\(\int \frac {e^{\frac {-18 x+3 x^2+3 \log (3)+(-5 x+x^2) \log (e^{e^x x^2}-x)}{-5 x+x^2}} (-25 x^2+7 x^3-x^4+(-15 x+6 x^2) \log (3)+e^{e^x x^2} (3 x^2+e^x (50 x^3+5 x^4-8 x^5+x^6)+(15-6 x) \log (3)))}{-25 x^3+10 x^4-x^5+e^{e^x x^2} (25 x^2-10 x^3+x^4)} \, dx\) [1805]

Optimal result

Integrand size = 165, antiderivative size = 34 \[ \int \frac {e^{\frac {-18 x+3 x^2+3 \log (3)+\left (-5 x+x^2\right ) \log \left (e^{e^x x^2}-x\right )}{-5 x+x^2}} \left (-25 x^2+7 x^3-x^4+\left (-15 x+6 x^2\right ) \log (3)+e^{e^x x^2} \left (3 x^2+e^x \left (50 x^3+5 x^4-8 x^5+x^6\right )+(15-6 x) \log (3)\right )\right )}{-25 x^3+10 x^4-x^5+e^{e^x x^2} \left (25 x^2-10 x^3+x^4\right )} \, dx=e^{3+\frac {3 (-x+\log (3))}{(-5+x) x}} \left (e^{e^x x^2}-x\right ) \] Output:

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

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.09 \[ \int \frac {e^{\frac {-18 x+3 x^2+3 \log (3)+\left (-5 x+x^2\right ) \log \left (e^{e^x x^2}-x\right )}{-5 x+x^2}} \left (-25 x^2+7 x^3-x^4+\left (-15 x+6 x^2\right ) \log (3)+e^{e^x x^2} \left (3 x^2+e^x \left (50 x^3+5 x^4-8 x^5+x^6\right )+(15-6 x) \log (3)\right )\right )}{-25 x^3+10 x^4-x^5+e^{e^x x^2} \left (25 x^2-10 x^3+x^4\right )} \, dx=3^{\frac {3}{(-5+x) x}} e^{3-\frac {3}{-5+x}} \left (e^{e^x x^2}-x\right ) \] Input:

Integrate[(E^((-18*x + 3*x^2 + 3*Log[3] + (-5*x + x^2)*Log[E^(E^x*x^2) - x 
])/(-5*x + x^2))*(-25*x^2 + 7*x^3 - x^4 + (-15*x + 6*x^2)*Log[3] + E^(E^x* 
x^2)*(3*x^2 + E^x*(50*x^3 + 5*x^4 - 8*x^5 + x^6) + (15 - 6*x)*Log[3])))/(- 
25*x^3 + 10*x^4 - x^5 + E^(E^x*x^2)*(25*x^2 - 10*x^3 + x^4)),x]
 

Output:

3^(3/((-5 + x)*x))*E^(3 - 3/(-5 + x))*(E^(E^x*x^2) - 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.

\(\displaystyle \int \frac {\left (-x^4+7 x^3-25 x^2+\left (6 x^2-15 x\right ) \log (3)+e^{e^x x^2} \left (3 x^2+e^x \left (x^6-8 x^5+5 x^4+50 x^3\right )+(15-6 x) \log (3)\right )\right ) \exp \left (\frac {3 x^2+\left (x^2-5 x\right ) \log \left (e^{e^x x^2}-x\right )-18 x+3 \log (3)}{x^2-5 x}\right )}{-x^5+10 x^4-25 x^3+e^{e^x x^2} \left (x^4-10 x^3+25 x^2\right )} \, dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {\left (-x^4+7 x^3-25 x^2+\left (6 x^2-15 x\right ) \log (3)+e^{e^x x^2} \left (3 x^2+e^x \left (x^6-8 x^5+5 x^4+50 x^3\right )+(15-6 x) \log (3)\right )\right ) \exp \left (\frac {3 x^2+\left (x^2-5 x\right ) \log \left (e^{e^x x^2}-x\right )-18 x+3 \log (3)}{(x-5) x}\right )}{(5-x)^2 \left (e^{e^x x^2}-x\right ) x^2}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (\frac {\left (e^x x^3+2 e^x x^2-1\right ) \exp \left (\frac {3 x^2+\left (x^2-5 x\right ) \log \left (e^{e^x x^2}-x\right )-18 x+3 \log (3)}{(x-5) x}\right )}{e^{e^x x^2}-x}+\frac {\left (e^x x^6-8 e^x x^5+5 e^x x^4+50 e^x x^3+3 x^2-6 x \log (3)+15 \log (3)\right ) \exp \left (\frac {3 x^2+\left (x^2-5 x\right ) \log \left (e^{e^x x^2}-x\right )-18 x+3 \log (3)}{(x-5) x}\right )}{(x-5)^2 x^2}\right )dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (3^{\frac {3}{(x-5) x}} e^{\frac {3 (x-6)}{x-5}} \left (e^x x^3+2 e^x x^2-1\right )+\frac {3^{\frac {3}{(x-5) x}} e^{\frac {3 (x-6)}{x-5}} \left (e^{e^x x^2}-x\right ) \left (e^x x^6-8 e^x x^5+5 e^x x^4+50 e^x x^3+3 x^2-6 x \log (3)+15 \log (3)\right )}{(5-x)^2 x^2}\right )dx\)

\(\Big \downarrow \) 7299

\(\displaystyle \int \left (3^{\frac {3}{(x-5) x}} e^{\frac {3 (x-6)}{x-5}} \left (e^x x^3+2 e^x x^2-1\right )+\frac {3^{\frac {3}{(x-5) x}} e^{\frac {3 (x-6)}{x-5}} \left (e^{e^x x^2}-x\right ) \left (e^x x^6-8 e^x x^5+5 e^x x^4+50 e^x x^3+3 x^2-6 x \log (3)+15 \log (3)\right )}{(5-x)^2 x^2}\right )dx\)

Input:

Int[(E^((-18*x + 3*x^2 + 3*Log[3] + (-5*x + x^2)*Log[E^(E^x*x^2) - x])/(-5 
*x + x^2))*(-25*x^2 + 7*x^3 - x^4 + (-15*x + 6*x^2)*Log[3] + E^(E^x*x^2)*( 
3*x^2 + E^x*(50*x^3 + 5*x^4 - 8*x^5 + x^6) + (15 - 6*x)*Log[3])))/(-25*x^3 
 + 10*x^4 - x^5 + E^(E^x*x^2)*(25*x^2 - 10*x^3 + x^4)),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.62

\[{\mathrm e}^{\frac {\ln \left ({\mathrm e}^{{\mathrm e}^{x} x^{2}}-x \right ) x^{2}-5 \ln \left ({\mathrm e}^{{\mathrm e}^{x} x^{2}}-x \right ) x +3 x^{2}+3 \ln \left (3\right )-18 x}{\left (-5+x \right ) x}}\]

Input:

int((((x^6-8*x^5+5*x^4+50*x^3)*exp(x)+(-6*x+15)*ln(3)+3*x^2)*exp(exp(x)*x^ 
2)+(6*x^2-15*x)*ln(3)-x^4+7*x^3-25*x^2)*exp(((x^2-5*x)*ln(exp(exp(x)*x^2)- 
x)+3*ln(3)+3*x^2-18*x)/(x^2-5*x))/((x^4-10*x^3+25*x^2)*exp(exp(x)*x^2)-x^5 
+10*x^4-25*x^3),x)
 

Output:

exp((ln(exp(exp(x)*x^2)-x)*x^2-5*ln(exp(exp(x)*x^2)-x)*x+3*x^2+3*ln(3)-18* 
x)/(-5+x)/x)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.29 \[ \int \frac {e^{\frac {-18 x+3 x^2+3 \log (3)+\left (-5 x+x^2\right ) \log \left (e^{e^x x^2}-x\right )}{-5 x+x^2}} \left (-25 x^2+7 x^3-x^4+\left (-15 x+6 x^2\right ) \log (3)+e^{e^x x^2} \left (3 x^2+e^x \left (50 x^3+5 x^4-8 x^5+x^6\right )+(15-6 x) \log (3)\right )\right )}{-25 x^3+10 x^4-x^5+e^{e^x x^2} \left (25 x^2-10 x^3+x^4\right )} \, dx=e^{\left (\frac {3 \, x^{2} + {\left (x^{2} - 5 \, x\right )} \log \left (-x + e^{\left (x^{2} e^{x}\right )}\right ) - 18 \, x + 3 \, \log \left (3\right )}{x^{2} - 5 \, x}\right )} \] Input:

integrate((((x^6-8*x^5+5*x^4+50*x^3)*exp(x)+(-6*x+15)*log(3)+3*x^2)*exp(ex 
p(x)*x^2)+(6*x^2-15*x)*log(3)-x^4+7*x^3-25*x^2)*exp(((x^2-5*x)*log(exp(exp 
(x)*x^2)-x)+3*log(3)+3*x^2-18*x)/(x^2-5*x))/((x^4-10*x^3+25*x^2)*exp(exp(x 
)*x^2)-x^5+10*x^4-25*x^3),x, algorithm="fricas")
 

Output:

e^((3*x^2 + (x^2 - 5*x)*log(-x + e^(x^2*e^x)) - 18*x + 3*log(3))/(x^2 - 5* 
x))
 

Sympy [F(-1)]

Timed out. \[ \int \frac {e^{\frac {-18 x+3 x^2+3 \log (3)+\left (-5 x+x^2\right ) \log \left (e^{e^x x^2}-x\right )}{-5 x+x^2}} \left (-25 x^2+7 x^3-x^4+\left (-15 x+6 x^2\right ) \log (3)+e^{e^x x^2} \left (3 x^2+e^x \left (50 x^3+5 x^4-8 x^5+x^6\right )+(15-6 x) \log (3)\right )\right )}{-25 x^3+10 x^4-x^5+e^{e^x x^2} \left (25 x^2-10 x^3+x^4\right )} \, dx=\text {Timed out} \] Input:

integrate((((x**6-8*x**5+5*x**4+50*x**3)*exp(x)+(-6*x+15)*ln(3)+3*x**2)*ex 
p(exp(x)*x**2)+(6*x**2-15*x)*ln(3)-x**4+7*x**3-25*x**2)*exp(((x**2-5*x)*ln 
(exp(exp(x)*x**2)-x)+3*ln(3)+3*x**2-18*x)/(x**2-5*x))/((x**4-10*x**3+25*x* 
*2)*exp(exp(x)*x**2)-x**5+10*x**4-25*x**3),x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.26 \[ \int \frac {e^{\frac {-18 x+3 x^2+3 \log (3)+\left (-5 x+x^2\right ) \log \left (e^{e^x x^2}-x\right )}{-5 x+x^2}} \left (-25 x^2+7 x^3-x^4+\left (-15 x+6 x^2\right ) \log (3)+e^{e^x x^2} \left (3 x^2+e^x \left (50 x^3+5 x^4-8 x^5+x^6\right )+(15-6 x) \log (3)\right )\right )}{-25 x^3+10 x^4-x^5+e^{e^x x^2} \left (25 x^2-10 x^3+x^4\right )} \, dx=-{\left (x e^{3} - e^{\left (x^{2} e^{x} + 3\right )}\right )} e^{\left (\frac {3 \, \log \left (3\right )}{5 \, {\left (x - 5\right )}} - \frac {3 \, \log \left (3\right )}{5 \, x} - \frac {3}{x - 5}\right )} \] Input:

integrate((((x^6-8*x^5+5*x^4+50*x^3)*exp(x)+(-6*x+15)*log(3)+3*x^2)*exp(ex 
p(x)*x^2)+(6*x^2-15*x)*log(3)-x^4+7*x^3-25*x^2)*exp(((x^2-5*x)*log(exp(exp 
(x)*x^2)-x)+3*log(3)+3*x^2-18*x)/(x^2-5*x))/((x^4-10*x^3+25*x^2)*exp(exp(x 
)*x^2)-x^5+10*x^4-25*x^3),x, algorithm="maxima")
 

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 90 vs. \(2 (31) = 62\).

Time = 0.31 (sec) , antiderivative size = 90, normalized size of antiderivative = 2.65 \[ \int \frac {e^{\frac {-18 x+3 x^2+3 \log (3)+\left (-5 x+x^2\right ) \log \left (e^{e^x x^2}-x\right )}{-5 x+x^2}} \left (-25 x^2+7 x^3-x^4+\left (-15 x+6 x^2\right ) \log (3)+e^{e^x x^2} \left (3 x^2+e^x \left (50 x^3+5 x^4-8 x^5+x^6\right )+(15-6 x) \log (3)\right )\right )}{-25 x^3+10 x^4-x^5+e^{e^x x^2} \left (25 x^2-10 x^3+x^4\right )} \, dx=e^{\left (\frac {x^{2} \log \left (-x + e^{\left (x^{2} e^{x}\right )}\right )}{x^{2} - 5 \, x} + \frac {3 \, x^{2}}{x^{2} - 5 \, x} - \frac {5 \, x \log \left (-x + e^{\left (x^{2} e^{x}\right )}\right )}{x^{2} - 5 \, x} - \frac {18 \, x}{x^{2} - 5 \, x} + \frac {3 \, \log \left (3\right )}{x^{2} - 5 \, x}\right )} \] Input:

integrate((((x^6-8*x^5+5*x^4+50*x^3)*exp(x)+(-6*x+15)*log(3)+3*x^2)*exp(ex 
p(x)*x^2)+(6*x^2-15*x)*log(3)-x^4+7*x^3-25*x^2)*exp(((x^2-5*x)*log(exp(exp 
(x)*x^2)-x)+3*log(3)+3*x^2-18*x)/(x^2-5*x))/((x^4-10*x^3+25*x^2)*exp(exp(x 
)*x^2)-x^5+10*x^4-25*x^3),x, algorithm="giac")
 

Output:

e^(x^2*log(-x + e^(x^2*e^x))/(x^2 - 5*x) + 3*x^2/(x^2 - 5*x) - 5*x*log(-x 
+ e^(x^2*e^x))/(x^2 - 5*x) - 18*x/(x^2 - 5*x) + 3*log(3)/(x^2 - 5*x))
 

Mupad [F(-1)]

Timed out. \[ \int \frac {e^{\frac {-18 x+3 x^2+3 \log (3)+\left (-5 x+x^2\right ) \log \left (e^{e^x x^2}-x\right )}{-5 x+x^2}} \left (-25 x^2+7 x^3-x^4+\left (-15 x+6 x^2\right ) \log (3)+e^{e^x x^2} \left (3 x^2+e^x \left (50 x^3+5 x^4-8 x^5+x^6\right )+(15-6 x) \log (3)\right )\right )}{-25 x^3+10 x^4-x^5+e^{e^x x^2} \left (25 x^2-10 x^3+x^4\right )} \, dx=-\int \frac {{\mathrm {e}}^{\frac {18\,x-3\,\ln \left (3\right )-3\,x^2+\ln \left ({\mathrm {e}}^{x^2\,{\mathrm {e}}^x}-x\right )\,\left (5\,x-x^2\right )}{5\,x-x^2}}\,\left (\ln \left (3\right )\,\left (15\,x-6\,x^2\right )-{\mathrm {e}}^{x^2\,{\mathrm {e}}^x}\,\left ({\mathrm {e}}^x\,\left (x^6-8\,x^5+5\,x^4+50\,x^3\right )-\ln \left (3\right )\,\left (6\,x-15\right )+3\,x^2\right )+25\,x^2-7\,x^3+x^4\right )}{{\mathrm {e}}^{x^2\,{\mathrm {e}}^x}\,\left (x^4-10\,x^3+25\,x^2\right )-25\,x^3+10\,x^4-x^5} \,d x \] Input:

int(-(exp((18*x - 3*log(3) - 3*x^2 + log(exp(x^2*exp(x)) - x)*(5*x - x^2)) 
/(5*x - x^2))*(log(3)*(15*x - 6*x^2) - exp(x^2*exp(x))*(exp(x)*(50*x^3 + 5 
*x^4 - 8*x^5 + x^6) - log(3)*(6*x - 15) + 3*x^2) + 25*x^2 - 7*x^3 + x^4))/ 
(exp(x^2*exp(x))*(25*x^2 - 10*x^3 + x^4) - 25*x^3 + 10*x^4 - x^5),x)
                                                                                    
                                                                                    
 

Output:

-int((exp((18*x - 3*log(3) - 3*x^2 + log(exp(x^2*exp(x)) - x)*(5*x - x^2)) 
/(5*x - x^2))*(log(3)*(15*x - 6*x^2) - exp(x^2*exp(x))*(exp(x)*(50*x^3 + 5 
*x^4 - 8*x^5 + x^6) - log(3)*(6*x - 15) + 3*x^2) + 25*x^2 - 7*x^3 + x^4))/ 
(exp(x^2*exp(x))*(25*x^2 - 10*x^3 + x^4) - 25*x^3 + 10*x^4 - x^5), x)
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.26 \[ \int \frac {e^{\frac {-18 x+3 x^2+3 \log (3)+\left (-5 x+x^2\right ) \log \left (e^{e^x x^2}-x\right )}{-5 x+x^2}} \left (-25 x^2+7 x^3-x^4+\left (-15 x+6 x^2\right ) \log (3)+e^{e^x x^2} \left (3 x^2+e^x \left (50 x^3+5 x^4-8 x^5+x^6\right )+(15-6 x) \log (3)\right )\right )}{-25 x^3+10 x^4-x^5+e^{e^x x^2} \left (25 x^2-10 x^3+x^4\right )} \, dx=\frac {e^{\frac {3 \,\mathrm {log}\left (3\right )}{x^{2}-5 x}} e^{3} \left (e^{e^{x} x^{2}}-x \right )}{e^{\frac {3}{-5+x}}} \] Input:

int((((x^6-8*x^5+5*x^4+50*x^3)*exp(x)+(-6*x+15)*log(3)+3*x^2)*exp(exp(x)*x 
^2)+(6*x^2-15*x)*log(3)-x^4+7*x^3-25*x^2)*exp(((x^2-5*x)*log(exp(exp(x)*x^ 
2)-x)+3*log(3)+3*x^2-18*x)/(x^2-5*x))/((x^4-10*x^3+25*x^2)*exp(exp(x)*x^2) 
-x^5+10*x^4-25*x^3),x)
 

Output:

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