3.42.58 \(\int \frac {5}{12} e^{-3+\frac {5}{12} e^{e^{-3+x}}+e^{-3+x}+x} \, dx\)

Optimal. Leaf size=15 \[ 6+e^{\frac {5}{12} e^{e^{-3+x}}} \]

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Rubi [A]  time = 0.04, antiderivative size = 13, normalized size of antiderivative = 0.87, number of steps used = 4, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {12, 2282, 2194} \begin {gather*} e^{\frac {5 e^{e^{x-3}}}{12}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(5*E^(-3 + (5*E^E^(-3 + x))/12 + E^(-3 + x) + x))/12,x]

[Out]

E^((5*E^E^(-3 + x))/12)

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} &=\frac {5}{12} \int e^{-3+\frac {5}{12} e^{e^{-3+x}}+e^{-3+x}+x} \, dx\\ &=\frac {5}{12} \operatorname {Subst}\left (\int e^{-3+\frac {5 e^{\frac {x}{e^3}}}{12}+\frac {x}{e^3}} \, dx,x,e^x\right )\\ &=\frac {1}{12} \left (5 e^3\right ) \operatorname {Subst}\left (\int e^{-3+\frac {5 x}{12}} \, dx,x,e^{e^{-3+x}}\right )\\ &=e^{\frac {5}{12} e^{e^{-3+x}}}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.04, size = 13, normalized size = 0.87 \begin {gather*} e^{\frac {5}{12} e^{e^{-3+x}}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(5*E^(-3 + (5*E^E^(-3 + x))/12 + E^(-3 + x) + x))/12,x]

[Out]

E^((5*E^E^(-3 + x))/12)

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fricas [A]  time = 0.71, size = 8, normalized size = 0.53 \begin {gather*} e^{\left (\frac {5}{12} \, e^{\left (e^{\left (x - 3\right )}\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(5/12*exp(x-3)*exp(exp(x-3))*exp(5/12*exp(exp(x-3))),x, algorithm="fricas")

[Out]

e^(5/12*e^(e^(x - 3)))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {5}{12} \, e^{\left (x + e^{\left (x - 3\right )} + \frac {5}{12} \, e^{\left (e^{\left (x - 3\right )}\right )} - 3\right )}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(5/12*exp(x-3)*exp(exp(x-3))*exp(5/12*exp(exp(x-3))),x, algorithm="giac")

[Out]

integrate(5/12*e^(x + e^(x - 3) + 5/12*e^(e^(x - 3)) - 3), x)

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maple [A]  time = 0.03, size = 9, normalized size = 0.60

method	result	size
derivativedivides	\({\mathrm e}^{\frac {5 \,{\mathrm e}^{{\mathrm e}^{x -3}}}{12}}\)	\(9\)
default	\({\mathrm e}^{\frac {5 \,{\mathrm e}^{{\mathrm e}^{x -3}}}{12}}\)	\(9\)
norman	\({\mathrm e}^{\frac {5 \,{\mathrm e}^{{\mathrm e}^{x -3}}}{12}}\)	\(9\)
risch	\({\mathrm e}^{\frac {5 \,{\mathrm e}^{{\mathrm e}^{x -3}}}{12}}\)	\(9\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(5/12*exp(x-3)*exp(exp(x-3))*exp(5/12*exp(exp(x-3))),x,method=_RETURNVERBOSE)

[Out]

exp(5/12*exp(exp(x-3)))

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maxima [A]  time = 0.42, size = 8, normalized size = 0.53 \begin {gather*} e^{\left (\frac {5}{12} \, e^{\left (e^{\left (x - 3\right )}\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(5/12*exp(x-3)*exp(exp(x-3))*exp(5/12*exp(exp(x-3))),x, algorithm="maxima")

[Out]

e^(5/12*e^(e^(x - 3)))

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mupad [B]  time = 0.08, size = 8, normalized size = 0.53 \begin {gather*} {\mathrm {e}}^{\frac {5\,{\mathrm {e}}^{{\mathrm {e}}^{x-3}}}{12}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5*exp(x - 3)*exp((5*exp(exp(x - 3)))/12)*exp(exp(x - 3)))/12,x)

[Out]

exp((5*exp(exp(x - 3)))/12)

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sympy [A]  time = 0.18, size = 10, normalized size = 0.67 \begin {gather*} e^{\frac {5 e^{e^{x - 3}}}{12}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(5/12*exp(x-3)*exp(exp(x-3))*exp(5/12*exp(exp(x-3))),x)

[Out]

exp(5*exp(exp(x - 3))/12)

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