\(\int \frac {e^x (-220-20 x+40 \log (2))+e^x (44+4 x) \log (2) \log (11+x)+(e^x (-220-240 x-20 x^2+(-44 x-48 x^2-4 x^3) \log (2))+e^x (44+48 x+4 x^2) \log (2) \log (11+x)) \log (\frac {(1+x) \log (2)}{-5-x \log (2)+\log (2) \log (11+x)})}{-55-60 x-5 x^2+(-11 x-12 x^2-x^3) \log (2)+(11+12 x+x^2) \log (2) \log (11+x)} \, dx\) [429]

Optimal result

Integrand size = 150, antiderivative size = 26 \[ \int \frac {e^x (-220-20 x+40 \log (2))+e^x (44+4 x) \log (2) \log (11+x)+\left (e^x \left (-220-240 x-20 x^2+\left (-44 x-48 x^2-4 x^3\right ) \log (2)\right )+e^x \left (44+48 x+4 x^2\right ) \log (2) \log (11+x)\right ) \log \left (\frac {(1+x) \log (2)}{-5-x \log (2)+\log (2) \log (11+x)}\right )}{-55-60 x-5 x^2+\left (-11 x-12 x^2-x^3\right ) \log (2)+\left (11+12 x+x^2\right ) \log (2) \log (11+x)} \, dx=4 e^x \log \left (\frac {1+x}{-x-\frac {5}{\log (2)}+\log (11+x)}\right ) \] Output:

4*ln((1+x)/(ln(11+x)-x-5/ln(2)))*exp(x)
 

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08 \[ \int \frac {e^x (-220-20 x+40 \log (2))+e^x (44+4 x) \log (2) \log (11+x)+\left (e^x \left (-220-240 x-20 x^2+\left (-44 x-48 x^2-4 x^3\right ) \log (2)\right )+e^x \left (44+48 x+4 x^2\right ) \log (2) \log (11+x)\right ) \log \left (\frac {(1+x) \log (2)}{-5-x \log (2)+\log (2) \log (11+x)}\right )}{-55-60 x-5 x^2+\left (-11 x-12 x^2-x^3\right ) \log (2)+\left (11+12 x+x^2\right ) \log (2) \log (11+x)} \, dx=4 e^x \log \left (\frac {(1+x) \log (2)}{-5-x \log (2)+\log (2) \log (11+x)}\right ) \] Input:

Integrate[(E^x*(-220 - 20*x + 40*Log[2]) + E^x*(44 + 4*x)*Log[2]*Log[11 + 
x] + (E^x*(-220 - 240*x - 20*x^2 + (-44*x - 48*x^2 - 4*x^3)*Log[2]) + E^x* 
(44 + 48*x + 4*x^2)*Log[2]*Log[11 + x])*Log[((1 + x)*Log[2])/(-5 - x*Log[2 
] + Log[2]*Log[11 + x])])/(-55 - 60*x - 5*x^2 + (-11*x - 12*x^2 - x^3)*Log 
[2] + (11 + 12*x + x^2)*Log[2]*Log[11 + x]),x]
 

Output:

4*E^x*Log[((1 + x)*Log[2])/(-5 - x*Log[2] + Log[2]*Log[11 + x])]
 

Rubi [A] (verified)

Time = 19.37 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.12, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.040, Rules used = {7292, 7279, 7239, 27, 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.

\(\displaystyle \int \frac {\left (e^x \left (4 x^2+48 x+44\right ) \log (2) \log (x+11)+e^x \left (-20 x^2+\left (-4 x^3-48 x^2-44 x\right ) \log (2)-240 x-220\right )\right ) \log \left (\frac {(x+1) \log (2)}{x (-\log (2))+\log (2) \log (x+11)-5}\right )+e^x (-20 x-220+40 \log (2))+e^x (4 x+44) \log (2) \log (x+11)}{-5 x^2+\left (x^2+12 x+11\right ) \log (2) \log (x+11)+\left (-x^3-12 x^2-11 x\right ) \log (2)-60 x-55} \, dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {-\left (e^x \left (4 x^2+48 x+44\right ) \log (2) \log (x+11)+e^x \left (-20 x^2+\left (-4 x^3-48 x^2-44 x\right ) \log (2)-240 x-220\right )\right ) \log \left (\frac {(x+1) \log (2)}{x (-\log (2))+\log (2) \log (x+11)-5}\right )-e^x (-20 x-220+40 \log (2))-e^x (4 x+44) \log (2) \log (x+11)}{\left (x^2+12 x+11\right ) (x \log (2)-\log (2) \log (x+11)+5)}dx\)

\(\Big \downarrow \) 7279

\(\displaystyle \int \left (\frac {2 e^x \left (x^3 \log (2) \log \left (\frac {(x+1) \log (2)}{x (-\log (2))+\log (2) \log (x+11)-5}\right )-x^2 \log (2) \log (x+11) \log \left (\frac {(x+1) \log (2)}{x (-\log (2))+\log (2) \log (x+11)-5}\right )+5 x^2 \left (1+\frac {12 \log (2)}{5}\right ) \log \left (\frac {(x+1) \log (2)}{x (-\log (2))+\log (2) \log (x+11)-5}\right )+5 x-x \log (2) \log (x+11)-12 x \log (2) \log (x+11) \log \left (\frac {(x+1) \log (2)}{x (-\log (2))+\log (2) \log (x+11)-5}\right )+60 x \left (1+\frac {11 \log (2)}{60}\right ) \log \left (\frac {(x+1) \log (2)}{x (-\log (2))+\log (2) \log (x+11)-5}\right )-11 \log (2) \log (x+11)-11 \log (2) \log (x+11) \log \left (\frac {(x+1) \log (2)}{x (-\log (2))+\log (2) \log (x+11)-5}\right )+55 \log \left (\frac {(x+1) \log (2)}{x (-\log (2))+\log (2) \log (x+11)-5}\right )+55 \left (1-\frac {2 \log (2)}{11}\right )\right )}{5 (x+1) (x \log (2)-\log (2) \log (x+11)+5)}+\frac {2 e^x \left (x^3 (-\log (2)) \log \left (\frac {(x+1) \log (2)}{x (-\log (2))+\log (2) \log (x+11)-5}\right )+x^2 \log (2) \log (x+11) \log \left (\frac {(x+1) \log (2)}{x (-\log (2))+\log (2) \log (x+11)-5}\right )-5 x^2 \left (1+\frac {12 \log (2)}{5}\right ) \log \left (\frac {(x+1) \log (2)}{x (-\log (2))+\log (2) \log (x+11)-5}\right )-5 x+x \log (2) \log (x+11)+12 x \log (2) \log (x+11) \log \left (\frac {(x+1) \log (2)}{x (-\log (2))+\log (2) \log (x+11)-5}\right )-60 x \left (1+\frac {11 \log (2)}{60}\right ) \log \left (\frac {(x+1) \log (2)}{x (-\log (2))+\log (2) \log (x+11)-5}\right )+11 \log (2) \log (x+11)+11 \log (2) \log (x+11) \log \left (\frac {(x+1) \log (2)}{x (-\log (2))+\log (2) \log (x+11)-5}\right )-55 \log \left (\frac {(x+1) \log (2)}{x (-\log (2))+\log (2) \log (x+11)-5}\right )-55 \left (1-\frac {2 \log (2)}{11}\right )\right )}{5 (x+11) (x \log (2)-\log (2) \log (x+11)+5)}\right )dx\)

\(\Big \downarrow \) 7239

\(\displaystyle \int \frac {4 e^x \left (\left (x^2+12 x+11\right ) (x \log (2)+5) \log \left (\frac {(x+1) \log (2)}{x (-\log (2))+\log (2) \log (x+11)-5}\right )+5 (x+11-\log (4))-(x+11) \log (2) \log (x+11) \left ((x+1) \log \left (\frac {(x+1) \log (2)}{x (-\log (2))+\log (2) \log (x+11)-5}\right )+1\right )\right )}{(x+1) (x+11) (x \log (2)-\log (2) \log (x+11)+5)}dx\)

\(\Big \downarrow \) 27

\(\displaystyle 4 \int \frac {e^x \left (5 (x-\log (4)+11)+\left (x^2+12 x+11\right ) (\log (2) x+5) \log \left (-\frac {(x+1) \log (2)}{\log (2) x-\log (2) \log (x+11)+5}\right )-(x+11) \log (2) \log (x+11) \left ((x+1) \log \left (-\frac {(x+1) \log (2)}{\log (2) x-\log (2) \log (x+11)+5}\right )+1\right )\right )}{(x+1) (x+11) (\log (2) x-\log (2) \log (x+11)+5)}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle 4 \int \left (\frac {e^x \left (-\log (2) \log (x+11) x+5 x-11 \log (2) \log (x+11)+55 \left (1-\frac {2 \log (2)}{11}\right )\right )}{(x+1) (x+11) (\log (2) x-\log (2) \log (x+11)+5)}+e^x \log \left (-\frac {(x+1) \log (2)}{\log (2) x-\log (2) \log (x+11)+5}\right )\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle 4 e^x \log \left (-\frac {(x+1) \log (2)}{x \log (2)-\log (2) \log (x+11)+5}\right )\)

Input:

Int[(E^x*(-220 - 20*x + 40*Log[2]) + E^x*(44 + 4*x)*Log[2]*Log[11 + x] + ( 
E^x*(-220 - 240*x - 20*x^2 + (-44*x - 48*x^2 - 4*x^3)*Log[2]) + E^x*(44 + 
48*x + 4*x^2)*Log[2]*Log[11 + x])*Log[((1 + x)*Log[2])/(-5 - x*Log[2] + Lo 
g[2]*Log[11 + x])])/(-55 - 60*x - 5*x^2 + (-11*x - 12*x^2 - x^3)*Log[2] + 
(11 + 12*x + x^2)*Log[2]*Log[11 + x]),x]
 

Output:

4*E^x*Log[-(((1 + x)*Log[2])/(5 + x*Log[2] - Log[2]*Log[11 + x]))]
 

Defintions of rubi rules used

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 7239

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl 
erIntegrandQ[v, u, x]]
 

rule 7279

Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[ 
{v = RationalFunctionExpand[u/(a + b*x^n + c*x^(2*n)), x]}, Int[v, x] /; Su 
mQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]
 

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 = 19.81 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08

method	result	size
parallelrisch	\(4 \,{\mathrm e}^{x} \ln \left (\frac {\left (1+x \right ) \ln \left (2\right )}{\ln \left (2\right ) \ln \left (11+x \right )-x \ln \left (2\right )-5}\right )\)	\(28\)
risch	\(-4 \,{\mathrm e}^{x} \ln \left (5+\left (x -\ln \left (11+x \right )\right ) \ln \left (2\right )\right )-4 i \pi \operatorname {csgn}\left (\frac {i \left (1+x \right )}{5+\left (x -\ln \left (11+x \right )\right ) \ln \left (2\right )}\right )^{2} {\mathrm e}^{x}+2 i \pi \,\operatorname {csgn}\left (\frac {i}{5+\left (x -\ln \left (11+x \right )\right ) \ln \left (2\right )}\right ) \operatorname {csgn}\left (\frac {i \left (1+x \right )}{5+\left (x -\ln \left (11+x \right )\right ) \ln \left (2\right )}\right )^{2} {\mathrm e}^{x}-2 i \pi \,\operatorname {csgn}\left (\frac {i}{5+\left (x -\ln \left (11+x \right )\right ) \ln \left (2\right )}\right ) \operatorname {csgn}\left (\frac {i \left (1+x \right )}{5+\left (x -\ln \left (11+x \right )\right ) \ln \left (2\right )}\right ) \operatorname {csgn}\left (i \left (1+x \right )\right ) {\mathrm e}^{x}+2 i \pi \operatorname {csgn}\left (\frac {i \left (1+x \right )}{5+\left (x -\ln \left (11+x \right )\right ) \ln \left (2\right )}\right )^{2} \operatorname {csgn}\left (i \left (1+x \right )\right ) {\mathrm e}^{x}+2 i \pi \operatorname {csgn}\left (\frac {i \left (1+x \right )}{5+\left (x -\ln \left (11+x \right )\right ) \ln \left (2\right )}\right )^{3} {\mathrm e}^{x}+4 i \pi \,{\mathrm e}^{x}+4 \,{\mathrm e}^{x} \ln \left (\ln \left (2\right )\right )+4 \,{\mathrm e}^{x} \ln \left (1+x \right )\)	\(241\)

Input:

int((((4*x^2+48*x+44)*ln(2)*exp(x)*ln(11+x)+((-4*x^3-48*x^2-44*x)*ln(2)-20 
*x^2-240*x-220)*exp(x))*ln((1+x)*ln(2)/(ln(2)*ln(11+x)-x*ln(2)-5))+(4*x+44 
)*ln(2)*exp(x)*ln(11+x)+(40*ln(2)-20*x-220)*exp(x))/((x^2+12*x+11)*ln(2)*l 
n(11+x)+(-x^3-12*x^2-11*x)*ln(2)-5*x^2-60*x-55),x,method=_RETURNVERBOSE)
 

Output:

4*exp(x)*ln((1+x)*ln(2)/(ln(2)*ln(11+x)-x*ln(2)-5))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08 \[ \int \frac {e^x (-220-20 x+40 \log (2))+e^x (44+4 x) \log (2) \log (11+x)+\left (e^x \left (-220-240 x-20 x^2+\left (-44 x-48 x^2-4 x^3\right ) \log (2)\right )+e^x \left (44+48 x+4 x^2\right ) \log (2) \log (11+x)\right ) \log \left (\frac {(1+x) \log (2)}{-5-x \log (2)+\log (2) \log (11+x)}\right )}{-55-60 x-5 x^2+\left (-11 x-12 x^2-x^3\right ) \log (2)+\left (11+12 x+x^2\right ) \log (2) \log (11+x)} \, dx=4 \, e^{x} \log \left (-\frac {{\left (x + 1\right )} \log \left (2\right )}{x \log \left (2\right ) - \log \left (2\right ) \log \left (x + 11\right ) + 5}\right ) \] Input:

integrate((((4*x^2+48*x+44)*log(2)*exp(x)*log(11+x)+((-4*x^3-48*x^2-44*x)* 
log(2)-20*x^2-240*x-220)*exp(x))*log((1+x)*log(2)/(log(2)*log(11+x)-x*log( 
2)-5))+(4*x+44)*log(2)*exp(x)*log(11+x)+(40*log(2)-20*x-220)*exp(x))/((x^2 
+12*x+11)*log(2)*log(11+x)+(-x^3-12*x^2-11*x)*log(2)-5*x^2-60*x-55),x, alg 
orithm="fricas")
 

Output:

4*e^x*log(-(x + 1)*log(2)/(x*log(2) - log(2)*log(x + 11) + 5))
 

Sympy [A] (verification not implemented)

Time = 13.79 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.04 \[ \int \frac {e^x (-220-20 x+40 \log (2))+e^x (44+4 x) \log (2) \log (11+x)+\left (e^x \left (-220-240 x-20 x^2+\left (-44 x-48 x^2-4 x^3\right ) \log (2)\right )+e^x \left (44+48 x+4 x^2\right ) \log (2) \log (11+x)\right ) \log \left (\frac {(1+x) \log (2)}{-5-x \log (2)+\log (2) \log (11+x)}\right )}{-55-60 x-5 x^2+\left (-11 x-12 x^2-x^3\right ) \log (2)+\left (11+12 x+x^2\right ) \log (2) \log (11+x)} \, dx=4 e^{x} \log {\left (\frac {\left (x + 1\right ) \log {\left (2 \right )}}{- x \log {\left (2 \right )} + \log {\left (2 \right )} \log {\left (x + 11 \right )} - 5} \right )} \] Input:

integrate((((4*x**2+48*x+44)*ln(2)*exp(x)*ln(11+x)+((-4*x**3-48*x**2-44*x) 
*ln(2)-20*x**2-240*x-220)*exp(x))*ln((1+x)*ln(2)/(ln(2)*ln(11+x)-x*ln(2)-5 
))+(4*x+44)*ln(2)*exp(x)*ln(11+x)+(40*ln(2)-20*x-220)*exp(x))/((x**2+12*x+ 
11)*ln(2)*ln(11+x)+(-x**3-12*x**2-11*x)*ln(2)-5*x**2-60*x-55),x)
 

Output:

4*exp(x)*log((x + 1)*log(2)/(-x*log(2) + log(2)*log(x + 11) - 5))
 

Maxima [A] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.35 \[ \int \frac {e^x (-220-20 x+40 \log (2))+e^x (44+4 x) \log (2) \log (11+x)+\left (e^x \left (-220-240 x-20 x^2+\left (-44 x-48 x^2-4 x^3\right ) \log (2)\right )+e^x \left (44+48 x+4 x^2\right ) \log (2) \log (11+x)\right ) \log \left (\frac {(1+x) \log (2)}{-5-x \log (2)+\log (2) \log (11+x)}\right )}{-55-60 x-5 x^2+\left (-11 x-12 x^2-x^3\right ) \log (2)+\left (11+12 x+x^2\right ) \log (2) \log (11+x)} \, dx=-4 \, e^{x} \log \left (-x \log \left (2\right ) + \log \left (2\right ) \log \left (x + 11\right ) - 5\right ) + 4 \, e^{x} \log \left (x + 1\right ) + 4 \, e^{x} \log \left (\log \left (2\right )\right ) \] Input:

integrate((((4*x^2+48*x+44)*log(2)*exp(x)*log(11+x)+((-4*x^3-48*x^2-44*x)* 
log(2)-20*x^2-240*x-220)*exp(x))*log((1+x)*log(2)/(log(2)*log(11+x)-x*log( 
2)-5))+(4*x+44)*log(2)*exp(x)*log(11+x)+(40*log(2)-20*x-220)*exp(x))/((x^2 
+12*x+11)*log(2)*log(11+x)+(-x^3-12*x^2-11*x)*log(2)-5*x^2-60*x-55),x, alg 
orithm="maxima")
 

Output:

-4*e^x*log(-x*log(2) + log(2)*log(x + 11) - 5) + 4*e^x*log(x + 1) + 4*e^x* 
log(log(2))
 

Giac [A] (verification not implemented)

Time = 0.35 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.35 \[ \int \frac {e^x (-220-20 x+40 \log (2))+e^x (44+4 x) \log (2) \log (11+x)+\left (e^x \left (-220-240 x-20 x^2+\left (-44 x-48 x^2-4 x^3\right ) \log (2)\right )+e^x \left (44+48 x+4 x^2\right ) \log (2) \log (11+x)\right ) \log \left (\frac {(1+x) \log (2)}{-5-x \log (2)+\log (2) \log (11+x)}\right )}{-55-60 x-5 x^2+\left (-11 x-12 x^2-x^3\right ) \log (2)+\left (11+12 x+x^2\right ) \log (2) \log (11+x)} \, dx=-4 \, e^{x} \log \left (x \log \left (2\right ) - \log \left (2\right ) \log \left (x + 11\right ) + 5\right ) + 4 \, e^{x} \log \left (-x \log \left (2\right ) - \log \left (2\right )\right ) \] Input:

integrate((((4*x^2+48*x+44)*log(2)*exp(x)*log(11+x)+((-4*x^3-48*x^2-44*x)* 
log(2)-20*x^2-240*x-220)*exp(x))*log((1+x)*log(2)/(log(2)*log(11+x)-x*log( 
2)-5))+(4*x+44)*log(2)*exp(x)*log(11+x)+(40*log(2)-20*x-220)*exp(x))/((x^2 
+12*x+11)*log(2)*log(11+x)+(-x^3-12*x^2-11*x)*log(2)-5*x^2-60*x-55),x, alg 
orithm="giac")
 

Output:

-4*e^x*log(x*log(2) - log(2)*log(x + 11) + 5) + 4*e^x*log(-x*log(2) - log( 
2))
 

Mupad [B] (verification not implemented)

Time = 3.22 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08 \[ \int \frac {e^x (-220-20 x+40 \log (2))+e^x (44+4 x) \log (2) \log (11+x)+\left (e^x \left (-220-240 x-20 x^2+\left (-44 x-48 x^2-4 x^3\right ) \log (2)\right )+e^x \left (44+48 x+4 x^2\right ) \log (2) \log (11+x)\right ) \log \left (\frac {(1+x) \log (2)}{-5-x \log (2)+\log (2) \log (11+x)}\right )}{-55-60 x-5 x^2+\left (-11 x-12 x^2-x^3\right ) \log (2)+\left (11+12 x+x^2\right ) \log (2) \log (11+x)} \, dx=4\,\ln \left (-\frac {\ln \left (2\right )\,\left (x+1\right )}{x\,\ln \left (2\right )-\ln \left (x+11\right )\,\ln \left (2\right )+5}\right )\,{\mathrm {e}}^x \] Input:

int((exp(x)*(20*x - 40*log(2) + 220) + log(-(log(2)*(x + 1))/(x*log(2) - l 
og(x + 11)*log(2) + 5))*(exp(x)*(240*x + log(2)*(44*x + 48*x^2 + 4*x^3) + 
20*x^2 + 220) - log(x + 11)*exp(x)*log(2)*(48*x + 4*x^2 + 44)) - log(x + 1 
1)*exp(x)*log(2)*(4*x + 44))/(60*x + log(2)*(11*x + 12*x^2 + x^3) + 5*x^2 
- log(x + 11)*log(2)*(12*x + x^2 + 11) + 55),x)
 

Output:

4*log(-(log(2)*(x + 1))/(x*log(2) - log(x + 11)*log(2) + 5))*exp(x)
 

Reduce [B] (verification not implemented)

Time = 0.51 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.15 \[ \int \frac {e^x (-220-20 x+40 \log (2))+e^x (44+4 x) \log (2) \log (11+x)+\left (e^x \left (-220-240 x-20 x^2+\left (-44 x-48 x^2-4 x^3\right ) \log (2)\right )+e^x \left (44+48 x+4 x^2\right ) \log (2) \log (11+x)\right ) \log \left (\frac {(1+x) \log (2)}{-5-x \log (2)+\log (2) \log (11+x)}\right )}{-55-60 x-5 x^2+\left (-11 x-12 x^2-x^3\right ) \log (2)+\left (11+12 x+x^2\right ) \log (2) \log (11+x)} \, dx=4 e^{x} \mathrm {log}\left (\frac {\mathrm {log}\left (2\right ) x +\mathrm {log}\left (2\right )}{\mathrm {log}\left (x +11\right ) \mathrm {log}\left (2\right )-\mathrm {log}\left (2\right ) x -5}\right ) \] Input:

int((((4*x^2+48*x+44)*log(2)*exp(x)*log(11+x)+((-4*x^3-48*x^2-44*x)*log(2) 
-20*x^2-240*x-220)*exp(x))*log((1+x)*log(2)/(log(2)*log(11+x)-x*log(2)-5)) 
+(4*x+44)*log(2)*exp(x)*log(11+x)+(40*log(2)-20*x-220)*exp(x))/((x^2+12*x+ 
11)*log(2)*log(11+x)+(-x^3-12*x^2-11*x)*log(2)-5*x^2-60*x-55),x)
 

Output:

4*e**x*log((log(2)*x + log(2))/(log(x + 11)*log(2) - log(2)*x - 5))