\(\int \frac {x^2+e^{5+x} (4+x^2)+(-4-4 e^{5+x}-x^3) \log (x)+(-e^{5+x} x^2+(x^2+e^{5+x} x^2) \log (x)) \log (e^{-5-x} (-e^{5+x}+(1+e^{5+x}) \log (x)))}{-e^{5+x} x^2+(x^2+e^{5+x} x^2) \log (x)} \, dx\) [1374]

Optimal result

Integrand size = 117, antiderivative size = 24 \[ \int \frac {x^2+e^{5+x} \left (4+x^2\right )+\left (-4-4 e^{5+x}-x^3\right ) \log (x)+\left (-e^{5+x} x^2+\left (x^2+e^{5+x} x^2\right ) \log (x)\right ) \log \left (e^{-5-x} \left (-e^{5+x}+\left (1+e^{5+x}\right ) \log (x)\right )\right )}{-e^{5+x} x^2+\left (x^2+e^{5+x} x^2\right ) \log (x)} \, dx=-1+\frac {4}{x}+x \log \left (-1+\log (x)+e^{-5-x} \log (x)\right ) \] Output:

x*ln(-1+ln(x)+ln(x)/exp(5)/exp(x))+4/x-1
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96 \[ \int \frac {x^2+e^{5+x} \left (4+x^2\right )+\left (-4-4 e^{5+x}-x^3\right ) \log (x)+\left (-e^{5+x} x^2+\left (x^2+e^{5+x} x^2\right ) \log (x)\right ) \log \left (e^{-5-x} \left (-e^{5+x}+\left (1+e^{5+x}\right ) \log (x)\right )\right )}{-e^{5+x} x^2+\left (x^2+e^{5+x} x^2\right ) \log (x)} \, dx=\frac {4}{x}+x \log \left (-1+\left (1+e^{-5-x}\right ) \log (x)\right ) \] Input:

Integrate[(x^2 + E^(5 + x)*(4 + x^2) + (-4 - 4*E^(5 + x) - x^3)*Log[x] + ( 
-(E^(5 + x)*x^2) + (x^2 + E^(5 + x)*x^2)*Log[x])*Log[E^(-5 - x)*(-E^(5 + x 
) + (1 + E^(5 + x))*Log[x])])/(-(E^(5 + x)*x^2) + (x^2 + E^(5 + x)*x^2)*Lo 
g[x]),x]
 

Output:

4/x + x*Log[-1 + (1 + E^(-5 - x))*Log[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.

\(\displaystyle \int \frac {\left (-x^3-4 e^{x+5}-4\right ) \log (x)+x^2+e^{x+5} \left (x^2+4\right )+\left (\left (e^{x+5} x^2+x^2\right ) \log (x)-e^{x+5} x^2\right ) \log \left (e^{-x-5} \left (\left (e^{x+5}+1\right ) \log (x)-e^{x+5}\right )\right )}{\left (e^{x+5} x^2+x^2\right ) \log (x)-e^{x+5} x^2} \, dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (\frac {x^2+x^2 \log (x) \log \left (\left (e^{-x-5}+1\right ) \log (x)-1\right )-x^2 \log \left (\left (e^{-x-5}+1\right ) \log (x)-1\right )-4 \log (x)+4}{x^2 (\log (x)-1)}-\frac {x \log ^2(x)-x \log (x)+1}{(\log (x)-1) \left (-e^{x+5}+e^{x+5} \log (x)+\log (x)\right )}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\int \frac {x \log ^2(x)}{(\log (x)-1) \left (e^{x+5} \log (x)+\log (x)-e^{x+5}\right )}dx-\int \frac {1}{(\log (x)-1) \left (e^{x+5} \log (x)+\log (x)-e^{x+5}\right )}dx+\int \frac {x \log (x)}{(\log (x)-1) \left (e^{x+5} \log (x)+\log (x)-e^{x+5}\right )}dx+\int \log \left (\left (1+e^{-x-5}\right ) \log (x)-1\right )dx+e \operatorname {ExpIntegralEi}(\log (x)-1)+\frac {4}{x}\)

Input:

Int[(x^2 + E^(5 + x)*(4 + x^2) + (-4 - 4*E^(5 + x) - x^3)*Log[x] + (-(E^(5 
 + x)*x^2) + (x^2 + E^(5 + x)*x^2)*Log[x])*Log[E^(-5 - x)*(-E^(5 + x) + (1 
 + E^(5 + x))*Log[x])])/(-(E^(5 + x)*x^2) + (x^2 + E^(5 + x)*x^2)*Log[x]), 
x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 10.40 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.54

method	result	size
parallelrisch	\(\frac {4+\ln \left (\left ({\mathrm e}^{5} {\mathrm e}^{x} \ln \left (x \right )-{\mathrm e}^{5} {\mathrm e}^{x}+\ln \left (x \right )\right ) {\mathrm e}^{-5} {\mathrm e}^{-x}\right ) x^{2}}{x}\)	\(37\)
risch	\(-x \ln \left ({\mathrm e}^{x}\right )+\frac {-i \pi \,x^{2} \operatorname {csgn}\left (i {\mathrm e}^{-x}\right ) \operatorname {csgn}\left (i \left (\left (\ln \left (x \right )-1\right ) {\mathrm e}^{5+x}+\ln \left (x \right )\right )\right ) \operatorname {csgn}\left (i {\mathrm e}^{-x} \left (\left (\ln \left (x \right )-1\right ) {\mathrm e}^{5+x}+\ln \left (x \right )\right )\right )+i \pi \,x^{2} \operatorname {csgn}\left (i {\mathrm e}^{-x}\right ) \operatorname {csgn}\left (i {\mathrm e}^{-x} \left (\left (\ln \left (x \right )-1\right ) {\mathrm e}^{5+x}+\ln \left (x \right )\right )\right )^{2}+i \pi \,x^{2} \operatorname {csgn}\left (i \left (\left (\ln \left (x \right )-1\right ) {\mathrm e}^{5+x}+\ln \left (x \right )\right )\right ) \operatorname {csgn}\left (i {\mathrm e}^{-x} \left (\left (\ln \left (x \right )-1\right ) {\mathrm e}^{5+x}+\ln \left (x \right )\right )\right )^{2}+8-i \pi \,x^{2} \operatorname {csgn}\left (i {\mathrm e}^{-x} \left (\left (\ln \left (x \right )-1\right ) {\mathrm e}^{5+x}+\ln \left (x \right )\right )\right )^{3}+2 x^{2} \ln \left (\left (\ln \left (x \right )-1\right ) {\mathrm e}^{5+x}+\ln \left (x \right )\right )-10 x^{2}}{2 x}\)	\(200\)

Input:

int((((x^2*exp(5)*exp(x)+x^2)*ln(x)-x^2*exp(5)*exp(x))*ln(((exp(5)*exp(x)+ 
1)*ln(x)-exp(5)*exp(x))/exp(5)/exp(x))+(-4*exp(5)*exp(x)-x^3-4)*ln(x)+(x^2 
+4)*exp(5)*exp(x)+x^2)/((x^2*exp(5)*exp(x)+x^2)*ln(x)-x^2*exp(5)*exp(x)),x 
,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.42 \[ \int \frac {x^2+e^{5+x} \left (4+x^2\right )+\left (-4-4 e^{5+x}-x^3\right ) \log (x)+\left (-e^{5+x} x^2+\left (x^2+e^{5+x} x^2\right ) \log (x)\right ) \log \left (e^{-5-x} \left (-e^{5+x}+\left (1+e^{5+x}\right ) \log (x)\right )\right )}{-e^{5+x} x^2+\left (x^2+e^{5+x} x^2\right ) \log (x)} \, dx=\frac {x^{2} \log \left ({\left ({\left (e^{\left (x + 5\right )} + 1\right )} \log \left (x\right ) - e^{\left (x + 5\right )}\right )} e^{\left (-x - 5\right )}\right ) + 4}{x} \] Input:

integrate((((x^2*exp(5)*exp(x)+x^2)*log(x)-x^2*exp(5)*exp(x))*log(((exp(5) 
*exp(x)+1)*log(x)-exp(5)*exp(x))/exp(5)/exp(x))+(-4*exp(5)*exp(x)-x^3-4)*l 
og(x)+(x^2+4)*exp(5)*exp(x)+x^2)/((x^2*exp(5)*exp(x)+x^2)*log(x)-x^2*exp(5 
)*exp(x)),x, algorithm="fricas")
 

Output:

(x^2*log(((e^(x + 5) + 1)*log(x) - e^(x + 5))*e^(-x - 5)) + 4)/x
 

Sympy [A] (verification not implemented)

Time = 0.44 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.29 \[ \int \frac {x^2+e^{5+x} \left (4+x^2\right )+\left (-4-4 e^{5+x}-x^3\right ) \log (x)+\left (-e^{5+x} x^2+\left (x^2+e^{5+x} x^2\right ) \log (x)\right ) \log \left (e^{-5-x} \left (-e^{5+x}+\left (1+e^{5+x}\right ) \log (x)\right )\right )}{-e^{5+x} x^2+\left (x^2+e^{5+x} x^2\right ) \log (x)} \, dx=x \log {\left (\frac {\left (\left (e^{5} e^{x} + 1\right ) \log {\left (x \right )} - e^{5} e^{x}\right ) e^{- x}}{e^{5}} \right )} + \frac {4}{x} \] Input:

integrate((((x**2*exp(5)*exp(x)+x**2)*ln(x)-x**2*exp(5)*exp(x))*ln(((exp(5 
)*exp(x)+1)*ln(x)-exp(5)*exp(x))/exp(5)/exp(x))+(-4*exp(5)*exp(x)-x**3-4)* 
ln(x)+(x**2+4)*exp(5)*exp(x)+x**2)/((x**2*exp(5)*exp(x)+x**2)*ln(x)-x**2*e 
xp(5)*exp(x)),x)
 

Output:

x*log(((exp(5)*exp(x) + 1)*log(x) - exp(5)*exp(x))*exp(-5)*exp(-x)) + 4/x
 

Maxima [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.54 \[ \int \frac {x^2+e^{5+x} \left (4+x^2\right )+\left (-4-4 e^{5+x}-x^3\right ) \log (x)+\left (-e^{5+x} x^2+\left (x^2+e^{5+x} x^2\right ) \log (x)\right ) \log \left (e^{-5-x} \left (-e^{5+x}+\left (1+e^{5+x}\right ) \log (x)\right )\right )}{-e^{5+x} x^2+\left (x^2+e^{5+x} x^2\right ) \log (x)} \, dx=-\frac {x^{3} - x^{2} \log \left ({\left (e^{\left (x + 5\right )} + 1\right )} \log \left (x\right ) - e^{\left (x + 5\right )}\right ) + 5 \, x^{2} - 4}{x} \] Input:

integrate((((x^2*exp(5)*exp(x)+x^2)*log(x)-x^2*exp(5)*exp(x))*log(((exp(5) 
*exp(x)+1)*log(x)-exp(5)*exp(x))/exp(5)/exp(x))+(-4*exp(5)*exp(x)-x^3-4)*l 
og(x)+(x^2+4)*exp(5)*exp(x)+x^2)/((x^2*exp(5)*exp(x)+x^2)*log(x)-x^2*exp(5 
)*exp(x)),x, algorithm="maxima")
 

Output:

-(x^3 - x^2*log((e^(x + 5) + 1)*log(x) - e^(x + 5)) + 5*x^2 - 4)/x
 

Giac [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.54 \[ \int \frac {x^2+e^{5+x} \left (4+x^2\right )+\left (-4-4 e^{5+x}-x^3\right ) \log (x)+\left (-e^{5+x} x^2+\left (x^2+e^{5+x} x^2\right ) \log (x)\right ) \log \left (e^{-5-x} \left (-e^{5+x}+\left (1+e^{5+x}\right ) \log (x)\right )\right )}{-e^{5+x} x^2+\left (x^2+e^{5+x} x^2\right ) \log (x)} \, dx=\frac {x^{2} \log \left ({\left (e^{\left (x + 5\right )} \log \left (x\right ) - e^{\left (x + 5\right )} + \log \left (x\right )\right )} e^{\left (-x\right )}\right ) - 5 \, x^{2} + 4}{x} \] Input:

integrate((((x^2*exp(5)*exp(x)+x^2)*log(x)-x^2*exp(5)*exp(x))*log(((exp(5) 
*exp(x)+1)*log(x)-exp(5)*exp(x))/exp(5)/exp(x))+(-4*exp(5)*exp(x)-x^3-4)*l 
og(x)+(x^2+4)*exp(5)*exp(x)+x^2)/((x^2*exp(5)*exp(x)+x^2)*log(x)-x^2*exp(5 
)*exp(x)),x, algorithm="giac")
 

Output:

(x^2*log((e^(x + 5)*log(x) - e^(x + 5) + log(x))*e^(-x)) - 5*x^2 + 4)/x
 

Mupad [B] (verification not implemented)

Time = 2.17 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \frac {x^2+e^{5+x} \left (4+x^2\right )+\left (-4-4 e^{5+x}-x^3\right ) \log (x)+\left (-e^{5+x} x^2+\left (x^2+e^{5+x} x^2\right ) \log (x)\right ) \log \left (e^{-5-x} \left (-e^{5+x}+\left (1+e^{5+x}\right ) \log (x)\right )\right )}{-e^{5+x} x^2+\left (x^2+e^{5+x} x^2\right ) \log (x)} \, dx=x\,\ln \left (\ln \left (x\right )+{\mathrm {e}}^{-x}\,{\mathrm {e}}^{-5}\,\ln \left (x\right )-1\right )+\frac {4}{x} \] Input:

int((x^2 - log(x)*(4*exp(5)*exp(x) + x^3 + 4) + log(exp(-x)*exp(-5)*(log(x 
)*(exp(5)*exp(x) + 1) - exp(5)*exp(x)))*(log(x)*(x^2 + x^2*exp(5)*exp(x)) 
- x^2*exp(5)*exp(x)) + exp(5)*exp(x)*(x^2 + 4))/(log(x)*(x^2 + x^2*exp(5)* 
exp(x)) - x^2*exp(5)*exp(x)),x)
 

Output:

x*log(log(x) + exp(-x)*exp(-5)*log(x) - 1) + 4/x
 

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.67 \[ \int \frac {x^2+e^{5+x} \left (4+x^2\right )+\left (-4-4 e^{5+x}-x^3\right ) \log (x)+\left (-e^{5+x} x^2+\left (x^2+e^{5+x} x^2\right ) \log (x)\right ) \log \left (e^{-5-x} \left (-e^{5+x}+\left (1+e^{5+x}\right ) \log (x)\right )\right )}{-e^{5+x} x^2+\left (x^2+e^{5+x} x^2\right ) \log (x)} \, dx=\frac {\mathrm {log}\left (\frac {e^{x} \mathrm {log}\left (x \right ) e^{5}-e^{x} e^{5}+\mathrm {log}\left (x \right )}{e^{x} e^{5}}\right ) x^{2}+4}{x} \] Input:

int((((x^2*exp(5)*exp(x)+x^2)*log(x)-x^2*exp(5)*exp(x))*log(((exp(5)*exp(x 
)+1)*log(x)-exp(5)*exp(x))/exp(5)/exp(x))+(-4*exp(5)*exp(x)-x^3-4)*log(x)+ 
(x^2+4)*exp(5)*exp(x)+x^2)/((x^2*exp(5)*exp(x)+x^2)*log(x)-x^2*exp(5)*exp( 
x)),x)
 

Output:

(log((e**x*log(x)*e**5 - e**x*e**5 + log(x))/(e**x*e**5))*x**2 + 4)/x