\(\int \frac {1-2 x+x^2+e^3 (-1+2 x-x^2)+(-x+x^2+e^3 (x-x^2)) \log (x)+(-4+e^3 (4-2 x)+2 x+(-2+e^3 (2-2 x)+2 x) \log (x)) \log (2+\log (2))+(-x+e^3 x+(-2+2 e^3) \log (2+\log (2))) \log (\frac {1}{2} (x+2 \log (2+\log (2))))}{-x^2-x^2 \log (x)+(-2 x-2 x \log (x)) \log (2+\log (2))+(x+2 \log (2+\log (2))) \log (\frac {1}{2} (x+2 \log (2+\log (2))))} \, dx\) [585]

Optimal result

Integrand size = 172, antiderivative size = 33 \[ \int \frac {1-2 x+x^2+e^3 \left (-1+2 x-x^2\right )+\left (-x+x^2+e^3 \left (x-x^2\right )\right ) \log (x)+\left (-4+e^3 (4-2 x)+2 x+\left (-2+e^3 (2-2 x)+2 x\right ) \log (x)\right ) \log (2+\log (2))+\left (-x+e^3 x+\left (-2+2 e^3\right ) \log (2+\log (2))\right ) \log \left (\frac {1}{2} (x+2 \log (2+\log (2)))\right )}{-x^2-x^2 \log (x)+(-2 x-2 x \log (x)) \log (2+\log (2))+(x+2 \log (2+\log (2))) \log \left (\frac {1}{2} (x+2 \log (2+\log (2)))\right )} \, dx=\left (1-e^3\right ) \left (-x+\log \left (x+x \log (x)-\log \left (\frac {x}{2}+\log (2+\log (2))\right )\right )\right ) \] Output:

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.94 \[ \int \frac {1-2 x+x^2+e^3 \left (-1+2 x-x^2\right )+\left (-x+x^2+e^3 \left (x-x^2\right )\right ) \log (x)+\left (-4+e^3 (4-2 x)+2 x+\left (-2+e^3 (2-2 x)+2 x\right ) \log (x)\right ) \log (2+\log (2))+\left (-x+e^3 x+\left (-2+2 e^3\right ) \log (2+\log (2))\right ) \log \left (\frac {1}{2} (x+2 \log (2+\log (2)))\right )}{-x^2-x^2 \log (x)+(-2 x-2 x \log (x)) \log (2+\log (2))+(x+2 \log (2+\log (2))) \log \left (\frac {1}{2} (x+2 \log (2+\log (2)))\right )} \, dx=\left (-1+e^3\right ) \left (x-\log \left (x+x \log (x)-\log \left (\frac {x}{2}+\log (2+\log (2))\right )\right )\right ) \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.92 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.03, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.029, Rules used = {7292, 27, 25, 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.

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

\(\Big \downarrow \) 7292

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 7293

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

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 = 2.62 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.03

method	result	size
norman	\(\left ({\mathrm e}^{3}-1\right ) x +\left (-{\mathrm e}^{3}+1\right ) \ln \left (x +x \ln \left (x \right )-\ln \left (\ln \left (\ln \left (2\right )+2\right )+\frac {x}{2}\right )\right )\)	\(34\)
default	\(\left ({\mathrm e}^{3}-1\right ) x +\left (-{\mathrm e}^{3}+1\right ) \ln \left (x \ln \left (x \right )+x -\ln \left (2 \ln \left (\ln \left (2\right )+2\right )+x \right )+\ln \left (2\right )\right )\)	\(36\)
risch	\(-\ln \left (\ln \left (\ln \left (\ln \left (2\right )+2\right )+\frac {x}{2}\right )-\left (\ln \left (x \right )+1\right ) x \right ) {\mathrm e}^{3}+x \,{\mathrm e}^{3}+\ln \left (\ln \left (\ln \left (\ln \left (2\right )+2\right )+\frac {x}{2}\right )-\left (\ln \left (x \right )+1\right ) x \right )-x\)	\(51\)
parallelrisch	\(-\ln \left (x +x \ln \left (x \right )-\ln \left (\ln \left (\ln \left (2\right )+2\right )+\frac {x}{2}\right )\right ) {\mathrm e}^{3}-4 \ln \left (\ln \left (2\right )+2\right ) {\mathrm e}^{3}+x \,{\mathrm e}^{3}+\ln \left (x +x \ln \left (x \right )-\ln \left (\ln \left (\ln \left (2\right )+2\right )+\frac {x}{2}\right )\right )+4 \ln \left (\ln \left (2\right )+2\right )-x\)	\(67\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.03 \[ \int \frac {1-2 x+x^2+e^3 \left (-1+2 x-x^2\right )+\left (-x+x^2+e^3 \left (x-x^2\right )\right ) \log (x)+\left (-4+e^3 (4-2 x)+2 x+\left (-2+e^3 (2-2 x)+2 x\right ) \log (x)\right ) \log (2+\log (2))+\left (-x+e^3 x+\left (-2+2 e^3\right ) \log (2+\log (2))\right ) \log \left (\frac {1}{2} (x+2 \log (2+\log (2)))\right )}{-x^2-x^2 \log (x)+(-2 x-2 x \log (x)) \log (2+\log (2))+(x+2 \log (2+\log (2))) \log \left (\frac {1}{2} (x+2 \log (2+\log (2)))\right )} \, dx=x e^{3} - {\left (e^{3} - 1\right )} \log \left (-x \log \left (x\right ) - x + \log \left (\frac {1}{2} \, x + \log \left (\log \left (2\right ) + 2\right )\right )\right ) - x \] Input:

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

Output:

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

Sympy [A] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.18 \[ \int \frac {1-2 x+x^2+e^3 \left (-1+2 x-x^2\right )+\left (-x+x^2+e^3 \left (x-x^2\right )\right ) \log (x)+\left (-4+e^3 (4-2 x)+2 x+\left (-2+e^3 (2-2 x)+2 x\right ) \log (x)\right ) \log (2+\log (2))+\left (-x+e^3 x+\left (-2+2 e^3\right ) \log (2+\log (2))\right ) \log \left (\frac {1}{2} (x+2 \log (2+\log (2)))\right )}{-x^2-x^2 \log (x)+(-2 x-2 x \log (x)) \log (2+\log (2))+(x+2 \log (2+\log (2))) \log \left (\frac {1}{2} (x+2 \log (2+\log (2)))\right )} \, dx=x \left (-1 + e^{3}\right ) - \left (-1 + e\right ) \left (1 + e + e^{2}\right ) \log {\left (- x \log {\left (x \right )} - x + \log {\left (\frac {x}{2} + \log {\left (\log {\left (2 \right )} + 2 \right )} \right )} \right )} \] Input:

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

Output:

x*(-1 + exp(3)) - (-1 + E)*(1 + E + exp(2))*log(-x*log(x) - x + log(x/2 + 
log(log(2) + 2)))
 

Maxima [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.12 \[ \int \frac {1-2 x+x^2+e^3 \left (-1+2 x-x^2\right )+\left (-x+x^2+e^3 \left (x-x^2\right )\right ) \log (x)+\left (-4+e^3 (4-2 x)+2 x+\left (-2+e^3 (2-2 x)+2 x\right ) \log (x)\right ) \log (2+\log (2))+\left (-x+e^3 x+\left (-2+2 e^3\right ) \log (2+\log (2))\right ) \log \left (\frac {1}{2} (x+2 \log (2+\log (2)))\right )}{-x^2-x^2 \log (x)+(-2 x-2 x \log (x)) \log (2+\log (2))+(x+2 \log (2+\log (2))) \log \left (\frac {1}{2} (x+2 \log (2+\log (2)))\right )} \, dx=x {\left (e^{3} - 1\right )} - {\left (e^{3} - 1\right )} \log \left (-x \log \left (x\right ) - x - \log \left (2\right ) + \log \left (x + 2 \, \log \left (\log \left (2\right ) + 2\right )\right )\right ) \] Input:

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

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 153 vs. \(2 (28) = 56\).

Time = 0.16 (sec) , antiderivative size = 153, normalized size of antiderivative = 4.64 \[ \int \frac {1-2 x+x^2+e^3 \left (-1+2 x-x^2\right )+\left (-x+x^2+e^3 \left (x-x^2\right )\right ) \log (x)+\left (-4+e^3 (4-2 x)+2 x+\left (-2+e^3 (2-2 x)+2 x\right ) \log (x)\right ) \log (2+\log (2))+\left (-x+e^3 x+\left (-2+2 e^3\right ) \log (2+\log (2))\right ) \log \left (\frac {1}{2} (x+2 \log (2+\log (2)))\right )}{-x^2-x^2 \log (x)+(-2 x-2 x \log (x)) \log (2+\log (2))+(x+2 \log (2+\log (2))) \log \left (\frac {1}{2} (x+2 \log (2+\log (2)))\right )} \, dx={\left (x + 2 \, \log \left (\log \left (2\right ) + 2\right )\right )} e^{3} - e^{3} \log \left (-{\left (x + 2 \, \log \left (\log \left (2\right ) + 2\right )\right )} \log \left (2\right ) - {\left (x + 2 \, \log \left (\log \left (2\right ) + 2\right )\right )} \log \left (\frac {1}{2} \, x\right ) + 2 \, \log \left (2\right ) \log \left (\log \left (2\right ) + 2\right ) + 2 \, \log \left (\frac {1}{2} \, x\right ) \log \left (\log \left (2\right ) + 2\right ) - x + \log \left (\frac {1}{2} \, x + \log \left (\log \left (2\right ) + 2\right )\right )\right ) - x + \log \left (-{\left (x + 2 \, \log \left (\log \left (2\right ) + 2\right )\right )} \log \left (2\right ) - {\left (x + 2 \, \log \left (\log \left (2\right ) + 2\right )\right )} \log \left (\frac {1}{2} \, x\right ) + 2 \, \log \left (2\right ) \log \left (\log \left (2\right ) + 2\right ) + 2 \, \log \left (\frac {1}{2} \, x\right ) \log \left (\log \left (2\right ) + 2\right ) - x + \log \left (\frac {1}{2} \, x + \log \left (\log \left (2\right ) + 2\right )\right )\right ) - 2 \, \log \left (\log \left (2\right ) + 2\right ) \] Input:

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

Output:

(x + 2*log(log(2) + 2))*e^3 - e^3*log(-(x + 2*log(log(2) + 2))*log(2) - (x 
 + 2*log(log(2) + 2))*log(1/2*x) + 2*log(2)*log(log(2) + 2) + 2*log(1/2*x) 
*log(log(2) + 2) - x + log(1/2*x + log(log(2) + 2))) - x + log(-(x + 2*log 
(log(2) + 2))*log(2) - (x + 2*log(log(2) + 2))*log(1/2*x) + 2*log(2)*log(l 
og(2) + 2) + 2*log(1/2*x)*log(log(2) + 2) - x + log(1/2*x + log(log(2) + 2 
))) - 2*log(log(2) + 2)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1-2 x+x^2+e^3 \left (-1+2 x-x^2\right )+\left (-x+x^2+e^3 \left (x-x^2\right )\right ) \log (x)+\left (-4+e^3 (4-2 x)+2 x+\left (-2+e^3 (2-2 x)+2 x\right ) \log (x)\right ) \log (2+\log (2))+\left (-x+e^3 x+\left (-2+2 e^3\right ) \log (2+\log (2))\right ) \log \left (\frac {1}{2} (x+2 \log (2+\log (2)))\right )}{-x^2-x^2 \log (x)+(-2 x-2 x \log (x)) \log (2+\log (2))+(x+2 \log (2+\log (2))) \log \left (\frac {1}{2} (x+2 \log (2+\log (2)))\right )} \, dx=\int -\frac {\ln \left (x\right )\,\left ({\mathrm {e}}^3\,\left (x-x^2\right )-x+x^2\right )-2\,x-\ln \left (\ln \left (2\right )+2\right )\,\left (\ln \left (x\right )\,\left ({\mathrm {e}}^3\,\left (2\,x-2\right )-2\,x+2\right )-2\,x+{\mathrm {e}}^3\,\left (2\,x-4\right )+4\right )+\ln \left (\frac {x}{2}+\ln \left (\ln \left (2\right )+2\right )\right )\,\left (x\,{\mathrm {e}}^3-x+\ln \left (\ln \left (2\right )+2\right )\,\left (2\,{\mathrm {e}}^3-2\right )\right )-{\mathrm {e}}^3\,\left (x^2-2\,x+1\right )+x^2+1}{x^2\,\ln \left (x\right )+\ln \left (\ln \left (2\right )+2\right )\,\left (2\,x+2\,x\,\ln \left (x\right )\right )-\ln \left (\frac {x}{2}+\ln \left (\ln \left (2\right )+2\right )\right )\,\left (x+2\,\ln \left (\ln \left (2\right )+2\right )\right )+x^2} \,d x \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.64 \[ \int \frac {1-2 x+x^2+e^3 \left (-1+2 x-x^2\right )+\left (-x+x^2+e^3 \left (x-x^2\right )\right ) \log (x)+\left (-4+e^3 (4-2 x)+2 x+\left (-2+e^3 (2-2 x)+2 x\right ) \log (x)\right ) \log (2+\log (2))+\left (-x+e^3 x+\left (-2+2 e^3\right ) \log (2+\log (2))\right ) \log \left (\frac {1}{2} (x+2 \log (2+\log (2)))\right )}{-x^2-x^2 \log (x)+(-2 x-2 x \log (x)) \log (2+\log (2))+(x+2 \log (2+\log (2))) \log \left (\frac {1}{2} (x+2 \log (2+\log (2)))\right )} \, dx=-\mathrm {log}\left (\mathrm {log}\left (\mathrm {log}\left (\mathrm {log}\left (2\right )+2\right )+\frac {x}{2}\right )-\mathrm {log}\left (x \right ) x -x \right ) e^{3}+\mathrm {log}\left (\mathrm {log}\left (\mathrm {log}\left (\mathrm {log}\left (2\right )+2\right )+\frac {x}{2}\right )-\mathrm {log}\left (x \right ) x -x \right )+e^{3} x -x \] Input:

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

Output:

 - log(log((2*log(log(2) + 2) + x)/2) - log(x)*x - x)*e**3 + log(log((2*lo 
g(log(2) + 2) + x)/2) - log(x)*x - x) + e**3*x - x