\(\int \frac {1}{2} e^{-2+\frac {e^{x^2}}{2}+x^2} (x+4 e^2 x) \, dx\) [7480]

Optimal result

Integrand size = 28, antiderivative size = 21 \[ \int \frac {1}{2} e^{-2+\frac {e^{x^2}}{2}+x^2} \left (x+4 e^2 x\right ) \, dx=\left (2+\frac {1}{2 e^2}\right ) e^{\frac {e^{x^2}}{2}} \]

[Out]

(2+1/2/exp(1)^2)*exp(1/2*exp(x^2))

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.14, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {6, 12, 6847, 2320, 2225} \[ \int \frac {1}{2} e^{-2+\frac {e^{x^2}}{2}+x^2} \left (x+4 e^2 x\right ) \, dx=\frac {1}{2} \left (1+4 e^2\right ) e^{\frac {e^{x^2}}{2}-2} \]

[In]

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

[Out]

(E^(-2 + E^x^2/2)*(1 + 4*E^2))/2

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2225

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2320

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 6847

Int[(u_)*(x_)^(m_.), x_Symbol] :> Dist[1/(m + 1), Subst[Int[SubstFor[x^(m + 1), u, x], x], x, x^(m + 1)], x] /
; FreeQ[m, x] && NeQ[m, -1] && FunctionOfQ[x^(m + 1), u, x]

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.14 \[ \int \frac {1}{2} e^{-2+\frac {e^{x^2}}{2}+x^2} \left (x+4 e^2 x\right ) \, dx=\frac {1}{2} e^{-2+\frac {e^{x^2}}{2}} \left (1+4 e^2\right ) \]

[In]

Integrate[(E^(-2 + E^x^2/2 + x^2)*(x + 4*E^2*x))/2,x]

[Out]

(E^(-2 + E^x^2/2)*(1 + 4*E^2))/2

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.05

method	result	size
norman	\(\frac {\left (4 \,{\mathrm e}^{2}+1\right ) {\mathrm e}^{-2} {\mathrm e}^{\frac {{\mathrm e}^{x^{2}}}{2}}}{2}\)	\(22\)
risch	\(2 \,{\mathrm e}^{-2+\frac {{\mathrm e}^{x^{2}}}{2}} {\mathrm e}^{2}+\frac {{\mathrm e}^{-2+\frac {{\mathrm e}^{x^{2}}}{2}}}{2}\)	\(26\)
default	\(\frac {{\mathrm e}^{-2} \left ({\mathrm e}^{\frac {{\mathrm e}^{x^{2}}}{2}}+4 \,{\mathrm e}^{2} {\mathrm e}^{\frac {{\mathrm e}^{x^{2}}}{2}}\right )}{2}\)	\(28\)
parallelrisch	\(\frac {{\mathrm e}^{-2} \left ({\mathrm e}^{\frac {{\mathrm e}^{x^{2}}}{2}}+4 \,{\mathrm e}^{2} {\mathrm e}^{\frac {{\mathrm e}^{x^{2}}}{2}}\right )}{2}\)	\(28\)

[In]

int(1/2*(4*x*exp(1)^2+x)*exp(x^2)*exp(1/2*exp(x^2))/exp(1)^2,x,method=_RETURNVERBOSE)

[Out]

1/2*(4*exp(1)^2+1)/exp(1)^2*exp(1/2*exp(x^2))

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81 \[ \int \frac {1}{2} e^{-2+\frac {e^{x^2}}{2}+x^2} \left (x+4 e^2 x\right ) \, dx=\frac {1}{2} \, {\left (4 \, e^{2} + 1\right )} e^{\left (\frac {1}{2} \, e^{\left (x^{2}\right )} - 2\right )} \]

[In]

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

[Out]

1/2*(4*e^2 + 1)*e^(1/2*e^(x^2) - 2)

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {1}{2} e^{-2+\frac {e^{x^2}}{2}+x^2} \left (x+4 e^2 x\right ) \, dx=\frac {\left (1 + 4 e^{2}\right ) e^{\frac {e^{x^{2}}}{2}}}{2 e^{2}} \]

[In]

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

[Out]

(1 + 4*exp(2))*exp(-2)*exp(exp(x**2)/2)/2

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int \frac {1}{2} e^{-2+\frac {e^{x^2}}{2}+x^2} \left (x+4 e^2 x\right ) \, dx=2 \, e^{\left (\frac {1}{2} \, e^{\left (x^{2}\right )}\right )} + \frac {1}{2} \, e^{\left (\frac {1}{2} \, e^{\left (x^{2}\right )} - 2\right )} \]

[In]

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

[Out]

2*e^(1/2*e^(x^2)) + 1/2*e^(1/2*e^(x^2) - 2)

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.10 \[ \int \frac {1}{2} e^{-2+\frac {e^{x^2}}{2}+x^2} \left (x+4 e^2 x\right ) \, dx=\frac {1}{2} \, {\left (e^{\left (\frac {1}{2} \, e^{\left (x^{2}\right )}\right )} + 4 \, e^{\left (\frac {1}{2} \, e^{\left (x^{2}\right )} + 2\right )}\right )} e^{\left (-2\right )} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 12.99 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.76 \[ \int \frac {1}{2} e^{-2+\frac {e^{x^2}}{2}+x^2} \left (x+4 e^2 x\right ) \, dx={\mathrm {e}}^{\frac {{\mathrm {e}}^{x^2}}{2}-2}\,\left (2\,{\mathrm {e}}^2+\frac {1}{2}\right ) \]

[In]

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

[Out]

exp(exp(x^2)/2 - 2)*(2*exp(2) + 1/2)