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

Optimal. Leaf size=16 \[ 1+e^{1-e^x}+x^2+\log (3) \]

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Rubi [A]  time = 0.06, antiderivative size = 13, normalized size of antiderivative = 0.81, number of steps used = 4, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {6688, 2282, 2194} \begin {gather*} x^2+e^{1-e^x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-E^(1 + x) + 2*E^E^x*x)/E^E^x,x]

[Out]

E^(1 - E^x) + x^2

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]]]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-e^{1-e^x+x}+2 x\right ) \, dx\\ &=x^2-\int e^{1-e^x+x} \, dx\\ &=x^2-\operatorname {Subst}\left (\int e^{1-x} \, dx,x,e^x\right )\\ &=e^{1-e^x}+x^2\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 13, normalized size = 0.81 \begin {gather*} e^{1-e^x}+x^2 \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

E^(1 - E^x) + x^2

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fricas [A]  time = 0.53, size = 16, normalized size = 1.00 \begin {gather*} {\left (x^{2} e^{\left (e^{x}\right )} + e\right )} e^{\left (-e^{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x^2*e^(e^x) + e)*e^(-e^x)

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giac [A]  time = 0.15, size = 11, normalized size = 0.69 \begin {gather*} x^{2} + e^{\left (-e^{x} + 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x^2 + e^(-e^x + 1)

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maple [A]  time = 0.03, size = 12, normalized size = 0.75

method	result	size
risch	\(x^{2}+{\mathrm e}^{1-{\mathrm e}^{x}}\)	\(12\)
default	\(x^{2}+{\mathrm e} \,{\mathrm e}^{-{\mathrm e}^{x}}\)	\(13\)
norman	\(\left ({\mathrm e}^{{\mathrm e}^{x}} x^{2}+{\mathrm e}\right ) {\mathrm e}^{-{\mathrm e}^{x}}\)	\(17\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x^2+exp(1-exp(x))

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maxima [A]  time = 0.37, size = 11, normalized size = 0.69 \begin {gather*} x^{2} + e^{\left (-e^{x} + 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x^2 + e^(-e^x + 1)

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mupad [B]  time = 0.07, size = 12, normalized size = 0.75 \begin {gather*} \mathrm {e}\,{\mathrm {e}}^{-{\mathrm {e}}^x}+x^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

exp(1)*exp(-exp(x)) + x^2

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sympy [A]  time = 0.12, size = 10, normalized size = 0.62 \begin {gather*} x^{2} + e e^{- e^{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x*exp(exp(x))-exp(1)*exp(x))/exp(exp(x)),x)

[Out]

x**2 + E*exp(-exp(x))

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