\(\int \frac {-16 x^2+(-2 e^4 x^2+6 x^4) \log (x)+(16+4 x^2 \log (x)) \log (\log ^2(x))-10 \log (x) \log ^2(\log ^2(x))}{(e^4 x^3+x^5) \log (x)-2 x^3 \log (x) \log (\log ^2(x))+x \log (x) \log ^2(\log ^2(x))} \, dx\) [732]

Optimal result

Integrand size = 90, antiderivative size = 27 \[ \int \frac {-16 x^2+\left (-2 e^4 x^2+6 x^4\right ) \log (x)+\left (16+4 x^2 \log (x)\right ) \log \left (\log ^2(x)\right )-10 \log (x) \log ^2\left (\log ^2(x)\right )}{\left (e^4 x^3+x^5\right ) \log (x)-2 x^3 \log (x) \log \left (\log ^2(x)\right )+x \log (x) \log ^2\left (\log ^2(x)\right )} \, dx=\log \left (\frac {8 \left (e^4+\left (-x+\frac {\log \left (\log ^2(x)\right )}{x}\right )^2\right )^4}{x^2}\right ) \] Output:

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

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.41 \[ \int \frac {-16 x^2+\left (-2 e^4 x^2+6 x^4\right ) \log (x)+\left (16+4 x^2 \log (x)\right ) \log \left (\log ^2(x)\right )-10 \log (x) \log ^2\left (\log ^2(x)\right )}{\left (e^4 x^3+x^5\right ) \log (x)-2 x^3 \log (x) \log \left (\log ^2(x)\right )+x \log (x) \log ^2\left (\log ^2(x)\right )} \, dx=-2 \left (5 \log (x)-2 \log \left (e^4 x^2+x^4-2 x^2 \log \left (\log ^2(x)\right )+\log ^2\left (\log ^2(x)\right )\right )\right ) \] Input:

Integrate[(-16*x^2 + (-2*E^4*x^2 + 6*x^4)*Log[x] + (16 + 4*x^2*Log[x])*Log 
[Log[x]^2] - 10*Log[x]*Log[Log[x]^2]^2)/((E^4*x^3 + x^5)*Log[x] - 2*x^3*Lo 
g[x]*Log[Log[x]^2] + x*Log[x]*Log[Log[x]^2]^2),x]
 

Output:

-2*(5*Log[x] - 2*Log[E^4*x^2 + x^4 - 2*x^2*Log[Log[x]^2] + Log[Log[x]^2]^2 
])
 

Rubi [A] (verified)

Time = 0.98 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.33, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.033, Rules used = {7292, 7293, 2009}

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.

\(\displaystyle \int \frac {-16 x^2+\left (4 x^2 \log (x)+16\right ) \log \left (\log ^2(x)\right )+\left (6 x^4-2 e^4 x^2\right ) \log (x)-10 \log (x) \log ^2\left (\log ^2(x)\right )}{-2 x^3 \log (x) \log \left (\log ^2(x)\right )+\left (x^5+e^4 x^3\right ) \log (x)+x \log (x) \log ^2\left (\log ^2(x)\right )} \, dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {-16 x^2+\left (4 x^2 \log (x)+16\right ) \log \left (\log ^2(x)\right )+\left (6 x^4-2 e^4 x^2\right ) \log (x)-10 \log (x) \log ^2\left (\log ^2(x)\right )}{x \log (x) \left (x^4+e^4 x^2-2 x^2 \log \left (\log ^2(x)\right )+\log ^2\left (\log ^2(x)\right )\right )}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (\frac {8 \left (2 x^4 \log (x)-2 x^2-2 x^2 \log (x) \log \left (\log ^2(x)\right )+e^4 x^2 \log (x)+2 \log \left (\log ^2(x)\right )\right )}{x \log (x) \left (x^4+e^4 x^2-2 x^2 \log \left (\log ^2(x)\right )+\log ^2\left (\log ^2(x)\right )\right )}-\frac {10}{x}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle 4 \log \left (x^4+e^4 x^2-2 x^2 \log \left (\log ^2(x)\right )+\log ^2\left (\log ^2(x)\right )\right )-10 \log (x)\)

Input:

Int[(-16*x^2 + (-2*E^4*x^2 + 6*x^4)*Log[x] + (16 + 4*x^2*Log[x])*Log[Log[x 
]^2] - 10*Log[x]*Log[Log[x]^2]^2)/((E^4*x^3 + x^5)*Log[x] - 2*x^3*Log[x]*L 
og[Log[x]^2] + x*Log[x]*Log[Log[x]^2]^2),x]
 

Output:

-10*Log[x] + 4*Log[E^4*x^2 + x^4 - 2*x^2*Log[Log[x]^2] + Log[Log[x]^2]^2]
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 7292

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =! 
= u]
 

rule 7293

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v] 
]
 

Maple [A] (verified)

Time = 2.15 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.33

method	result	size
parallelrisch	\(-10 \ln \left (x \right )+4 \ln \left (x^{4}-2 x^{2} \ln \left (\ln \left (x \right )^{2}\right )+x^{2} {\mathrm e}^{4}+\ln \left (\ln \left (x \right )^{2}\right )^{2}\right )\)	\(36\)
risch	\(-10 \ln \left (x \right )+4 \ln \left (-\frac {\pi ^{2} \operatorname {csgn}\left (i \ln \left (x \right )\right )^{4} \operatorname {csgn}\left (i \ln \left (x \right )^{2}\right )^{2}}{16}+\frac {\pi ^{2} \operatorname {csgn}\left (i \ln \left (x \right )\right )^{3} \operatorname {csgn}\left (i \ln \left (x \right )^{2}\right )^{3}}{4}-\frac {3 \pi ^{2} \operatorname {csgn}\left (i \ln \left (x \right )\right )^{2} \operatorname {csgn}\left (i \ln \left (x \right )^{2}\right )^{4}}{8}+\frac {\pi ^{2} \operatorname {csgn}\left (i \ln \left (x \right )\right ) \operatorname {csgn}\left (i \ln \left (x \right )^{2}\right )^{5}}{4}-\frac {\pi ^{2} \operatorname {csgn}\left (i \ln \left (x \right )^{2}\right )^{6}}{16}+\frac {i \pi \,x^{2} \operatorname {csgn}\left (i \ln \left (x \right )\right )^{2} \operatorname {csgn}\left (i \ln \left (x \right )^{2}\right )}{4}-\frac {i \pi \,x^{2} \operatorname {csgn}\left (i \ln \left (x \right )\right ) \operatorname {csgn}\left (i \ln \left (x \right )^{2}\right )^{2}}{2}+\frac {i \pi \,x^{2} \operatorname {csgn}\left (i \ln \left (x \right )^{2}\right )^{3}}{4}+\frac {x^{4}}{4}+\frac {x^{2} {\mathrm e}^{4}}{4}+\left (-\frac {i \pi \operatorname {csgn}\left (i \ln \left (x \right )\right )^{2} \operatorname {csgn}\left (i \ln \left (x \right )^{2}\right )}{2}+i \pi \,\operatorname {csgn}\left (i \ln \left (x \right )\right ) \operatorname {csgn}\left (i \ln \left (x \right )^{2}\right )^{2}-\frac {i \pi \operatorname {csgn}\left (i \ln \left (x \right )^{2}\right )^{3}}{2}-x^{2}\right ) \ln \left (\ln \left (x \right )\right )+\ln \left (\ln \left (x \right )\right )^{2}\right )\)	\(259\)

Input:

int((-10*ln(x)*ln(ln(x)^2)^2+(4*x^2*ln(x)+16)*ln(ln(x)^2)+(-2*x^2*exp(4)+6 
*x^4)*ln(x)-16*x^2)/(x*ln(x)*ln(ln(x)^2)^2-2*x^3*ln(x)*ln(ln(x)^2)+(x^3*ex 
p(4)+x^5)*ln(x)),x,method=_RETURNVERBOSE)
 

Output:

-10*ln(x)+4*ln(x^4-2*x^2*ln(ln(x)^2)+x^2*exp(4)+ln(ln(x)^2)^2)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.30 \[ \int \frac {-16 x^2+\left (-2 e^4 x^2+6 x^4\right ) \log (x)+\left (16+4 x^2 \log (x)\right ) \log \left (\log ^2(x)\right )-10 \log (x) \log ^2\left (\log ^2(x)\right )}{\left (e^4 x^3+x^5\right ) \log (x)-2 x^3 \log (x) \log \left (\log ^2(x)\right )+x \log (x) \log ^2\left (\log ^2(x)\right )} \, dx=4 \, \log \left (x^{4} + x^{2} e^{4} - 2 \, x^{2} \log \left (\log \left (x\right )^{2}\right ) + \log \left (\log \left (x\right )^{2}\right )^{2}\right ) - 10 \, \log \left (x\right ) \] Input:

integrate((-10*log(x)*log(log(x)^2)^2+(4*x^2*log(x)+16)*log(log(x)^2)+(-2* 
x^2*exp(4)+6*x^4)*log(x)-16*x^2)/(x*log(x)*log(log(x)^2)^2-2*x^3*log(x)*lo 
g(log(x)^2)+(x^3*exp(4)+x^5)*log(x)),x, algorithm="fricas")
 

Output:

4*log(x^4 + x^2*e^4 - 2*x^2*log(log(x)^2) + log(log(x)^2)^2) - 10*log(x)
 

Sympy [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.37 \[ \int \frac {-16 x^2+\left (-2 e^4 x^2+6 x^4\right ) \log (x)+\left (16+4 x^2 \log (x)\right ) \log \left (\log ^2(x)\right )-10 \log (x) \log ^2\left (\log ^2(x)\right )}{\left (e^4 x^3+x^5\right ) \log (x)-2 x^3 \log (x) \log \left (\log ^2(x)\right )+x \log (x) \log ^2\left (\log ^2(x)\right )} \, dx=- 10 \log {\left (x \right )} + 4 \log {\left (x^{4} - 2 x^{2} \log {\left (\log {\left (x \right )}^{2} \right )} + x^{2} e^{4} + \log {\left (\log {\left (x \right )}^{2} \right )}^{2} \right )} \] Input:

integrate((-10*ln(x)*ln(ln(x)**2)**2+(4*x**2*ln(x)+16)*ln(ln(x)**2)+(-2*x* 
*2*exp(4)+6*x**4)*ln(x)-16*x**2)/(x*ln(x)*ln(ln(x)**2)**2-2*x**3*ln(x)*ln( 
ln(x)**2)+(x**3*exp(4)+x**5)*ln(x)),x)
 

Output:

-10*log(x) + 4*log(x**4 - 2*x**2*log(log(x)**2) + x**2*exp(4) + log(log(x) 
**2)**2)
 

Maxima [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.26 \[ \int \frac {-16 x^2+\left (-2 e^4 x^2+6 x^4\right ) \log (x)+\left (16+4 x^2 \log (x)\right ) \log \left (\log ^2(x)\right )-10 \log (x) \log ^2\left (\log ^2(x)\right )}{\left (e^4 x^3+x^5\right ) \log (x)-2 x^3 \log (x) \log \left (\log ^2(x)\right )+x \log (x) \log ^2\left (\log ^2(x)\right )} \, dx=4 \, \log \left (\frac {1}{4} \, x^{4} + \frac {1}{4} \, x^{2} e^{4} - x^{2} \log \left (\log \left (x\right )\right ) + \log \left (\log \left (x\right )\right )^{2}\right ) - 10 \, \log \left (x\right ) \] Input:

integrate((-10*log(x)*log(log(x)^2)^2+(4*x^2*log(x)+16)*log(log(x)^2)+(-2* 
x^2*exp(4)+6*x^4)*log(x)-16*x^2)/(x*log(x)*log(log(x)^2)^2-2*x^3*log(x)*lo 
g(log(x)^2)+(x^3*exp(4)+x^5)*log(x)),x, algorithm="maxima")
 

Output:

4*log(1/4*x^4 + 1/4*x^2*e^4 - x^2*log(log(x)) + log(log(x))^2) - 10*log(x)
 

Giac [A] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.30 \[ \int \frac {-16 x^2+\left (-2 e^4 x^2+6 x^4\right ) \log (x)+\left (16+4 x^2 \log (x)\right ) \log \left (\log ^2(x)\right )-10 \log (x) \log ^2\left (\log ^2(x)\right )}{\left (e^4 x^3+x^5\right ) \log (x)-2 x^3 \log (x) \log \left (\log ^2(x)\right )+x \log (x) \log ^2\left (\log ^2(x)\right )} \, dx=4 \, \log \left (x^{4} + x^{2} e^{4} - 2 \, x^{2} \log \left (\log \left (x\right )^{2}\right ) + \log \left (\log \left (x\right )^{2}\right )^{2}\right ) - 10 \, \log \left (x\right ) \] Input:

integrate((-10*log(x)*log(log(x)^2)^2+(4*x^2*log(x)+16)*log(log(x)^2)+(-2* 
x^2*exp(4)+6*x^4)*log(x)-16*x^2)/(x*log(x)*log(log(x)^2)^2-2*x^3*log(x)*lo 
g(log(x)^2)+(x^3*exp(4)+x^5)*log(x)),x, algorithm="giac")
 

Output:

4*log(x^4 + x^2*e^4 - 2*x^2*log(log(x)^2) + log(log(x)^2)^2) - 10*log(x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {-16 x^2+\left (-2 e^4 x^2+6 x^4\right ) \log (x)+\left (16+4 x^2 \log (x)\right ) \log \left (\log ^2(x)\right )-10 \log (x) \log ^2\left (\log ^2(x)\right )}{\left (e^4 x^3+x^5\right ) \log (x)-2 x^3 \log (x) \log \left (\log ^2(x)\right )+x \log (x) \log ^2\left (\log ^2(x)\right )} \, dx=\int -\frac {10\,{\ln \left ({\ln \left (x\right )}^2\right )}^2\,\ln \left (x\right )+\ln \left (x\right )\,\left (2\,x^2\,{\mathrm {e}}^4-6\,x^4\right )-\ln \left ({\ln \left (x\right )}^2\right )\,\left (4\,x^2\,\ln \left (x\right )+16\right )+16\,x^2}{\ln \left (x\right )\,\left (x^5+{\mathrm {e}}^4\,x^3\right )+x\,{\ln \left ({\ln \left (x\right )}^2\right )}^2\,\ln \left (x\right )-2\,x^3\,\ln \left ({\ln \left (x\right )}^2\right )\,\ln \left (x\right )} \,d x \] Input:

int(-(10*log(log(x)^2)^2*log(x) + log(x)*(2*x^2*exp(4) - 6*x^4) - log(log( 
x)^2)*(4*x^2*log(x) + 16) + 16*x^2)/(log(x)*(x^3*exp(4) + x^5) + x*log(log 
(x)^2)^2*log(x) - 2*x^3*log(log(x)^2)*log(x)),x)
 

Output:

int(-(10*log(log(x)^2)^2*log(x) + log(x)*(2*x^2*exp(4) - 6*x^4) - log(log( 
x)^2)*(4*x^2*log(x) + 16) + 16*x^2)/(log(x)*(x^3*exp(4) + x^5) + x*log(log 
(x)^2)^2*log(x) - 2*x^3*log(log(x)^2)*log(x)), x)
 

Reduce [F]

\[ \int \frac {-16 x^2+\left (-2 e^4 x^2+6 x^4\right ) \log (x)+\left (16+4 x^2 \log (x)\right ) \log \left (\log ^2(x)\right )-10 \log (x) \log ^2\left (\log ^2(x)\right )}{\left (e^4 x^3+x^5\right ) \log (x)-2 x^3 \log (x) \log \left (\log ^2(x)\right )+x \log (x) \log ^2\left (\log ^2(x)\right )} \, dx=-10 \left (\int \frac {\mathrm {log}\left (\mathrm {log}\left (x \right )^{2}\right )^{2}}{\mathrm {log}\left (\mathrm {log}\left (x \right )^{2}\right )^{2} x -2 \,\mathrm {log}\left (\mathrm {log}\left (x \right )^{2}\right ) x^{3}+e^{4} x^{3}+x^{5}}d x \right )+6 \left (\int \frac {x^{3}}{\mathrm {log}\left (\mathrm {log}\left (x \right )^{2}\right )^{2}-2 \,\mathrm {log}\left (\mathrm {log}\left (x \right )^{2}\right ) x^{2}+e^{4} x^{2}+x^{4}}d x \right )+16 \left (\int \frac {\mathrm {log}\left (\mathrm {log}\left (x \right )^{2}\right )}{\mathrm {log}\left (\mathrm {log}\left (x \right )^{2}\right )^{2} \mathrm {log}\left (x \right ) x -2 \,\mathrm {log}\left (\mathrm {log}\left (x \right )^{2}\right ) \mathrm {log}\left (x \right ) x^{3}+\mathrm {log}\left (x \right ) e^{4} x^{3}+\mathrm {log}\left (x \right ) x^{5}}d x \right )+4 \left (\int \frac {\mathrm {log}\left (\mathrm {log}\left (x \right )^{2}\right ) x}{\mathrm {log}\left (\mathrm {log}\left (x \right )^{2}\right )^{2}-2 \,\mathrm {log}\left (\mathrm {log}\left (x \right )^{2}\right ) x^{2}+e^{4} x^{2}+x^{4}}d x \right )-2 \left (\int \frac {x}{\mathrm {log}\left (\mathrm {log}\left (x \right )^{2}\right )^{2}-2 \,\mathrm {log}\left (\mathrm {log}\left (x \right )^{2}\right ) x^{2}+e^{4} x^{2}+x^{4}}d x \right ) e^{4}-16 \left (\int \frac {x}{\mathrm {log}\left (\mathrm {log}\left (x \right )^{2}\right )^{2} \mathrm {log}\left (x \right )-2 \,\mathrm {log}\left (\mathrm {log}\left (x \right )^{2}\right ) \mathrm {log}\left (x \right ) x^{2}+\mathrm {log}\left (x \right ) e^{4} x^{2}+\mathrm {log}\left (x \right ) x^{4}}d x \right ) \] Input:

int((-10*log(x)*log(log(x)^2)^2+(4*x^2*log(x)+16)*log(log(x)^2)+(-2*x^2*ex 
p(4)+6*x^4)*log(x)-16*x^2)/(x*log(x)*log(log(x)^2)^2-2*x^3*log(x)*log(log( 
x)^2)+(x^3*exp(4)+x^5)*log(x)),x)
 

Output:

2*( - 5*int(log(log(x)**2)**2/(log(log(x)**2)**2*x - 2*log(log(x)**2)*x**3 
 + e**4*x**3 + x**5),x) + 3*int(x**3/(log(log(x)**2)**2 - 2*log(log(x)**2) 
*x**2 + e**4*x**2 + x**4),x) + 8*int(log(log(x)**2)/(log(log(x)**2)**2*log 
(x)*x - 2*log(log(x)**2)*log(x)*x**3 + log(x)*e**4*x**3 + log(x)*x**5),x) 
+ 2*int((log(log(x)**2)*x)/(log(log(x)**2)**2 - 2*log(log(x)**2)*x**2 + e* 
*4*x**2 + x**4),x) - int(x/(log(log(x)**2)**2 - 2*log(log(x)**2)*x**2 + e* 
*4*x**2 + x**4),x)*e**4 - 8*int(x/(log(log(x)**2)**2*log(x) - 2*log(log(x) 
**2)*log(x)*x**2 + log(x)*e**4*x**2 + log(x)*x**4),x))