\(\int 10 e^{e^{2 x}+2 x} \, dx\) [5271]

Optimal result

Integrand size = 13, antiderivative size = 9 \[ \int 10 e^{e^{2 x}+2 x} \, dx=5 e^{e^{2 x}} \]

[Out]

5*exp(exp(x)^2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 9, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {12, 2320, 2225} \[ \int 10 e^{e^{2 x}+2 x} \, dx=5 e^{e^{2 x}} \]

[In]

Int[10*E^(E^(2*x) + 2*x),x]

[Out]

5*E^E^(2*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]]]

Rubi steps \begin{align*} \text {integral}& = 10 \int e^{e^{2 x}+2 x} \, dx \\ & = 5 \text {Subst}\left (\int e^x \, dx,x,e^{2 x}\right ) \\ & = 5 e^{e^{2 x}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 9, normalized size of antiderivative = 1.00 \[ \int 10 e^{e^{2 x}+2 x} \, dx=5 e^{e^{2 x}} \]

[In]

Integrate[10*E^(E^(2*x) + 2*x),x]

[Out]

5*E^E^(2*x)

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.89

method	result	size
derivativedivides	\(5 \,{\mathrm e}^{{\mathrm e}^{2 x}}\)	\(8\)
default	\(5 \,{\mathrm e}^{{\mathrm e}^{2 x}}\)	\(8\)
norman	\(5 \,{\mathrm e}^{{\mathrm e}^{2 x}}\)	\(8\)
risch	\(5 \,{\mathrm e}^{{\mathrm e}^{2 x}}\)	\(8\)
parallelrisch	\(5 \,{\mathrm e}^{{\mathrm e}^{2 x}}\)	\(8\)

[In]

int(10*exp(x)^2*exp(exp(x)^2),x,method=_RETURNVERBOSE)

[Out]

5*exp(exp(x)^2)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.78 \[ \int 10 e^{e^{2 x}+2 x} \, dx=5 \, e^{\left (e^{\left (2 \, x\right )}\right )} \]

[In]

integrate(10*exp(x)^2*exp(exp(x)^2),x, algorithm="fricas")

[Out]

5*e^(e^(2*x))

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.78 \[ \int 10 e^{e^{2 x}+2 x} \, dx=5 e^{e^{2 x}} \]

[In]

integrate(10*exp(x)**2*exp(exp(x)**2),x)

[Out]

5*exp(exp(2*x))

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.78 \[ \int 10 e^{e^{2 x}+2 x} \, dx=5 \, e^{\left (e^{\left (2 \, x\right )}\right )} \]

[In]

integrate(10*exp(x)^2*exp(exp(x)^2),x, algorithm="maxima")

[Out]

5*e^(e^(2*x))

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.78 \[ \int 10 e^{e^{2 x}+2 x} \, dx=5 \, e^{\left (e^{\left (2 \, x\right )}\right )} \]

[In]

integrate(10*exp(x)^2*exp(exp(x)^2),x, algorithm="giac")

[Out]

5*e^(e^(2*x))

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.78 \[ \int 10 e^{e^{2 x}+2 x} \, dx=5\,{\mathrm {e}}^{{\mathrm {e}}^{2\,x}} \]

[In]

int(10*exp(2*x)*exp(exp(2*x)),x)

[Out]

5*exp(exp(2*x))