\(\int (-1+12 x^3+e^{e^x} (4 x+2 e^x x^2)) \, dx\) [3185]

Optimal result

Integrand size = 25, antiderivative size = 22 \[ \int \left (-1+12 x^3+e^{e^x} \left (4 x+2 e^x x^2\right )\right ) \, dx=-x+3 x^4+2 \left (4+e^{e^x} x^2\right ) \]

[Out]

8+2*exp(exp(x))*x^2+3*x^4-x

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.86, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.040, Rules used = {2326} \[ \int \left (-1+12 x^3+e^{e^x} \left (4 x+2 e^x x^2\right )\right ) \, dx=3 x^4+2 e^{e^x} x^2-x \]

[In]

Int[-1 + 12*x^3 + E^E^x*(4*x + 2*E^x*x^2),x]

[Out]

-x + 2*E^E^x*x^2 + 3*x^4

Rule 2326

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = v*(y/(Log[F]*D[u, x]))}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.86 \[ \int \left (-1+12 x^3+e^{e^x} \left (4 x+2 e^x x^2\right )\right ) \, dx=-x+2 e^{e^x} x^2+3 x^4 \]

[In]

Integrate[-1 + 12*x^3 + E^E^x*(4*x + 2*E^x*x^2),x]

[Out]

-x + 2*E^E^x*x^2 + 3*x^4

Maple [A] (verified)

Time = 0.24 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.82

method	result	size
default	\(-x +2 \,{\mathrm e}^{{\mathrm e}^{x}} x^{2}+3 x^{4}\)	\(18\)
norman	\(-x +2 \,{\mathrm e}^{{\mathrm e}^{x}} x^{2}+3 x^{4}\)	\(18\)
risch	\(-x +2 \,{\mathrm e}^{{\mathrm e}^{x}} x^{2}+3 x^{4}\)	\(18\)
parallelrisch	\(-x +2 \,{\mathrm e}^{{\mathrm e}^{x}} x^{2}+3 x^{4}\)	\(18\)

[In]

int((2*exp(x)*x^2+4*x)*exp(exp(x))+12*x^3-1,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.77 \[ \int \left (-1+12 x^3+e^{e^x} \left (4 x+2 e^x x^2\right )\right ) \, dx=3 \, x^{4} + 2 \, x^{2} e^{\left (e^{x}\right )} - x \]

[In]

integrate((2*exp(x)*x^2+4*x)*exp(exp(x))+12*x^3-1,x, algorithm="fricas")

[Out]

3*x^4 + 2*x^2*e^(e^x) - x

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.68 \[ \int \left (-1+12 x^3+e^{e^x} \left (4 x+2 e^x x^2\right )\right ) \, dx=3 x^{4} + 2 x^{2} e^{e^{x}} - x \]

[In]

integrate((2*exp(x)*x**2+4*x)*exp(exp(x))+12*x**3-1,x)

[Out]

3*x**4 + 2*x**2*exp(exp(x)) - x

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.77 \[ \int \left (-1+12 x^3+e^{e^x} \left (4 x+2 e^x x^2\right )\right ) \, dx=3 \, x^{4} + 2 \, x^{2} e^{\left (e^{x}\right )} - x \]

[In]

integrate((2*exp(x)*x^2+4*x)*exp(exp(x))+12*x^3-1,x, algorithm="maxima")

[Out]

3*x^4 + 2*x^2*e^(e^x) - x

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.77 \[ \int \left (-1+12 x^3+e^{e^x} \left (4 x+2 e^x x^2\right )\right ) \, dx=3 \, x^{4} + 2 \, x^{2} e^{\left (e^{x}\right )} - x \]

[In]

integrate((2*exp(x)*x^2+4*x)*exp(exp(x))+12*x^3-1,x, algorithm="giac")

[Out]

3*x^4 + 2*x^2*e^(e^x) - x

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.77 \[ \int \left (-1+12 x^3+e^{e^x} \left (4 x+2 e^x x^2\right )\right ) \, dx=2\,x^2\,{\mathrm {e}}^{{\mathrm {e}}^x}-x+3\,x^4 \]

[In]

int(12*x^3 + exp(exp(x))*(4*x + 2*x^2*exp(x)) - 1,x)

[Out]

2*x^2*exp(exp(x)) - x + 3*x^4