3.42.51 \(\int 2 e^{1+e^{1+2 x}+2 x} \, dx\)

Optimal. Leaf size=14 \[ e^{e^{1+2 x}}-\log (3) \]

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Rubi [A]  time = 0.01, antiderivative size = 9, normalized size of antiderivative = 0.64, number of steps used = 3, number of rules used = 3, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {12, 2282, 2194} \begin {gather*} e^{e^{2 x+1}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[2*E^(1 + E^(1 + 2*x) + 2*x),x]

[Out]

E^E^(1 + 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 2194

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 2282

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 {gather*} \begin {aligned} \text {integral} &=2 \int e^{1+e^{1+2 x}+2 x} \, dx\\ &=\operatorname {Subst}\left (\int e^{1+e x} \, dx,x,e^{2 x}\right )\\ &=e^{e^{1+2 x}}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 9, normalized size = 0.64 \begin {gather*} e^{e^{1+2 x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[2*E^(1 + E^(1 + 2*x) + 2*x),x]

[Out]

E^E^(1 + 2*x)

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fricas [A]  time = 0.67, size = 7, normalized size = 0.50 \begin {gather*} e^{\left (e^{\left (2 \, x + 1\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e^(e^(2*x + 1))

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giac [A]  time = 0.16, size = 7, normalized size = 0.50 \begin {gather*} e^{\left (e^{\left (2 \, x + 1\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e^(e^(2*x + 1))

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maple [A]  time = 0.01, size = 8, normalized size = 0.57

method	result	size
derivativedivides	\({\mathrm e}^{{\mathrm e}^{2 x +1}}\)	\(8\)
default	\({\mathrm e}^{{\mathrm e}^{2 x +1}}\)	\(8\)
norman	\({\mathrm e}^{{\mathrm e}^{2 x +1}}\)	\(8\)
risch	\({\mathrm e}^{{\mathrm e}^{2 x +1}}\)	\(8\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

exp(exp(2*x+1))

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maxima [A]  time = 0.38, size = 7, normalized size = 0.50 \begin {gather*} e^{\left (e^{\left (2 \, x + 1\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e^(e^(2*x + 1))

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mupad [B]  time = 3.39, size = 7, normalized size = 0.50 \begin {gather*} {\mathrm {e}}^{{\mathrm {e}}^{2\,x+1}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(2*exp(exp(2*x + 1))*exp(2*x + 1),x)

[Out]

exp(exp(2*x + 1))

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sympy [A]  time = 0.10, size = 7, normalized size = 0.50 \begin {gather*} e^{e^{2 x + 1}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

exp(exp(2*x + 1))

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