3.78.32 \(\int \frac {1}{15} e^{\frac {1}{15} (-2 x+x^2)} (-2+2 x) \, dx\)

Optimal. Leaf size=10 \[ e^{\frac {1}{15} (-2+x) x} \]

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Rubi [A]  time = 0.03, antiderivative size = 15, normalized size of antiderivative = 1.50, number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {12, 2244, 2236} \begin {gather*} e^{\frac {x^2}{15}-\frac {2 x}{15}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^((-2*x + x^2)/15)*(-2 + 2*x))/15,x]

[Out]

E^((-2*x)/15 + x^2/15)

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}{15} \int e^{\frac {1}{15} \left (-2 x+x^2\right )} (-2+2 x) \, dx\\ &=\frac {1}{15} \int e^{-\frac {2 x}{15}+\frac {x^2}{15}} (-2+2 x) \, dx\\ &=e^{-\frac {2 x}{15}+\frac {x^2}{15}}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 10, normalized size = 1.00 \begin {gather*} e^{\frac {1}{15} (-2+x) x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((-2*x + x^2)/15)*(-2 + 2*x))/15,x]

[Out]

E^(((-2 + x)*x)/15)

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fricas [A]  time = 0.58, size = 10, normalized size = 1.00 \begin {gather*} e^{\left (\frac {1}{15} \, x^{2} - \frac {2}{15} \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/15*(2*x-2)*exp(1/15*x^2-2/15*x),x, algorithm="fricas")

[Out]

e^(1/15*x^2 - 2/15*x)

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giac [A]  time = 0.13, size = 10, normalized size = 1.00 \begin {gather*} e^{\left (\frac {1}{15} \, x^{2} - \frac {2}{15} \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/15*(2*x-2)*exp(1/15*x^2-2/15*x),x, algorithm="giac")

[Out]

e^(1/15*x^2 - 2/15*x)

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maple [A]  time = 0.03, size = 8, normalized size = 0.80

method	result	size
risch	\({\mathrm e}^{\frac {\left (x -2\right ) x}{15}}\)	\(8\)
gosper	\({\mathrm e}^{\frac {1}{15} x^{2}-\frac {2}{15} x}\)	\(11\)
default	\({\mathrm e}^{\frac {1}{15} x^{2}-\frac {2}{15} x}\)	\(11\)
norman	\({\mathrm e}^{\frac {1}{15} x^{2}-\frac {2}{15} x}\)	\(11\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/15*(2*x-2)*exp(1/15*x^2-2/15*x),x,method=_RETURNVERBOSE)

[Out]

exp(1/15*(x-2)*x)

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maxima [A]  time = 0.36, size = 10, normalized size = 1.00 \begin {gather*} e^{\left (\frac {1}{15} \, x^{2} - \frac {2}{15} \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/15*(2*x-2)*exp(1/15*x^2-2/15*x),x, algorithm="maxima")

[Out]

e^(1/15*x^2 - 2/15*x)

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mupad [B]  time = 5.18, size = 7, normalized size = 0.70 \begin {gather*} {\mathrm {e}}^{\frac {x\,\left (x-2\right )}{15}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x^2/15 - (2*x)/15)*(2*x - 2))/15,x)

[Out]

exp((x*(x - 2))/15)

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sympy [A]  time = 0.10, size = 10, normalized size = 1.00 \begin {gather*} e^{\frac {x^{2}}{15} - \frac {2 x}{15}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/15*(2*x-2)*exp(1/15*x**2-2/15*x),x)

[Out]

exp(x**2/15 - 2*x/15)

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