3.54.31 \(\int \frac {1}{3} e^{\frac {1}{3} (8-10 x-3 x^2+e (4+x))} (-10+e-6 x) \, dx\)

Optimal. Leaf size=19 \[ e^{\frac {1}{3} (-4-x) (-2-e+3 x)} \]

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Rubi [A]  time = 0.11, antiderivative size = 25, normalized size of antiderivative = 1.32, number of steps used = 3, number of rules used = 3, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {12, 2244, 2236} \begin {gather*} e^{-x^2-\frac {1}{3} (10-e) x+\frac {4 (2+e)}{3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^((8 - 10*x - 3*x^2 + E*(4 + x))/3)*(-10 + E - 6*x))/3,x]

[Out]

E^((4*(2 + E))/3 - ((10 - E)*x)/3 - x^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 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}{3} \int e^{\frac {1}{3} \left (8-10 x-3 x^2+e (4+x)\right )} (-10+e-6 x) \, dx\\ &=\frac {1}{3} \int e^{\frac {4 (2+e)}{3}+\frac {1}{3} (-10+e) x-x^2} (-10+e-6 x) \, dx\\ &=e^{\frac {4 (2+e)}{3}-\frac {1}{3} (10-e) x-x^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 15, normalized size = 0.79 \begin {gather*} e^{\frac {1}{3} (2+e-3 x) (4+x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((8 - 10*x - 3*x^2 + E*(4 + x))/3)*(-10 + E - 6*x))/3,x]

[Out]

E^(((2 + E - 3*x)*(4 + x))/3)

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fricas [A]  time = 0.69, size = 18, normalized size = 0.95 \begin {gather*} e^{\left (-x^{2} + \frac {1}{3} \, {\left (x + 4\right )} e - \frac {10}{3} \, x + \frac {8}{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(exp(1)-6*x-10)*exp(1/3*(4+x)*exp(1)-x^2-10/3*x+8/3),x, algorithm="fricas")

[Out]

e^(-x^2 + 1/3*(x + 4)*e - 10/3*x + 8/3)

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giac [A]  time = 0.13, size = 20, normalized size = 1.05 \begin {gather*} e^{\left (-x^{2} + \frac {1}{3} \, x e - \frac {10}{3} \, x + \frac {4}{3} \, e + \frac {8}{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(exp(1)-6*x-10)*exp(1/3*(4+x)*exp(1)-x^2-10/3*x+8/3),x, algorithm="giac")

[Out]

e^(-x^2 + 1/3*x*e - 10/3*x + 4/3*e + 8/3)

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maple [A]  time = 0.07, size = 14, normalized size = 0.74

method	result	size
risch	\({\mathrm e}^{\frac {\left (4+x \right ) \left (-3 x +{\mathrm e}+2\right )}{3}}\)	\(14\)
norman	\({\mathrm e}^{\frac {\left (4+x \right ) {\mathrm e}}{3}-x^{2}-\frac {10 x}{3}+\frac {8}{3}}\)	\(19\)
gosper	\({\mathrm e}^{\frac {x \,{\mathrm e}}{3}+\frac {4 \,{\mathrm e}}{3}-x^{2}-\frac {10 x}{3}+\frac {8}{3}}\)	\(21\)
default	\(\frac {\sqrt {\pi }\, {\mathrm e}^{\frac {4 \,{\mathrm e}}{3}+\frac {11}{3}+\frac {\left (\frac {{\mathrm e}}{3}-\frac {10}{3}\right )^{2}}{4}} \erf \left (x -\frac {{\mathrm e}}{6}+\frac {5}{3}\right )}{6}-\frac {5 \sqrt {\pi }\, {\mathrm e}^{\frac {4 \,{\mathrm e}}{3}+\frac {8}{3}+\frac {\left (\frac {{\mathrm e}}{3}-\frac {10}{3}\right )^{2}}{4}} \erf \left (x -\frac {{\mathrm e}}{6}+\frac {5}{3}\right )}{3}+{\mathrm e}^{-x^{2}+\left (\frac {{\mathrm e}}{3}-\frac {10}{3}\right ) x +\frac {4 \,{\mathrm e}}{3}+\frac {8}{3}}-\frac {\left (\frac {{\mathrm e}}{3}-\frac {10}{3}\right ) \sqrt {\pi }\, {\mathrm e}^{\frac {4 \,{\mathrm e}}{3}+\frac {8}{3}+\frac {\left (\frac {{\mathrm e}}{3}-\frac {10}{3}\right )^{2}}{4}} \erf \left (x -\frac {{\mathrm e}}{6}+\frac {5}{3}\right )}{2}\)	\(118\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/3*(exp(1)-6*x-10)*exp(1/3*(4+x)*exp(1)-x^2-10/3*x+8/3),x,method=_RETURNVERBOSE)

[Out]

exp(1/3*(4+x)*(-3*x+exp(1)+2))

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maxima [A]  time = 0.45, size = 18, normalized size = 0.95 \begin {gather*} e^{\left (-x^{2} + \frac {1}{3} \, {\left (x + 4\right )} e - \frac {10}{3} \, x + \frac {8}{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(exp(1)-6*x-10)*exp(1/3*(4+x)*exp(1)-x^2-10/3*x+8/3),x, algorithm="maxima")

[Out]

e^(-x^2 + 1/3*(x + 4)*e - 10/3*x + 8/3)

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mupad [B]  time = 0.16, size = 24, normalized size = 1.26 \begin {gather*} {\mathrm {e}}^{\frac {4\,\mathrm {e}}{3}}\,{\mathrm {e}}^{-\frac {10\,x}{3}}\,{\mathrm {e}}^{8/3}\,{\mathrm {e}}^{-x^2}\,{\mathrm {e}}^{\frac {x\,\mathrm {e}}{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp((exp(1)*(x + 4))/3 - (10*x)/3 - x^2 + 8/3)*(6*x - exp(1) + 10))/3,x)

[Out]

exp((4*exp(1))/3)*exp(-(10*x)/3)*exp(8/3)*exp(-x^2)*exp((x*exp(1))/3)

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sympy [A]  time = 0.12, size = 22, normalized size = 1.16 \begin {gather*} e^{- x^{2} - \frac {10 x}{3} + e \left (\frac {x}{3} + \frac {4}{3}\right ) + \frac {8}{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(exp(1)-6*x-10)*exp(1/3*(4+x)*exp(1)-x**2-10/3*x+8/3),x)

[Out]

exp(-x**2 - 10*x/3 + E*(x/3 + 4/3) + 8/3)

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