3.44.85 \(\int (5+e^{e^{2 e^{5-x}}+x} (1-2 e^{5+2 e^{5-x}-x})+2 x) \, dx\) [4385]

Optimal. Leaf size=24 \[ -\frac {7}{3}+e^{e^{2 e^{5-x}}+x}+x (5+x) \]

[Out]

-7/3+exp(exp(2*exp(5-x))+x)+(5+x)*x

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Rubi [A]
time = 0.12, antiderivative size = 22, normalized size of antiderivative = 0.92, number of steps used = 3, number of rules used = 2, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.049, Rules used = {2320, 2326} \begin {gather*} x^2+5 x+e^{x+e^{2 e^{5-x}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

E^(E^(2*E^(5 - x)) + x) + 5*x + x^2

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]]]

Rule 2326

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = v*(y/(Log[F]*D[u, x]))}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rubi steps

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

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Mathematica [A]
time = 0.05, size = 22, normalized size = 0.92 \begin {gather*} e^{e^{2 e^{5-x}}+x}+5 x+x^2 \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

E^(E^(2*E^(5 - x)) + x) + 5*x + x^2

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Maple [A]
time = 0.17, size = 20, normalized size = 0.83

method	result	size
default	\(5 x +{\mathrm e}^{{\mathrm e}^{2 \,{\mathrm e}^{5-x}}+x}+x^{2}\)	\(20\)
norman	\(5 x +{\mathrm e}^{{\mathrm e}^{2 \,{\mathrm e}^{5-x}}+x}+x^{2}\)	\(20\)
risch	\(5 x +{\mathrm e}^{{\mathrm e}^{2 \,{\mathrm e}^{5-x}}+x}+x^{2}\)	\(20\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.30, size = 19, normalized size = 0.79 \begin {gather*} x^{2} + 5 \, x + e^{\left (x + e^{\left (2 \, e^{\left (-x + 5\right )}\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x^2 + 5*x + e^(x + e^(2*e^(-x + 5)))

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Fricas [A]
time = 0.40, size = 19, normalized size = 0.79 \begin {gather*} x^{2} + 5 \, x + e^{\left (x + e^{\left (2 \, e^{\left (-x + 5\right )}\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x^2 + 5*x + e^(x + e^(2*e^(-x + 5)))

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Sympy [A]
time = 0.12, size = 17, normalized size = 0.71 \begin {gather*} x^{2} + 5 x + e^{x + e^{2 e^{5 - x}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 44 vs. \(2 (19) = 38\).
time = 0.41, size = 44, normalized size = 1.83 \begin {gather*} x^{2} + 5 \, x + e^{\left ({\left (2 \, e^{5} + e^{\left (x + 2 \, e^{\left (-x + 5\right )}\right )} + 5 \, e^{x}\right )} e^{\left (-x\right )} + x - 2 \, e^{\left (-x + 5\right )} - 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x^2 + 5*x + e^((2*e^5 + e^(x + 2*e^(-x + 5)) + 5*e^x)*e^(-x) + x - 2*e^(-x + 5) - 5)

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Mupad [B]
time = 3.16, size = 20, normalized size = 0.83 \begin {gather*} 5\,x+{\mathrm {e}}^{{\mathrm {e}}^{2\,{\mathrm {e}}^{-x}\,{\mathrm {e}}^5}}\,{\mathrm {e}}^x+x^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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