\(\int \frac {1}{5} e^{-x} (4+e^x (30-10 x)-4 x+e^{e^{e^{-x} (4+x)}} (5 e^x+e^{e^{-x} (4+x)} (-15 x-5 x^2))) \, dx\) [3778]

Optimal result

Integrand size = 64, antiderivative size = 30 \[ \int \frac {1}{5} e^{-x} \left (4+e^x (30-10 x)-4 x+e^{e^{e^{-x} (4+x)}} \left (5 e^x+e^{e^{-x} (4+x)} \left (-15 x-5 x^2\right )\right )\right ) \, dx=-x \left (-6-e^{e^{e^{-x} (4+x)}}-\frac {4 e^{-x}}{5}+x\right ) \]

[Out]

-x*(x-exp(exp((4+x)/exp(x)))-6-4/5/exp(x))

Rubi [F]

\[ \int \frac {1}{5} e^{-x} \left (4+e^x (30-10 x)-4 x+e^{e^{e^{-x} (4+x)}} \left (5 e^x+e^{e^{-x} (4+x)} \left (-15 x-5 x^2\right )\right )\right ) \, dx=\int \frac {1}{5} e^{-x} \left (4+e^x (30-10 x)-4 x+e^{e^{e^{-x} (4+x)}} \left (5 e^x+e^{e^{-x} (4+x)} \left (-15 x-5 x^2\right )\right )\right ) \, dx \]

[In]

Int[(4 + E^x*(30 - 10*x) - 4*x + E^E^((4 + x)/E^x)*(5*E^x + E^((4 + x)/E^x)*(-15*x - 5*x^2)))/(5*E^x),x]

[Out]

-(3 - x)^2 + (4*x)/(5*E^x) + Defer[Int][E^E^((4 + x)/E^x), x] - 3*Defer[Int][E^(E^((4 + x)/E^x) - x + (4 + x)/
E^x)*x, x] - Defer[Int][E^(E^((4 + x)/E^x) - x + (4 + x)/E^x)*x^2, x]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{5} \int e^{-x} \left (4+e^x (30-10 x)-4 x+e^{e^{e^{-x} (4+x)}} \left (5 e^x+e^{e^{-x} (4+x)} \left (-15 x-5 x^2\right )\right )\right ) \, dx \\ & = \frac {1}{5} \int \left (4 e^{-x}-10 (-3+x)-4 e^{-x} x+5 e^{e^{e^{-x} (4+x)}-x} \left (e^x-3 e^{4 e^{-x}+e^{-x} x} x-e^{4 e^{-x}+e^{-x} x} x^2\right )\right ) \, dx \\ & = -(3-x)^2+\frac {4}{5} \int e^{-x} \, dx-\frac {4}{5} \int e^{-x} x \, dx+\int e^{e^{e^{-x} (4+x)}-x} \left (e^x-3 e^{4 e^{-x}+e^{-x} x} x-e^{4 e^{-x}+e^{-x} x} x^2\right ) \, dx \\ & = -\frac {4 e^{-x}}{5}-(3-x)^2+\frac {4 e^{-x} x}{5}-\frac {4}{5} \int e^{-x} \, dx+\int e^{e^{e^{-x} (4+x)}-x} \left (e^x-e^{e^{-x} (4+x)} x (3+x)\right ) \, dx \\ & = -(3-x)^2+\frac {4 e^{-x} x}{5}+\int \left (e^{e^{e^{-x} (4+x)}}+e^{e^{e^{-x} (4+x)}-x+e^{-x} (4+x)} (-3-x) x\right ) \, dx \\ & = -(3-x)^2+\frac {4 e^{-x} x}{5}+\int e^{e^{e^{-x} (4+x)}} \, dx+\int e^{e^{e^{-x} (4+x)}-x+e^{-x} (4+x)} (-3-x) x \, dx \\ & = -(3-x)^2+\frac {4 e^{-x} x}{5}+\int e^{e^{e^{-x} (4+x)}} \, dx+\int \left (-3 e^{e^{e^{-x} (4+x)}-x+e^{-x} (4+x)} x-e^{e^{e^{-x} (4+x)}-x+e^{-x} (4+x)} x^2\right ) \, dx \\ & = -(3-x)^2+\frac {4 e^{-x} x}{5}-3 \int e^{e^{e^{-x} (4+x)}-x+e^{-x} (4+x)} x \, dx+\int e^{e^{e^{-x} (4+x)}} \, dx-\int e^{e^{e^{-x} (4+x)}-x+e^{-x} (4+x)} x^2 \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 3.37 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.23 \[ \int \frac {1}{5} e^{-x} \left (4+e^x (30-10 x)-4 x+e^{e^{e^{-x} (4+x)}} \left (5 e^x+e^{e^{-x} (4+x)} \left (-15 x-5 x^2\right )\right )\right ) \, dx=\frac {1}{5} \left (30 x+5 e^{e^{e^{-x} (4+x)}} x+4 e^{-x} x-5 x^2\right ) \]

[In]

Integrate[(4 + E^x*(30 - 10*x) - 4*x + E^E^((4 + x)/E^x)*(5*E^x + E^((4 + x)/E^x)*(-15*x - 5*x^2)))/(5*E^x),x]

[Out]

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

Maple [A] (verified)

Time = 1.18 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.97

method	result	size
risch	\(-x^{2}+6 x +\frac {4 x \,{\mathrm e}^{-x}}{5}+x \,{\mathrm e}^{{\mathrm e}^{\left (4+x \right ) {\mathrm e}^{-x}}}\)	\(29\)
norman	\(\left ({\mathrm e}^{x} x \,{\mathrm e}^{{\mathrm e}^{\left (4+x \right ) {\mathrm e}^{-x}}}+\frac {4 x}{5}+6 \,{\mathrm e}^{x} x -{\mathrm e}^{x} x^{2}\right ) {\mathrm e}^{-x}\)	\(36\)
parallelrisch	\(\frac {{\mathrm e}^{-x} \left (-5 \,{\mathrm e}^{x} x^{2}+5 \,{\mathrm e}^{x} x \,{\mathrm e}^{{\mathrm e}^{\left (4+x \right ) {\mathrm e}^{-x}}}+30 \,{\mathrm e}^{x} x +4 x \right )}{5}\)	\(38\)

[In]

int(1/5*(((-5*x^2-15*x)*exp((4+x)/exp(x))+5*exp(x))*exp(exp((4+x)/exp(x)))+(-10*x+30)*exp(x)-4*x+4)/exp(x),x,m
ethod=_RETURNVERBOSE)

[Out]

-x^2+6*x+4/5*x*exp(-x)+x*exp(exp((4+x)*exp(-x)))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.20 \[ \int \frac {1}{5} e^{-x} \left (4+e^x (30-10 x)-4 x+e^{e^{e^{-x} (4+x)}} \left (5 e^x+e^{e^{-x} (4+x)} \left (-15 x-5 x^2\right )\right )\right ) \, dx=\frac {1}{5} \, {\left (5 \, x e^{\left (x + e^{\left ({\left (x + 4\right )} e^{\left (-x\right )}\right )}\right )} - 5 \, {\left (x^{2} - 6 \, x\right )} e^{x} + 4 \, x\right )} e^{\left (-x\right )} \]

[In]

integrate(1/5*(((-5*x^2-15*x)*exp((4+x)/exp(x))+5*exp(x))*exp(exp((4+x)/exp(x)))+(-10*x+30)*exp(x)-4*x+4)/exp(
x),x, algorithm="fricas")

[Out]

1/5*(5*x*e^(x + e^((x + 4)*e^(-x))) - 5*(x^2 - 6*x)*e^x + 4*x)*e^(-x)

Sympy [A] (verification not implemented)

Time = 43.03 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.87 \[ \int \frac {1}{5} e^{-x} \left (4+e^x (30-10 x)-4 x+e^{e^{e^{-x} (4+x)}} \left (5 e^x+e^{e^{-x} (4+x)} \left (-15 x-5 x^2\right )\right )\right ) \, dx=- x^{2} + x e^{e^{\left (x + 4\right ) e^{- x}}} + 6 x + \frac {4 x e^{- x}}{5} \]

[In]

integrate(1/5*(((-5*x**2-15*x)*exp((4+x)/exp(x))+5*exp(x))*exp(exp((4+x)/exp(x)))+(-10*x+30)*exp(x)-4*x+4)/exp
(x),x)

[Out]

-x**2 + x*exp(exp((x + 4)*exp(-x))) + 6*x + 4*x*exp(-x)/5

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.37 \[ \int \frac {1}{5} e^{-x} \left (4+e^x (30-10 x)-4 x+e^{e^{e^{-x} (4+x)}} \left (5 e^x+e^{e^{-x} (4+x)} \left (-15 x-5 x^2\right )\right )\right ) \, dx=-x^{2} + \frac {4}{5} \, {\left (x + 1\right )} e^{\left (-x\right )} + x e^{\left (e^{\left (x e^{\left (-x\right )} + 4 \, e^{\left (-x\right )}\right )}\right )} + 6 \, x - \frac {4}{5} \, e^{\left (-x\right )} \]

[In]

integrate(1/5*(((-5*x^2-15*x)*exp((4+x)/exp(x))+5*exp(x))*exp(exp((4+x)/exp(x)))+(-10*x+30)*exp(x)-4*x+4)/exp(
x),x, algorithm="maxima")

[Out]

-x^2 + 4/5*(x + 1)*e^(-x) + x*e^(e^(x*e^(-x) + 4*e^(-x))) + 6*x - 4/5*e^(-x)

Giac [F]

\[ \int \frac {1}{5} e^{-x} \left (4+e^x (30-10 x)-4 x+e^{e^{e^{-x} (4+x)}} \left (5 e^x+e^{e^{-x} (4+x)} \left (-15 x-5 x^2\right )\right )\right ) \, dx=\int { -\frac {1}{5} \, {\left (10 \, {\left (x - 3\right )} e^{x} + 5 \, {\left ({\left (x^{2} + 3 \, x\right )} e^{\left ({\left (x + 4\right )} e^{\left (-x\right )}\right )} - e^{x}\right )} e^{\left (e^{\left ({\left (x + 4\right )} e^{\left (-x\right )}\right )}\right )} + 4 \, x - 4\right )} e^{\left (-x\right )} \,d x } \]

[In]

integrate(1/5*(((-5*x^2-15*x)*exp((4+x)/exp(x))+5*exp(x))*exp(exp((4+x)/exp(x)))+(-10*x+30)*exp(x)-4*x+4)/exp(
x),x, algorithm="giac")

[Out]

integrate(-1/5*(10*(x - 3)*e^x + 5*((x^2 + 3*x)*e^((x + 4)*e^(-x)) - e^x)*e^(e^((x + 4)*e^(-x))) + 4*x - 4)*e^
(-x), x)

Mupad [B] (verification not implemented)

Time = 9.29 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.13 \[ \int \frac {1}{5} e^{-x} \left (4+e^x (30-10 x)-4 x+e^{e^{e^{-x} (4+x)}} \left (5 e^x+e^{e^{-x} (4+x)} \left (-15 x-5 x^2\right )\right )\right ) \, dx=6\,x+\frac {4\,x\,{\mathrm {e}}^{-x}}{5}+x\,{\mathrm {e}}^{{\mathrm {e}}^{4\,{\mathrm {e}}^{-x}}\,{\mathrm {e}}^{x\,{\mathrm {e}}^{-x}}}-x^2 \]

[In]

int(-exp(-x)*((4*x)/5 + (exp(x)*(10*x - 30))/5 - (exp(exp(exp(-x)*(x + 4)))*(5*exp(x) - exp(exp(-x)*(x + 4))*(
15*x + 5*x^2)))/5 - 4/5),x)

[Out]

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