3.7.53 \(\int 2 e^{e^{\frac {1}{81} (812+162 x-36 \log (4)+162 \log ^2(4))}+\frac {1}{81} (812+162 x-36 \log (4)+162 \log ^2(4))} \, dx\)

Optimal. Leaf size=23 \[ e^{e^{-2 \left (-5-x-\left (\frac {1}{9}-\log (4)\right )^2\right )}} \]

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Rubi [A]  time = 0.04, antiderivative size = 25, normalized size of antiderivative = 1.09, number of steps used = 3, number of rules used = 3, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {12, 2282, 2194} \begin {gather*} e^{\frac {e^{\frac {2}{81} \left (81 x+406+81 \log ^2(4)\right )}}{2^{8/9}}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[2*E^(E^((812 + 162*x - 36*Log[4] + 162*Log[4]^2)/81) + (812 + 162*x - 36*Log[4] + 162*Log[4]^2)/81),x]

[Out]

E^(E^((2*(406 + 81*x + 81*Log[4]^2))/81)/2^(8/9))

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 \exp \left (e^{\frac {1}{81} \left (812+162 x-36 \log (4)+162 \log ^2(4)\right )}+\frac {1}{81} \left (812+162 x-36 \log (4)+162 \log ^2(4)\right )\right ) \, dx\\ &=\operatorname {Subst}\left (\int e^x \, dx,x,e^{\frac {1}{81} \left (812+162 x-36 \log (4)+162 \log ^2(4)\right )}\right )\\ &=e^{e^{\frac {1}{81} \left (812+162 x-36 \log (4)+162 \log ^2(4)\right )}}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.07, size = 23, normalized size = 1.00 \begin {gather*} e^{\frac {e^{\frac {812}{81}+2 x+2 \log ^2(4)}}{2^{8/9}}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[2*E^(E^((812 + 162*x - 36*Log[4] + 162*Log[4]^2)/81) + (812 + 162*x - 36*Log[4] + 162*Log[4]^2)/81),
x]

[Out]

E^(E^(812/81 + 2*x + 2*Log[4]^2)/2^(8/9))

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fricas [A]  time = 0.82, size = 17, normalized size = 0.74 \begin {gather*} e^{\left (e^{\left (8 \, \log \relax (2)^{2} + 2 \, x - \frac {8}{9} \, \log \relax (2) + \frac {812}{81}\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2*exp(8*log(2)^2-8/9*log(2)+2*x+812/81)*exp(exp(8*log(2)^2-8/9*log(2)+2*x+812/81)),x, algorithm="fri
cas")

[Out]

e^(e^(8*log(2)^2 + 2*x - 8/9*log(2) + 812/81))

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giac [A]  time = 0.50, size = 18, normalized size = 0.78 \begin {gather*} e^{\left (\frac {1}{2} \cdot 2^{\frac {1}{9}} e^{\left (8 \, \log \relax (2)^{2} + 2 \, x + \frac {812}{81}\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2*exp(8*log(2)^2-8/9*log(2)+2*x+812/81)*exp(exp(8*log(2)^2-8/9*log(2)+2*x+812/81)),x, algorithm="gia
c")

[Out]

e^(1/2*2^(1/9)*e^(8*log(2)^2 + 2*x + 812/81))

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maple [A]  time = 0.02, size = 18, normalized size = 0.78

method	result	size
derivativedivides	\({\mathrm e}^{{\mathrm e}^{8 \ln \relax (2)^{2}-\frac {8 \ln \relax (2)}{9}+2 x +\frac {812}{81}}}\)	\(18\)
default	\({\mathrm e}^{{\mathrm e}^{8 \ln \relax (2)^{2}-\frac {8 \ln \relax (2)}{9}+2 x +\frac {812}{81}}}\)	\(18\)
norman	\({\mathrm e}^{{\mathrm e}^{8 \ln \relax (2)^{2}-\frac {8 \ln \relax (2)}{9}+2 x +\frac {812}{81}}}\)	\(18\)
risch	\({\mathrm e}^{\frac {2^{\frac {1}{9}} {\mathrm e}^{8 \ln \relax (2)^{2}+\frac {812}{81}+2 x}}{2}}\)	\(19\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(2*exp(8*ln(2)^2-8/9*ln(2)+2*x+812/81)*exp(exp(8*ln(2)^2-8/9*ln(2)+2*x+812/81)),x,method=_RETURNVERBOSE)

[Out]

exp(exp(8*ln(2)^2-8/9*ln(2)+2*x+812/81))

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maxima [A]  time = 0.81, size = 18, normalized size = 0.78 \begin {gather*} e^{\left (\frac {1}{2} \cdot 2^{\frac {1}{9}} e^{\left (8 \, \log \relax (2)^{2} + 2 \, x + \frac {812}{81}\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2*exp(8*log(2)^2-8/9*log(2)+2*x+812/81)*exp(exp(8*log(2)^2-8/9*log(2)+2*x+812/81)),x, algorithm="max
ima")

[Out]

e^(1/2*2^(1/9)*e^(8*log(2)^2 + 2*x + 812/81))

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mupad [B]  time = 0.10, size = 19, normalized size = 0.83 \begin {gather*} {\mathrm {e}}^{\frac {2^{1/9}\,{\mathrm {e}}^{2\,x}\,{\mathrm {e}}^{812/81}\,{\mathrm {e}}^{8\,{\ln \relax (2)}^2}}{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(2*exp(2*x - (8*log(2))/9 + 8*log(2)^2 + 812/81)*exp(exp(2*x - (8*log(2))/9 + 8*log(2)^2 + 812/81)),x)

[Out]

exp((2^(1/9)*exp(2*x)*exp(812/81)*exp(8*log(2)^2))/2)

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sympy [A]  time = 0.16, size = 22, normalized size = 0.96 \begin {gather*} e^{\frac {\sqrt [9]{2} e^{2 x + 8 \log {\relax (2 )}^{2} + \frac {812}{81}}}{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2*exp(8*ln(2)**2-8/9*ln(2)+2*x+812/81)*exp(exp(8*ln(2)**2-8/9*ln(2)+2*x+812/81)),x)

[Out]

exp(2**(1/9)*exp(2*x + 8*log(2)**2 + 812/81)/2)

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