\(\int \frac {1}{5} (10 x+e^{e^{\frac {1}{5} (10 x-2 e^{e^x} x-2 e^x x+2 x^2)}} (-5+e^{\frac {1}{5} (10 x-2 e^{e^x} x-2 e^x x+2 x^2)} (-10 x-4 x^2+e^x (2 x+2 x^2)+e^{e^x} (2 x+2 e^x x^2)))) \, dx\) [2402]

Optimal result

Integrand size = 112, antiderivative size = 30 \[ \int \frac {1}{5} \left (10 x+e^{e^{\frac {1}{5} \left (10 x-2 e^{e^x} x-2 e^x x+2 x^2\right )}} \left (-5+e^{\frac {1}{5} \left (10 x-2 e^{e^x} x-2 e^x x+2 x^2\right )} \left (-10 x-4 x^2+e^x \left (2 x+2 x^2\right )+e^{e^x} \left (2 x+2 e^x x^2\right )\right )\right )\right ) \, dx=x \left (-e^{e^{\frac {2}{5} x \left (5-e^{e^x}-e^x+x\right )}}+x\right ) \] Output:

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

Mathematica [A] (verified)

Time = 0.26 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00 \[ \int \frac {1}{5} \left (10 x+e^{e^{\frac {1}{5} \left (10 x-2 e^{e^x} x-2 e^x x+2 x^2\right )}} \left (-5+e^{\frac {1}{5} \left (10 x-2 e^{e^x} x-2 e^x x+2 x^2\right )} \left (-10 x-4 x^2+e^x \left (2 x+2 x^2\right )+e^{e^x} \left (2 x+2 e^x x^2\right )\right )\right )\right ) \, dx=x \left (-e^{e^{\frac {2}{5} x \left (5-e^{e^x}-e^x+x\right )}}+x\right ) \] Input:

Integrate[(10*x + E^E^((10*x - 2*E^E^x*x - 2*E^x*x + 2*x^2)/5)*(-5 + E^((1 
0*x - 2*E^E^x*x - 2*E^x*x + 2*x^2)/5)*(-10*x - 4*x^2 + E^x*(2*x + 2*x^2) + 
 E^E^x*(2*x + 2*E^x*x^2))))/5,x]
 

Output:

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

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(111\) vs. \(2(30)=60\).

Time = 0.43 (sec) , antiderivative size = 111, normalized size of antiderivative = 3.70, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.018, Rules used = {27, 2009}

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.

\(\displaystyle \int \frac {1}{5} \left (e^{e^{\frac {1}{5} \left (2 x^2-2 e^{e^x} x-2 e^x x+10 x\right )}} \left (e^{\frac {1}{5} \left (2 x^2-2 e^{e^x} x-2 e^x x+10 x\right )} \left (-4 x^2+e^x \left (2 x^2+2 x\right )+e^{e^x} \left (2 e^x x^2+2 x\right )-10 x\right )-5\right )+10 x\right ) \, dx\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{5} \int \left (10 x-e^{e^{\frac {2}{5} \left (x^2-e^{e^x} x-e^x x+5 x\right )}} \left (2 e^{\frac {2}{5} \left (x^2-e^{e^x} x-e^x x+5 x\right )} \left (2 x^2+5 x-e^x \left (x^2+x\right )-e^{e^x} \left (e^x x^2+x\right )\right )+5\right )\right )dx\)

\(\Big \downarrow \) 2009

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

Input:

Int[(10*x + E^E^((10*x - 2*E^E^x*x - 2*E^x*x + 2*x^2)/5)*(-5 + E^((10*x - 
2*E^E^x*x - 2*E^x*x + 2*x^2)/5)*(-10*x - 4*x^2 + E^x*(2*x + 2*x^2) + E^E^x 
*(2*x + 2*E^x*x^2))))/5,x]
 

Output:

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

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

Maple [A] (verified)

Time = 0.68 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.77

method	result	size
risch	\(-x \,{\mathrm e}^{{\mathrm e}^{-\frac {2 x \left ({\mathrm e}^{{\mathrm e}^{x}}+{\mathrm e}^{x}-x -5\right )}{5}}}+x^{2}\)	\(23\)
parallelrisch	\(-x \,{\mathrm e}^{{\mathrm e}^{-\frac {2 x \left ({\mathrm e}^{{\mathrm e}^{x}}+{\mathrm e}^{x}-x -5\right )}{5}}}+x^{2}\)	\(23\)

Input:

int(1/5*(((2*exp(x)*x^2+2*x)*exp(exp(x))+(2*x^2+2*x)*exp(x)-4*x^2-10*x)*ex 
p(-2/5*x*exp(exp(x))-2/5*exp(x)*x+2/5*x^2+2*x)-5)*exp(exp(-2/5*x*exp(exp(x 
))-2/5*exp(x)*x+2/5*x^2+2*x))+2*x,x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.97 \[ \int \frac {1}{5} \left (10 x+e^{e^{\frac {1}{5} \left (10 x-2 e^{e^x} x-2 e^x x+2 x^2\right )}} \left (-5+e^{\frac {1}{5} \left (10 x-2 e^{e^x} x-2 e^x x+2 x^2\right )} \left (-10 x-4 x^2+e^x \left (2 x+2 x^2\right )+e^{e^x} \left (2 x+2 e^x x^2\right )\right )\right )\right ) \, dx=x^{2} - x e^{\left (e^{\left (\frac {2}{5} \, x^{2} - \frac {2}{5} \, x e^{x} - \frac {2}{5} \, x e^{\left (e^{x}\right )} + 2 \, x\right )}\right )} \] Input:

integrate(1/5*(((2*exp(x)*x^2+2*x)*exp(exp(x))+(2*x^2+2*x)*exp(x)-4*x^2-10 
*x)*exp(-2/5*x*exp(exp(x))-2/5*exp(x)*x+2/5*x^2+2*x)-5)*exp(exp(-2/5*x*exp 
(exp(x))-2/5*exp(x)*x+2/5*x^2+2*x))+2*x,x, algorithm="fricas")
 

Output:

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

Sympy [A] (verification not implemented)

Time = 4.41 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.20 \[ \int \frac {1}{5} \left (10 x+e^{e^{\frac {1}{5} \left (10 x-2 e^{e^x} x-2 e^x x+2 x^2\right )}} \left (-5+e^{\frac {1}{5} \left (10 x-2 e^{e^x} x-2 e^x x+2 x^2\right )} \left (-10 x-4 x^2+e^x \left (2 x+2 x^2\right )+e^{e^x} \left (2 x+2 e^x x^2\right )\right )\right )\right ) \, dx=x^{2} - x e^{e^{\frac {2 x^{2}}{5} - \frac {2 x e^{x}}{5} - \frac {2 x e^{e^{x}}}{5} + 2 x}} \] Input:

integrate(1/5*(((2*exp(x)*x**2+2*x)*exp(exp(x))+(2*x**2+2*x)*exp(x)-4*x**2 
-10*x)*exp(-2/5*x*exp(exp(x))-2/5*exp(x)*x+2/5*x**2+2*x)-5)*exp(exp(-2/5*x 
*exp(exp(x))-2/5*exp(x)*x+2/5*x**2+2*x))+2*x,x)
 

Output:

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

Maxima [F]

\[ \int \frac {1}{5} \left (10 x+e^{e^{\frac {1}{5} \left (10 x-2 e^{e^x} x-2 e^x x+2 x^2\right )}} \left (-5+e^{\frac {1}{5} \left (10 x-2 e^{e^x} x-2 e^x x+2 x^2\right )} \left (-10 x-4 x^2+e^x \left (2 x+2 x^2\right )+e^{e^x} \left (2 x+2 e^x x^2\right )\right )\right )\right ) \, dx=\int { -\frac {1}{5} \, {\left (2 \, {\left (2 \, x^{2} - {\left (x^{2} + x\right )} e^{x} - {\left (x^{2} e^{x} + x\right )} e^{\left (e^{x}\right )} + 5 \, x\right )} e^{\left (\frac {2}{5} \, x^{2} - \frac {2}{5} \, x e^{x} - \frac {2}{5} \, x e^{\left (e^{x}\right )} + 2 \, x\right )} + 5\right )} e^{\left (e^{\left (\frac {2}{5} \, x^{2} - \frac {2}{5} \, x e^{x} - \frac {2}{5} \, x e^{\left (e^{x}\right )} + 2 \, x\right )}\right )} + 2 \, x \,d x } \] Input:

integrate(1/5*(((2*exp(x)*x^2+2*x)*exp(exp(x))+(2*x^2+2*x)*exp(x)-4*x^2-10 
*x)*exp(-2/5*x*exp(exp(x))-2/5*exp(x)*x+2/5*x^2+2*x)-5)*exp(exp(-2/5*x*exp 
(exp(x))-2/5*exp(x)*x+2/5*x^2+2*x))+2*x,x, algorithm="maxima")
 

Output:

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

Giac [F]

\[ \int \frac {1}{5} \left (10 x+e^{e^{\frac {1}{5} \left (10 x-2 e^{e^x} x-2 e^x x+2 x^2\right )}} \left (-5+e^{\frac {1}{5} \left (10 x-2 e^{e^x} x-2 e^x x+2 x^2\right )} \left (-10 x-4 x^2+e^x \left (2 x+2 x^2\right )+e^{e^x} \left (2 x+2 e^x x^2\right )\right )\right )\right ) \, dx=\int { -\frac {1}{5} \, {\left (2 \, {\left (2 \, x^{2} - {\left (x^{2} + x\right )} e^{x} - {\left (x^{2} e^{x} + x\right )} e^{\left (e^{x}\right )} + 5 \, x\right )} e^{\left (\frac {2}{5} \, x^{2} - \frac {2}{5} \, x e^{x} - \frac {2}{5} \, x e^{\left (e^{x}\right )} + 2 \, x\right )} + 5\right )} e^{\left (e^{\left (\frac {2}{5} \, x^{2} - \frac {2}{5} \, x e^{x} - \frac {2}{5} \, x e^{\left (e^{x}\right )} + 2 \, x\right )}\right )} + 2 \, x \,d x } \] Input:

integrate(1/5*(((2*exp(x)*x^2+2*x)*exp(exp(x))+(2*x^2+2*x)*exp(x)-4*x^2-10 
*x)*exp(-2/5*x*exp(exp(x))-2/5*exp(x)*x+2/5*x^2+2*x)-5)*exp(exp(-2/5*x*exp 
(exp(x))-2/5*exp(x)*x+2/5*x^2+2*x))+2*x,x, algorithm="giac")
 

Output:

integrate(-1/5*(2*(2*x^2 - (x^2 + x)*e^x - (x^2*e^x + x)*e^(e^x) + 5*x)*e^ 
(2/5*x^2 - 2/5*x*e^x - 2/5*x*e^(e^x) + 2*x) + 5)*e^(e^(2/5*x^2 - 2/5*x*e^x 
 - 2/5*x*e^(e^x) + 2*x)) + 2*x, x)
                                                                                    
                                                                                    
 

Mupad [B] (verification not implemented)

Time = 3.32 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.03 \[ \int \frac {1}{5} \left (10 x+e^{e^{\frac {1}{5} \left (10 x-2 e^{e^x} x-2 e^x x+2 x^2\right )}} \left (-5+e^{\frac {1}{5} \left (10 x-2 e^{e^x} x-2 e^x x+2 x^2\right )} \left (-10 x-4 x^2+e^x \left (2 x+2 x^2\right )+e^{e^x} \left (2 x+2 e^x x^2\right )\right )\right )\right ) \, dx=x\,\left (x-{\mathrm {e}}^{{\mathrm {e}}^{-\frac {2\,x\,{\mathrm {e}}^x}{5}}\,{\mathrm {e}}^{2\,x}\,{\mathrm {e}}^{-\frac {2\,x\,{\mathrm {e}}^{{\mathrm {e}}^x}}{5}}\,{\mathrm {e}}^{\frac {2\,x^2}{5}}}\right ) \] Input:

int(2*x - (exp(exp(2*x - (2*x*exp(exp(x)))/5 - (2*x*exp(x))/5 + (2*x^2)/5) 
)*(exp(2*x - (2*x*exp(exp(x)))/5 - (2*x*exp(x))/5 + (2*x^2)/5)*(10*x - exp 
(x)*(2*x + 2*x^2) + 4*x^2 - exp(exp(x))*(2*x + 2*x^2*exp(x))) + 5))/5,x)
 

Output:

x*(x - exp(exp(-(2*x*exp(x))/5)*exp(2*x)*exp(-(2*x*exp(exp(x)))/5)*exp((2* 
x^2)/5)))
 

Reduce [F]

\[ \int \frac {1}{5} \left (10 x+e^{e^{\frac {1}{5} \left (10 x-2 e^{e^x} x-2 e^x x+2 x^2\right )}} \left (-5+e^{\frac {1}{5} \left (10 x-2 e^{e^x} x-2 e^x x+2 x^2\right )} \left (-10 x-4 x^2+e^x \left (2 x+2 x^2\right )+e^{e^x} \left (2 x+2 e^x x^2\right )\right )\right )\right ) \, dx=-\left (\int e^{\frac {e^{\frac {2}{5} x^{2}+2 x}}{e^{\frac {2 e^{e^{x}} x}{5}+\frac {2 e^{x} x}{5}}}}d x \right )+\frac {2 \left (\int \frac {e^{\frac {5 e^{\frac {2}{5} x^{2}+2 x}+5 e^{\frac {2 e^{e^{x}} x}{5}+\frac {2 e^{x} x}{5}+x}+2 e^{\frac {2 e^{e^{x}} x}{5}+\frac {2 e^{x} x}{5}} x^{2}+15 e^{\frac {2 e^{e^{x}} x}{5}+\frac {2 e^{x} x}{5}} x}{5 e^{\frac {2 e^{e^{x}} x}{5}+\frac {2 e^{x} x}{5}}}} x^{2}}{e^{\frac {2 e^{e^{x}} x}{5}+\frac {2 e^{x} x}{5}}}d x \right )}{5}+\frac {2 \left (\int \frac {e^{\frac {5 e^{\frac {2}{5} x^{2}+2 x}+5 e^{\frac {2 e^{e^{x}} x}{5}+\frac {2 e^{x} x}{5}+x}+2 e^{\frac {2 e^{e^{x}} x}{5}+\frac {2 e^{x} x}{5}} x^{2}+10 e^{\frac {2 e^{e^{x}} x}{5}+\frac {2 e^{x} x}{5}} x}{5 e^{\frac {2 e^{e^{x}} x}{5}+\frac {2 e^{x} x}{5}}}} x}{e^{\frac {2 e^{e^{x}} x}{5}+\frac {2 e^{x} x}{5}}}d x \right )}{5}+\frac {2 \left (\int \frac {e^{\frac {5 e^{\frac {2}{5} x^{2}+2 x}+2 e^{\frac {2 e^{e^{x}} x}{5}+\frac {2 e^{x} x}{5}} x^{2}+15 e^{\frac {2 e^{e^{x}} x}{5}+\frac {2 e^{x} x}{5}} x}{5 e^{\frac {2 e^{e^{x}} x}{5}+\frac {2 e^{x} x}{5}}}} x^{2}}{e^{\frac {2 e^{e^{x}} x}{5}+\frac {2 e^{x} x}{5}}}d x \right )}{5}+\frac {2 \left (\int \frac {e^{\frac {5 e^{\frac {2}{5} x^{2}+2 x}+2 e^{\frac {2 e^{e^{x}} x}{5}+\frac {2 e^{x} x}{5}} x^{2}+15 e^{\frac {2 e^{e^{x}} x}{5}+\frac {2 e^{x} x}{5}} x}{5 e^{\frac {2 e^{e^{x}} x}{5}+\frac {2 e^{x} x}{5}}}} x}{e^{\frac {2 e^{e^{x}} x}{5}+\frac {2 e^{x} x}{5}}}d x \right )}{5}-\frac {4 \left (\int \frac {e^{\frac {5 e^{\frac {2}{5} x^{2}+2 x}+2 e^{\frac {2 e^{e^{x}} x}{5}+\frac {2 e^{x} x}{5}} x^{2}+10 e^{\frac {2 e^{e^{x}} x}{5}+\frac {2 e^{x} x}{5}} x}{5 e^{\frac {2 e^{e^{x}} x}{5}+\frac {2 e^{x} x}{5}}}} x^{2}}{e^{\frac {2 e^{e^{x}} x}{5}+\frac {2 e^{x} x}{5}}}d x \right )}{5}-2 \left (\int \frac {e^{\frac {5 e^{\frac {2}{5} x^{2}+2 x}+2 e^{\frac {2 e^{e^{x}} x}{5}+\frac {2 e^{x} x}{5}} x^{2}+10 e^{\frac {2 e^{e^{x}} x}{5}+\frac {2 e^{x} x}{5}} x}{5 e^{\frac {2 e^{e^{x}} x}{5}+\frac {2 e^{x} x}{5}}}} x}{e^{\frac {2 e^{e^{x}} x}{5}+\frac {2 e^{x} x}{5}}}d x \right )+x^{2} \] Input:

int(1/5*(((2*exp(x)*x^2+2*x)*exp(exp(x))+(2*x^2+2*x)*exp(x)-4*x^2-10*x)*ex 
p(-2/5*x*exp(exp(x))-2/5*exp(x)*x+2/5*x^2+2*x)-5)*exp(exp(-2/5*x*exp(exp(x 
))-2/5*exp(x)*x+2/5*x^2+2*x))+2*x,x)
 

Output:

( - 5*int(e**(e**((2*x**2 + 10*x)/5)/e**((2*e**(e**x)*x + 2*e**x*x)/5)),x) 
 + 2*int((e**((5*e**((2*x**2 + 10*x)/5) + 5*e**((2*e**(e**x)*x + 2*e**x*x 
+ 5*x)/5) + 2*e**((2*e**(e**x)*x + 2*e**x*x)/5)*x**2 + 15*e**((2*e**(e**x) 
*x + 2*e**x*x)/5)*x)/(5*e**((2*e**(e**x)*x + 2*e**x*x)/5)))*x**2)/e**((2*e 
**(e**x)*x + 2*e**x*x)/5),x) + 2*int((e**((5*e**((2*x**2 + 10*x)/5) + 5*e* 
*((2*e**(e**x)*x + 2*e**x*x + 5*x)/5) + 2*e**((2*e**(e**x)*x + 2*e**x*x)/5 
)*x**2 + 10*e**((2*e**(e**x)*x + 2*e**x*x)/5)*x)/(5*e**((2*e**(e**x)*x + 2 
*e**x*x)/5)))*x)/e**((2*e**(e**x)*x + 2*e**x*x)/5),x) + 2*int((e**((5*e**( 
(2*x**2 + 10*x)/5) + 2*e**((2*e**(e**x)*x + 2*e**x*x)/5)*x**2 + 15*e**((2* 
e**(e**x)*x + 2*e**x*x)/5)*x)/(5*e**((2*e**(e**x)*x + 2*e**x*x)/5)))*x**2) 
/e**((2*e**(e**x)*x + 2*e**x*x)/5),x) + 2*int((e**((5*e**((2*x**2 + 10*x)/ 
5) + 2*e**((2*e**(e**x)*x + 2*e**x*x)/5)*x**2 + 15*e**((2*e**(e**x)*x + 2* 
e**x*x)/5)*x)/(5*e**((2*e**(e**x)*x + 2*e**x*x)/5)))*x)/e**((2*e**(e**x)*x 
 + 2*e**x*x)/5),x) - 4*int((e**((5*e**((2*x**2 + 10*x)/5) + 2*e**((2*e**(e 
**x)*x + 2*e**x*x)/5)*x**2 + 10*e**((2*e**(e**x)*x + 2*e**x*x)/5)*x)/(5*e* 
*((2*e**(e**x)*x + 2*e**x*x)/5)))*x**2)/e**((2*e**(e**x)*x + 2*e**x*x)/5), 
x) - 10*int((e**((5*e**((2*x**2 + 10*x)/5) + 2*e**((2*e**(e**x)*x + 2*e**x 
*x)/5)*x**2 + 10*e**((2*e**(e**x)*x + 2*e**x*x)/5)*x)/(5*e**((2*e**(e**x)* 
x + 2*e**x*x)/5)))*x)/e**((2*e**(e**x)*x + 2*e**x*x)/5),x) + 5*x**2)/5