3.62.29 \(\int -e^{6+e^4-e^{5+e^4+x}+x} \, dx\)

Optimal. Leaf size=14 \[ e^{1-e^{5+e^4+x}} \]

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Rubi [A]  time = 0.02, antiderivative size = 14, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {2282, 2194} \begin {gather*} e^{1-e^{x+e^4+5}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[-E^(6 + E^4 - E^(5 + E^4 + x) + x),x]

[Out]

E^(1 - E^(5 + E^4 + 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} &=-\operatorname {Subst}\left (\int e^{6+e^4-e^{5+e^4} x} \, dx,x,e^x\right )\\ &=e^{1-e^{5+e^4+x}}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 14, normalized size = 1.00 \begin {gather*} e^{1-e^{5+e^4+x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[-E^(6 + E^4 - E^(5 + E^4 + x) + x),x]

[Out]

E^(1 - E^(5 + E^4 + x))

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fricas [A]  time = 0.71, size = 11, normalized size = 0.79 \begin {gather*} e^{\left (-e^{\left (x + e^{4} + 5\right )} + 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-exp(exp(1)^4+5+x)*exp(-exp(exp(1)^4+5+x)+1),x, algorithm="fricas")

[Out]

e^(-e^(x + e^4 + 5) + 1)

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giac [A]  time = 0.17, size = 11, normalized size = 0.79 \begin {gather*} e^{\left (-e^{\left (x + e^{4} + 5\right )} + 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-exp(exp(1)^4+5+x)*exp(-exp(exp(1)^4+5+x)+1),x, algorithm="giac")

[Out]

e^(-e^(x + e^4 + 5) + 1)

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maple [A]  time = 0.02, size = 12, normalized size = 0.86

method	result	size
risch	\({\mathrm e}^{-{\mathrm e}^{{\mathrm e}^{4}+x +5}+1}\)	\(12\)
derivativedivides	\({\mathrm e}^{-{\mathrm e}^{{\mathrm e}^{4}+x +5}+1}\)	\(14\)
default	\({\mathrm e}^{-{\mathrm e}^{{\mathrm e}^{4}+x +5}+1}\)	\(14\)
norman	\({\mathrm e}^{-{\mathrm e}^{{\mathrm e}^{4}+x +5}+1}\)	\(14\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-exp(exp(1)^4+5+x)*exp(-exp(exp(1)^4+5+x)+1),x,method=_RETURNVERBOSE)

[Out]

exp(-exp(exp(4)+x+5)+1)

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maxima [A]  time = 0.35, size = 11, normalized size = 0.79 \begin {gather*} e^{\left (-e^{\left (x + e^{4} + 5\right )} + 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-exp(exp(1)^4+5+x)*exp(-exp(exp(1)^4+5+x)+1),x, algorithm="maxima")

[Out]

e^(-e^(x + e^4 + 5) + 1)

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mupad [B]  time = 4.30, size = 12, normalized size = 0.86 \begin {gather*} \mathrm {e}\,{\mathrm {e}}^{-{\mathrm {e}}^{x+{\mathrm {e}}^4+5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-exp(x + exp(4) + 5)*exp(1 - exp(x + exp(4) + 5)),x)

[Out]

exp(1)*exp(-exp(x + exp(4) + 5))

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sympy [A]  time = 0.11, size = 10, normalized size = 0.71 \begin {gather*} e^{1 - e^{x + 5 + e^{4}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-exp(exp(1)**4+5+x)*exp(-exp(exp(1)**4+5+x)+1),x)

[Out]

exp(1 - exp(x + 5 + exp(4)))

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