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

Optimal. Leaf size=13 \[ e^{-2-e^5+e^x+x} \]

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Rubi [B]  time = 0.04, antiderivative size = 33, normalized size of antiderivative = 2.54, number of steps used = 3, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {2282, 2176, 2194} \begin {gather*} e^{e^x-2-e^5} \left (e^x+1\right )-e^{e^x-2-e^5} \end {gather*}

Antiderivative was successfully verified.

[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 2176

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] &&  !$UseGamma === True

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^{-2-e^5+x} (1+x) \, dx,x,e^x\right )\\ &=e^{-2-e^5+e^x} \left (1+e^x\right )-\operatorname {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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 13, normalized size = 1.00 \begin {gather*} e^{-2-e^5+e^x+x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

E^(-2 - E^5 + E^x + x)

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fricas [A]  time = 0.85, size = 10, normalized size = 0.77 \begin {gather*} e^{\left (x - e^{5} + e^{x} - 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e^(x - e^5 + e^x - 2)

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giac [A]  time = 0.24, size = 10, normalized size = 0.77 \begin {gather*} e^{\left (x - e^{5} + e^{x} - 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e^(x - e^5 + e^x - 2)

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maple [A]  time = 0.02, size = 11, normalized size = 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	\({\mathrm e}^{{\mathrm e}^{x}-{\mathrm e}^{5}+x -2}\)	\(11\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.74, size = 10, normalized size = 0.77 \begin {gather*} e^{\left (x - e^{5} + e^{x} - 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e^(x - e^5 + e^x - 2)

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mupad [B]  time = 1.99, size = 13, normalized size = 1.00 \begin {gather*} {\mathrm {e}}^{-{\mathrm {e}}^5}\,{\mathrm {e}}^{{\mathrm {e}}^x}\,{\mathrm {e}}^{-2}\,{\mathrm {e}}^x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.14, size = 10, normalized size = 0.77 \begin {gather*} e^{x + e^{x} - e^{5} - 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

exp(x + exp(x) - exp(5) - 2)

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