\(\int e^{16+e^{16+8 e^x+e^{2 x}}+8 e^x+e^{2 x}} (8 e^x+2 e^{2 x}) \, dx\) [6576]

Optimal result

Integrand size = 42, antiderivative size = 13 \[ \int e^{16+e^{16+8 e^x+e^{2 x}}+8 e^x+e^{2 x}} \left (8 e^x+2 e^{2 x}\right ) \, dx=e^{e^{\left (-4-e^x\right )^2}} \]

[Out]

exp(exp((-4-exp(x))^2))

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.85, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2320, 12, 6847, 2225} \[ \int e^{16+e^{16+8 e^x+e^{2 x}}+8 e^x+e^{2 x}} \left (8 e^x+2 e^{2 x}\right ) \, dx=e^{e^{\left (e^x+4\right )^2}} \]

[In]

Int[E^(16 + E^(16 + 8*E^x + E^(2*x)) + 8*E^x + E^(2*x))*(8*E^x + 2*E^(2*x)),x]

[Out]

E^E^(4 + E^x)^2

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

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.85 \[ \int e^{16+e^{16+8 e^x+e^{2 x}}+8 e^x+e^{2 x}} \left (8 e^x+2 e^{2 x}\right ) \, dx=e^{e^{\left (4+e^x\right )^2}} \]

[In]

Integrate[E^(16 + E^(16 + 8*E^x + E^(2*x)) + 8*E^x + E^(2*x))*(8*E^x + 2*E^(2*x)),x]

[Out]

E^E^(4 + E^x)^2

Maple [A] (verified)

Time = 0.32 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00

method	result	size
derivativedivides	\({\mathrm e}^{{\mathrm e}^{{\mathrm e}^{2 x}+8 \,{\mathrm e}^{x}+16}}\)	\(13\)
norman	\({\mathrm e}^{{\mathrm e}^{{\mathrm e}^{2 x}+8 \,{\mathrm e}^{x}+16}}\)	\(13\)
risch	\({\mathrm e}^{{\mathrm e}^{{\mathrm e}^{2 x}+8 \,{\mathrm e}^{x}+16}}\)	\(13\)
parallelrisch	\({\mathrm e}^{{\mathrm e}^{{\mathrm e}^{2 x}+8 \,{\mathrm e}^{x}+16}}\)	\(13\)

[In]

int((2*exp(x)^2+8*exp(x))*exp(exp(x)^2+8*exp(x)+16)*exp(exp(exp(x)^2+8*exp(x)+16)),x,method=_RETURNVERBOSE)

[Out]

exp(exp(exp(x)^2+8*exp(x)+16))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.92 \[ \int e^{16+e^{16+8 e^x+e^{2 x}}+8 e^x+e^{2 x}} \left (8 e^x+2 e^{2 x}\right ) \, dx=e^{\left (e^{\left (e^{\left (2 \, x\right )} + 8 \, e^{x} + 16\right )}\right )} \]

[In]

integrate((2*exp(x)^2+8*exp(x))*exp(exp(x)^2+8*exp(x)+16)*exp(exp(exp(x)^2+8*exp(x)+16)),x, algorithm="fricas"
)

[Out]

e^(e^(e^(2*x) + 8*e^x + 16))

Sympy [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.08 \[ \int e^{16+e^{16+8 e^x+e^{2 x}}+8 e^x+e^{2 x}} \left (8 e^x+2 e^{2 x}\right ) \, dx=e^{e^{e^{2 x} + 8 e^{x} + 16}} \]

[In]

integrate((2*exp(x)**2+8*exp(x))*exp(exp(x)**2+8*exp(x)+16)*exp(exp(exp(x)**2+8*exp(x)+16)),x)

[Out]

exp(exp(exp(2*x) + 8*exp(x) + 16))

Maxima [A] (verification not implemented)

none

Time = 0.39 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.92 \[ \int e^{16+e^{16+8 e^x+e^{2 x}}+8 e^x+e^{2 x}} \left (8 e^x+2 e^{2 x}\right ) \, dx=e^{\left (e^{\left (e^{\left (2 \, x\right )} + 8 \, e^{x} + 16\right )}\right )} \]

[In]

integrate((2*exp(x)^2+8*exp(x))*exp(exp(x)^2+8*exp(x)+16)*exp(exp(exp(x)^2+8*exp(x)+16)),x, algorithm="maxima"
)

[Out]

e^(e^(e^(2*x) + 8*e^x + 16))

Giac [F]

\[ \int e^{16+e^{16+8 e^x+e^{2 x}}+8 e^x+e^{2 x}} \left (8 e^x+2 e^{2 x}\right ) \, dx=\int { 2 \, {\left (e^{\left (2 \, x\right )} + 4 \, e^{x}\right )} e^{\left (e^{\left (2 \, x\right )} + 8 \, e^{x} + e^{\left (e^{\left (2 \, x\right )} + 8 \, e^{x} + 16\right )} + 16\right )} \,d x } \]

[In]

integrate((2*exp(x)^2+8*exp(x))*exp(exp(x)^2+8*exp(x)+16)*exp(exp(exp(x)^2+8*exp(x)+16)),x, algorithm="giac")

[Out]

integrate(2*(e^(2*x) + 4*e^x)*e^(e^(2*x) + 8*e^x + e^(e^(2*x) + 8*e^x + 16) + 16), x)

Mupad [B] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.08 \[ \int e^{16+e^{16+8 e^x+e^{2 x}}+8 e^x+e^{2 x}} \left (8 e^x+2 e^{2 x}\right ) \, dx={\mathrm {e}}^{{\mathrm {e}}^{16}\,{\mathrm {e}}^{{\mathrm {e}}^{2\,x}}\,{\mathrm {e}}^{8\,{\mathrm {e}}^x}} \]

[In]

int(exp(exp(2*x) + 8*exp(x) + 16)*exp(exp(exp(2*x) + 8*exp(x) + 16))*(2*exp(2*x) + 8*exp(x)),x)

[Out]

exp(exp(16)*exp(exp(2*x))*exp(8*exp(x)))