\(\int e^{7-e^{2+x}+x} \, dx\) [6785]

Optimal result

Integrand size = 12, antiderivative size = 19 \[ \int e^{7-e^{2+x}+x} \, dx=e^{16+e}-e^{5-e^{2+x}} \]

[Out]

exp(16+exp(1))-exp(-exp(2+x)+5)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.68, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2320, 2225} \[ \int e^{7-e^{2+x}+x} \, dx=-e^{5-e^{x+2}} \]

[In]

Int[E^(7 - E^(2 + x) + x),x]

[Out]

-E^(5 - E^(2 + x))

Rule 2225

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 2320

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{align*} \text {integral}& = \text {Subst}\left (\int e^{7-e^2 x} \, dx,x,e^x\right ) \\ & = -e^{5-e^{2+x}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.68 \[ \int e^{7-e^{2+x}+x} \, dx=-e^{5-e^{2+x}} \]

[In]

Integrate[E^(7 - E^(2 + x) + x),x]

[Out]

-E^(5 - E^(2 + x))

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.63

method	result	size
derivativedivides	\(-{\mathrm e}^{-{\mathrm e}^{2+x}+5}\)	\(12\)
default	\(-{\mathrm e}^{-{\mathrm e}^{2+x}+5}\)	\(12\)
norman	\(-{\mathrm e}^{-{\mathrm e}^{2+x}+5}\)	\(12\)
risch	\(-{\mathrm e}^{-{\mathrm e}^{2+x}+5}\)	\(12\)
parallelrisch	\(-{\mathrm e}^{-{\mathrm e}^{2+x}+5}\)	\(12\)

[In]

int(exp(2+x)*exp(-exp(2+x)+5),x,method=_RETURNVERBOSE)

[Out]

-exp(-exp(2+x)+5)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.58 \[ \int e^{7-e^{2+x}+x} \, dx=-e^{\left (-e^{\left (x + 2\right )} + 5\right )} \]

[In]

integrate(exp(2+x)*exp(-exp(2+x)+5),x, algorithm="fricas")

[Out]

-e^(-e^(x + 2) + 5)

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.42 \[ \int e^{7-e^{2+x}+x} \, dx=- e^{5 - e^{x + 2}} \]

[In]

integrate(exp(2+x)*exp(-exp(2+x)+5),x)

[Out]

-exp(5 - exp(x + 2))

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.58 \[ \int e^{7-e^{2+x}+x} \, dx=-e^{\left (-e^{\left (x + 2\right )} + 5\right )} \]

[In]

integrate(exp(2+x)*exp(-exp(2+x)+5),x, algorithm="maxima")

[Out]

-e^(-e^(x + 2) + 5)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.58 \[ \int e^{7-e^{2+x}+x} \, dx=-e^{\left (-e^{\left (x + 2\right )} + 5\right )} \]

[In]

integrate(exp(2+x)*exp(-exp(2+x)+5),x, algorithm="giac")

[Out]

-e^(-e^(x + 2) + 5)

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.58 \[ \int e^{7-e^{2+x}+x} \, dx=-{\mathrm {e}}^{-{\mathrm {e}}^{x+2}}\,{\mathrm {e}}^5 \]

[In]

int(exp(x + 2)*exp(5 - exp(x + 2)),x)

[Out]

-exp(-exp(x + 2))*exp(5)