\(\int \frac {12+e^{1-x} (-12-4 x)+4 x+(-12-4 e^{1-x}-4 x+4 \log (2)) \log (3+e^{1-x}+x-\log (2))}{27+27 x+9 x^2+x^3+e^{1-x} (9+6 x+x^2)+(-9-6 x-x^2) \log (2)+(-72+e^{1-x} (-24-8 x)-48 x-8 x^2+(24+8 x) \log (2)) \log (3+e^{1-x}+x-\log (2))+(48+16 e^{1-x}+16 x-16 \log (2)) \log ^2(3+e^{1-x}+x-\log (2))} \, dx\) [1302]

Optimal result

Integrand size = 180, antiderivative size = 30 \[ \int \frac {12+e^{1-x} (-12-4 x)+4 x+\left (-12-4 e^{1-x}-4 x+4 \log (2)\right ) \log \left (3+e^{1-x}+x-\log (2)\right )}{27+27 x+9 x^2+x^3+e^{1-x} \left (9+6 x+x^2\right )+\left (-9-6 x-x^2\right ) \log (2)+\left (-72+e^{1-x} (-24-8 x)-48 x-8 x^2+(24+8 x) \log (2)\right ) \log \left (3+e^{1-x}+x-\log (2)\right )+\left (48+16 e^{1-x}+16 x-16 \log (2)\right ) \log ^2\left (3+e^{1-x}+x-\log (2)\right )} \, dx=\frac {2}{-2+\frac {3+x}{2 \log \left (3+e^{1-x}+x-\log (2)\right )}} \] Output:

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

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.87 \[ \int \frac {12+e^{1-x} (-12-4 x)+4 x+\left (-12-4 e^{1-x}-4 x+4 \log (2)\right ) \log \left (3+e^{1-x}+x-\log (2)\right )}{27+27 x+9 x^2+x^3+e^{1-x} \left (9+6 x+x^2\right )+\left (-9-6 x-x^2\right ) \log (2)+\left (-72+e^{1-x} (-24-8 x)-48 x-8 x^2+(24+8 x) \log (2)\right ) \log \left (3+e^{1-x}+x-\log (2)\right )+\left (48+16 e^{1-x}+16 x-16 \log (2)\right ) \log ^2\left (3+e^{1-x}+x-\log (2)\right )} \, dx=\frac {3+x}{3+x-4 \log \left (3+e^{1-x}+x-\log (2)\right )} \] Input:

Integrate[(12 + E^(1 - x)*(-12 - 4*x) + 4*x + (-12 - 4*E^(1 - x) - 4*x + 4 
*Log[2])*Log[3 + E^(1 - x) + x - Log[2]])/(27 + 27*x + 9*x^2 + x^3 + E^(1 
- x)*(9 + 6*x + x^2) + (-9 - 6*x - x^2)*Log[2] + (-72 + E^(1 - x)*(-24 - 8 
*x) - 48*x - 8*x^2 + (24 + 8*x)*Log[2])*Log[3 + E^(1 - x) + x - Log[2]] + 
(48 + 16*E^(1 - x) + 16*x - 16*Log[2])*Log[3 + E^(1 - x) + x - Log[2]]^2), 
x]
 

Output:

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

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.

\(\displaystyle \int \frac {e^{1-x} (-4 x-12)+4 x+\left (-4 x-4 e^{1-x}-12+4 \log (2)\right ) \log \left (x+e^{1-x}+3-\log (2)\right )+12}{x^3+9 x^2+e^{1-x} \left (x^2+6 x+9\right )+\left (-8 x^2-48 x+e^{1-x} (-8 x-24)+(8 x+24) \log (2)-72\right ) \log \left (x+e^{1-x}+3-\log (2)\right )+\left (-x^2-6 x-9\right ) \log (2)+27 x+\left (16 x+16 e^{1-x}+48-16 \log (2)\right ) \log ^2\left (x+e^{1-x}+3-\log (2)\right )+27} \, dx\)

\(\Big \downarrow \) 7239

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

\(\Big \downarrow \) 7293

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

\(\Big \downarrow \) 2009

\(\displaystyle \int \frac {1}{\left (x-4 \log \left (x+e^{1-x}-\log (2)+3\right )+3\right )^2}dx-\int \frac {x}{\left (x-4 \log \left (x+e^{1-x}-\log (2)+3\right )+3\right )^2}dx+16 e \int \frac {1}{\left (-e^x x-3 e^x \left (1-\frac {\log (2)}{3}\right )-e\right ) \left (x-4 \log \left (x+e^{1-x}-\log (2)+3\right )+3\right )^2}dx+4 e \int \frac {x}{\left (-e^x x-3 e^x \left (1-\frac {\log (2)}{3}\right )-e\right ) \left (x-4 \log \left (x+e^{1-x}-\log (2)+3\right )+3\right )^2}dx+4 \log (2) \int \frac {1}{(x-\log (2)+3) \left (x-4 \log \left (x+e^{1-x}-\log (2)+3\right )+3\right )^2}dx+4 e \log (2) \int \frac {1}{\left (-e^x x-3 e^x \left (1-\frac {\log (2)}{3}\right )-e\right ) (x-\log (2)+3) \left (x-4 \log \left (x+e^{1-x}-\log (2)+3\right )+3\right )^2}dx+\int \frac {1}{x-4 \log \left (x+e^{1-x}-\log (2)+3\right )+3}dx\)

Input:

Int[(12 + E^(1 - x)*(-12 - 4*x) + 4*x + (-12 - 4*E^(1 - x) - 4*x + 4*Log[2 
])*Log[3 + E^(1 - x) + x - Log[2]])/(27 + 27*x + 9*x^2 + x^3 + E^(1 - x)*( 
9 + 6*x + x^2) + (-9 - 6*x - x^2)*Log[2] + (-72 + E^(1 - x)*(-24 - 8*x) - 
48*x - 8*x^2 + (24 + 8*x)*Log[2])*Log[3 + E^(1 - x) + x - Log[2]] + (48 + 
16*E^(1 - x) + 16*x - 16*Log[2])*Log[3 + E^(1 - x) + x - Log[2]]^2),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 0.34 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.87

method	result	size
risch	\(\frac {3+x}{x -4 \ln \left ({\mathrm e}^{1-x}-\ln \left (2\right )+3+x \right )+3}\)	\(26\)
parallelrisch	\(\frac {4 x +12}{4 x -16 \ln \left ({\mathrm e}^{1-x}-\ln \left (2\right )+3+x \right )+12}\)	\(29\)

Input:

int(((-4*exp(1-x)+4*ln(2)-4*x-12)*ln(exp(1-x)-ln(2)+3+x)+(-4*x-12)*exp(1-x 
)+4*x+12)/((16*exp(1-x)-16*ln(2)+16*x+48)*ln(exp(1-x)-ln(2)+3+x)^2+((-8*x- 
24)*exp(1-x)+(8*x+24)*ln(2)-8*x^2-48*x-72)*ln(exp(1-x)-ln(2)+3+x)+(x^2+6*x 
+9)*exp(1-x)+(-x^2-6*x-9)*ln(2)+x^3+9*x^2+27*x+27),x,method=_RETURNVERBOSE 
)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.83 \[ \int \frac {12+e^{1-x} (-12-4 x)+4 x+\left (-12-4 e^{1-x}-4 x+4 \log (2)\right ) \log \left (3+e^{1-x}+x-\log (2)\right )}{27+27 x+9 x^2+x^3+e^{1-x} \left (9+6 x+x^2\right )+\left (-9-6 x-x^2\right ) \log (2)+\left (-72+e^{1-x} (-24-8 x)-48 x-8 x^2+(24+8 x) \log (2)\right ) \log \left (3+e^{1-x}+x-\log (2)\right )+\left (48+16 e^{1-x}+16 x-16 \log (2)\right ) \log ^2\left (3+e^{1-x}+x-\log (2)\right )} \, dx=\frac {x + 3}{x - 4 \, \log \left (x + e^{\left (-x + 1\right )} - \log \left (2\right ) + 3\right ) + 3} \] Input:

integrate(((-4*exp(1-x)+4*log(2)-4*x-12)*log(exp(1-x)-log(2)+3+x)+(-4*x-12 
)*exp(1-x)+4*x+12)/((16*exp(1-x)-16*log(2)+16*x+48)*log(exp(1-x)-log(2)+3+ 
x)^2+((-8*x-24)*exp(1-x)+(8*x+24)*log(2)-8*x^2-48*x-72)*log(exp(1-x)-log(2 
)+3+x)+(x^2+6*x+9)*exp(1-x)+(-x^2-6*x-9)*log(2)+x^3+9*x^2+27*x+27),x, algo 
rithm="fricas")
 

Output:

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

Sympy [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.73 \[ \int \frac {12+e^{1-x} (-12-4 x)+4 x+\left (-12-4 e^{1-x}-4 x+4 \log (2)\right ) \log \left (3+e^{1-x}+x-\log (2)\right )}{27+27 x+9 x^2+x^3+e^{1-x} \left (9+6 x+x^2\right )+\left (-9-6 x-x^2\right ) \log (2)+\left (-72+e^{1-x} (-24-8 x)-48 x-8 x^2+(24+8 x) \log (2)\right ) \log \left (3+e^{1-x}+x-\log (2)\right )+\left (48+16 e^{1-x}+16 x-16 \log (2)\right ) \log ^2\left (3+e^{1-x}+x-\log (2)\right )} \, dx=\frac {- x - 3}{- x + 4 \log {\left (x + e^{1 - x} - \log {\left (2 \right )} + 3 \right )} - 3} \] Input:

integrate(((-4*exp(1-x)+4*ln(2)-4*x-12)*ln(exp(1-x)-ln(2)+3+x)+(-4*x-12)*e 
xp(1-x)+4*x+12)/((16*exp(1-x)-16*ln(2)+16*x+48)*ln(exp(1-x)-ln(2)+3+x)**2+ 
((-8*x-24)*exp(1-x)+(8*x+24)*ln(2)-8*x**2-48*x-72)*ln(exp(1-x)-ln(2)+3+x)+ 
(x**2+6*x+9)*exp(1-x)+(-x**2-6*x-9)*ln(2)+x**3+9*x**2+27*x+27),x)
 

Output:

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

Maxima [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.90 \[ \int \frac {12+e^{1-x} (-12-4 x)+4 x+\left (-12-4 e^{1-x}-4 x+4 \log (2)\right ) \log \left (3+e^{1-x}+x-\log (2)\right )}{27+27 x+9 x^2+x^3+e^{1-x} \left (9+6 x+x^2\right )+\left (-9-6 x-x^2\right ) \log (2)+\left (-72+e^{1-x} (-24-8 x)-48 x-8 x^2+(24+8 x) \log (2)\right ) \log \left (3+e^{1-x}+x-\log (2)\right )+\left (48+16 e^{1-x}+16 x-16 \log (2)\right ) \log ^2\left (3+e^{1-x}+x-\log (2)\right )} \, dx=\frac {x + 3}{5 \, x - 4 \, \log \left ({\left (x - \log \left (2\right ) + 3\right )} e^{x} + e\right ) + 3} \] Input:

integrate(((-4*exp(1-x)+4*log(2)-4*x-12)*log(exp(1-x)-log(2)+3+x)+(-4*x-12 
)*exp(1-x)+4*x+12)/((16*exp(1-x)-16*log(2)+16*x+48)*log(exp(1-x)-log(2)+3+ 
x)^2+((-8*x-24)*exp(1-x)+(8*x+24)*log(2)-8*x^2-48*x-72)*log(exp(1-x)-log(2 
)+3+x)+(x^2+6*x+9)*exp(1-x)+(-x^2-6*x-9)*log(2)+x^3+9*x^2+27*x+27),x, algo 
rithm="maxima")
 

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 298 vs. \(2 (26) = 52\).

Time = 0.34 (sec) , antiderivative size = 298, normalized size of antiderivative = 9.93 \[ \int \frac {12+e^{1-x} (-12-4 x)+4 x+\left (-12-4 e^{1-x}-4 x+4 \log (2)\right ) \log \left (3+e^{1-x}+x-\log (2)\right )}{27+27 x+9 x^2+x^3+e^{1-x} \left (9+6 x+x^2\right )+\left (-9-6 x-x^2\right ) \log (2)+\left (-72+e^{1-x} (-24-8 x)-48 x-8 x^2+(24+8 x) \log (2)\right ) \log \left (3+e^{1-x}+x-\log (2)\right )+\left (48+16 e^{1-x}+16 x-16 \log (2)\right ) \log ^2\left (3+e^{1-x}+x-\log (2)\right )} \, dx=\frac {4 \, {\left (x e^{x} \log \left ({\left (x e^{x} - e^{x} \log \left (2\right ) + e + 3 \, e^{x}\right )} e^{\left (-x\right )}\right ) - e^{x} \log \left (2\right ) \log \left ({\left (x e^{x} - e^{x} \log \left (2\right ) + e + 3 \, e^{x}\right )} e^{\left (-x\right )}\right ) - 4 \, x e + 4 \, x e^{x} + e \log \left ({\left (x e^{x} - e^{x} \log \left (2\right ) + e + 3 \, e^{x}\right )} e^{\left (-x\right )}\right ) + 3 \, e^{x} \log \left ({\left (x e^{x} - e^{x} \log \left (2\right ) + e + 3 \, e^{x}\right )} e^{\left (-x\right )}\right ) + 4 \, e \log \left (x e^{x} - e^{x} \log \left (2\right ) + e + 3 \, e^{x}\right ) - 4 \, e^{x} \log \left (x e^{x} - e^{x} \log \left (2\right ) + e + 3 \, e^{x}\right )\right )}}{5 \, x^{2} e^{x} - 5 \, x e^{x} \log \left (2\right ) - 4 \, x e^{x} \log \left (x e^{x} - e^{x} \log \left (2\right ) + e + 3 \, e^{x}\right ) + 4 \, e^{x} \log \left (2\right ) \log \left (x e^{x} - e^{x} \log \left (2\right ) + e + 3 \, e^{x}\right ) + 25 \, x e - 2 \, x e^{x} - 3 \, e^{x} \log \left (2\right ) - 20 \, e \log \left (x e^{x} - e^{x} \log \left (2\right ) + e + 3 \, e^{x}\right ) + 4 \, e^{x} \log \left (x e^{x} - e^{x} \log \left (2\right ) + e + 3 \, e^{x}\right ) + 15 \, e - 3 \, e^{x}} \] Input:

integrate(((-4*exp(1-x)+4*log(2)-4*x-12)*log(exp(1-x)-log(2)+3+x)+(-4*x-12 
)*exp(1-x)+4*x+12)/((16*exp(1-x)-16*log(2)+16*x+48)*log(exp(1-x)-log(2)+3+ 
x)^2+((-8*x-24)*exp(1-x)+(8*x+24)*log(2)-8*x^2-48*x-72)*log(exp(1-x)-log(2 
)+3+x)+(x^2+6*x+9)*exp(1-x)+(-x^2-6*x-9)*log(2)+x^3+9*x^2+27*x+27),x, algo 
rithm="giac")
 

Output:

4*(x*e^x*log((x*e^x - e^x*log(2) + e + 3*e^x)*e^(-x)) - e^x*log(2)*log((x* 
e^x - e^x*log(2) + e + 3*e^x)*e^(-x)) - 4*x*e + 4*x*e^x + e*log((x*e^x - e 
^x*log(2) + e + 3*e^x)*e^(-x)) + 3*e^x*log((x*e^x - e^x*log(2) + e + 3*e^x 
)*e^(-x)) + 4*e*log(x*e^x - e^x*log(2) + e + 3*e^x) - 4*e^x*log(x*e^x - e^ 
x*log(2) + e + 3*e^x))/(5*x^2*e^x - 5*x*e^x*log(2) - 4*x*e^x*log(x*e^x - e 
^x*log(2) + e + 3*e^x) + 4*e^x*log(2)*log(x*e^x - e^x*log(2) + e + 3*e^x) 
+ 25*x*e - 2*x*e^x - 3*e^x*log(2) - 20*e*log(x*e^x - e^x*log(2) + e + 3*e^ 
x) + 4*e^x*log(x*e^x - e^x*log(2) + e + 3*e^x) + 15*e - 3*e^x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {12+e^{1-x} (-12-4 x)+4 x+\left (-12-4 e^{1-x}-4 x+4 \log (2)\right ) \log \left (3+e^{1-x}+x-\log (2)\right )}{27+27 x+9 x^2+x^3+e^{1-x} \left (9+6 x+x^2\right )+\left (-9-6 x-x^2\right ) \log (2)+\left (-72+e^{1-x} (-24-8 x)-48 x-8 x^2+(24+8 x) \log (2)\right ) \log \left (3+e^{1-x}+x-\log (2)\right )+\left (48+16 e^{1-x}+16 x-16 \log (2)\right ) \log ^2\left (3+e^{1-x}+x-\log (2)\right )} \, dx=\int \frac {4\,x-\ln \left (x-\ln \left (2\right )+{\mathrm {e}}^{1-x}+3\right )\,\left (4\,x-4\,\ln \left (2\right )+4\,{\mathrm {e}}^{1-x}+12\right )-{\mathrm {e}}^{1-x}\,\left (4\,x+12\right )+12}{27\,x+{\mathrm {e}}^{1-x}\,\left (x^2+6\,x+9\right )+{\ln \left (x-\ln \left (2\right )+{\mathrm {e}}^{1-x}+3\right )}^2\,\left (16\,x-16\,\ln \left (2\right )+16\,{\mathrm {e}}^{1-x}+48\right )-\ln \left (x-\ln \left (2\right )+{\mathrm {e}}^{1-x}+3\right )\,\left (48\,x-\ln \left (2\right )\,\left (8\,x+24\right )+{\mathrm {e}}^{1-x}\,\left (8\,x+24\right )+8\,x^2+72\right )+9\,x^2+x^3-\ln \left (2\right )\,\left (x^2+6\,x+9\right )+27} \,d x \] Input:

int((4*x - log(x - log(2) + exp(1 - x) + 3)*(4*x - 4*log(2) + 4*exp(1 - x) 
 + 12) - exp(1 - x)*(4*x + 12) + 12)/(27*x + exp(1 - x)*(6*x + x^2 + 9) + 
log(x - log(2) + exp(1 - x) + 3)^2*(16*x - 16*log(2) + 16*exp(1 - x) + 48) 
 - log(x - log(2) + exp(1 - x) + 3)*(48*x - log(2)*(8*x + 24) + exp(1 - x) 
*(8*x + 24) + 8*x^2 + 72) + 9*x^2 + x^3 - log(2)*(6*x + x^2 + 9) + 27),x)
 

Output:

int((4*x - log(x - log(2) + exp(1 - x) + 3)*(4*x - 4*log(2) + 4*exp(1 - x) 
 + 12) - exp(1 - x)*(4*x + 12) + 12)/(27*x + exp(1 - x)*(6*x + x^2 + 9) + 
log(x - log(2) + exp(1 - x) + 3)^2*(16*x - 16*log(2) + 16*exp(1 - x) + 48) 
 - log(x - log(2) + exp(1 - x) + 3)*(48*x - log(2)*(8*x + 24) + exp(1 - x) 
*(8*x + 24) + 8*x^2 + 72) + 9*x^2 + x^3 - log(2)*(6*x + x^2 + 9) + 27), x)
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 200, normalized size of antiderivative = 6.67 \[ \int \frac {12+e^{1-x} (-12-4 x)+4 x+\left (-12-4 e^{1-x}-4 x+4 \log (2)\right ) \log \left (3+e^{1-x}+x-\log (2)\right )}{27+27 x+9 x^2+x^3+e^{1-x} \left (9+6 x+x^2\right )+\left (-9-6 x-x^2\right ) \log (2)+\left (-72+e^{1-x} (-24-8 x)-48 x-8 x^2+(24+8 x) \log (2)\right ) \log \left (3+e^{1-x}+x-\log (2)\right )+\left (48+16 e^{1-x}+16 x-16 \log (2)\right ) \log ^2\left (3+e^{1-x}+x-\log (2)\right )} \, dx=\frac {16 \,\mathrm {log}\left (e^{x} \mathrm {log}\left (2\right )-e^{x} x -3 e^{x}-e \right ) \mathrm {log}\left (\frac {-e^{x} \mathrm {log}\left (2\right )+e^{x} x +3 e^{x}+e}{e^{x}}\right )-4 \,\mathrm {log}\left (e^{x} \mathrm {log}\left (2\right )-e^{x} x -3 e^{x}-e \right ) x -12 \,\mathrm {log}\left (e^{x} \mathrm {log}\left (2\right )-e^{x} x -3 e^{x}-e \right )-16 \mathrm {log}\left (\frac {-e^{x} \mathrm {log}\left (2\right )+e^{x} x +3 e^{x}+e}{e^{x}}\right )^{2}-12 \,\mathrm {log}\left (\frac {-e^{x} \mathrm {log}\left (2\right )+e^{x} x +3 e^{x}+e}{e^{x}}\right ) x +4 x^{2}-36}{60 \,\mathrm {log}\left (\frac {-e^{x} \mathrm {log}\left (2\right )+e^{x} x +3 e^{x}+e}{e^{x}}\right )-15 x -45} \] Input:

int(((-4*exp(1-x)+4*log(2)-4*x-12)*log(exp(1-x)-log(2)+3+x)+(-4*x-12)*exp( 
1-x)+4*x+12)/((16*exp(1-x)-16*log(2)+16*x+48)*log(exp(1-x)-log(2)+3+x)^2+( 
(-8*x-24)*exp(1-x)+(8*x+24)*log(2)-8*x^2-48*x-72)*log(exp(1-x)-log(2)+3+x) 
+(x^2+6*x+9)*exp(1-x)+(-x^2-6*x-9)*log(2)+x^3+9*x^2+27*x+27),x)
 

Output:

(4*(4*log(e**x*log(2) - e**x*x - 3*e**x - e)*log(( - e**x*log(2) + e**x*x 
+ 3*e**x + e)/e**x) - log(e**x*log(2) - e**x*x - 3*e**x - e)*x - 3*log(e** 
x*log(2) - e**x*x - 3*e**x - e) - 4*log(( - e**x*log(2) + e**x*x + 3*e**x 
+ e)/e**x)**2 - 3*log(( - e**x*log(2) + e**x*x + 3*e**x + e)/e**x)*x + x** 
2 - 9))/(15*(4*log(( - e**x*log(2) + e**x*x + 3*e**x + e)/e**x) - x - 3))