*exp(4*log(2)-25)+x^2)*log(x)*log(exp(4*log(2)-25)+x)^2+(ex 
p(4*log(2)-25)+x)*log(exp(4*log(2)-25)+x))/(x^2*exp(4*log(2)-25)+x^3)/log( 
x)/log(exp(4*log(2)-25)+x)^2,x, algorithm="fricas")
 

Output:

(x*log(x + e^(4*log(2) - 25))*log(x) + log(log(x)))/(x*log(x + e^(4*log(2) 
 - 25)))
 

Sympy [F(-2)]

Exception generated. \[ \int \frac {\left (\frac {16}{e^{25}}+x\right ) \log \left (\frac {16}{e^{25}}+x\right )+\left (\frac {16 x}{e^{25}}+x^2\right ) \log (x) \log ^2\left (\frac {16}{e^{25}}+x\right )+\left (-x \log (x)+\left (-\frac {16}{e^{25}}-x\right ) \log (x) \log \left (\frac {16}{e^{25}}+x\right )\right ) \log (\log (x))}{\left (\frac {16 x^2}{e^{25}}+x^3\right ) \log (x) \log ^2\left (\frac {16}{e^{25}}+x\right )} \, dx=\text {Exception raised: TypeError} \] Input:

integrate((((-exp(4*ln(2)-25)-x)*ln(x)*ln(exp(4*ln(2)-25)+x)-x*ln(x))*ln(l 
n(x))+(x*exp(4*ln(2)-25)+x**2)*ln(x)*ln(exp(4*ln(2)-25)+x)**2+(exp(4*ln(2) 
-25)+x)*ln(exp(4*ln(2)-25)+x))/(x**2*exp(4*ln(2)-25)+x**3)/ln(x)/ln(exp(4* 
ln(2)-25)+x)**2,x)
 

Output:

Exception raised: TypeError >> '>' not supported between instances of 'Pol 
y' and 'int'
 

Maxima [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.10 \[ \int \frac {\left (\frac {16}{e^{25}}+x\right ) \log \left (\frac {16}{e^{25}}+x\right )+\left (\frac {16 x}{e^{25}}+x^2\right ) \log (x) \log ^2\left (\frac {16}{e^{25}}+x\right )+\left (-x \log (x)+\left (-\frac {16}{e^{25}}-x\right ) \log (x) \log \left (\frac {16}{e^{25}}+x\right )\right ) \log (\log (x))}{\left (\frac {16 x^2}{e^{25}}+x^3\right ) \log (x) \log ^2\left (\frac {16}{e^{25}}+x\right )} \, dx=\frac {\log \left (\log \left (x\right )\right )}{x \log \left (x e^{25} + 16\right ) - 25 \, x} + \log \left (x\right ) \] Input:

integrate((((-exp(4*log(2)-25)-x)*log(x)*log(exp(4*log(2)-25)+x)-x*log(x)) 
*log(log(x))+(x*exp(4*log(2)-25)+x^2)*log(x)*log(exp(4*log(2)-25)+x)^2+(ex 
p(4*log(2)-25)+x)*log(exp(4*log(2)-25)+x))/(x^2*exp(4*log(2)-25)+x^3)/log( 
x)/log(exp(4*log(2)-25)+x)^2,x, algorithm="maxima")
 

Output:

log(log(x))/(x*log(x*e^25 + 16) - 25*x) + log(x)
 

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.80 \[ \int \frac {\left (\frac {16}{e^{25}}+x\right ) \log \left (\frac {16}{e^{25}}+x\right )+\left (\frac {16 x}{e^{25}}+x^2\right ) \log (x) \log ^2\left (\frac {16}{e^{25}}+x\right )+\left (-x \log (x)+\left (-\frac {16}{e^{25}}-x\right ) \log (x) \log \left (\frac {16}{e^{25}}+x\right )\right ) \log (\log (x))}{\left (\frac {16 x^2}{e^{25}}+x^3\right ) \log (x) \log ^2\left (\frac {16}{e^{25}}+x\right )} \, dx=\frac {x \log \left (x e^{25} + 16\right ) \log \left (x\right ) - 25 \, x \log \left (x\right ) + \log \left (\log \left (x\right )\right )}{x \log \left (x e^{25} + 16\right ) - 25 \, x} \] Input:

integrate((((-exp(4*log(2)-25)-x)*log(x)*log(exp(4*log(2)-25)+x)-x*log(x)) 
*log(log(x))+(x*exp(4*log(2)-25)+x^2)*log(x)*log(exp(4*log(2)-25)+x)^2+(ex 
p(4*log(2)-25)+x)*log(exp(4*log(2)-25)+x))/(x^2*exp(4*log(2)-25)+x^3)/log( 
x)/log(exp(4*log(2)-25)+x)^2,x, algorithm="giac")
 

Output:

(x*log(x*e^25 + 16)*log(x) - 25*x*log(x) + log(log(x)))/(x*log(x*e^25 + 16 
) - 25*x)
 

Mupad [B] (verification not implemented)

Time = 3.37 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95 \[ \int \frac {\left (\frac {16}{e^{25}}+x\right ) \log \left (\frac {16}{e^{25}}+x\right )+\left (\frac {16 x}{e^{25}}+x^2\right ) \log (x) \log ^2\left (\frac {16}{e^{25}}+x\right )+\left (-x \log (x)+\left (-\frac {16}{e^{25}}-x\right ) \log (x) \log \left (\frac {16}{e^{25}}+x\right )\right ) \log (\log (x))}{\left (\frac {16 x^2}{e^{25}}+x^3\right ) \log (x) \log ^2\left (\frac {16}{e^{25}}+x\right )} \, dx=\ln \left (x\right )+\frac {\ln \left (\ln \left (x\right )\right )}{x\,\ln \left (x+16\,{\mathrm {e}}^{-25}\right )} \] Input:

int((log(x + exp(4*log(2) - 25))*(x + exp(4*log(2) - 25)) - log(log(x))*(x 
*log(x) + log(x + exp(4*log(2) - 25))*log(x)*(x + exp(4*log(2) - 25))) + l 
og(x + exp(4*log(2) - 25))^2*log(x)*(x*exp(4*log(2) - 25) + x^2))/(log(x + 
 exp(4*log(2) - 25))^2*log(x)*(x^2*exp(4*log(2) - 25) + x^3)),x)
 

Output:

log(x) + log(log(x))/(x*log(x + 16*exp(-25)))
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.90 \[ \int \frac {\left (\frac {16}{e^{25}}+x\right ) \log \left (\frac {16}{e^{25}}+x\right )+\left (\frac {16 x}{e^{25}}+x^2\right ) \log (x) \log ^2\left (\frac {16}{e^{25}}+x\right )+\left (-x \log (x)+\left (-\frac {16}{e^{25}}-x\right ) \log (x) \log \left (\frac {16}{e^{25}}+x\right )\right ) \log (\log (x))}{\left (\frac {16 x^2}{e^{25}}+x^3\right ) \log (x) \log ^2\left (\frac {16}{e^{25}}+x\right )} \, dx=\frac {\mathrm {log}\left (\mathrm {log}\left (x \right )\right )+\mathrm {log}\left (\frac {e^{25} x +16}{e^{25}}\right ) \mathrm {log}\left (x \right ) x}{\mathrm {log}\left (\frac {e^{25} x +16}{e^{25}}\right ) x} \] Input:

int((((-exp(4*log(2)-25)-x)*log(x)*log(exp(4*log(2)-25)+x)-x*log(x))*log(l 
og(x))+(x*exp(4*log(2)-25)+x^2)*log(x)*log(exp(4*log(2)-25)+x)^2+(exp(4*lo 
g(2)-25)+x)*log(exp(4*log(2)-25)+x))/(x^2*exp(4*log(2)-25)+x^3)/log(x)/log 
(exp(4*log(2)-25)+x)^2,x)
 

Output:

(log(log(x)) + log((e**25*x + 16)/e**25)*log(x)*x)/(log((e**25*x + 16)/e** 
25)*x)