\(\int \frac {(-3 x-2 x^2) \log (x^2)+(4+6 x+2 x^2) \log (2+3 x+x^2)+(-2 x-3 x^2-x^3) \log ^2(2+3 x+x^2)+e^{e^x} (3 x+2 x^2+e^x (-2 x-3 x^2-x^3) \log (2+3 x+x^2))}{(2 x+3 x^2+x^3) \log ^2(2+3 x+x^2)} \, dx\) [2027]

Optimal result

Integrand size = 129, antiderivative size = 32 \[ \int \frac {\left (-3 x-2 x^2\right ) \log \left (x^2\right )+\left (4+6 x+2 x^2\right ) \log \left (2+3 x+x^2\right )+\left (-2 x-3 x^2-x^3\right ) \log ^2\left (2+3 x+x^2\right )+e^{e^x} \left (3 x+2 x^2+e^x \left (-2 x-3 x^2-x^3\right ) \log \left (2+3 x+x^2\right )\right )}{\left (2 x+3 x^2+x^3\right ) \log ^2\left (2+3 x+x^2\right )} \, dx=-x+\frac {-e^{e^x}+\log \left (x^2\right )}{\log \left (\frac {(2+x) \left (x+x^2\right )}{x}\right )} \] Output:

(ln(x^2)-exp(exp(x)))/ln((2+x)/x*(x^2+x))-x
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.28 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.19 \[ \int \frac {\left (-3 x-2 x^2\right ) \log \left (x^2\right )+\left (4+6 x+2 x^2\right ) \log \left (2+3 x+x^2\right )+\left (-2 x-3 x^2-x^3\right ) \log ^2\left (2+3 x+x^2\right )+e^{e^x} \left (3 x+2 x^2+e^x \left (-2 x-3 x^2-x^3\right ) \log \left (2+3 x+x^2\right )\right )}{\left (2 x+3 x^2+x^3\right ) \log ^2\left (2+3 x+x^2\right )} \, dx=-x-\frac {e^{e^x}}{\log \left (2+3 x+x^2\right )}+\frac {\log \left (x^2\right )}{\log \left (2+3 x+x^2\right )} \] Input:

Integrate[((-3*x - 2*x^2)*Log[x^2] + (4 + 6*x + 2*x^2)*Log[2 + 3*x + x^2] 
+ (-2*x - 3*x^2 - x^3)*Log[2 + 3*x + x^2]^2 + E^E^x*(3*x + 2*x^2 + E^x*(-2 
*x - 3*x^2 - x^3)*Log[2 + 3*x + x^2]))/((2*x + 3*x^2 + x^3)*Log[2 + 3*x + 
x^2]^2),x]
 

Output:

-x - E^E^x/Log[2 + 3*x + x^2] + Log[x^2]/Log[2 + 3*x + x^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.

\(\displaystyle \int \frac {\left (2 x^2+6 x+4\right ) \log \left (x^2+3 x+2\right )+\left (-2 x^2-3 x\right ) \log \left (x^2\right )+\left (-x^3-3 x^2-2 x\right ) \log ^2\left (x^2+3 x+2\right )+e^{e^x} \left (2 x^2+e^x \left (-x^3-3 x^2-2 x\right ) \log \left (x^2+3 x+2\right )+3 x\right )}{\left (x^3+3 x^2+2 x\right ) \log ^2\left (x^2+3 x+2\right )} \, dx\)

\(\Big \downarrow \) 2026

\(\displaystyle \int \frac {\left (2 x^2+6 x+4\right ) \log \left (x^2+3 x+2\right )+\left (-2 x^2-3 x\right ) \log \left (x^2\right )+\left (-x^3-3 x^2-2 x\right ) \log ^2\left (x^2+3 x+2\right )+e^{e^x} \left (2 x^2+e^x \left (-x^3-3 x^2-2 x\right ) \log \left (x^2+3 x+2\right )+3 x\right )}{x \left (x^2+3 x+2\right ) \log ^2\left (x^2+3 x+2\right )}dx\)

\(\Big \downarrow \) 7239

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

\(\Big \downarrow \) 2009

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

Input:

Int[((-3*x - 2*x^2)*Log[x^2] + (4 + 6*x + 2*x^2)*Log[2 + 3*x + x^2] + (-2* 
x - 3*x^2 - x^3)*Log[2 + 3*x + x^2]^2 + E^E^x*(3*x + 2*x^2 + E^x*(-2*x - 3 
*x^2 - x^3)*Log[2 + 3*x + x^2]))/((2*x + 3*x^2 + x^3)*Log[2 + 3*x + x^2]^2 
),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 60.84 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.53

method	result	size
parallelrisch	\(\frac {-6 x \ln \left (x^{2}+3 x +2\right )+6 \ln \left (x^{2}\right )-6 \,{\mathrm e}^{{\mathrm e}^{x}}+13 \ln \left (x^{2}+3 x +2\right )}{6 \ln \left (x^{2}+3 x +2\right )}\)	\(49\)
risch	\(-x +\frac {4 \ln \left (x \right )-i \pi \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )+2 i \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2}-i \pi \operatorname {csgn}\left (i x^{2}\right )^{3}}{2 \ln \left (x^{2}+3 x +2\right )}-\frac {{\mathrm e}^{{\mathrm e}^{x}}}{\ln \left (x^{2}+3 x +2\right )}\)	\(88\)

Input:

int((((-x^3-3*x^2-2*x)*exp(x)*ln(x^2+3*x+2)+2*x^2+3*x)*exp(exp(x))+(-x^3-3 
*x^2-2*x)*ln(x^2+3*x+2)^2+(2*x^2+6*x+4)*ln(x^2+3*x+2)+(-2*x^2-3*x)*ln(x^2) 
)/(x^3+3*x^2+2*x)/ln(x^2+3*x+2)^2,x,method=_RETURNVERBOSE)
 

Output:

1/6*(-6*x*ln(x^2+3*x+2)+6*ln(x^2)-6*exp(exp(x))+13*ln(x^2+3*x+2))/ln(x^2+3 
*x+2)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.06 \[ \int \frac {\left (-3 x-2 x^2\right ) \log \left (x^2\right )+\left (4+6 x+2 x^2\right ) \log \left (2+3 x+x^2\right )+\left (-2 x-3 x^2-x^3\right ) \log ^2\left (2+3 x+x^2\right )+e^{e^x} \left (3 x+2 x^2+e^x \left (-2 x-3 x^2-x^3\right ) \log \left (2+3 x+x^2\right )\right )}{\left (2 x+3 x^2+x^3\right ) \log ^2\left (2+3 x+x^2\right )} \, dx=-\frac {x \log \left (x^{2} + 3 \, x + 2\right ) + e^{\left (e^{x}\right )} - \log \left (x^{2}\right )}{\log \left (x^{2} + 3 \, x + 2\right )} \] Input:

integrate((((-x^3-3*x^2-2*x)*exp(x)*log(x^2+3*x+2)+2*x^2+3*x)*exp(exp(x))+ 
(-x^3-3*x^2-2*x)*log(x^2+3*x+2)^2+(2*x^2+6*x+4)*log(x^2+3*x+2)+(-2*x^2-3*x 
)*log(x^2))/(x^3+3*x^2+2*x)/log(x^2+3*x+2)^2,x, algorithm="fricas")
 

Output:

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

Sympy [A] (verification not implemented)

Time = 0.34 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.97 \[ \int \frac {\left (-3 x-2 x^2\right ) \log \left (x^2\right )+\left (4+6 x+2 x^2\right ) \log \left (2+3 x+x^2\right )+\left (-2 x-3 x^2-x^3\right ) \log ^2\left (2+3 x+x^2\right )+e^{e^x} \left (3 x+2 x^2+e^x \left (-2 x-3 x^2-x^3\right ) \log \left (2+3 x+x^2\right )\right )}{\left (2 x+3 x^2+x^3\right ) \log ^2\left (2+3 x+x^2\right )} \, dx=- x - \frac {e^{e^{x}}}{\log {\left (x^{2} + 3 x + 2 \right )}} + \frac {\log {\left (x^{2} \right )}}{\log {\left (x^{2} + 3 x + 2 \right )}} \] Input:

integrate((((-x**3-3*x**2-2*x)*exp(x)*ln(x**2+3*x+2)+2*x**2+3*x)*exp(exp(x 
))+(-x**3-3*x**2-2*x)*ln(x**2+3*x+2)**2+(2*x**2+6*x+4)*ln(x**2+3*x+2)+(-2* 
x**2-3*x)*ln(x**2))/(x**3+3*x**2+2*x)/ln(x**2+3*x+2)**2,x)
 

Output:

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

Maxima [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.03 \[ \int \frac {\left (-3 x-2 x^2\right ) \log \left (x^2\right )+\left (4+6 x+2 x^2\right ) \log \left (2+3 x+x^2\right )+\left (-2 x-3 x^2-x^3\right ) \log ^2\left (2+3 x+x^2\right )+e^{e^x} \left (3 x+2 x^2+e^x \left (-2 x-3 x^2-x^3\right ) \log \left (2+3 x+x^2\right )\right )}{\left (2 x+3 x^2+x^3\right ) \log ^2\left (2+3 x+x^2\right )} \, dx=-\frac {x \log \left (x + 2\right ) + x \log \left (x + 1\right ) + e^{\left (e^{x}\right )} - 2 \, \log \left (x\right )}{\log \left (x + 2\right ) + \log \left (x + 1\right )} \] Input:

integrate((((-x^3-3*x^2-2*x)*exp(x)*log(x^2+3*x+2)+2*x^2+3*x)*exp(exp(x))+ 
(-x^3-3*x^2-2*x)*log(x^2+3*x+2)^2+(2*x^2+6*x+4)*log(x^2+3*x+2)+(-2*x^2-3*x 
)*log(x^2))/(x^3+3*x^2+2*x)/log(x^2+3*x+2)^2,x, algorithm="maxima")
 

Output:

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

Giac [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.06 \[ \int \frac {\left (-3 x-2 x^2\right ) \log \left (x^2\right )+\left (4+6 x+2 x^2\right ) \log \left (2+3 x+x^2\right )+\left (-2 x-3 x^2-x^3\right ) \log ^2\left (2+3 x+x^2\right )+e^{e^x} \left (3 x+2 x^2+e^x \left (-2 x-3 x^2-x^3\right ) \log \left (2+3 x+x^2\right )\right )}{\left (2 x+3 x^2+x^3\right ) \log ^2\left (2+3 x+x^2\right )} \, dx=-\frac {x \log \left (x^{2} + 3 \, x + 2\right ) + e^{\left (e^{x}\right )} - \log \left (x^{2}\right )}{\log \left (x^{2} + 3 \, x + 2\right )} \] Input:

integrate((((-x^3-3*x^2-2*x)*exp(x)*log(x^2+3*x+2)+2*x^2+3*x)*exp(exp(x))+ 
(-x^3-3*x^2-2*x)*log(x^2+3*x+2)^2+(2*x^2+6*x+4)*log(x^2+3*x+2)+(-2*x^2-3*x 
)*log(x^2))/(x^3+3*x^2+2*x)/log(x^2+3*x+2)^2,x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (-3 x-2 x^2\right ) \log \left (x^2\right )+\left (4+6 x+2 x^2\right ) \log \left (2+3 x+x^2\right )+\left (-2 x-3 x^2-x^3\right ) \log ^2\left (2+3 x+x^2\right )+e^{e^x} \left (3 x+2 x^2+e^x \left (-2 x-3 x^2-x^3\right ) \log \left (2+3 x+x^2\right )\right )}{\left (2 x+3 x^2+x^3\right ) \log ^2\left (2+3 x+x^2\right )} \, dx=\int \frac {{\mathrm {e}}^{{\mathrm {e}}^x}\,\left (3\,x+2\,x^2-{\mathrm {e}}^x\,\ln \left (x^2+3\,x+2\right )\,\left (x^3+3\,x^2+2\,x\right )\right )-\ln \left (x^2\right )\,\left (2\,x^2+3\,x\right )+\ln \left (x^2+3\,x+2\right )\,\left (2\,x^2+6\,x+4\right )-{\ln \left (x^2+3\,x+2\right )}^2\,\left (x^3+3\,x^2+2\,x\right )}{{\ln \left (x^2+3\,x+2\right )}^2\,\left (x^3+3\,x^2+2\,x\right )} \,d x \] Input:

int((exp(exp(x))*(3*x + 2*x^2 - exp(x)*log(3*x + x^2 + 2)*(2*x + 3*x^2 + x 
^3)) - log(x^2)*(3*x + 2*x^2) + log(3*x + x^2 + 2)*(6*x + 2*x^2 + 4) - log 
(3*x + x^2 + 2)^2*(2*x + 3*x^2 + x^3))/(log(3*x + x^2 + 2)^2*(2*x + 3*x^2 
+ x^3)),x)
 

Output:

int((exp(exp(x))*(3*x + 2*x^2 - exp(x)*log(3*x + x^2 + 2)*(2*x + 3*x^2 + x 
^3)) - log(x^2)*(3*x + 2*x^2) + log(3*x + x^2 + 2)*(6*x + 2*x^2 + 4) - log 
(3*x + x^2 + 2)^2*(2*x + 3*x^2 + x^3))/(log(3*x + x^2 + 2)^2*(2*x + 3*x^2 
+ x^3)), x)
 

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.12 \[ \int \frac {\left (-3 x-2 x^2\right ) \log \left (x^2\right )+\left (4+6 x+2 x^2\right ) \log \left (2+3 x+x^2\right )+\left (-2 x-3 x^2-x^3\right ) \log ^2\left (2+3 x+x^2\right )+e^{e^x} \left (3 x+2 x^2+e^x \left (-2 x-3 x^2-x^3\right ) \log \left (2+3 x+x^2\right )\right )}{\left (2 x+3 x^2+x^3\right ) \log ^2\left (2+3 x+x^2\right )} \, dx=\frac {-e^{e^{x}}+\mathrm {log}\left (x^{2}\right )-\mathrm {log}\left (x^{2}+3 x +2\right ) x}{\mathrm {log}\left (x^{2}+3 x +2\right )} \] Input:

int((((-x^3-3*x^2-2*x)*exp(x)*log(x^2+3*x+2)+2*x^2+3*x)*exp(exp(x))+(-x^3- 
3*x^2-2*x)*log(x^2+3*x+2)^2+(2*x^2+6*x+4)*log(x^2+3*x+2)+(-2*x^2-3*x)*log( 
x^2))/(x^3+3*x^2+2*x)/log(x^2+3*x+2)^2,x)
 

Output:

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