\(\int \frac {e^{-x^2} (e^{e^{e^{-x^2} x}} (e^{x^2} (-15+12 x+3 x^2)+e^{e^{-x^2} x} (-15 x+12 x^2+33 x^3-24 x^4-6 x^5))+(\frac {-5+4 x+x^2}{x})^{5 x/3} (e^{x^2} (25+5 x^2)+e^{x^2} (-25+20 x+5 x^2) \log (\frac {-5+4 x+x^2}{x})))}{-15+12 x+3 x^2} \, dx\) [1097]

Optimal result

Integrand size = 150, antiderivative size = 30 \[ \int \frac {e^{-x^2} \left (e^{e^{e^{-x^2} x}} \left (e^{x^2} \left (-15+12 x+3 x^2\right )+e^{e^{-x^2} x} \left (-15 x+12 x^2+33 x^3-24 x^4-6 x^5\right )\right )+\left (\frac {-5+4 x+x^2}{x}\right )^{5 x/3} \left (e^{x^2} \left (25+5 x^2\right )+e^{x^2} \left (-25+20 x+5 x^2\right ) \log \left (\frac {-5+4 x+x^2}{x}\right )\right )\right )}{-15+12 x+3 x^2} \, dx=e^{e^{e^{-x^2} x}} x+\left (4-\frac {5}{x}+x\right )^{5 x/3} \] Output:

exp(5/3*ln(4+x-5/x)*x)+x*exp(exp(x/exp(x^2)))
 

Mathematica [A] (verified)

Time = 0.71 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-x^2} \left (e^{e^{e^{-x^2} x}} \left (e^{x^2} \left (-15+12 x+3 x^2\right )+e^{e^{-x^2} x} \left (-15 x+12 x^2+33 x^3-24 x^4-6 x^5\right )\right )+\left (\frac {-5+4 x+x^2}{x}\right )^{5 x/3} \left (e^{x^2} \left (25+5 x^2\right )+e^{x^2} \left (-25+20 x+5 x^2\right ) \log \left (\frac {-5+4 x+x^2}{x}\right )\right )\right )}{-15+12 x+3 x^2} \, dx=e^{e^{e^{-x^2} x}} x+\left (4-\frac {5}{x}+x\right )^{5 x/3} \] Input:

Integrate[(E^E^(x/E^x^2)*(E^x^2*(-15 + 12*x + 3*x^2) + E^(x/E^x^2)*(-15*x 
+ 12*x^2 + 33*x^3 - 24*x^4 - 6*x^5)) + ((-5 + 4*x + x^2)/x)^((5*x)/3)*(E^x 
^2*(25 + 5*x^2) + E^x^2*(-25 + 20*x + 5*x^2)*Log[(-5 + 4*x + x^2)/x]))/(E^ 
x^2*(-15 + 12*x + 3*x^2)),x]
 

Output:

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

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.

\(\displaystyle \int \frac {e^{-x^2} \left (\left (\frac {x^2+4 x-5}{x}\right )^{5 x/3} \left (e^{x^2} \left (5 x^2+25\right )+e^{x^2} \left (5 x^2+20 x-25\right ) \log \left (\frac {x^2+4 x-5}{x}\right )\right )+e^{e^{e^{-x^2} x}} \left (e^{x^2} \left (3 x^2+12 x-15\right )+e^{e^{-x^2} x} \left (-6 x^5-24 x^4+33 x^3+12 x^2-15 x\right )\right )\right )}{3 x^2+12 x-15} \, dx\)

\(\Big \downarrow \) 7279

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

\(\Big \downarrow \) 2009

\(\displaystyle \int e^{e^{e^{-x^2} x}}dx+\int e^{-x^2+e^{-x^2} x+e^{e^{-x^2} x}} xdx-2 \int e^{-x^2+e^{-x^2} x+e^{e^{-x^2} x}} x^3dx+\frac {5}{3} \int \left (x+4-\frac {5}{x}\right )^{5 x/3}dx+\frac {5}{3} \int \frac {\left (x+4-\frac {5}{x}\right )^{5 x/3}}{x-1}dx-\frac {25}{3} \int \frac {\left (x+4-\frac {5}{x}\right )^{5 x/3}}{x+5}dx-\frac {25}{3} \int \frac {\left (x+4-\frac {5}{x}\right )^{\frac {5 x}{3}-1}}{x}dx-\frac {5}{3} \int x \left (x+4-\frac {5}{x}\right )^{\frac {5 x}{3}-1}dx+\left (x-\frac {5}{x}+4\right )^{5 x/3}\)

Input:

Int[(E^E^(x/E^x^2)*(E^x^2*(-15 + 12*x + 3*x^2) + E^(x/E^x^2)*(-15*x + 12*x 
^2 + 33*x^3 - 24*x^4 - 6*x^5)) + ((-5 + 4*x + x^2)/x)^((5*x)/3)*(E^x^2*(25 
 + 5*x^2) + E^x^2*(-25 + 20*x + 5*x^2)*Log[(-5 + 4*x + x^2)/x]))/(E^x^2*(- 
15 + 12*x + 3*x^2)),x]
 

Output:

$Aborted
 

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.07 (sec) , antiderivative size = 108, normalized size of antiderivative = 3.60

\[x \,{\mathrm e}^{{\mathrm e}^{{\mathrm e}^{-x^{2}} x}}+x^{-\frac {5 x}{3}} \left (x^{2}+4 x -5\right )^{\frac {5 x}{3}} {\mathrm e}^{-\frac {5 i \pi \,\operatorname {csgn}\left (\frac {i \left (x^{2}+4 x -5\right )}{x}\right ) x \left (-\operatorname {csgn}\left (\frac {i \left (x^{2}+4 x -5\right )}{x}\right )+\operatorname {csgn}\left (i \left (x^{2}+4 x -5\right )\right )\right ) \left (-\operatorname {csgn}\left (\frac {i \left (x^{2}+4 x -5\right )}{x}\right )+\operatorname {csgn}\left (\frac {i}{x}\right )\right )}{6}}\]

Input:

int((((-6*x^5-24*x^4+33*x^3+12*x^2-15*x)*exp(x/exp(x^2))+(3*x^2+12*x-15)*e 
xp(x^2))*exp(exp(x/exp(x^2)))+((5*x^2+20*x-25)*exp(x^2)*ln((x^2+4*x-5)/x)+ 
(5*x^2+25)*exp(x^2))*exp(5/3*x*ln((x^2+4*x-5)/x)))/(3*x^2+12*x-15)/exp(x^2 
),x)
 

Output:

x*exp(exp(exp(-x^2)*x))+x^(-5/3*x)*(x^2+4*x-5)^(5/3*x)*exp(-5/6*I*Pi*csgn( 
I/x*(x^2+4*x-5))*x*(-csgn(I/x*(x^2+4*x-5))+csgn(I*(x^2+4*x-5)))*(-csgn(I/x 
*(x^2+4*x-5))+csgn(I/x)))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.97 \[ \int \frac {e^{-x^2} \left (e^{e^{e^{-x^2} x}} \left (e^{x^2} \left (-15+12 x+3 x^2\right )+e^{e^{-x^2} x} \left (-15 x+12 x^2+33 x^3-24 x^4-6 x^5\right )\right )+\left (\frac {-5+4 x+x^2}{x}\right )^{5 x/3} \left (e^{x^2} \left (25+5 x^2\right )+e^{x^2} \left (-25+20 x+5 x^2\right ) \log \left (\frac {-5+4 x+x^2}{x}\right )\right )\right )}{-15+12 x+3 x^2} \, dx=x e^{\left (e^{\left (x e^{\left (-x^{2}\right )}\right )}\right )} + \left (\frac {x^{2} + 4 \, x - 5}{x}\right )^{\frac {5}{3} \, x} \] Input:

integrate((((-6*x^5-24*x^4+33*x^3+12*x^2-15*x)*exp(x/exp(x^2))+(3*x^2+12*x 
-15)*exp(x^2))*exp(exp(x/exp(x^2)))+((5*x^2+20*x-25)*exp(x^2)*log((x^2+4*x 
-5)/x)+(5*x^2+25)*exp(x^2))*exp(5/3*x*log((x^2+4*x-5)/x)))/(3*x^2+12*x-15) 
/exp(x^2),x, algorithm="fricas")
 

Output:

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

Sympy [F(-1)]

Timed out. \[ \int \frac {e^{-x^2} \left (e^{e^{e^{-x^2} x}} \left (e^{x^2} \left (-15+12 x+3 x^2\right )+e^{e^{-x^2} x} \left (-15 x+12 x^2+33 x^3-24 x^4-6 x^5\right )\right )+\left (\frac {-5+4 x+x^2}{x}\right )^{5 x/3} \left (e^{x^2} \left (25+5 x^2\right )+e^{x^2} \left (-25+20 x+5 x^2\right ) \log \left (\frac {-5+4 x+x^2}{x}\right )\right )\right )}{-15+12 x+3 x^2} \, dx=\text {Timed out} \] Input:

integrate((((-6*x**5-24*x**4+33*x**3+12*x**2-15*x)*exp(x/exp(x**2))+(3*x** 
2+12*x-15)*exp(x**2))*exp(exp(x/exp(x**2)))+((5*x**2+20*x-25)*exp(x**2)*ln 
((x**2+4*x-5)/x)+(5*x**2+25)*exp(x**2))*exp(5/3*x*ln((x**2+4*x-5)/x)))/(3* 
x**2+12*x-15)/exp(x**2),x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.43 \[ \int \frac {e^{-x^2} \left (e^{e^{e^{-x^2} x}} \left (e^{x^2} \left (-15+12 x+3 x^2\right )+e^{e^{-x^2} x} \left (-15 x+12 x^2+33 x^3-24 x^4-6 x^5\right )\right )+\left (\frac {-5+4 x+x^2}{x}\right )^{5 x/3} \left (e^{x^2} \left (25+5 x^2\right )+e^{x^2} \left (-25+20 x+5 x^2\right ) \log \left (\frac {-5+4 x+x^2}{x}\right )\right )\right )}{-15+12 x+3 x^2} \, dx=\frac {x e^{\left (\frac {5}{3} \, x \log \left (x\right ) + e^{\left (x e^{\left (-x^{2}\right )}\right )}\right )} + e^{\left (\frac {5}{3} \, x \log \left (x + 5\right ) + \frac {5}{3} \, x \log \left (x - 1\right )\right )}}{x^{\frac {5}{3} \, x}} \] Input:

integrate((((-6*x^5-24*x^4+33*x^3+12*x^2-15*x)*exp(x/exp(x^2))+(3*x^2+12*x 
-15)*exp(x^2))*exp(exp(x/exp(x^2)))+((5*x^2+20*x-25)*exp(x^2)*log((x^2+4*x 
-5)/x)+(5*x^2+25)*exp(x^2))*exp(5/3*x*log((x^2+4*x-5)/x)))/(3*x^2+12*x-15) 
/exp(x^2),x, algorithm="maxima")
 

Output:

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

Giac [F]

\[ \int \frac {e^{-x^2} \left (e^{e^{e^{-x^2} x}} \left (e^{x^2} \left (-15+12 x+3 x^2\right )+e^{e^{-x^2} x} \left (-15 x+12 x^2+33 x^3-24 x^4-6 x^5\right )\right )+\left (\frac {-5+4 x+x^2}{x}\right )^{5 x/3} \left (e^{x^2} \left (25+5 x^2\right )+e^{x^2} \left (-25+20 x+5 x^2\right ) \log \left (\frac {-5+4 x+x^2}{x}\right )\right )\right )}{-15+12 x+3 x^2} \, dx=\int { \frac {{\left (5 \, {\left ({\left (x^{2} + 4 \, x - 5\right )} e^{\left (x^{2}\right )} \log \left (\frac {x^{2} + 4 \, x - 5}{x}\right ) + {\left (x^{2} + 5\right )} e^{\left (x^{2}\right )}\right )} \left (\frac {x^{2} + 4 \, x - 5}{x}\right )^{\frac {5}{3} \, x} + 3 \, {\left ({\left (x^{2} + 4 \, x - 5\right )} e^{\left (x^{2}\right )} - {\left (2 \, x^{5} + 8 \, x^{4} - 11 \, x^{3} - 4 \, x^{2} + 5 \, x\right )} e^{\left (x e^{\left (-x^{2}\right )}\right )}\right )} e^{\left (e^{\left (x e^{\left (-x^{2}\right )}\right )}\right )}\right )} e^{\left (-x^{2}\right )}}{3 \, {\left (x^{2} + 4 \, x - 5\right )}} \,d x } \] Input:

integrate((((-6*x^5-24*x^4+33*x^3+12*x^2-15*x)*exp(x/exp(x^2))+(3*x^2+12*x 
-15)*exp(x^2))*exp(exp(x/exp(x^2)))+((5*x^2+20*x-25)*exp(x^2)*log((x^2+4*x 
-5)/x)+(5*x^2+25)*exp(x^2))*exp(5/3*x*log((x^2+4*x-5)/x)))/(3*x^2+12*x-15) 
/exp(x^2),x, algorithm="giac")
 

Output:

integrate(1/3*(5*((x^2 + 4*x - 5)*e^(x^2)*log((x^2 + 4*x - 5)/x) + (x^2 + 
5)*e^(x^2))*((x^2 + 4*x - 5)/x)^(5/3*x) + 3*((x^2 + 4*x - 5)*e^(x^2) - (2* 
x^5 + 8*x^4 - 11*x^3 - 4*x^2 + 5*x)*e^(x*e^(-x^2)))*e^(e^(x*e^(-x^2))))*e^ 
(-x^2)/(x^2 + 4*x - 5), x)
 

Mupad [B] (verification not implemented)

Time = 2.06 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.83 \[ \int \frac {e^{-x^2} \left (e^{e^{e^{-x^2} x}} \left (e^{x^2} \left (-15+12 x+3 x^2\right )+e^{e^{-x^2} x} \left (-15 x+12 x^2+33 x^3-24 x^4-6 x^5\right )\right )+\left (\frac {-5+4 x+x^2}{x}\right )^{5 x/3} \left (e^{x^2} \left (25+5 x^2\right )+e^{x^2} \left (-25+20 x+5 x^2\right ) \log \left (\frac {-5+4 x+x^2}{x}\right )\right )\right )}{-15+12 x+3 x^2} \, dx=x\,{\mathrm {e}}^{{\mathrm {e}}^{x\,{\mathrm {e}}^{-x^2}}}+{\left (x-\frac {5}{x}+4\right )}^{\frac {5\,x}{3}} \] Input:

int((exp(-x^2)*(exp(exp(x*exp(-x^2)))*(exp(x^2)*(12*x + 3*x^2 - 15) - exp( 
x*exp(-x^2))*(15*x - 12*x^2 - 33*x^3 + 24*x^4 + 6*x^5)) + exp((5*x*log((4* 
x + x^2 - 5)/x))/3)*(exp(x^2)*(5*x^2 + 25) + exp(x^2)*log((4*x + x^2 - 5)/ 
x)*(20*x + 5*x^2 - 25))))/(12*x + 3*x^2 - 15),x)
 

Output:

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

Reduce [F]

\[ \int \frac {e^{-x^2} \left (e^{e^{e^{-x^2} x}} \left (e^{x^2} \left (-15+12 x+3 x^2\right )+e^{e^{-x^2} x} \left (-15 x+12 x^2+33 x^3-24 x^4-6 x^5\right )\right )+\left (\frac {-5+4 x+x^2}{x}\right )^{5 x/3} \left (e^{x^2} \left (25+5 x^2\right )+e^{x^2} \left (-25+20 x+5 x^2\right ) \log \left (\frac {-5+4 x+x^2}{x}\right )\right )\right )}{-15+12 x+3 x^2} \, dx=e^{e^{\frac {x}{e^{x^{2}}}}} x +\frac {25 \left (\int \frac {\left (x^{2}+4 x -5\right )^{\frac {5 x}{3}}}{x^{\frac {5 x}{3}} x^{2}+4 x^{\frac {5 x}{3}} x -5 x^{\frac {5 x}{3}}}d x \right )}{3}+\frac {5 \left (\int \frac {\left (x^{2}+4 x -5\right )^{\frac {5 x}{3}} x^{2}}{x^{\frac {5 x}{3}} x^{2}+4 x^{\frac {5 x}{3}} x -5 x^{\frac {5 x}{3}}}d x \right )}{3}+\frac {5 \left (\int \frac {\left (x^{2}+4 x -5\right )^{\frac {5 x}{3}} \mathrm {log}\left (\frac {x^{2}+4 x -5}{x}\right ) x^{2}}{x^{\frac {5 x}{3}} x^{2}+4 x^{\frac {5 x}{3}} x -5 x^{\frac {5 x}{3}}}d x \right )}{3}+\frac {20 \left (\int \frac {\left (x^{2}+4 x -5\right )^{\frac {5 x}{3}} \mathrm {log}\left (\frac {x^{2}+4 x -5}{x}\right ) x}{x^{\frac {5 x}{3}} x^{2}+4 x^{\frac {5 x}{3}} x -5 x^{\frac {5 x}{3}}}d x \right )}{3}-\frac {25 \left (\int \frac {\left (x^{2}+4 x -5\right )^{\frac {5 x}{3}} \mathrm {log}\left (\frac {x^{2}+4 x -5}{x}\right )}{x^{\frac {5 x}{3}} x^{2}+4 x^{\frac {5 x}{3}} x -5 x^{\frac {5 x}{3}}}d x \right )}{3} \] Input:

int((((-6*x^5-24*x^4+33*x^3+12*x^2-15*x)*exp(x/exp(x^2))+(3*x^2+12*x-15)*e 
xp(x^2))*exp(exp(x/exp(x^2)))+((5*x^2+20*x-25)*exp(x^2)*log((x^2+4*x-5)/x) 
+(5*x^2+25)*exp(x^2))*exp(5/3*x*log((x^2+4*x-5)/x)))/(3*x^2+12*x-15)/exp(x 
^2),x)
 

Output:

(3*e**(e**(x/e**(x**2)))*x + 25*int((x**2 + 4*x - 5)**((5*x)/3)/(x**((5*x) 
/3)*x**2 + 4*x**((5*x)/3)*x - 5*x**((5*x)/3)),x) + 5*int(((x**2 + 4*x - 5) 
**((5*x)/3)*x**2)/(x**((5*x)/3)*x**2 + 4*x**((5*x)/3)*x - 5*x**((5*x)/3)), 
x) + 5*int(((x**2 + 4*x - 5)**((5*x)/3)*log((x**2 + 4*x - 5)/x)*x**2)/(x** 
((5*x)/3)*x**2 + 4*x**((5*x)/3)*x - 5*x**((5*x)/3)),x) + 20*int(((x**2 + 4 
*x - 5)**((5*x)/3)*log((x**2 + 4*x - 5)/x)*x)/(x**((5*x)/3)*x**2 + 4*x**(( 
5*x)/3)*x - 5*x**((5*x)/3)),x) - 25*int(((x**2 + 4*x - 5)**((5*x)/3)*log(( 
x**2 + 4*x - 5)/x))/(x**((5*x)/3)*x**2 + 4*x**((5*x)/3)*x - 5*x**((5*x)/3) 
),x))/3