3.18.42 \(\int \frac {1}{6} e^{-1+\frac {1}{16} (-5 x+x^2)} (-5+2 x) \, dx\)

Optimal. Leaf size=16 \[ \frac {8}{3} e^{-1+\frac {1}{16} (-5+x) x} \]

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Rubi [A]  time = 0.03, antiderivative size = 20, normalized size of antiderivative = 1.25, number of steps used = 3, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {12, 2244, 2236} \begin {gather*} \frac {8}{3} e^{\frac {x^2}{16}-\frac {5 x}{16}-1} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(-1 + (-5*x + x^2)/16)*(-5 + 2*x))/6,x]

[Out]

(8*E^(-1 - (5*x)/16 + x^2/16))/3

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]

Rule 2244

Int[(F_)^(v_)*(u_)^(m_.), x_Symbol] :> Int[ExpandToSum[u, x]^m*F^ExpandToSum[v, x], x] /; FreeQ[{F, m}, x] &&
LinearQ[u, x] && QuadraticQ[v, x] &&  !(LinearMatchQ[u, x] && QuadraticMatchQ[v, x])

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{6} \int e^{-1+\frac {1}{16} \left (-5 x+x^2\right )} (-5+2 x) \, dx\\ &=\frac {1}{6} \int e^{-1-\frac {5 x}{16}+\frac {x^2}{16}} (-5+2 x) \, dx\\ &=\frac {8}{3} e^{-1-\frac {5 x}{16}+\frac {x^2}{16}}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.04, size = 20, normalized size = 1.25 \begin {gather*} \frac {8}{3} e^{-1-\frac {5 x}{16}+\frac {x^2}{16}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(-1 + (-5*x + x^2)/16)*(-5 + 2*x))/6,x]

[Out]

(8*E^(-1 - (5*x)/16 + x^2/16))/3

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fricas [A]  time = 0.65, size = 13, normalized size = 0.81 \begin {gather*} \frac {8}{3} \, e^{\left (\frac {1}{16} \, x^{2} - \frac {5}{16} \, x - 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/6*(2*x-5)*exp(1/16*x^2-5/16*x)/exp(1),x, algorithm="fricas")

[Out]

8/3*e^(1/16*x^2 - 5/16*x - 1)

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giac [A]  time = 0.21, size = 13, normalized size = 0.81 \begin {gather*} \frac {8}{3} \, e^{\left (\frac {1}{16} \, x^{2} - \frac {5}{16} \, x - 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/6*(2*x-5)*exp(1/16*x^2-5/16*x)/exp(1),x, algorithm="giac")

[Out]

8/3*e^(1/16*x^2 - 5/16*x - 1)

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maple [A]  time = 0.04, size = 14, normalized size = 0.88

method	result	size
risch	\(\frac {8 \,{\mathrm e}^{\frac {1}{16} x^{2}-\frac {5}{16} x -1}}{3}\)	\(14\)
gosper	\(\frac {8 \,{\mathrm e}^{\frac {1}{16} x^{2}-\frac {5}{16} x} {\mathrm e}^{-1}}{3}\)	\(17\)
default	\(\frac {8 \,{\mathrm e}^{\frac {1}{16} x^{2}-\frac {5}{16} x} {\mathrm e}^{-1}}{3}\)	\(17\)
norman	\(\frac {8 \,{\mathrm e}^{\frac {1}{16} x^{2}-\frac {5}{16} x} {\mathrm e}^{-1}}{3}\)	\(17\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/6*(2*x-5)*exp(1/16*x^2-5/16*x)/exp(1),x,method=_RETURNVERBOSE)

[Out]

8/3*exp(1/16*x^2-5/16*x-1)

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maxima [A]  time = 0.34, size = 13, normalized size = 0.81 \begin {gather*} \frac {8}{3} \, e^{\left (\frac {1}{16} \, x^{2} - \frac {5}{16} \, x - 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/6*(2*x-5)*exp(1/16*x^2-5/16*x)/exp(1),x, algorithm="maxima")

[Out]

8/3*e^(1/16*x^2 - 5/16*x - 1)

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mupad [B]  time = 1.09, size = 14, normalized size = 0.88 \begin {gather*} \frac {8\,{\mathrm {e}}^{-\frac {5\,x}{16}}\,{\mathrm {e}}^{-1}\,{\mathrm {e}}^{\frac {x^2}{16}}}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-1)*exp(x^2/16 - (5*x)/16)*(2*x - 5))/6,x)

[Out]

(8*exp(-(5*x)/16)*exp(-1)*exp(x^2/16))/3

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sympy [A]  time = 0.11, size = 17, normalized size = 1.06 \begin {gather*} \frac {8 e^{\frac {x^{2}}{16} - \frac {5 x}{16}}}{3 e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/6*(2*x-5)*exp(1/16*x**2-5/16*x)/exp(1),x)

[Out]

8*exp(-1)*exp(x**2/16 - 5*x/16)/3

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