\(\int \frac {x \log (4)+\log ^2(4)+e^x (-x^2 \log (4)-x \log ^2(4))+(-x \log (4)+\log ^2(4)+e^x (x^3 \log (4)+x^2 \log ^2(4))) \log (\frac {x}{4})+(x \log (4)+\log ^2(4)+(x \log (4)+\log ^2(4)) \log (\frac {x}{4})) \log (x^2+2 x \log (4)+\log ^2(4))}{(x^3+x^2 \log (4)) \log ^2(\frac {x}{4})} \, dx\) [2793]

Optimal result

Integrand size = 125, antiderivative size = 29 \[ \int \frac {x \log (4)+\log ^2(4)+e^x \left (-x^2 \log (4)-x \log ^2(4)\right )+\left (-x \log (4)+\log ^2(4)+e^x \left (x^3 \log (4)+x^2 \log ^2(4)\right )\right ) \log \left (\frac {x}{4}\right )+\left (x \log (4)+\log ^2(4)+\left (x \log (4)+\log ^2(4)\right ) \log \left (\frac {x}{4}\right )\right ) \log \left (x^2+2 x \log (4)+\log ^2(4)\right )}{\left (x^3+x^2 \log (4)\right ) \log ^2\left (\frac {x}{4}\right )} \, dx=\frac {\log (4) \left (e^x-\frac {1+\log \left ((x+\log (4))^2\right )}{x}\right )}{\log \left (\frac {x}{4}\right )} \] Output:

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

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.03 \[ \int \frac {x \log (4)+\log ^2(4)+e^x \left (-x^2 \log (4)-x \log ^2(4)\right )+\left (-x \log (4)+\log ^2(4)+e^x \left (x^3 \log (4)+x^2 \log ^2(4)\right )\right ) \log \left (\frac {x}{4}\right )+\left (x \log (4)+\log ^2(4)+\left (x \log (4)+\log ^2(4)\right ) \log \left (\frac {x}{4}\right )\right ) \log \left (x^2+2 x \log (4)+\log ^2(4)\right )}{\left (x^3+x^2 \log (4)\right ) \log ^2\left (\frac {x}{4}\right )} \, dx=\frac {\log (4) \left (-1+e^x x-\log \left ((x+\log (4))^2\right )\right )}{x \log \left (\frac {x}{4}\right )} \] Input:

Integrate[(x*Log[4] + Log[4]^2 + E^x*(-(x^2*Log[4]) - x*Log[4]^2) + (-(x*L 
og[4]) + Log[4]^2 + E^x*(x^3*Log[4] + x^2*Log[4]^2))*Log[x/4] + (x*Log[4] 
+ Log[4]^2 + (x*Log[4] + Log[4]^2)*Log[x/4])*Log[x^2 + 2*x*Log[4] + Log[4] 
^2])/((x^3 + x^2*Log[4])*Log[x/4]^2),x]
 

Output:

(Log[4]*(-1 + E^x*x - Log[(x + Log[4])^2]))/(x*Log[x/4])
 

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.

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

\(\Big \downarrow \) 2026

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

\(\Big \downarrow \) 7293

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

$Aborted
 

Maple [C] (warning: unable to verify)

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

Time = 2.14 (sec) , antiderivative size = 115, normalized size of antiderivative = 3.97

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

Input:

int((((4*ln(2)^2+2*x*ln(2))*ln(1/4*x)+4*ln(2)^2+2*x*ln(2))*ln(4*ln(2)^2+4* 
x*ln(2)+x^2)+((4*x^2*ln(2)^2+2*x^3*ln(2))*exp(x)+4*ln(2)^2-2*x*ln(2))*ln(1 
/4*x)+(-4*x*ln(2)^2-2*x^2*ln(2))*exp(x)+4*ln(2)^2+2*x*ln(2))/(2*x^2*ln(2)+ 
x^3)/ln(1/4*x)^2,x)
 

Output:

-4/x*ln(2)/ln(1/4*x)*ln(ln(2)+1/2*x)+1/x*ln(2)*(I*Pi*csgn(I*(ln(2)+1/2*x)) 
^2*csgn(I*(ln(2)+1/2*x)^2)-2*I*Pi*csgn(I*(ln(2)+1/2*x))*csgn(I*(ln(2)+1/2* 
x)^2)^2+I*Pi*csgn(I*(ln(2)+1/2*x)^2)^3+2*exp(x)*x-2)/ln(1/4*x)
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.45 \[ \int \frac {x \log (4)+\log ^2(4)+e^x \left (-x^2 \log (4)-x \log ^2(4)\right )+\left (-x \log (4)+\log ^2(4)+e^x \left (x^3 \log (4)+x^2 \log ^2(4)\right )\right ) \log \left (\frac {x}{4}\right )+\left (x \log (4)+\log ^2(4)+\left (x \log (4)+\log ^2(4)\right ) \log \left (\frac {x}{4}\right )\right ) \log \left (x^2+2 x \log (4)+\log ^2(4)\right )}{\left (x^3+x^2 \log (4)\right ) \log ^2\left (\frac {x}{4}\right )} \, dx=\frac {2 \, {\left (x e^{x} \log \left (2\right ) - \log \left (2\right ) \log \left (x^{2} + 4 \, x \log \left (2\right ) + 4 \, \log \left (2\right )^{2}\right ) - \log \left (2\right )\right )}}{x \log \left (\frac {1}{4} \, x\right )} \] Input:

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

Output:

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

Sympy [F(-2)]

Exception generated. \[ \int \frac {x \log (4)+\log ^2(4)+e^x \left (-x^2 \log (4)-x \log ^2(4)\right )+\left (-x \log (4)+\log ^2(4)+e^x \left (x^3 \log (4)+x^2 \log ^2(4)\right )\right ) \log \left (\frac {x}{4}\right )+\left (x \log (4)+\log ^2(4)+\left (x \log (4)+\log ^2(4)\right ) \log \left (\frac {x}{4}\right )\right ) \log \left (x^2+2 x \log (4)+\log ^2(4)\right )}{\left (x^3+x^2 \log (4)\right ) \log ^2\left (\frac {x}{4}\right )} \, dx=\text {Exception raised: TypeError} \] Input:

integrate((((4*ln(2)**2+2*x*ln(2))*ln(1/4*x)+4*ln(2)**2+2*x*ln(2))*ln(4*ln 
(2)**2+4*x*ln(2)+x**2)+((4*x**2*ln(2)**2+2*x**3*ln(2))*exp(x)+4*ln(2)**2-2 
*x*ln(2))*ln(1/4*x)+(-4*x*ln(2)**2-2*x**2*ln(2))*exp(x)+4*ln(2)**2+2*x*ln( 
2))/(2*x**2*ln(2)+x**3)/ln(1/4*x)**2,x)
 

Output:

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

Maxima [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.28 \[ \int \frac {x \log (4)+\log ^2(4)+e^x \left (-x^2 \log (4)-x \log ^2(4)\right )+\left (-x \log (4)+\log ^2(4)+e^x \left (x^3 \log (4)+x^2 \log ^2(4)\right )\right ) \log \left (\frac {x}{4}\right )+\left (x \log (4)+\log ^2(4)+\left (x \log (4)+\log ^2(4)\right ) \log \left (\frac {x}{4}\right )\right ) \log \left (x^2+2 x \log (4)+\log ^2(4)\right )}{\left (x^3+x^2 \log (4)\right ) \log ^2\left (\frac {x}{4}\right )} \, dx=-\frac {2 \, {\left (x e^{x} \log \left (2\right ) - 2 \, \log \left (2\right ) \log \left (x + 2 \, \log \left (2\right )\right ) - \log \left (2\right )\right )}}{2 \, x \log \left (2\right ) - x \log \left (x\right )} \] Input:

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

Output:

-2*(x*e^x*log(2) - 2*log(2)*log(x + 2*log(2)) - log(2))/(2*x*log(2) - x*lo 
g(x))
 

Giac [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.59 \[ \int \frac {x \log (4)+\log ^2(4)+e^x \left (-x^2 \log (4)-x \log ^2(4)\right )+\left (-x \log (4)+\log ^2(4)+e^x \left (x^3 \log (4)+x^2 \log ^2(4)\right )\right ) \log \left (\frac {x}{4}\right )+\left (x \log (4)+\log ^2(4)+\left (x \log (4)+\log ^2(4)\right ) \log \left (\frac {x}{4}\right )\right ) \log \left (x^2+2 x \log (4)+\log ^2(4)\right )}{\left (x^3+x^2 \log (4)\right ) \log ^2\left (\frac {x}{4}\right )} \, dx=-\frac {2 \, {\left (x e^{x} \log \left (2\right ) - \log \left (2\right ) \log \left (x^{2} + 4 \, x \log \left (2\right ) + 4 \, \log \left (2\right )^{2}\right ) - \log \left (2\right )\right )}}{2 \, x \log \left (2\right ) - x \log \left (x\right )} \] Input:

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.31 \[ \int \frac {x \log (4)+\log ^2(4)+e^x \left (-x^2 \log (4)-x \log ^2(4)\right )+\left (-x \log (4)+\log ^2(4)+e^x \left (x^3 \log (4)+x^2 \log ^2(4)\right )\right ) \log \left (\frac {x}{4}\right )+\left (x \log (4)+\log ^2(4)+\left (x \log (4)+\log ^2(4)\right ) \log \left (\frac {x}{4}\right )\right ) \log \left (x^2+2 x \log (4)+\log ^2(4)\right )}{\left (x^3+x^2 \log (4)\right ) \log ^2\left (\frac {x}{4}\right )} \, dx=\frac {2 \,\mathrm {log}\left (2\right ) \left (e^{x} x -\mathrm {log}\left (4 \mathrm {log}\left (2\right )^{2}+4 \,\mathrm {log}\left (2\right ) x +x^{2}\right )-1\right )}{\mathrm {log}\left (\frac {x}{4}\right ) x} \] Input:

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

Output:

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