\(\int \frac {-32+2 e^4+4 x-8 e^2 x+6 x^2+(16-e^4-x+2 e^2 x-x^2) \log (-16 x+e^4 x+x^2-2 e^2 x^2+x^3) \log (\log ^2(-16 x+e^4 x+x^2-2 e^2 x^2+x^3))}{(-16 x^2+e^4 x^2+x^3-2 e^2 x^3+x^4) \log (-16 x+e^4 x+x^2-2 e^2 x^2+x^3)} \, dx\) [915]

Optimal result

Integrand size = 150, antiderivative size = 22 \[ \int \frac {-32+2 e^4+4 x-8 e^2 x+6 x^2+\left (16-e^4-x+2 e^2 x-x^2\right ) \log \left (-16 x+e^4 x+x^2-2 e^2 x^2+x^3\right ) \log \left (\log ^2\left (-16 x+e^4 x+x^2-2 e^2 x^2+x^3\right )\right )}{\left (-16 x^2+e^4 x^2+x^3-2 e^2 x^3+x^4\right ) \log \left (-16 x+e^4 x+x^2-2 e^2 x^2+x^3\right )} \, dx=\frac {\log \left (\log ^2\left (x \left (-16+\left (e^2-x\right )^2+x\right )\right )\right )}{x} \] Output:

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

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.14 \[ \int \frac {-32+2 e^4+4 x-8 e^2 x+6 x^2+\left (16-e^4-x+2 e^2 x-x^2\right ) \log \left (-16 x+e^4 x+x^2-2 e^2 x^2+x^3\right ) \log \left (\log ^2\left (-16 x+e^4 x+x^2-2 e^2 x^2+x^3\right )\right )}{\left (-16 x^2+e^4 x^2+x^3-2 e^2 x^3+x^4\right ) \log \left (-16 x+e^4 x+x^2-2 e^2 x^2+x^3\right )} \, dx=\frac {\log \left (\log ^2\left (x \left (-16+e^4+x-2 e^2 x+x^2\right )\right )\right )}{x} \] Input:

Integrate[(-32 + 2*E^4 + 4*x - 8*E^2*x + 6*x^2 + (16 - E^4 - x + 2*E^2*x - 
 x^2)*Log[-16*x + E^4*x + x^2 - 2*E^2*x^2 + x^3]*Log[Log[-16*x + E^4*x + x 
^2 - 2*E^2*x^2 + x^3]^2])/((-16*x^2 + E^4*x^2 + x^3 - 2*E^2*x^3 + x^4)*Log 
[-16*x + E^4*x + x^2 - 2*E^2*x^2 + x^3]),x]
 

Output:

Log[Log[x*(-16 + E^4 + x - 2*E^2*x + 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.

\(\displaystyle \int \frac {6 x^2+\left (-x^2+2 e^2 x-x-e^4+16\right ) \log \left (x^3-2 e^2 x^2+x^2+e^4 x-16 x\right ) \log \left (\log ^2\left (x^3-2 e^2 x^2+x^2+e^4 x-16 x\right )\right )-8 e^2 x+4 x+2 e^4-32}{\left (x^4-2 e^2 x^3+x^3+e^4 x^2-16 x^2\right ) \log \left (x^3-2 e^2 x^2+x^2+e^4 x-16 x\right )} \, dx\)

\(\Big \downarrow \) 6

\(\displaystyle \int \frac {6 x^2+\left (-x^2+2 e^2 x-x-e^4+16\right ) \log \left (x^3-2 e^2 x^2+x^2+e^4 x-16 x\right ) \log \left (\log ^2\left (x^3-2 e^2 x^2+x^2+e^4 x-16 x\right )\right )-8 e^2 x+4 x+2 e^4-32}{\left (x^4-2 e^2 x^3+x^3+\left (e^4-16\right ) x^2\right ) \log \left (x^3-2 e^2 x^2+x^2+e^4 x-16 x\right )}dx\)

\(\Big \downarrow \) 6

\(\displaystyle \int \frac {6 x^2+\left (-x^2+2 e^2 x-x-e^4+16\right ) \log \left (x^3-2 e^2 x^2+x^2+e^4 x-16 x\right ) \log \left (\log ^2\left (x^3-2 e^2 x^2+x^2+e^4 x-16 x\right )\right )-8 e^2 x+4 x+2 e^4-32}{\left (x^4+\left (1-2 e^2\right ) x^3+\left (e^4-16\right ) x^2\right ) \log \left (x^3-2 e^2 x^2+x^2+e^4 x-16 x\right )}dx\)

\(\Big \downarrow \) 6

\(\displaystyle \int \frac {6 x^2+\left (-x^2+2 e^2 x-x-e^4+16\right ) \log \left (x^3-2 e^2 x^2+x^2+e^4 x-16 x\right ) \log \left (\log ^2\left (x^3-2 e^2 x^2+x^2+e^4 x-16 x\right )\right )+\left (4-8 e^2\right ) x+2 e^4-32}{\left (x^4+\left (1-2 e^2\right ) x^3+\left (e^4-16\right ) x^2\right ) \log \left (x^3-2 e^2 x^2+x^2+e^4 x-16 x\right )}dx\)

\(\Big \downarrow \) 2026

\(\displaystyle \int \frac {6 x^2+\left (-x^2+2 e^2 x-x-e^4+16\right ) \log \left (x^3-2 e^2 x^2+x^2+e^4 x-16 x\right ) \log \left (\log ^2\left (x^3-2 e^2 x^2+x^2+e^4 x-16 x\right )\right )+\left (4-8 e^2\right ) x+2 e^4-32}{x^2 \left (x^2+\left (1-2 e^2\right ) x+e^4-16\right ) \log \left (x^3-2 e^2 x^2+x^2+e^4 x-16 x\right )}dx\)

\(\Big \downarrow \) 7279

\(\displaystyle \int \left (\frac {2 \left (-3 x^2-2 \left (1-2 e^2\right ) x-e^4+16\right )}{x^2 \left (-x^2-\left (1-2 e^2\right ) x-e^4+16\right ) \log \left (x \left (x^2+\left (1-2 e^2\right ) x+e^4-16\right )\right )}-\frac {\log \left (\log ^2\left (x \left (x^2+\left (1-2 e^2\right ) x+e^4-16\right )\right )\right )}{x^2}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\int \frac {\log \left (\log ^2\left (x \left (x^2+\left (1-2 e^2\right ) x+e^4-16\right )\right )\right )}{x^2}dx-\frac {2 \left (1-2 e^2\right ) \left (1+\frac {1-2 e^2}{\sqrt {65-4 e^2}}\right ) \int \frac {1}{\left (-2 x-\sqrt {65-4 e^2}+2 e^2-1\right ) \log \left (x \left (x^2+\left (1-2 e^2\right ) x+e^4-16\right )\right )}dx}{16-e^4}-\frac {2 \left (1-2 e^2\right ) \left (1-\frac {1-2 e^2}{\sqrt {65-4 e^2}}\right ) \int \frac {1}{\left (-2 x+\sqrt {65-4 e^2}+2 e^2-1\right ) \log \left (x \left (x^2+\left (1-2 e^2\right ) x+e^4-16\right )\right )}dx}{16-e^4}-\frac {4 \left (33-4 e^2+2 e^4\right ) \int \frac {1}{\left (-2 x+\sqrt {65-4 e^2}+2 e^2-1\right ) \log \left (x \left (x^2+\left (1-2 e^2\right ) x+e^4-16\right )\right )}dx}{\sqrt {65-4 e^2} \left (16-e^4\right )}+2 \int \frac {1}{x^2 \log \left (x \left (x^2+\left (1-2 e^2\right ) x+e^4-16\right )\right )}dx-\frac {2 \left (1-2 e^2\right ) \int \frac {1}{x \log \left (x \left (x^2+\left (1-2 e^2\right ) x+e^4-16\right )\right )}dx}{16-e^4}-\frac {4 \left (33-4 e^2+2 e^4\right ) \int \frac {1}{\left (2 x+\sqrt {65-4 e^2}-2 e^2+1\right ) \log \left (x \left (x^2+\left (1-2 e^2\right ) x+e^4-16\right )\right )}dx}{\sqrt {65-4 e^2} \left (16-e^4\right )}\)

Input:

Int[(-32 + 2*E^4 + 4*x - 8*E^2*x + 6*x^2 + (16 - E^4 - x + 2*E^2*x - x^2)* 
Log[-16*x + E^4*x + x^2 - 2*E^2*x^2 + x^3]*Log[Log[-16*x + E^4*x + x^2 - 2 
*E^2*x^2 + x^3]^2])/((-16*x^2 + E^4*x^2 + x^3 - 2*E^2*x^3 + x^4)*Log[-16*x 
 + E^4*x + x^2 - 2*E^2*x^2 + x^3]),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 3.20 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.27

method	result	size
parallelrisch	\(\frac {\ln \left ({\ln \left (x \left (-16+{\mathrm e}^{4}-2 \,{\mathrm e}^{2} x +x^{2}+x \right )\right )}^{2}\right )}{x}\)	\(28\)

Input:

int(((-exp(1)^4+2*x*exp(1)^2-x^2-x+16)*ln(x*exp(1)^4-2*x^2*exp(1)^2+x^3+x^ 
2-16*x)*ln(ln(x*exp(1)^4-2*x^2*exp(1)^2+x^3+x^2-16*x)^2)+2*exp(1)^4-8*x*ex 
p(1)^2+6*x^2+4*x-32)/(x^2*exp(1)^4-2*x^3*exp(1)^2+x^4+x^3-16*x^2)/ln(x*exp 
(1)^4-2*x^2*exp(1)^2+x^3+x^2-16*x),x,method=_RETURNVERBOSE)
 

Output:

ln(ln(x*(-16+exp(1)^4-2*x*exp(1)^2+x^2+x))^2)/x
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.32 \[ \int \frac {-32+2 e^4+4 x-8 e^2 x+6 x^2+\left (16-e^4-x+2 e^2 x-x^2\right ) \log \left (-16 x+e^4 x+x^2-2 e^2 x^2+x^3\right ) \log \left (\log ^2\left (-16 x+e^4 x+x^2-2 e^2 x^2+x^3\right )\right )}{\left (-16 x^2+e^4 x^2+x^3-2 e^2 x^3+x^4\right ) \log \left (-16 x+e^4 x+x^2-2 e^2 x^2+x^3\right )} \, dx=\frac {\log \left (\log \left (x^{3} - 2 \, x^{2} e^{2} + x^{2} + x e^{4} - 16 \, x\right )^{2}\right )}{x} \] Input:

integrate(((-exp(1)^4+2*x*exp(1)^2-x^2-x+16)*log(x*exp(1)^4-2*x^2*exp(1)^2 
+x^3+x^2-16*x)*log(log(x*exp(1)^4-2*x^2*exp(1)^2+x^3+x^2-16*x)^2)+2*exp(1) 
^4-8*x*exp(1)^2+6*x^2+4*x-32)/(x^2*exp(1)^4-2*x^3*exp(1)^2+x^4+x^3-16*x^2) 
/log(x*exp(1)^4-2*x^2*exp(1)^2+x^3+x^2-16*x),x, algorithm="fricas")
 

Output:

log(log(x^3 - 2*x^2*e^2 + x^2 + x*e^4 - 16*x)^2)/x
 

Sympy [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.32 \[ \int \frac {-32+2 e^4+4 x-8 e^2 x+6 x^2+\left (16-e^4-x+2 e^2 x-x^2\right ) \log \left (-16 x+e^4 x+x^2-2 e^2 x^2+x^3\right ) \log \left (\log ^2\left (-16 x+e^4 x+x^2-2 e^2 x^2+x^3\right )\right )}{\left (-16 x^2+e^4 x^2+x^3-2 e^2 x^3+x^4\right ) \log \left (-16 x+e^4 x+x^2-2 e^2 x^2+x^3\right )} \, dx=\frac {\log {\left (\log {\left (x^{3} - 2 x^{2} e^{2} + x^{2} - 16 x + x e^{4} \right )}^{2} \right )}}{x} \] Input:

integrate(((-exp(1)**4+2*x*exp(1)**2-x**2-x+16)*ln(x*exp(1)**4-2*x**2*exp( 
1)**2+x**3+x**2-16*x)*ln(ln(x*exp(1)**4-2*x**2*exp(1)**2+x**3+x**2-16*x)** 
2)+2*exp(1)**4-8*x*exp(1)**2+6*x**2+4*x-32)/(x**2*exp(1)**4-2*x**3*exp(1)* 
*2+x**4+x**3-16*x**2)/ln(x*exp(1)**4-2*x**2*exp(1)**2+x**3+x**2-16*x),x)
 

Output:

log(log(x**3 - 2*x**2*exp(2) + x**2 - 16*x + x*exp(4))**2)/x
 

Maxima [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.18 \[ \int \frac {-32+2 e^4+4 x-8 e^2 x+6 x^2+\left (16-e^4-x+2 e^2 x-x^2\right ) \log \left (-16 x+e^4 x+x^2-2 e^2 x^2+x^3\right ) \log \left (\log ^2\left (-16 x+e^4 x+x^2-2 e^2 x^2+x^3\right )\right )}{\left (-16 x^2+e^4 x^2+x^3-2 e^2 x^3+x^4\right ) \log \left (-16 x+e^4 x+x^2-2 e^2 x^2+x^3\right )} \, dx=\frac {2 \, \log \left (\log \left (x^{2} - x {\left (2 \, e^{2} - 1\right )} + e^{4} - 16\right ) + \log \left (x\right )\right )}{x} \] Input:

integrate(((-exp(1)^4+2*x*exp(1)^2-x^2-x+16)*log(x*exp(1)^4-2*x^2*exp(1)^2 
+x^3+x^2-16*x)*log(log(x*exp(1)^4-2*x^2*exp(1)^2+x^3+x^2-16*x)^2)+2*exp(1) 
^4-8*x*exp(1)^2+6*x^2+4*x-32)/(x^2*exp(1)^4-2*x^3*exp(1)^2+x^4+x^3-16*x^2) 
/log(x*exp(1)^4-2*x^2*exp(1)^2+x^3+x^2-16*x),x, algorithm="maxima")
 

Output:

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

Giac [A] (verification not implemented)

Time = 12.82 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.32 \[ \int \frac {-32+2 e^4+4 x-8 e^2 x+6 x^2+\left (16-e^4-x+2 e^2 x-x^2\right ) \log \left (-16 x+e^4 x+x^2-2 e^2 x^2+x^3\right ) \log \left (\log ^2\left (-16 x+e^4 x+x^2-2 e^2 x^2+x^3\right )\right )}{\left (-16 x^2+e^4 x^2+x^3-2 e^2 x^3+x^4\right ) \log \left (-16 x+e^4 x+x^2-2 e^2 x^2+x^3\right )} \, dx=\frac {\log \left (\log \left (x^{3} - 2 \, x^{2} e^{2} + x^{2} + x e^{4} - 16 \, x\right )^{2}\right )}{x} \] Input:

integrate(((-exp(1)^4+2*x*exp(1)^2-x^2-x+16)*log(x*exp(1)^4-2*x^2*exp(1)^2 
+x^3+x^2-16*x)*log(log(x*exp(1)^4-2*x^2*exp(1)^2+x^3+x^2-16*x)^2)+2*exp(1) 
^4-8*x*exp(1)^2+6*x^2+4*x-32)/(x^2*exp(1)^4-2*x^3*exp(1)^2+x^4+x^3-16*x^2) 
/log(x*exp(1)^4-2*x^2*exp(1)^2+x^3+x^2-16*x),x, algorithm="giac")
 

Output:

log(log(x^3 - 2*x^2*e^2 + x^2 + x*e^4 - 16*x)^2)/x
 

Mupad [B] (verification not implemented)

Time = 1.37 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.32 \[ \int \frac {-32+2 e^4+4 x-8 e^2 x+6 x^2+\left (16-e^4-x+2 e^2 x-x^2\right ) \log \left (-16 x+e^4 x+x^2-2 e^2 x^2+x^3\right ) \log \left (\log ^2\left (-16 x+e^4 x+x^2-2 e^2 x^2+x^3\right )\right )}{\left (-16 x^2+e^4 x^2+x^3-2 e^2 x^3+x^4\right ) \log \left (-16 x+e^4 x+x^2-2 e^2 x^2+x^3\right )} \, dx=\frac {\ln \left ({\ln \left (x\,{\mathrm {e}}^4-16\,x-2\,x^2\,{\mathrm {e}}^2+x^2+x^3\right )}^2\right )}{x} \] Input:

int((4*x + 2*exp(4) - 8*x*exp(2) + 6*x^2 - log(x*exp(4) - 16*x - 2*x^2*exp 
(2) + x^2 + x^3)*log(log(x*exp(4) - 16*x - 2*x^2*exp(2) + x^2 + x^3)^2)*(x 
 + exp(4) - 2*x*exp(2) + x^2 - 16) - 32)/(log(x*exp(4) - 16*x - 2*x^2*exp( 
2) + x^2 + x^3)*(x^2*exp(4) - 2*x^3*exp(2) - 16*x^2 + x^3 + x^4)),x)
 

Output:

log(log(x*exp(4) - 16*x - 2*x^2*exp(2) + x^2 + x^3)^2)/x
 

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.41 \[ \int \frac {-32+2 e^4+4 x-8 e^2 x+6 x^2+\left (16-e^4-x+2 e^2 x-x^2\right ) \log \left (-16 x+e^4 x+x^2-2 e^2 x^2+x^3\right ) \log \left (\log ^2\left (-16 x+e^4 x+x^2-2 e^2 x^2+x^3\right )\right )}{\left (-16 x^2+e^4 x^2+x^3-2 e^2 x^3+x^4\right ) \log \left (-16 x+e^4 x+x^2-2 e^2 x^2+x^3\right )} \, dx=\frac {\mathrm {log}\left (\mathrm {log}\left (e^{4} x -2 e^{2} x^{2}+x^{3}+x^{2}-16 x \right )^{2}\right )}{x} \] Input:

int(((-exp(1)^4+2*x*exp(1)^2-x^2-x+16)*log(x*exp(1)^4-2*x^2*exp(1)^2+x^3+x 
^2-16*x)*log(log(x*exp(1)^4-2*x^2*exp(1)^2+x^3+x^2-16*x)^2)+2*exp(1)^4-8*x 
*exp(1)^2+6*x^2+4*x-32)/(x^2*exp(1)^4-2*x^3*exp(1)^2+x^4+x^3-16*x^2)/log(x 
*exp(1)^4-2*x^2*exp(1)^2+x^3+x^2-16*x),x)
 

Output:

log(log(e**4*x - 2*e**2*x**2 + x**3 + x**2 - 16*x)**2)/x