\(\int \frac {2 x \log (x)+(-x+(x+x^2-x^3) \log (x)) \log (2 x^2)+(-2 \log (x)+(-1+x^2) \log (x) \log (2 x^2)) \log (\log (x))+(2 x^2 \log (x)-2 x \log (x) \log (\log (x))) \log (\frac {e^x}{-x+\log (\log (x))})}{-x^3 \log (5) \log (x) \log ^2(2 x^2)+x^2 \log (5) \log (x) \log ^2(2 x^2) \log (\log (x))} \, dx\) [101]

Optimal result

Integrand size = 124, antiderivative size = 31 \[ \int \frac {2 x \log (x)+\left (-x+\left (x+x^2-x^3\right ) \log (x)\right ) \log \left (2 x^2\right )+\left (-2 \log (x)+\left (-1+x^2\right ) \log (x) \log \left (2 x^2\right )\right ) \log (\log (x))+\left (2 x^2 \log (x)-2 x \log (x) \log (\log (x))\right ) \log \left (\frac {e^x}{-x+\log (\log (x))}\right )}{-x^3 \log (5) \log (x) \log ^2\left (2 x^2\right )+x^2 \log (5) \log (x) \log ^2\left (2 x^2\right ) \log (\log (x))} \, dx=\frac {\frac {1}{x}+\log \left (\frac {e^x}{-x+\log (\log (x))}\right )}{\log (5) \log \left (2 x^2\right )} \] Output:

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

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.10 \[ \int \frac {2 x \log (x)+\left (-x+\left (x+x^2-x^3\right ) \log (x)\right ) \log \left (2 x^2\right )+\left (-2 \log (x)+\left (-1+x^2\right ) \log (x) \log \left (2 x^2\right )\right ) \log (\log (x))+\left (2 x^2 \log (x)-2 x \log (x) \log (\log (x))\right ) \log \left (\frac {e^x}{-x+\log (\log (x))}\right )}{-x^3 \log (5) \log (x) \log ^2\left (2 x^2\right )+x^2 \log (5) \log (x) \log ^2\left (2 x^2\right ) \log (\log (x))} \, dx=\frac {1+x \log \left (\frac {e^x}{-x+\log (\log (x))}\right )}{x \log (5) \log \left (2 x^2\right )} \] Input:

Integrate[(2*x*Log[x] + (-x + (x + x^2 - x^3)*Log[x])*Log[2*x^2] + (-2*Log 
[x] + (-1 + x^2)*Log[x]*Log[2*x^2])*Log[Log[x]] + (2*x^2*Log[x] - 2*x*Log[ 
x]*Log[Log[x]])*Log[E^x/(-x + Log[Log[x]])])/(-(x^3*Log[5]*Log[x]*Log[2*x^ 
2]^2) + x^2*Log[5]*Log[x]*Log[2*x^2]^2*Log[Log[x]]),x]
 

Output:

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

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.

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

\(\Big \downarrow \) 7292

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 7293

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

\(\Big \downarrow \) 2009

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

Input:

Int[(2*x*Log[x] + (-x + (x + x^2 - x^3)*Log[x])*Log[2*x^2] + (-2*Log[x] + 
(-1 + x^2)*Log[x]*Log[2*x^2])*Log[Log[x]] + (2*x^2*Log[x] - 2*x*Log[x]*Log 
[Log[x]])*Log[E^x/(-x + Log[Log[x]])])/(-(x^3*Log[5]*Log[x]*Log[2*x^2]^2) 
+ x^2*Log[5]*Log[x]*Log[2*x^2]^2*Log[Log[x]]),x]
 

Output:

$Aborted
 

Maple [C] (warning: unable to verify)

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

Time = 1.45 (sec) , antiderivative size = 293, normalized size of antiderivative = 9.45

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

Input:

int(((-2*x*ln(x)*ln(ln(x))+2*x^2*ln(x))*ln(exp(x)/(ln(ln(x))-x))+((x^2-1)* 
ln(x)*ln(2*x^2)-2*ln(x))*ln(ln(x))+((-x^3+x^2+x)*ln(x)-x)*ln(2*x^2)+2*x*ln 
(x))/(x^2*ln(5)*ln(x)*ln(2*x^2)^2*ln(ln(x))-x^3*ln(5)*ln(x)*ln(2*x^2)^2),x 
)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.13 \[ \int \frac {2 x \log (x)+\left (-x+\left (x+x^2-x^3\right ) \log (x)\right ) \log \left (2 x^2\right )+\left (-2 \log (x)+\left (-1+x^2\right ) \log (x) \log \left (2 x^2\right )\right ) \log (\log (x))+\left (2 x^2 \log (x)-2 x \log (x) \log (\log (x))\right ) \log \left (\frac {e^x}{-x+\log (\log (x))}\right )}{-x^3 \log (5) \log (x) \log ^2\left (2 x^2\right )+x^2 \log (5) \log (x) \log ^2\left (2 x^2\right ) \log (\log (x))} \, dx=\frac {x \log \left (-\frac {e^{x}}{x - \log \left (\log \left (x\right )\right )}\right ) + 1}{x \log \left (5\right ) \log \left (2\right ) + 2 \, x \log \left (5\right ) \log \left (x\right )} \] Input:

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

Output:

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

Sympy [F(-2)]

Exception generated. \[ \int \frac {2 x \log (x)+\left (-x+\left (x+x^2-x^3\right ) \log (x)\right ) \log \left (2 x^2\right )+\left (-2 \log (x)+\left (-1+x^2\right ) \log (x) \log \left (2 x^2\right )\right ) \log (\log (x))+\left (2 x^2 \log (x)-2 x \log (x) \log (\log (x))\right ) \log \left (\frac {e^x}{-x+\log (\log (x))}\right )}{-x^3 \log (5) \log (x) \log ^2\left (2 x^2\right )+x^2 \log (5) \log (x) \log ^2\left (2 x^2\right ) \log (\log (x))} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(((-2*x*ln(x)*ln(ln(x))+2*x**2*ln(x))*ln(exp(x)/(ln(ln(x))-x))+(( 
x**2-1)*ln(x)*ln(2*x**2)-2*ln(x))*ln(ln(x))+((-x**3+x**2+x)*ln(x)-x)*ln(2* 
x**2)+2*x*ln(x))/(x**2*ln(5)*ln(x)*ln(2*x**2)**2*ln(ln(x))-x**3*ln(5)*ln(x 
)*ln(2*x**2)**2),x)
 

Output:

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

Maxima [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.06 \[ \int \frac {2 x \log (x)+\left (-x+\left (x+x^2-x^3\right ) \log (x)\right ) \log \left (2 x^2\right )+\left (-2 \log (x)+\left (-1+x^2\right ) \log (x) \log \left (2 x^2\right )\right ) \log (\log (x))+\left (2 x^2 \log (x)-2 x \log (x) \log (\log (x))\right ) \log \left (\frac {e^x}{-x+\log (\log (x))}\right )}{-x^3 \log (5) \log (x) \log ^2\left (2 x^2\right )+x^2 \log (5) \log (x) \log ^2\left (2 x^2\right ) \log (\log (x))} \, dx=\frac {x^{2} - x \log \left (-x + \log \left (\log \left (x\right )\right )\right ) + 1}{x \log \left (5\right ) \log \left (2\right ) + 2 \, x \log \left (5\right ) \log \left (x\right )} \] Input:

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

Output:

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

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.20 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.65 \[ \int \frac {2 x \log (x)+\left (-x+\left (x+x^2-x^3\right ) \log (x)\right ) \log \left (2 x^2\right )+\left (-2 \log (x)+\left (-1+x^2\right ) \log (x) \log \left (2 x^2\right )\right ) \log (\log (x))+\left (2 x^2 \log (x)-2 x \log (x) \log (\log (x))\right ) \log \left (\frac {e^x}{-x+\log (\log (x))}\right )}{-x^3 \log (5) \log (x) \log ^2\left (2 x^2\right )+x^2 \log (5) \log (x) \log ^2\left (2 x^2\right ) \log (\log (x))} \, dx=\frac {i \, \pi x + x^{2} + 1}{x \log \left (5\right ) \log \left (2\right ) + 2 \, x \log \left (5\right ) \log \left (x\right )} - \frac {\log \left (x - \log \left (\log \left (x\right )\right )\right )}{\log \left (5\right ) \log \left (2\right ) + 2 \, \log \left (5\right ) \log \left (x\right )} \] Input:

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

Output:

(I*pi*x + x^2 + 1)/(x*log(5)*log(2) + 2*x*log(5)*log(x)) - log(x - log(log 
(x)))/(log(5)*log(2) + 2*log(5)*log(x))
 

Mupad [B] (verification not implemented)

Time = 4.49 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.10 \[ \int \frac {2 x \log (x)+\left (-x+\left (x+x^2-x^3\right ) \log (x)\right ) \log \left (2 x^2\right )+\left (-2 \log (x)+\left (-1+x^2\right ) \log (x) \log \left (2 x^2\right )\right ) \log (\log (x))+\left (2 x^2 \log (x)-2 x \log (x) \log (\log (x))\right ) \log \left (\frac {e^x}{-x+\log (\log (x))}\right )}{-x^3 \log (5) \log (x) \log ^2\left (2 x^2\right )+x^2 \log (5) \log (x) \log ^2\left (2 x^2\right ) \log (\log (x))} \, dx=\frac {x\,\ln \left (-\frac {{\mathrm {e}}^x}{x-\ln \left (\ln \left (x\right )\right )}\right )+1}{x\,\ln \left (5\right )\,\ln \left (2\,x^2\right )} \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.10 \[ \int \frac {2 x \log (x)+\left (-x+\left (x+x^2-x^3\right ) \log (x)\right ) \log \left (2 x^2\right )+\left (-2 \log (x)+\left (-1+x^2\right ) \log (x) \log \left (2 x^2\right )\right ) \log (\log (x))+\left (2 x^2 \log (x)-2 x \log (x) \log (\log (x))\right ) \log \left (\frac {e^x}{-x+\log (\log (x))}\right )}{-x^3 \log (5) \log (x) \log ^2\left (2 x^2\right )+x^2 \log (5) \log (x) \log ^2\left (2 x^2\right ) \log (\log (x))} \, dx=\frac {\mathrm {log}\left (\frac {e^{x}}{\mathrm {log}\left (\mathrm {log}\left (x \right )\right )-x}\right ) x +1}{\mathrm {log}\left (2 x^{2}\right ) \mathrm {log}\left (5\right ) x} \] Input:

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

Output:

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