\(\int (5 e^x+e^{e^{-x^2-x^3+e^{2 x} (1+x)}-x^2-x^3+e^{2 x} (1+x)} (-2 x-3 x^2+e^{2 x} (3+2 x))) \, dx\) [8864]

   Optimal result
   Rubi [F]
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [B] (verification not implemented)
   Sympy [A] (verification not implemented)
   Maxima [A] (verification not implemented)
   Giac [F]
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 71, antiderivative size = 29 \[ \int \left (5 e^x+e^{e^{-x^2-x^3+e^{2 x} (1+x)}-x^2-x^3+e^{2 x} (1+x)} \left (-2 x-3 x^2+e^{2 x} (3+2 x)\right )\right ) \, dx=e^{e^{\left (\frac {e^{2 x}}{x}-x\right ) \left (x+x^2\right )}}+5 e^x \]

[Out]

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

Rubi [F]

\[ \int \left (5 e^x+e^{e^{-x^2-x^3+e^{2 x} (1+x)}-x^2-x^3+e^{2 x} (1+x)} \left (-2 x-3 x^2+e^{2 x} (3+2 x)\right )\right ) \, dx=\int \left (5 e^x+\exp \left (e^{-x^2-x^3+e^{2 x} (1+x)}-x^2-x^3+e^{2 x} (1+x)\right ) \left (-2 x-3 x^2+e^{2 x} (3+2 x)\right )\right ) \, dx \]

[In]

Int[5*E^x + E^(E^(-x^2 - x^3 + E^(2*x)*(1 + x)) - x^2 - x^3 + E^(2*x)*(1 + x))*(-2*x - 3*x^2 + E^(2*x)*(3 + 2*
x)),x]

[Out]

5*E^x + 3*Defer[Int][E^(E^(-x^2 - x^3 + E^(2*x)*(1 + x)) + 2*x - x^2 - x^3 + E^(2*x)*(1 + x)), x] - 2*Defer[In
t][E^(E^(-x^2 - x^3 + E^(2*x)*(1 + x)) - x^2 - x^3 + E^(2*x)*(1 + x))*x, x] + 2*Defer[Int][E^(E^(-x^2 - x^3 +
E^(2*x)*(1 + x)) + 2*x - x^2 - x^3 + E^(2*x)*(1 + x))*x, x] - 3*Defer[Int][E^(E^(-x^2 - x^3 + E^(2*x)*(1 + x))
 - x^2 - x^3 + E^(2*x)*(1 + x))*x^2, x]

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

Mathematica [A] (verified)

Time = 3.24 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.03 \[ \int \left (5 e^x+e^{e^{-x^2-x^3+e^{2 x} (1+x)}-x^2-x^3+e^{2 x} (1+x)} \left (-2 x-3 x^2+e^{2 x} (3+2 x)\right )\right ) \, dx=e^{e^{-x^2-x^3+e^{2 x} (1+x)}}+5 e^x \]

[In]

Integrate[5*E^x + E^(E^(-x^2 - x^3 + E^(2*x)*(1 + x)) - x^2 - x^3 + E^(2*x)*(1 + x))*(-2*x - 3*x^2 + E^(2*x)*(
3 + 2*x)),x]

[Out]

E^E^(-x^2 - x^3 + E^(2*x)*(1 + x)) + 5*E^x

Maple [A] (verified)

Time = 0.20 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.79

method result size
risch \({\mathrm e}^{{\mathrm e}^{-\left (1+x \right ) \left (x^{2}-{\mathrm e}^{2 x}\right )}}+5 \,{\mathrm e}^{x}\) \(23\)
default \({\mathrm e}^{{\mathrm e}^{\left (1+x \right ) {\mathrm e}^{2 x}-x^{3}-x^{2}}}+5 \,{\mathrm e}^{x}\) \(27\)
parallelrisch \({\mathrm e}^{{\mathrm e}^{\left (1+x \right ) {\mathrm e}^{2 x}-x^{3}-x^{2}}}+5 \,{\mathrm e}^{x}\) \(27\)

[In]

int(((3+2*x)*exp(2*x)-3*x^2-2*x)*exp((1+x)*exp(2*x)-x^3-x^2)*exp(exp((1+x)*exp(2*x)-x^3-x^2))+5*exp(x),x,metho
d=_RETURNVERBOSE)

[Out]

exp(exp(-(1+x)*(x^2-exp(2*x))))+5*exp(x)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 82 vs. \(2 (25) = 50\).

Time = 0.26 (sec) , antiderivative size = 82, normalized size of antiderivative = 2.83 \[ \int \left (5 e^x+e^{e^{-x^2-x^3+e^{2 x} (1+x)}-x^2-x^3+e^{2 x} (1+x)} \left (-2 x-3 x^2+e^{2 x} (3+2 x)\right )\right ) \, dx={\left (5 \, e^{\left (-x^{3} - x^{2} + {\left (x + 1\right )} e^{\left (2 \, x\right )} + x\right )} + e^{\left (-x^{3} - x^{2} + {\left (x + 1\right )} e^{\left (2 \, x\right )} + e^{\left (-x^{3} - x^{2} + {\left (x + 1\right )} e^{\left (2 \, x\right )}\right )}\right )}\right )} e^{\left (x^{3} + x^{2} - {\left (x + 1\right )} e^{\left (2 \, x\right )}\right )} \]

[In]

integrate(((3+2*x)*exp(2*x)-3*x^2-2*x)*exp((1+x)*exp(2*x)-x^3-x^2)*exp(exp((1+x)*exp(2*x)-x^3-x^2))+5*exp(x),x
, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.36 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.76 \[ \int \left (5 e^x+e^{e^{-x^2-x^3+e^{2 x} (1+x)}-x^2-x^3+e^{2 x} (1+x)} \left (-2 x-3 x^2+e^{2 x} (3+2 x)\right )\right ) \, dx=5 e^{x} + e^{e^{- x^{3} - x^{2} + \left (x + 1\right ) e^{2 x}}} \]

[In]

integrate(((3+2*x)*exp(2*x)-3*x**2-2*x)*exp((1+x)*exp(2*x)-x**3-x**2)*exp(exp((1+x)*exp(2*x)-x**3-x**2))+5*exp
(x),x)

[Out]

5*exp(x) + exp(exp(-x**3 - x**2 + (x + 1)*exp(2*x)))

Maxima [A] (verification not implemented)

none

Time = 0.52 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.97 \[ \int \left (5 e^x+e^{e^{-x^2-x^3+e^{2 x} (1+x)}-x^2-x^3+e^{2 x} (1+x)} \left (-2 x-3 x^2+e^{2 x} (3+2 x)\right )\right ) \, dx=5 \, e^{x} + e^{\left (e^{\left (-x^{3} - x^{2} + x e^{\left (2 \, x\right )} + e^{\left (2 \, x\right )}\right )}\right )} \]

[In]

integrate(((3+2*x)*exp(2*x)-3*x^2-2*x)*exp((1+x)*exp(2*x)-x^3-x^2)*exp(exp((1+x)*exp(2*x)-x^3-x^2))+5*exp(x),x
, algorithm="maxima")

[Out]

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

Giac [F]

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

[In]

integrate(((3+2*x)*exp(2*x)-3*x^2-2*x)*exp((1+x)*exp(2*x)-x^3-x^2)*exp(exp((1+x)*exp(2*x)-x^3-x^2))+5*exp(x),x
, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 12.45 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.07 \[ \int \left (5 e^x+e^{e^{-x^2-x^3+e^{2 x} (1+x)}-x^2-x^3+e^{2 x} (1+x)} \left (-2 x-3 x^2+e^{2 x} (3+2 x)\right )\right ) \, dx={\mathrm {e}}^{{\mathrm {e}}^{x\,{\mathrm {e}}^{2\,x}}\,{\mathrm {e}}^{-x^2}\,{\mathrm {e}}^{-x^3}\,{\mathrm {e}}^{{\mathrm {e}}^{2\,x}}}+5\,{\mathrm {e}}^x \]

[In]

int(5*exp(x) - exp(exp(exp(2*x)*(x + 1) - x^2 - x^3))*exp(exp(2*x)*(x + 1) - x^2 - x^3)*(2*x - exp(2*x)*(2*x +
 3) + 3*x^2),x)

[Out]

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