\(\int (2 e^{x^2} x+e^{4 e^x} (6 x^5+4 e^x x^6)) \, dx\) [508]

Optimal result

Integrand size = 31, antiderivative size = 19 \[ \int \left (2 e^{x^2} x+e^{4 e^x} \left (6 x^5+4 e^x x^6\right )\right ) \, dx=e^{x^2}+e^{4 e^x} x^6+\log (2) \]

[Out]

ln(2)+exp(x^2)+exp(exp(x))^4*x^6

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {2240, 2326} \[ \int \left (2 e^{x^2} x+e^{4 e^x} \left (6 x^5+4 e^x x^6\right )\right ) \, dx=e^{4 e^x} x^6+e^{x^2} \]

[In]

Int[2*E^x^2*x + E^(4*E^x)*(6*x^5 + 4*E^x*x^6),x]

[Out]

E^x^2 + E^(4*E^x)*x^6

Rule 2240

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[(e + f*x)^n*(
F^(a + b*(c + d*x)^n)/(b*f*n*(c + d*x)^n*Log[F])), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[m, n - 1] &
& EqQ[d*e - c*f, 0]

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}& = 2 \int e^{x^2} x \, dx+\int e^{4 e^x} \left (6 x^5+4 e^x x^6\right ) \, dx \\ & = e^{x^2}+e^{4 e^x} x^6 \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89 \[ \int \left (2 e^{x^2} x+e^{4 e^x} \left (6 x^5+4 e^x x^6\right )\right ) \, dx=e^{x^2}+e^{4 e^x} x^6 \]

[In]

Integrate[2*E^x^2*x + E^(4*E^x)*(6*x^5 + 4*E^x*x^6),x]

[Out]

E^x^2 + E^(4*E^x)*x^6

Maple [A] (verified)

Time = 0.53 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79

method	result	size
default	\(x^{6} {\mathrm e}^{4 \,{\mathrm e}^{x}}+{\mathrm e}^{x^{2}}\)	\(15\)
risch	\(x^{6} {\mathrm e}^{4 \,{\mathrm e}^{x}}+{\mathrm e}^{x^{2}}\)	\(15\)
parallelrisch	\(x^{6} {\mathrm e}^{4 \,{\mathrm e}^{x}}+{\mathrm e}^{x^{2}}\)	\(15\)

[In]

int((4*x^6*exp(x)+6*x^5)*exp(exp(x))^4+2*exp(x^2)*x,x,method=_RETURNVERBOSE)

[Out]

x^6*exp(4*exp(x))+exp(x^2)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.74 \[ \int \left (2 e^{x^2} x+e^{4 e^x} \left (6 x^5+4 e^x x^6\right )\right ) \, dx=x^{6} e^{\left (4 \, e^{x}\right )} + e^{\left (x^{2}\right )} \]

[In]

integrate((4*x^6*exp(x)+6*x^5)*exp(exp(x))^4+2*exp(x^2)*x,x, algorithm="fricas")

[Out]

x^6*e^(4*e^x) + e^(x^2)

Sympy [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.74 \[ \int \left (2 e^{x^2} x+e^{4 e^x} \left (6 x^5+4 e^x x^6\right )\right ) \, dx=x^{6} e^{4 e^{x}} + e^{x^{2}} \]

[In]

integrate((4*x**6*exp(x)+6*x**5)*exp(exp(x))**4+2*exp(x**2)*x,x)

[Out]

x**6*exp(4*exp(x)) + exp(x**2)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.74 \[ \int \left (2 e^{x^2} x+e^{4 e^x} \left (6 x^5+4 e^x x^6\right )\right ) \, dx=x^{6} e^{\left (4 \, e^{x}\right )} + e^{\left (x^{2}\right )} \]

[In]

integrate((4*x^6*exp(x)+6*x^5)*exp(exp(x))^4+2*exp(x^2)*x,x, algorithm="maxima")

[Out]

x^6*e^(4*e^x) + e^(x^2)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.74 \[ \int \left (2 e^{x^2} x+e^{4 e^x} \left (6 x^5+4 e^x x^6\right )\right ) \, dx=x^{6} e^{\left (4 \, e^{x}\right )} + e^{\left (x^{2}\right )} \]

[In]

integrate((4*x^6*exp(x)+6*x^5)*exp(exp(x))^4+2*exp(x^2)*x,x, algorithm="giac")

[Out]

x^6*e^(4*e^x) + e^(x^2)

Mupad [B] (verification not implemented)

Time = 7.46 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.74 \[ \int \left (2 e^{x^2} x+e^{4 e^x} \left (6 x^5+4 e^x x^6\right )\right ) \, dx={\mathrm {e}}^{x^2}+x^6\,{\mathrm {e}}^{4\,{\mathrm {e}}^x} \]

[In]

int(2*x*exp(x^2) + exp(4*exp(x))*(4*x^6*exp(x) + 6*x^5),x)

[Out]

exp(x^2) + x^6*exp(4*exp(x))