\(\int e^{-2-e^5+e^x+x} (1+e^x) \, dx\) [3300]

Optimal result

Integrand size = 19, antiderivative size = 13 \[ \int e^{-2-e^5+e^x+x} \left (1+e^x\right ) \, dx=e^{-2-e^5+e^x+x} \]

[Out]

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

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(33\) vs. \(2(13)=26\).

Time = 0.03 (sec) , antiderivative size = 33, normalized size of antiderivative = 2.54, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {2320, 2207, 2225} \[ \int e^{-2-e^5+e^x+x} \left (1+e^x\right ) \, dx=e^{e^x-2-e^5} \left (e^x+1\right )-e^{e^x-2-e^5} \]

[In]

Int[E^(-2 - E^5 + E^x + x)*(1 + E^x),x]

[Out]

-E^(-2 - E^5 + E^x) + E^(-2 - E^5 + E^x)*(1 + E^x)

Rule 2207

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(c + d*x)^m*
((b*F^(g*(e + f*x)))^n/(f*g*n*Log[F])), x] - Dist[d*(m/(f*g*n*Log[F])), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] &&  !TrueQ[$UseGamma]

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^{-2-e^5+x} (1+x) \, dx,x,e^x\right ) \\ & = e^{-2-e^5+e^x} \left (1+e^x\right )-\text {Subst}\left (\int e^{-2-e^5+x} \, dx,x,e^x\right ) \\ & = -e^{-2-e^5+e^x}+e^{-2-e^5+e^x} \left (1+e^x\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00 \[ \int e^{-2-e^5+e^x+x} \left (1+e^x\right ) \, dx=e^{-2-e^5+e^x+x} \]

[In]

Integrate[E^(-2 - E^5 + E^x + x)*(1 + E^x),x]

[Out]

E^(-2 - E^5 + E^x + x)

Maple [A] (verified)

Time = 0.60 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.85

method	result	size
derivativedivides	\({\mathrm e}^{{\mathrm e}^{x}-{\mathrm e}^{5}+x -2}\)	\(11\)
default	\({\mathrm e}^{{\mathrm e}^{x}-{\mathrm e}^{5}+x -2}\)	\(11\)
norman	\({\mathrm e}^{{\mathrm e}^{x}-{\mathrm e}^{5}+x -2}\)	\(11\)
risch	\({\mathrm e}^{{\mathrm e}^{x}-{\mathrm e}^{5}+x -2}\)	\(11\)
parallelrisch	\({\mathrm e}^{{\mathrm e}^{x}-{\mathrm e}^{5}+x -2}\)	\(11\)
parts	\({\mathrm e}^{-{\mathrm e}^{5}} {\mathrm e}^{-2} \left ({\mathrm e}^{x} {\mathrm e}^{{\mathrm e}^{x}}-{\mathrm e}^{{\mathrm e}^{x}}\right )+{\mathrm e}^{-{\mathrm e}^{5}} {\mathrm e}^{-2} {\mathrm e}^{{\mathrm e}^{x}}\)	\(33\)

[In]

int((exp(x)+1)*exp(exp(x)-exp(5)+x-2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.77 \[ \int e^{-2-e^5+e^x+x} \left (1+e^x\right ) \, dx=e^{\left (x - e^{5} + e^{x} - 2\right )} \]

[In]

integrate((exp(x)+1)*exp(exp(x)-exp(5)+x-2),x, algorithm="fricas")

[Out]

e^(x - e^5 + e^x - 2)

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.77 \[ \int e^{-2-e^5+e^x+x} \left (1+e^x\right ) \, dx=e^{x + e^{x} - e^{5} - 2} \]

[In]

integrate((exp(x)+1)*exp(exp(x)-exp(5)+x-2),x)

[Out]

exp(x + exp(x) - exp(5) - 2)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.77 \[ \int e^{-2-e^5+e^x+x} \left (1+e^x\right ) \, dx=e^{\left (x - e^{5} + e^{x} - 2\right )} \]

[In]

integrate((exp(x)+1)*exp(exp(x)-exp(5)+x-2),x, algorithm="maxima")

[Out]

e^(x - e^5 + e^x - 2)

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.77 \[ \int e^{-2-e^5+e^x+x} \left (1+e^x\right ) \, dx=e^{\left (x - e^{5} + e^{x} - 2\right )} \]

[In]

integrate((exp(x)+1)*exp(exp(x)-exp(5)+x-2),x, algorithm="giac")

[Out]

e^(x - e^5 + e^x - 2)

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00 \[ \int e^{-2-e^5+e^x+x} \left (1+e^x\right ) \, dx={\mathrm {e}}^{-{\mathrm {e}}^5}\,{\mathrm {e}}^{{\mathrm {e}}^x}\,{\mathrm {e}}^{-2}\,{\mathrm {e}}^x \]

[In]

int(exp(x - exp(5) + exp(x) - 2)*(exp(x) + 1),x)

[Out]

exp(-exp(5))*exp(exp(x))*exp(-2)*exp(x)