\(\int \frac {(2+e^x+x)^{\frac {1}{2 \log (\frac {1}{4} (-12+x))}} ((-12+e^x (-12+x)+x) \log (\frac {1}{4} (-12+x))+(-2-e^x-x) \log (2+e^x+x))}{(-48-20 x+2 x^2+e^x (-24+2 x)) \log ^2(\frac {1}{4} (-12+x))} \, dx\) [1866]

Optimal result

Integrand size = 91, antiderivative size = 21 \[ \int \frac {\left (2+e^x+x\right )^{\frac {1}{2 \log \left (\frac {1}{4} (-12+x)\right )}} \left (\left (-12+e^x (-12+x)+x\right ) \log \left (\frac {1}{4} (-12+x)\right )+\left (-2-e^x-x\right ) \log \left (2+e^x+x\right )\right )}{\left (-48-20 x+2 x^2+e^x (-24+2 x)\right ) \log ^2\left (\frac {1}{4} (-12+x)\right )} \, dx=\left (2+e^x+x\right )^{\frac {1}{2 \log \left (-3+\frac {x}{4}\right )}} \] Output:

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

Mathematica [A] (verified)

Time = 0.42 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int \frac {\left (2+e^x+x\right )^{\frac {1}{2 \log \left (\frac {1}{4} (-12+x)\right )}} \left (\left (-12+e^x (-12+x)+x\right ) \log \left (\frac {1}{4} (-12+x)\right )+\left (-2-e^x-x\right ) \log \left (2+e^x+x\right )\right )}{\left (-48-20 x+2 x^2+e^x (-24+2 x)\right ) \log ^2\left (\frac {1}{4} (-12+x)\right )} \, dx=\left (2+e^x+x\right )^{\frac {1}{2 \log \left (-3+\frac {x}{4}\right )}} \] Input:

Integrate[((2 + E^x + x)^(1/(2*Log[(-12 + x)/4]))*((-12 + E^x*(-12 + x) + 
x)*Log[(-12 + x)/4] + (-2 - E^x - x)*Log[2 + E^x + x]))/((-48 - 20*x + 2*x 
^2 + E^x*(-24 + 2*x))*Log[(-12 + x)/4]^2),x]
 

Output:

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

\(\displaystyle \int \frac {\left (x+e^x+2\right )^{\frac {1}{2 \log \left (\frac {x-12}{4}\right )}} \left (\left (e^x (x-12)+x-12\right ) \log \left (\frac {x-12}{4}\right )+\left (-x-e^x-2\right ) \log \left (x+e^x+2\right )\right )}{\left (2 x^2-20 x+e^x (2 x-24)-48\right ) \log ^2\left (\frac {x-12}{4}\right )} \, dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {\left (x+e^x+2\right )^{\frac {1}{2 \log \left (\frac {x}{4}-3\right )}-1} \left (-\left (\left (e^x (x-12)+x-12\right ) \log \left (\frac {x-12}{4}\right )\right )-\left (-x-e^x-2\right ) \log \left (x+e^x+2\right )\right )}{2 (12-x) \log ^2\left (\frac {x}{4}-3\right )}dx\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{2} \int \frac {\left (x+e^x+2\right )^{\frac {1}{2 \log \left (\frac {x}{4}-3\right )}-1} \left (\left (e^x (12-x)-x+12\right ) \log \left (\frac {x-12}{4}\right )+\left (x+e^x+2\right ) \log \left (x+e^x+2\right )\right )}{(12-x) \log ^2\left (\frac {x}{4}-3\right )}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \frac {1}{2} \int \left (\frac {e^x \left (x \log \left (\frac {x}{4}-3\right )-12 \log \left (\frac {x}{4}-3\right )-\log \left (x+e^x+2\right )\right ) \left (x+e^x+2\right )^{\frac {1}{2 \log \left (\frac {x}{4}-3\right )}-1}}{(x-12) \log ^2\left (\frac {x}{4}-3\right )}+\frac {\left (x \log \left (\frac {x}{4}-3\right )-12 \log \left (\frac {x}{4}-3\right )-x \log \left (x+e^x+2\right )-2 \log \left (x+e^x+2\right )\right ) \left (x+e^x+2\right )^{\frac {1}{2 \log \left (\frac {x}{4}-3\right )}-1}}{(x-12) \log ^2\left (\frac {x}{4}-3\right )}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{2} \left (-\int \frac {\left (x+e^x+2\right )^{\frac {1}{2 \log \left (\frac {x}{4}-3\right )}-1} \log \left (x+e^x+2\right )}{\log ^2\left (\frac {x}{4}-3\right )}dx-14 \int \frac {\left (x+e^x+2\right )^{\frac {1}{2 \log \left (\frac {x}{4}-3\right )}-1} \log \left (x+e^x+2\right )}{(x-12) \log ^2\left (\frac {x}{4}-3\right )}dx-\int \frac {e^x \left (x+e^x+2\right )^{\frac {1}{2 \log \left (\frac {x}{4}-3\right )}-1} \log \left (x+e^x+2\right )}{(x-12) \log ^2\left (\frac {x}{4}-3\right )}dx+\int \frac {\left (x+e^x+2\right )^{\frac {1}{2 \log \left (\frac {x}{4}-3\right )}-1}}{\log \left (\frac {x}{4}-3\right )}dx+\int \frac {e^x \left (x+e^x+2\right )^{\frac {1}{2 \log \left (\frac {x}{4}-3\right )}-1}}{\log \left (\frac {x}{4}-3\right )}dx\right )\)

Input:

Int[((2 + E^x + x)^(1/(2*Log[(-12 + x)/4]))*((-12 + E^x*(-12 + x) + x)*Log 
[(-12 + x)/4] + (-2 - E^x - x)*Log[2 + E^x + x]))/((-48 - 20*x + 2*x^2 + E 
^x*(-24 + 2*x))*Log[(-12 + x)/4]^2),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 25.04 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.86

method	result	size
parallelrisch	\({\mathrm e}^{\frac {\ln \left ({\mathrm e}^{x}+2+x \right )}{2 \ln \left (\frac {x}{4}-3\right )}}\)	\(18\)
risch	\(\left ({\mathrm e}^{x}+2+x \right )^{-\frac {1}{2 \left (2 \ln \left (2\right )-\ln \left (x -12\right )\right )}}\)	\(22\)

Input:

int(((-exp(x)-x-2)*ln(exp(x)+2+x)+((x-12)*exp(x)+x-12)*ln(1/4*x-3))*exp(1/ 
2*ln(exp(x)+2+x)/ln(1/4*x-3))/((2*x-24)*exp(x)+2*x^2-20*x-48)/ln(1/4*x-3)^ 
2,x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.76 \[ \int \frac {\left (2+e^x+x\right )^{\frac {1}{2 \log \left (\frac {1}{4} (-12+x)\right )}} \left (\left (-12+e^x (-12+x)+x\right ) \log \left (\frac {1}{4} (-12+x)\right )+\left (-2-e^x-x\right ) \log \left (2+e^x+x\right )\right )}{\left (-48-20 x+2 x^2+e^x (-24+2 x)\right ) \log ^2\left (\frac {1}{4} (-12+x)\right )} \, dx={\left (x + e^{x} + 2\right )}^{\frac {1}{2 \, \log \left (\frac {1}{4} \, x - 3\right )}} \] Input:

integrate(((-exp(x)-x-2)*log(exp(x)+2+x)+((x-12)*exp(x)+x-12)*log(1/4*x-3) 
)*exp(1/2*log(exp(x)+2+x)/log(1/4*x-3))/((2*x-24)*exp(x)+2*x^2-20*x-48)/lo 
g(1/4*x-3)^2,x, algorithm="fricas")
 

Output:

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

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (2+e^x+x\right )^{\frac {1}{2 \log \left (\frac {1}{4} (-12+x)\right )}} \left (\left (-12+e^x (-12+x)+x\right ) \log \left (\frac {1}{4} (-12+x)\right )+\left (-2-e^x-x\right ) \log \left (2+e^x+x\right )\right )}{\left (-48-20 x+2 x^2+e^x (-24+2 x)\right ) \log ^2\left (\frac {1}{4} (-12+x)\right )} \, dx=\text {Timed out} \] Input:

integrate(((-exp(x)-x-2)*ln(exp(x)+2+x)+((x-12)*exp(x)+x-12)*ln(1/4*x-3))* 
exp(1/2*ln(exp(x)+2+x)/ln(1/4*x-3))/((2*x-24)*exp(x)+2*x**2-20*x-48)/ln(1/ 
4*x-3)**2,x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.10 \[ \int \frac {\left (2+e^x+x\right )^{\frac {1}{2 \log \left (\frac {1}{4} (-12+x)\right )}} \left (\left (-12+e^x (-12+x)+x\right ) \log \left (\frac {1}{4} (-12+x)\right )+\left (-2-e^x-x\right ) \log \left (2+e^x+x\right )\right )}{\left (-48-20 x+2 x^2+e^x (-24+2 x)\right ) \log ^2\left (\frac {1}{4} (-12+x)\right )} \, dx=\frac {1}{{\left (x + e^{x} + 2\right )}^{\frac {1}{2 \, {\left (2 \, \log \left (2\right ) - \log \left (x - 12\right )\right )}}}} \] Input:

integrate(((-exp(x)-x-2)*log(exp(x)+2+x)+((x-12)*exp(x)+x-12)*log(1/4*x-3) 
)*exp(1/2*log(exp(x)+2+x)/log(1/4*x-3))/((2*x-24)*exp(x)+2*x^2-20*x-48)/lo 
g(1/4*x-3)^2,x, algorithm="maxima")
 

Output:

1/((x + e^x + 2)^(1/2/(2*log(2) - log(x - 12))))
 

Giac [F]

\[ \int \frac {\left (2+e^x+x\right )^{\frac {1}{2 \log \left (\frac {1}{4} (-12+x)\right )}} \left (\left (-12+e^x (-12+x)+x\right ) \log \left (\frac {1}{4} (-12+x)\right )+\left (-2-e^x-x\right ) \log \left (2+e^x+x\right )\right )}{\left (-48-20 x+2 x^2+e^x (-24+2 x)\right ) \log ^2\left (\frac {1}{4} (-12+x)\right )} \, dx=\int { -\frac {{\left ({\left (x + e^{x} + 2\right )} \log \left (x + e^{x} + 2\right ) - {\left ({\left (x - 12\right )} e^{x} + x - 12\right )} \log \left (\frac {1}{4} \, x - 3\right )\right )} {\left (x + e^{x} + 2\right )}^{\frac {1}{2 \, \log \left (\frac {1}{4} \, x - 3\right )}}}{2 \, {\left (x^{2} + {\left (x - 12\right )} e^{x} - 10 \, x - 24\right )} \log \left (\frac {1}{4} \, x - 3\right )^{2}} \,d x } \] Input:

integrate(((-exp(x)-x-2)*log(exp(x)+2+x)+((x-12)*exp(x)+x-12)*log(1/4*x-3) 
)*exp(1/2*log(exp(x)+2+x)/log(1/4*x-3))/((2*x-24)*exp(x)+2*x^2-20*x-48)/lo 
g(1/4*x-3)^2,x, algorithm="giac")
 

Output:

integrate(-1/2*((x + e^x + 2)*log(x + e^x + 2) - ((x - 12)*e^x + x - 12)*l 
og(1/4*x - 3))*(x + e^x + 2)^(1/2/log(1/4*x - 3))/((x^2 + (x - 12)*e^x - 1 
0*x - 24)*log(1/4*x - 3)^2), x)
 

Mupad [B] (verification not implemented)

Time = 3.95 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81 \[ \int \frac {\left (2+e^x+x\right )^{\frac {1}{2 \log \left (\frac {1}{4} (-12+x)\right )}} \left (\left (-12+e^x (-12+x)+x\right ) \log \left (\frac {1}{4} (-12+x)\right )+\left (-2-e^x-x\right ) \log \left (2+e^x+x\right )\right )}{\left (-48-20 x+2 x^2+e^x (-24+2 x)\right ) \log ^2\left (\frac {1}{4} (-12+x)\right )} \, dx={\mathrm {e}}^{\frac {\ln \left (x+{\mathrm {e}}^x+2\right )}{2\,\ln \left (\frac {x}{4}-3\right )}} \] Input:

int((exp(log(x + exp(x) + 2)/(2*log(x/4 - 3)))*(log(x + exp(x) + 2)*(x + e 
xp(x) + 2) - log(x/4 - 3)*(x + exp(x)*(x - 12) - 12)))/(log(x/4 - 3)^2*(20 
*x - exp(x)*(2*x - 24) - 2*x^2 + 48)),x)
 

Output:

exp(log(x + exp(x) + 2)/(2*log(x/4 - 3)))
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {\left (2+e^x+x\right )^{\frac {1}{2 \log \left (\frac {1}{4} (-12+x)\right )}} \left (\left (-12+e^x (-12+x)+x\right ) \log \left (\frac {1}{4} (-12+x)\right )+\left (-2-e^x-x\right ) \log \left (2+e^x+x\right )\right )}{\left (-48-20 x+2 x^2+e^x (-24+2 x)\right ) \log ^2\left (\frac {1}{4} (-12+x)\right )} \, dx=e^{\frac {\mathrm {log}\left (e^{x}+x +2\right )}{2 \,\mathrm {log}\left (\frac {x}{4}-3\right )}} \] Input:

int(((-exp(x)-x-2)*log(exp(x)+2+x)+((x-12)*exp(x)+x-12)*log(1/4*x-3))*exp( 
1/2*log(exp(x)+2+x)/log(1/4*x-3))/((2*x-24)*exp(x)+2*x^2-20*x-48)/log(1/4* 
x-3)^2,x)
 

Output:

e**(log(e**x + x + 2)/(2*log((x - 12)/4)))