3.17.83 \(\int \frac {1}{5} (10+e^{5-e^{10}-x-x^2} (-2-4 x)) \, dx\)

Optimal. Leaf size=26 \[ 2 \left (2+\frac {1}{5} e^{5-e^{10}-x-x^2}+x\right ) \]

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Rubi [A]  time = 0.02, antiderivative size = 25, normalized size of antiderivative = 0.96, number of steps used = 3, number of rules used = 2, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {12, 2236} \begin {gather*} \frac {2}{5} e^{-x^2-x-e^{10}+5}+2 x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(10 + E^(5 - E^10 - x - x^2)*(-2 - 4*x))/5,x]

[Out]

(2*E^(5 - E^10 - x - x^2))/5 + 2*x

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2236

Int[(F_)^((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)*((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[(e*F^(a + b*x + c*x^2))/(
2*c*Log[F]), x] /; FreeQ[{F, a, b, c, d, e}, x] && EqQ[b*e - 2*c*d, 0]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{5} \int \left (10+e^{5-e^{10}-x-x^2} (-2-4 x)\right ) \, dx\\ &=2 x+\frac {1}{5} \int e^{5-e^{10}-x-x^2} (-2-4 x) \, dx\\ &=\frac {2}{5} e^{5-e^{10}-x-x^2}+2 x\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.05, size = 25, normalized size = 0.96 \begin {gather*} \frac {2}{5} e^{5-e^{10}-x-x^2}+2 x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(10 + E^(5 - E^10 - x - x^2)*(-2 - 4*x))/5,x]

[Out]

(2*E^(5 - E^10 - x - x^2))/5 + 2*x

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fricas [A]  time = 0.75, size = 21, normalized size = 0.81 \begin {gather*} 2 \, x + \frac {2}{5} \, e^{\left (-x^{2} - x - e^{10} + 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(-4*x-2)*exp(-exp(5)^2-x^2-x+5)+2,x, algorithm="fricas")

[Out]

2*x + 2/5*e^(-x^2 - x - e^10 + 5)

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giac [A]  time = 0.17, size = 21, normalized size = 0.81 \begin {gather*} 2 \, x + \frac {2}{5} \, e^{\left (-x^{2} - x - e^{10} + 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(-4*x-2)*exp(-exp(5)^2-x^2-x+5)+2,x, algorithm="giac")

[Out]

2*x + 2/5*e^(-x^2 - x - e^10 + 5)

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maple [A]  time = 0.02, size = 22, normalized size = 0.85

method	result	size
risch	\(2 x +\frac {2 \,{\mathrm e}^{-{\mathrm e}^{10}-x^{2}-x +5}}{5}\)	\(22\)
default	\(2 x +\frac {2 \,{\mathrm e}^{-{\mathrm e}^{10}-x^{2}-x +5}}{5}\)	\(24\)
norman	\(2 x +\frac {2 \,{\mathrm e}^{-{\mathrm e}^{10}-x^{2}-x +5}}{5}\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/5*(-4*x-2)*exp(-exp(5)^2-x^2-x+5)+2,x,method=_RETURNVERBOSE)

[Out]

2*x+2/5*exp(-exp(10)-x^2-x+5)

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maxima [A]  time = 0.33, size = 21, normalized size = 0.81 \begin {gather*} 2 \, x + \frac {2}{5} \, e^{\left (-x^{2} - x - e^{10} + 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(-4*x-2)*exp(-exp(5)^2-x^2-x+5)+2,x, algorithm="maxima")

[Out]

2*x + 2/5*e^(-x^2 - x - e^10 + 5)

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mupad [B]  time = 0.12, size = 23, normalized size = 0.88 \begin {gather*} 2\,x+\frac {2\,{\mathrm {e}}^{-{\mathrm {e}}^{10}}\,{\mathrm {e}}^{-x}\,{\mathrm {e}}^5\,{\mathrm {e}}^{-x^2}}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(2 - (exp(5 - exp(10) - x^2 - x)*(4*x + 2))/5,x)

[Out]

2*x + (2*exp(-exp(10))*exp(-x)*exp(5)*exp(-x^2))/5

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sympy [A]  time = 0.11, size = 17, normalized size = 0.65 \begin {gather*} 2 x + \frac {2 e^{- x^{2} - x - e^{10} + 5}}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(-4*x-2)*exp(-exp(5)**2-x**2-x+5)+2,x)

[Out]

2*x + 2*exp(-x**2 - x - exp(10) + 5)/5

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