3.53.71 \(\int 10 e^{e^{2 x}+2 x} \, dx\)

Optimal. Leaf size=9 \[ 5 e^{e^{2 x}} \]

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Rubi [A]  time = 0.01, antiderivative size = 9, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {12, 2282, 2194} \begin {gather*} 5 e^{e^{2 x}} \end {gather*}

Antiderivative was successfully verified.

[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 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} &=10 \int e^{e^{2 x}+2 x} \, dx\\ &=5 \operatorname {Subst}\left (\int e^x \, dx,x,e^{2 x}\right )\\ &=5 e^{e^{2 x}}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 9, normalized size = 1.00 \begin {gather*} 5 e^{e^{2 x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[10*E^(E^(2*x) + 2*x),x]

[Out]

5*E^E^(2*x)

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fricas [A]  time = 0.54, size = 7, normalized size = 0.78 \begin {gather*} 5 \, e^{\left (e^{\left (2 \, x\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.20, size = 7, normalized size = 0.78 \begin {gather*} 5 \, e^{\left (e^{\left (2 \, x\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.03, size = 8, normalized size = 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	\(5 \,{\mathrm e}^{{\mathrm e}^{2 x}}\)	\(8\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5*exp(exp(x)^2)

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maxima [A]  time = 0.36, size = 7, normalized size = 0.78 \begin {gather*} 5 \, e^{\left (e^{\left (2 \, x\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.04, size = 7, normalized size = 0.78 \begin {gather*} 5\,{\mathrm {e}}^{{\mathrm {e}}^{2\,x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5*exp(exp(2*x))

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sympy [A]  time = 0.09, size = 7, normalized size = 0.78 \begin {gather*} 5 e^{e^{2 x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5*exp(exp(2*x))

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