\(\int \frac {2 x^3+e^{\frac {1}{2} (-e^{\frac {1}{x^2}+e^x x}+e^{\frac {1}{x^2}} (3+x))} (e^{\frac {1}{x^2}} (-6-2 x+x^3)+e^{e^x x} (2 e^{\frac {1}{x^2}}+e^{\frac {1}{x^2}+x} (-x^3-x^4)))}{2 x^3} \, dx\) [779]

Optimal result

Integrand size = 93, antiderivative size = 31 \[ \int \frac {2 x^3+e^{\frac {1}{2} \left (-e^{\frac {1}{x^2}+e^x x}+e^{\frac {1}{x^2}} (3+x)\right )} \left (e^{\frac {1}{x^2}} \left (-6-2 x+x^3\right )+e^{e^x x} \left (2 e^{\frac {1}{x^2}}+e^{\frac {1}{x^2}+x} \left (-x^3-x^4\right )\right )\right )}{2 x^3} \, dx=-3-e^3+e^{\frac {1}{2} e^{\frac {1}{x^2}} \left (3-e^{e^x x}+x\right )}+x \] Output:

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

Mathematica [A] (verified)

Time = 6.11 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.03 \[ \int \frac {2 x^3+e^{\frac {1}{2} \left (-e^{\frac {1}{x^2}+e^x x}+e^{\frac {1}{x^2}} (3+x)\right )} \left (e^{\frac {1}{x^2}} \left (-6-2 x+x^3\right )+e^{e^x x} \left (2 e^{\frac {1}{x^2}}+e^{\frac {1}{x^2}+x} \left (-x^3-x^4\right )\right )\right )}{2 x^3} \, dx=e^{-\frac {1}{2} e^{\frac {1}{x^2}+e^x x}+\frac {1}{2} e^{\frac {1}{x^2}} (3+x)}+x \] Input:

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

Output:

E^(-1/2*E^(x^(-2) + E^x*x) + (E^x^(-2)*(3 + 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.

\(\displaystyle \int \frac {2 x^3+e^{\frac {1}{2} \left (e^{\frac {1}{x^2}} (x+3)-e^{\frac {1}{x^2}+e^x x}\right )} \left (e^{\frac {1}{x^2}} \left (x^3-2 x-6\right )+e^{e^x x} \left (2 e^{\frac {1}{x^2}}+e^{\frac {1}{x^2}+x} \left (-x^4-x^3\right )\right )\right )}{2 x^3} \, dx\)

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 2010

\(\displaystyle \frac {1}{2} \int \left (2-\frac {\exp \left (\frac {1}{2} e^{\frac {1}{x^2}} (x+3)-\frac {1}{2} e^{e^x x+\frac {1}{x^2}}+\frac {1}{x^2}\right ) \left (e^{e^x x+x} x^4+e^{e^x x+x} x^3-x^3+2 x-2 e^{e^x x}+6\right )}{x^3}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{2} \left (\int \exp \left (\frac {1}{2} e^{\frac {1}{x^2}} (x+3)-\frac {1}{2} e^{e^x x+\frac {1}{x^2}}+\frac {1}{x^2}\right )dx-\int \exp \left (e^x x+x-\frac {1}{2} e^{e^x x+\frac {1}{x^2}}+\frac {1}{2} e^{\frac {1}{x^2}} (x+3)+\frac {1}{x^2}\right )dx-2 \int \frac {\exp \left (\frac {1}{2} e^{\frac {1}{x^2}} (x+3)-\frac {1}{2} e^{e^x x+\frac {1}{x^2}}+\frac {1}{x^2}\right )}{x^2}dx-\int \exp \left (e^x x+x-\frac {1}{2} e^{e^x x+\frac {1}{x^2}}+\frac {1}{2} e^{\frac {1}{x^2}} (x+3)+\frac {1}{x^2}\right ) xdx-6 \int \frac {\exp \left (\frac {1}{2} e^{\frac {1}{x^2}} (x+3)-\frac {1}{2} e^{e^x x+\frac {1}{x^2}}+\frac {1}{x^2}\right )}{x^3}dx+2 \int \frac {\exp \left (e^x x-\frac {1}{2} e^{e^x x+\frac {1}{x^2}}+\frac {1}{2} e^{\frac {1}{x^2}} (x+3)+\frac {1}{x^2}\right )}{x^3}dx+2 x\right )\)

Input:

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

Output:

$Aborted
 

Maple [A] (verified)

Time = 10.01 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.65

method	result	size
parallelrisch	\(x +{\mathrm e}^{-\frac {{\mathrm e}^{\frac {1}{x^{2}}} \left ({\mathrm e}^{{\mathrm e}^{x} x}-x -3\right )}{2}}\)	\(20\)
risch	\(x +{\mathrm e}^{-\frac {{\mathrm e}^{\frac {{\mathrm e}^{x} x^{3}+1}{x^{2}}}}{2}+\frac {{\mathrm e}^{\frac {1}{x^{2}}} x}{2}+\frac {3 \,{\mathrm e}^{\frac {1}{x^{2}}}}{2}}\)	\(33\)

Input:

int(1/2*((((-x^4-x^3)*exp(1/x^2)*exp(x)+2*exp(1/x^2))*exp(exp(x)*x)+(x^3-2 
*x-6)*exp(1/x^2))*exp(-1/2*exp(1/x^2)*exp(exp(x)*x)+1/2*(3+x)*exp(1/x^2))+ 
2*x^3)/x^3,x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.32 \[ \int \frac {2 x^3+e^{\frac {1}{2} \left (-e^{\frac {1}{x^2}+e^x x}+e^{\frac {1}{x^2}} (3+x)\right )} \left (e^{\frac {1}{x^2}} \left (-6-2 x+x^3\right )+e^{e^x x} \left (2 e^{\frac {1}{x^2}}+e^{\frac {1}{x^2}+x} \left (-x^3-x^4\right )\right )\right )}{2 x^3} \, dx=x + e^{\left (\frac {1}{2} \, {\left ({\left (x + 3\right )} e^{\left (\frac {x^{3} + 1}{x^{2}}\right )} - e^{\left (x + \frac {x^{3} e^{x} + 1}{x^{2}}\right )}\right )} e^{\left (-x\right )}\right )} \] Input:

integrate(1/2*((((-x^4-x^3)*exp(1/x^2)*exp(x)+2*exp(1/x^2))*exp(exp(x)*x)+ 
(x^3-2*x-6)*exp(1/x^2))*exp(-1/2*exp(1/x^2)*exp(exp(x)*x)+1/2*(3+x)*exp(1/ 
x^2))+2*x^3)/x^3,x, algorithm="fricas")
 

Output:

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

Sympy [A] (verification not implemented)

Time = 5.63 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00 \[ \int \frac {2 x^3+e^{\frac {1}{2} \left (-e^{\frac {1}{x^2}+e^x x}+e^{\frac {1}{x^2}} (3+x)\right )} \left (e^{\frac {1}{x^2}} \left (-6-2 x+x^3\right )+e^{e^x x} \left (2 e^{\frac {1}{x^2}}+e^{\frac {1}{x^2}+x} \left (-x^3-x^4\right )\right )\right )}{2 x^3} \, dx=x + e^{\left (\frac {x}{2} + \frac {3}{2}\right ) e^{\frac {1}{x^{2}}} - \frac {e^{\frac {1}{x^{2}}} e^{x e^{x}}}{2}} \] Input:

integrate(1/2*((((-x**4-x**3)*exp(1/x**2)*exp(x)+2*exp(1/x**2))*exp(exp(x) 
*x)+(x**3-2*x-6)*exp(1/x**2))*exp(-1/2*exp(1/x**2)*exp(exp(x)*x)+1/2*(3+x) 
*exp(1/x**2))+2*x**3)/x**3,x)
 

Output:

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

Maxima [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.90 \[ \int \frac {2 x^3+e^{\frac {1}{2} \left (-e^{\frac {1}{x^2}+e^x x}+e^{\frac {1}{x^2}} (3+x)\right )} \left (e^{\frac {1}{x^2}} \left (-6-2 x+x^3\right )+e^{e^x x} \left (2 e^{\frac {1}{x^2}}+e^{\frac {1}{x^2}+x} \left (-x^3-x^4\right )\right )\right )}{2 x^3} \, dx=x + e^{\left (\frac {1}{2} \, x e^{\left (\frac {1}{x^{2}}\right )} - \frac {1}{2} \, e^{\left (x e^{x} + \frac {1}{x^{2}}\right )} + \frac {3}{2} \, e^{\left (\frac {1}{x^{2}}\right )}\right )} \] Input:

integrate(1/2*((((-x^4-x^3)*exp(1/x^2)*exp(x)+2*exp(1/x^2))*exp(exp(x)*x)+ 
(x^3-2*x-6)*exp(1/x^2))*exp(-1/2*exp(1/x^2)*exp(exp(x)*x)+1/2*(3+x)*exp(1/ 
x^2))+2*x^3)/x^3,x, algorithm="maxima")
 

Output:

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

Giac [F]

\[ \int \frac {2 x^3+e^{\frac {1}{2} \left (-e^{\frac {1}{x^2}+e^x x}+e^{\frac {1}{x^2}} (3+x)\right )} \left (e^{\frac {1}{x^2}} \left (-6-2 x+x^3\right )+e^{e^x x} \left (2 e^{\frac {1}{x^2}}+e^{\frac {1}{x^2}+x} \left (-x^3-x^4\right )\right )\right )}{2 x^3} \, dx=\int { \frac {2 \, x^{3} - {\left ({\left ({\left (x^{4} + x^{3}\right )} e^{\left (x + \frac {1}{x^{2}}\right )} - 2 \, e^{\left (\frac {1}{x^{2}}\right )}\right )} e^{\left (x e^{x}\right )} - {\left (x^{3} - 2 \, x - 6\right )} e^{\left (\frac {1}{x^{2}}\right )}\right )} e^{\left (\frac {1}{2} \, {\left (x + 3\right )} e^{\left (\frac {1}{x^{2}}\right )} - \frac {1}{2} \, e^{\left (x e^{x} + \frac {1}{x^{2}}\right )}\right )}}{2 \, x^{3}} \,d x } \] Input:

integrate(1/2*((((-x^4-x^3)*exp(1/x^2)*exp(x)+2*exp(1/x^2))*exp(exp(x)*x)+ 
(x^3-2*x-6)*exp(1/x^2))*exp(-1/2*exp(1/x^2)*exp(exp(x)*x)+1/2*(3+x)*exp(1/ 
x^2))+2*x^3)/x^3,x, algorithm="giac")
 

Output:

integrate(1/2*(2*x^3 - (((x^4 + x^3)*e^(x + 1/x^2) - 2*e^(x^(-2)))*e^(x*e^ 
x) - (x^3 - 2*x - 6)*e^(x^(-2)))*e^(1/2*(x + 3)*e^(x^(-2)) - 1/2*e^(x*e^x 
+ 1/x^2)))/x^3, x)
 

Mupad [B] (verification not implemented)

Time = 7.20 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.90 \[ \int \frac {2 x^3+e^{\frac {1}{2} \left (-e^{\frac {1}{x^2}+e^x x}+e^{\frac {1}{x^2}} (3+x)\right )} \left (e^{\frac {1}{x^2}} \left (-6-2 x+x^3\right )+e^{e^x x} \left (2 e^{\frac {1}{x^2}}+e^{\frac {1}{x^2}+x} \left (-x^3-x^4\right )\right )\right )}{2 x^3} \, dx=x+{\mathrm {e}}^{\frac {3\,{\mathrm {e}}^{\frac {1}{x^2}}}{2}+\frac {x\,{\mathrm {e}}^{\frac {1}{x^2}}}{2}-\frac {{\mathrm {e}}^{x\,{\mathrm {e}}^x}\,{\mathrm {e}}^{\frac {1}{x^2}}}{2}} \] Input:

int(-((exp((exp(1/x^2)*(x + 3))/2 - (exp(x*exp(x))*exp(1/x^2))/2)*(exp(1/x 
^2)*(2*x - x^3 + 6) - exp(x*exp(x))*(2*exp(1/x^2) - exp(1/x^2)*exp(x)*(x^3 
 + x^4))))/2 - x^3)/x^3,x)
 

Output:

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

Reduce [B] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 62, normalized size of antiderivative = 2.00 \[ \int \frac {2 x^3+e^{\frac {1}{2} \left (-e^{\frac {1}{x^2}+e^x x}+e^{\frac {1}{x^2}} (3+x)\right )} \left (e^{\frac {1}{x^2}} \left (-6-2 x+x^3\right )+e^{e^x x} \left (2 e^{\frac {1}{x^2}}+e^{\frac {1}{x^2}+x} \left (-x^3-x^4\right )\right )\right )}{2 x^3} \, dx=\frac {e^{\frac {e^{\frac {e^{x} x^{3}+1}{x^{2}}}}{2}} x +e^{\frac {e^{\frac {1}{x^{2}}} x}{2}+\frac {3 e^{\frac {1}{x^{2}}}}{2}}}{e^{\frac {e^{\frac {e^{x} x^{3}+1}{x^{2}}}}{2}}} \] Input:

int(1/2*((((-x^4-x^3)*exp(1/x^2)*exp(x)+2*exp(1/x^2))*exp(exp(x)*x)+(x^3-2 
*x-6)*exp(1/x^2))*exp(-1/2*exp(1/x^2)*exp(exp(x)*x)+1/2*(3+x)*exp(1/x^2))+ 
2*x^3)/x^3,x)
 

Output:

(e**(e**((e**x*x**3 + 1)/x**2)/2)*x + e**((e**(1/x**2)*x + 3*e**(1/x**2))/ 
2))/e**(e**((e**x*x**3 + 1)/x**2)/2)