\(\int \frac {4 x^2-2 e^x x^2+3 x^3+x^4+(x^2+x^3) \log (9)+(-2 x^2-2 x \log (9)+e^x (2 x^2+2 x \log (9))) \log (x+\log (9))+\log (\frac {1}{5} (-2+e^x-x)) (-4 x+2 e^x x-2 x^2+(4 x+2 x^2+e^x (-2 x-2 \log (9))+(4+2 x) \log (9)) \log (x+\log (9)))}{-2 x^3-x^4+(-2 x^2-x^3) \log (9)+e^x (x^3+x^2 \log (9))} \, dx\) [604]

Optimal result

Integrand size = 170, antiderivative size = 33 \[ \int \frac {4 x^2-2 e^x x^2+3 x^3+x^4+\left (x^2+x^3\right ) \log (9)+\left (-2 x^2-2 x \log (9)+e^x \left (2 x^2+2 x \log (9)\right )\right ) \log (x+\log (9))+\log \left (\frac {1}{5} \left (-2+e^x-x\right )\right ) \left (-4 x+2 e^x x-2 x^2+\left (4 x+2 x^2+e^x (-2 x-2 \log (9))+(4+2 x) \log (9)\right ) \log (x+\log (9))\right )}{-2 x^3-x^4+\left (-2 x^2-x^3\right ) \log (9)+e^x \left (x^3+x^2 \log (9)\right )} \, dx=5-\frac {\left (x-\log \left (\frac {1}{5} \left (-2+e^x-x\right )\right )\right ) (x+2 \log (x+\log (9)))}{x} \] Output:

5-(x+2*ln(2*ln(3)+x))/x*(x-ln(1/5*exp(x)-2/5-1/5*x))
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.30 \[ \int \frac {4 x^2-2 e^x x^2+3 x^3+x^4+\left (x^2+x^3\right ) \log (9)+\left (-2 x^2-2 x \log (9)+e^x \left (2 x^2+2 x \log (9)\right )\right ) \log (x+\log (9))+\log \left (\frac {1}{5} \left (-2+e^x-x\right )\right ) \left (-4 x+2 e^x x-2 x^2+\left (4 x+2 x^2+e^x (-2 x-2 \log (9))+(4+2 x) \log (9)\right ) \log (x+\log (9))\right )}{-2 x^3-x^4+\left (-2 x^2-x^3\right ) \log (9)+e^x \left (x^3+x^2 \log (9)\right )} \, dx=-x+\log \left (2-e^x+x\right )-2 \log (x+\log (9))+\frac {2 \log \left (\frac {1}{5} \left (-2+e^x-x\right )\right ) \log (x+\log (9))}{x} \] Input:

Integrate[(4*x^2 - 2*E^x*x^2 + 3*x^3 + x^4 + (x^2 + x^3)*Log[9] + (-2*x^2 
- 2*x*Log[9] + E^x*(2*x^2 + 2*x*Log[9]))*Log[x + Log[9]] + Log[(-2 + E^x - 
 x)/5]*(-4*x + 2*E^x*x - 2*x^2 + (4*x + 2*x^2 + E^x*(-2*x - 2*Log[9]) + (4 
 + 2*x)*Log[9])*Log[x + Log[9]]))/(-2*x^3 - x^4 + (-2*x^2 - x^3)*Log[9] + 
E^x*(x^3 + x^2*Log[9])),x]
 

Output:

-x + Log[2 - E^x + x] - 2*Log[x + Log[9]] + (2*Log[(-2 + E^x - x)/5]*Log[x 
 + Log[9]])/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.

\(\displaystyle \int \frac {x^4+3 x^3-2 e^x x^2+4 x^2+\left (-2 x^2+e^x \left (2 x^2+2 x \log (9)\right )-2 x \log (9)\right ) \log (x+\log (9))+\log \left (\frac {1}{5} \left (-x+e^x-2\right )\right ) \left (-2 x^2+\left (2 x^2+4 x+e^x (-2 x-2 \log (9))+(2 x+4) \log (9)\right ) \log (x+\log (9))+2 e^x x-4 x\right )+\left (x^3+x^2\right ) \log (9)}{-x^4-2 x^3+e^x \left (x^3+x^2 \log (9)\right )+\left (-x^3-2 x^2\right ) \log (9)} \, dx\)

\(\Big \downarrow \) 7239

\(\displaystyle \int \frac {2 \left (-x+e^x-2\right ) \log \left (\frac {1}{5} \left (-x+e^x-2\right )\right ) ((x+\log (9)) \log (x+\log (9))-x)-x \left (x \left (x^2-2 e^x+x (3+\log (9))+4+\log (9)\right )+2 \left (e^x-1\right ) (x+\log (9)) \log (x+\log (9))\right )}{x^2 \left (x-e^x+2\right ) (x+\log (9))}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (\frac {2 \left (x-\log \left (-x+e^x-2\right )+\log (5)\right ) (-x+x \log (x+\log (9))+\log (9) \log (x+\log (9)))}{x^2 (x+\log (9))}+\frac {(x+1) (x+2 \log (x+\log (9)))}{\left (-x+e^x-2\right ) x}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \int \frac {1}{-x+e^x-2}dx+\int \frac {x}{-x+e^x-2}dx+\frac {2 \log \left (-x+e^x-2\right ) \log (x+\log (9))}{x}-\frac {2 \log (5) \log (x+\log (9))}{\log (9)}-\frac {2 \log \left (\frac {9}{5}\right ) \log (x+\log (9))}{\log (9)}-\frac {2 \log (5) \log (x+\log (9))}{x}\)

Input:

Int[(4*x^2 - 2*E^x*x^2 + 3*x^3 + x^4 + (x^2 + x^3)*Log[9] + (-2*x^2 - 2*x* 
Log[9] + E^x*(2*x^2 + 2*x*Log[9]))*Log[x + Log[9]] + Log[(-2 + E^x - x)/5] 
*(-4*x + 2*E^x*x - 2*x^2 + (4*x + 2*x^2 + E^x*(-2*x - 2*Log[9]) + (4 + 2*x 
)*Log[9])*Log[x + Log[9]]))/(-2*x^3 - x^4 + (-2*x^2 - x^3)*Log[9] + E^x*(x 
^3 + x^2*Log[9])),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 269.71 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.33

method	result	size
risch	\(\frac {2 \ln \left (2 \ln \left (3\right )+x \right ) \ln \left (\frac {{\mathrm e}^{x}}{5}-\frac {2}{5}-\frac {x}{5}\right )}{x}-2 \ln \left (2 \ln \left (3\right )+x \right )-x +\ln \left ({\mathrm e}^{x}-2-x \right )\)	\(44\)
parallelrisch	\(\frac {4 x \ln \left (3\right )-x^{2}-2 \ln \left (2 \ln \left (3\right )+x \right ) x +\ln \left (\frac {{\mathrm e}^{x}}{5}-\frac {2}{5}-\frac {x}{5}\right ) x +2 \ln \left (\frac {{\mathrm e}^{x}}{5}-\frac {2}{5}-\frac {x}{5}\right ) \ln \left (2 \ln \left (3\right )+x \right )}{x}\)	\(57\)

Input:

int(((((-4*ln(3)-2*x)*exp(x)+2*(4+2*x)*ln(3)+2*x^2+4*x)*ln(2*ln(3)+x)+2*ex 
p(x)*x-2*x^2-4*x)*ln(1/5*exp(x)-2/5-1/5*x)+((4*x*ln(3)+2*x^2)*exp(x)-4*x*l 
n(3)-2*x^2)*ln(2*ln(3)+x)-2*exp(x)*x^2+2*(x^3+x^2)*ln(3)+x^4+3*x^3+4*x^2)/ 
((2*x^2*ln(3)+x^3)*exp(x)+2*(-x^3-2*x^2)*ln(3)-x^4-2*x^3),x,method=_RETURN 
VERBOSE)
 

Output:

2/x*ln(2*ln(3)+x)*ln(1/5*exp(x)-2/5-1/5*x)-2*ln(2*ln(3)+x)-x+ln(exp(x)-2-x 
)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.27 \[ \int \frac {4 x^2-2 e^x x^2+3 x^3+x^4+\left (x^2+x^3\right ) \log (9)+\left (-2 x^2-2 x \log (9)+e^x \left (2 x^2+2 x \log (9)\right )\right ) \log (x+\log (9))+\log \left (\frac {1}{5} \left (-2+e^x-x\right )\right ) \left (-4 x+2 e^x x-2 x^2+\left (4 x+2 x^2+e^x (-2 x-2 \log (9))+(4+2 x) \log (9)\right ) \log (x+\log (9))\right )}{-2 x^3-x^4+\left (-2 x^2-x^3\right ) \log (9)+e^x \left (x^3+x^2 \log (9)\right )} \, dx=-\frac {x^{2} + 2 \, x \log \left (x + 2 \, \log \left (3\right )\right ) - {\left (x + 2 \, \log \left (x + 2 \, \log \left (3\right )\right )\right )} \log \left (-\frac {1}{5} \, x + \frac {1}{5} \, e^{x} - \frac {2}{5}\right )}{x} \] Input:

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

Output:

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

Sympy [A] (verification not implemented)

Time = 0.45 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.33 \[ \int \frac {4 x^2-2 e^x x^2+3 x^3+x^4+\left (x^2+x^3\right ) \log (9)+\left (-2 x^2-2 x \log (9)+e^x \left (2 x^2+2 x \log (9)\right )\right ) \log (x+\log (9))+\log \left (\frac {1}{5} \left (-2+e^x-x\right )\right ) \left (-4 x+2 e^x x-2 x^2+\left (4 x+2 x^2+e^x (-2 x-2 \log (9))+(4+2 x) \log (9)\right ) \log (x+\log (9))\right )}{-2 x^3-x^4+\left (-2 x^2-x^3\right ) \log (9)+e^x \left (x^3+x^2 \log (9)\right )} \, dx=- x - 2 \log {\left (x + 2 \log {\left (3 \right )} \right )} + \log {\left (- x + e^{x} - 2 \right )} + \frac {2 \log {\left (x + 2 \log {\left (3 \right )} \right )} \log {\left (- \frac {x}{5} + \frac {e^{x}}{5} - \frac {2}{5} \right )}}{x} \] Input:

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

Output:

-x - 2*log(x + 2*log(3)) + log(-x + exp(x) - 2) + 2*log(x + 2*log(3))*log( 
-x/5 + exp(x)/5 - 2/5)/x
 

Maxima [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.30 \[ \int \frac {4 x^2-2 e^x x^2+3 x^3+x^4+\left (x^2+x^3\right ) \log (9)+\left (-2 x^2-2 x \log (9)+e^x \left (2 x^2+2 x \log (9)\right )\right ) \log (x+\log (9))+\log \left (\frac {1}{5} \left (-2+e^x-x\right )\right ) \left (-4 x+2 e^x x-2 x^2+\left (4 x+2 x^2+e^x (-2 x-2 \log (9))+(4+2 x) \log (9)\right ) \log (x+\log (9))\right )}{-2 x^3-x^4+\left (-2 x^2-x^3\right ) \log (9)+e^x \left (x^3+x^2 \log (9)\right )} \, dx=-\frac {x^{2} + 2 \, {\left (x + \log \left (5\right )\right )} \log \left (x + 2 \, \log \left (3\right )\right ) - {\left (x + 2 \, \log \left (x + 2 \, \log \left (3\right )\right )\right )} \log \left (-x + e^{x} - 2\right )}{x} \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.76 \[ \int \frac {4 x^2-2 e^x x^2+3 x^3+x^4+\left (x^2+x^3\right ) \log (9)+\left (-2 x^2-2 x \log (9)+e^x \left (2 x^2+2 x \log (9)\right )\right ) \log (x+\log (9))+\log \left (\frac {1}{5} \left (-2+e^x-x\right )\right ) \left (-4 x+2 e^x x-2 x^2+\left (4 x+2 x^2+e^x (-2 x-2 \log (9))+(4+2 x) \log (9)\right ) \log (x+\log (9))\right )}{-2 x^3-x^4+\left (-2 x^2-x^3\right ) \log (9)+e^x \left (x^3+x^2 \log (9)\right )} \, dx=-\frac {x^{2} - x \log \left (x - e^{x} + 2\right ) + 2 \, x \log \left (x + 2 \, \log \left (3\right )\right ) + 2 \, \log \left (5\right ) \log \left (x + 2 \, \log \left (3\right )\right ) - 2 \, \log \left (x + 2 \, \log \left (3\right )\right ) \log \left (-x + e^{x} - 2\right )}{x} \] Input:

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {4 x^2-2 e^x x^2+3 x^3+x^4+\left (x^2+x^3\right ) \log (9)+\left (-2 x^2-2 x \log (9)+e^x \left (2 x^2+2 x \log (9)\right )\right ) \log (x+\log (9))+\log \left (\frac {1}{5} \left (-2+e^x-x\right )\right ) \left (-4 x+2 e^x x-2 x^2+\left (4 x+2 x^2+e^x (-2 x-2 \log (9))+(4+2 x) \log (9)\right ) \log (x+\log (9))\right )}{-2 x^3-x^4+\left (-2 x^2-x^3\right ) \log (9)+e^x \left (x^3+x^2 \log (9)\right )} \, dx=\int -\frac {4\,x^2-\ln \left (x+2\,\ln \left (3\right )\right )\,\left (4\,x\,\ln \left (3\right )-{\mathrm {e}}^x\,\left (2\,x^2+4\,\ln \left (3\right )\,x\right )+2\,x^2\right )-\ln \left (\frac {{\mathrm {e}}^x}{5}-\frac {x}{5}-\frac {2}{5}\right )\,\left (4\,x-\ln \left (x+2\,\ln \left (3\right )\right )\,\left (4\,x+2\,\ln \left (3\right )\,\left (2\,x+4\right )-{\mathrm {e}}^x\,\left (2\,x+4\,\ln \left (3\right )\right )+2\,x^2\right )-2\,x\,{\mathrm {e}}^x+2\,x^2\right )-2\,x^2\,{\mathrm {e}}^x+3\,x^3+x^4+2\,\ln \left (3\right )\,\left (x^3+x^2\right )}{2\,\ln \left (3\right )\,\left (x^3+2\,x^2\right )-{\mathrm {e}}^x\,\left (x^3+2\,\ln \left (3\right )\,x^2\right )+2\,x^3+x^4} \,d x \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.55 \[ \int \frac {4 x^2-2 e^x x^2+3 x^3+x^4+\left (x^2+x^3\right ) \log (9)+\left (-2 x^2-2 x \log (9)+e^x \left (2 x^2+2 x \log (9)\right )\right ) \log (x+\log (9))+\log \left (\frac {1}{5} \left (-2+e^x-x\right )\right ) \left (-4 x+2 e^x x-2 x^2+\left (4 x+2 x^2+e^x (-2 x-2 \log (9))+(4+2 x) \log (9)\right ) \log (x+\log (9))\right )}{-2 x^3-x^4+\left (-2 x^2-x^3\right ) \log (9)+e^x \left (x^3+x^2 \log (9)\right )} \, dx=\frac {\mathrm {log}\left (e^{x}-x -2\right ) x +2 \,\mathrm {log}\left (2 \,\mathrm {log}\left (3\right )+x \right ) \mathrm {log}\left (\frac {e^{x}}{5}-\frac {x}{5}-\frac {2}{5}\right )-2 \,\mathrm {log}\left (2 \,\mathrm {log}\left (3\right )+x \right ) x -x^{2}}{x} \] Input:

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

Output:

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