\(\int \frac {1}{2} e^{-6-x} (10-e-2 x+x^2) \, dx\) [10009]

Optimal result

Integrand size = 22, antiderivative size = 19 \[ \int \frac {1}{2} e^{-6-x} \left (10-e-2 x+x^2\right ) \, dx=\frac {1}{2} e^{-6-x} \left (-10+e-x^2\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.63, number of steps used = 9, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {12, 2227, 2225, 2207} \[ \int \frac {1}{2} e^{-6-x} \left (10-e-2 x+x^2\right ) \, dx=-\frac {1}{2} e^{-x-6} x^2-\frac {1}{2} (10-e) e^{-x-6} \]

[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 2207

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] &&  !TrueQ[$UseGamma]

Rule 2225

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 2227

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] &&  !TrueQ[$UseGamma]

Rubi steps \begin{align*} \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{align*}

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \frac {1}{2} e^{-6-x} \left (10-e-2 x+x^2\right ) \, dx=\frac {1}{2} e^{-6-x} \left (-10+e-x^2\right ) \]

[In]

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

[Out]

(E^(-6 - x)*(-10 + E - x^2))/2

Maple [A] (verified)

Time = 0.61 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.95

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

[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)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89 \[ \int \frac {1}{2} e^{-6-x} \left (10-e-2 x+x^2\right ) \, dx=-\frac {1}{2} \, {\left (x^{2} - e + 10\right )} e^{\left (-x - 6\right )} \]

[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)

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \frac {1}{2} e^{-6-x} \left (10-e-2 x+x^2\right ) \, dx=\frac {\left (- x^{2} - 10 + e\right ) e^{- x - 3}}{2 e^{3}} \]

[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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 43 vs. \(2 (17) = 34\).

Time = 0.21 (sec) , antiderivative size = 43, normalized size of antiderivative = 2.26 \[ \int \frac {1}{2} e^{-6-x} \left (10-e-2 x+x^2\right ) \, dx=-\frac {1}{2} \, {\left (x^{2} + 2 \, x + 2\right )} e^{\left (-x - 6\right )} + {\left (x + 1\right )} e^{\left (-x - 6\right )} + \frac {1}{2} \, e^{\left (-x - 5\right )} - 5 \, e^{\left (-x - 6\right )} \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.42 \[ \int \frac {1}{2} e^{-6-x} \left (10-e-2 x+x^2\right ) \, dx=-\frac {1}{2} \, {\left ({\left (x + 6\right )}^{2} - 12 \, x - 26\right )} e^{\left (-x - 6\right )} + \frac {1}{2} \, e^{\left (-x - 5\right )} \]

[In]

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

[Out]

-1/2*((x + 6)^2 - 12*x - 26)*e^(-x - 6) + 1/2*e^(-x - 5)

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \frac {1}{2} e^{-6-x} \left (10-e-2 x+x^2\right ) \, dx=-{\mathrm {e}}^{-x-6}\,\left (\frac {x^2}{2}-\frac {\mathrm {e}}{2}+5\right ) \]

[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)