3.101.9 \(\int \frac {1}{2} e^{-6-x} (10-e-2 x+x^2) \, dx\)

Optimal. Leaf size=19 \[ \frac {1}{2} e^{-6-x} \left (-10+e-x^2\right ) \]

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Rubi [A]  time = 0.07, antiderivative size = 31, normalized size of antiderivative = 1.63, number of steps used = 9, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {12, 2196, 2194, 2176} \begin {gather*} -\frac {1}{2} e^{-x-6} x^2-\frac {1}{2} (10-e) e^{-x-6} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(-6 - x)*(10 - E - 2*x + x^2))/2,x]

[Out]

-1/2*((10 - E)*E^(-6 - x)) - (E^(-6 - x)*x^2)/2

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 2196

Int[(F_)^((c_.)*(v_))*(u_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), u, x], x] /; FreeQ[{F, c
}, x] && PolynomialQ[u, x] && LinearQ[v, x] &&  !$UseGamma === True

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{2} \int e^{-6-x} \left (10-e-2 x+x^2\right ) \, dx\\ &=\frac {1}{2} \int \left (10 \left (1-\frac {e}{10}\right ) e^{-6-x}-2 e^{-6-x} x+e^{-6-x} x^2\right ) \, dx\\ &=\frac {1}{2} \int e^{-6-x} x^2 \, dx+\frac {1}{2} (10-e) \int e^{-6-x} \, dx-\int e^{-6-x} x \, dx\\ &=-\frac {1}{2} (10-e) e^{-6-x}+e^{-6-x} x-\frac {1}{2} e^{-6-x} x^2-\int e^{-6-x} \, dx+\int e^{-6-x} x \, dx\\ &=e^{-6-x}-\frac {1}{2} (10-e) e^{-6-x}-\frac {1}{2} e^{-6-x} x^2+\int e^{-6-x} \, dx\\ &=-\frac {1}{2} (10-e) e^{-6-x}-\frac {1}{2} e^{-6-x} x^2\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 19, normalized size = 1.00 \begin {gather*} \frac {1}{2} e^{-6-x} \left (-10+e-x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(-6 - x)*(10 - E - 2*x + x^2))/2,x]

[Out]

(E^(-6 - x)*(-10 + E - x^2))/2

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fricas [A]  time = 0.62, size = 17, normalized size = 0.89 \begin {gather*} -\frac {1}{2} \, {\left (x^{2} - e + 10\right )} e^{\left (-x - 6\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(-exp(1)+x^2-2*x+10)/exp(3)/exp(3+x),x, algorithm="fricas")

[Out]

-1/2*(x^2 - e + 10)*e^(-x - 6)

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giac [A]  time = 0.16, size = 22, normalized size = 1.16 \begin {gather*} -\frac {1}{2} \, {\left (x^{2} + 10\right )} e^{\left (-x - 6\right )} + \frac {1}{2} \, e^{\left (-x - 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(-exp(1)+x^2-2*x+10)/exp(3)/exp(3+x),x, algorithm="giac")

[Out]

-1/2*(x^2 + 10)*e^(-x - 6) + 1/2*e^(-x - 5)

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maple [A]  time = 0.05, size = 18, normalized size = 0.95

method	result	size
risch	\(\frac {\left ({\mathrm e}-x^{2}-10\right ) {\mathrm e}^{-x -6}}{2}\)	\(18\)
gosper	\(\frac {{\mathrm e}^{-3} \left ({\mathrm e}-x^{2}-10\right ) {\mathrm e}^{-3-x}}{2}\)	\(22\)
norman	\(\left (-\frac {x^{2} {\mathrm e}^{-3}}{2}+\frac {{\mathrm e}^{-3} \left ({\mathrm e}-10\right )}{2}\right ) {\mathrm e}^{-3-x}\)	\(28\)
derivativedivides	\(\frac {{\mathrm e}^{-3} \left (-{\mathrm e}^{-3-x} \left (3+x \right )^{2}+6 \,{\mathrm e}^{-3-x} \left (3+x \right )-19 \,{\mathrm e}^{-3-x}+{\mathrm e}^{-3-x} {\mathrm e}\right )}{2}\)	\(49\)
default	\(\frac {{\mathrm e}^{-3} \left (-{\mathrm e}^{-3-x} \left (3+x \right )^{2}+6 \,{\mathrm e}^{-3-x} \left (3+x \right )-19 \,{\mathrm e}^{-3-x}+{\mathrm e}^{-3-x} {\mathrm e}\right )}{2}\)	\(49\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/2*(-exp(1)+x^2-2*x+10)/exp(3)/exp(3+x),x,method=_RETURNVERBOSE)

[Out]

1/2*(exp(1)-x^2-10)*exp(-x-6)

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maxima [B]  time = 0.43, size = 43, normalized size = 2.26 \begin {gather*} -\frac {1}{2} \, {\left (x^{2} + 2 \, x + 2\right )} e^{\left (-x - 6\right )} + {\left (x + 1\right )} e^{\left (-x - 6\right )} + \frac {1}{2} \, e^{\left (-x - 5\right )} - 5 \, e^{\left (-x - 6\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(-exp(1)+x^2-2*x+10)/exp(3)/exp(3+x),x, algorithm="maxima")

[Out]

-1/2*(x^2 + 2*x + 2)*e^(-x - 6) + (x + 1)*e^(-x - 6) + 1/2*e^(-x - 5) - 5*e^(-x - 6)

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mupad [B]  time = 0.07, size = 19, normalized size = 1.00 \begin {gather*} -{\mathrm {e}}^{-x-6}\,\left (\frac {x^2}{2}-\frac {\mathrm {e}}{2}+5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-exp(-3)*exp(- x - 3)*(x + exp(1)/2 - x^2/2 - 5),x)

[Out]

-exp(- x - 6)*(x^2/2 - exp(1)/2 + 5)

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sympy [A]  time = 0.11, size = 19, normalized size = 1.00 \begin {gather*} \frac {\left (- x^{2} - 10 + e\right ) e^{- x - 3}}{2 e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(-exp(1)+x**2-2*x+10)/exp(3)/exp(3+x),x)

[Out]

(-x**2 - 10 + E)*exp(-3)*exp(-x - 3)/2

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