\(\int \frac {1}{15} e^{\frac {1}{15} (-2 x+x^2)} (-2+2 x) \, dx\) [7732]

Optimal result

Integrand size = 22, antiderivative size = 10 \[ \int \frac {1}{15} e^{\frac {1}{15} \left (-2 x+x^2\right )} (-2+2 x) \, dx=e^{\frac {1}{15} (-2+x) x} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.50, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {12, 2276, 2268} \[ \int \frac {1}{15} e^{\frac {1}{15} \left (-2 x+x^2\right )} (-2+2 x) \, dx=e^{\frac {x^2}{15}-\frac {2 x}{15}} \]

[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 2268

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 2276

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{align*} \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{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 10, normalized size of antiderivative = 1.00 \[ \int \frac {1}{15} e^{\frac {1}{15} \left (-2 x+x^2\right )} (-2+2 x) \, dx=e^{\frac {1}{15} (-2+x) x} \]

[In]

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

[Out]

E^(((-2 + x)*x)/15)

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.80

method	result	size
risch	\({\mathrm e}^{\frac {\left (-2+x \right ) x}{15}}\)	\(8\)
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	\({\mathrm e}^{\frac {1}{15} x^{2}-\frac {2}{15} x}\)	\(11\)
parallelrisch	\({\mathrm e}^{\frac {1}{15} x^{2}-\frac {2}{15} x}\)	\(11\)
parts	\(-\frac {i \sqrt {\pi }\, {\mathrm e}^{-\frac {1}{15}} \sqrt {15}\, \operatorname {erf}\left (\frac {i \sqrt {15}\, x}{15}-\frac {i \sqrt {15}}{15}\right ) x}{15}+\frac {i \sqrt {\pi }\, {\mathrm e}^{-\frac {1}{15}} \sqrt {15}\, \operatorname {erf}\left (\frac {i \sqrt {15}\, x}{15}-\frac {i \sqrt {15}}{15}\right )}{15}-\frac {i {\mathrm e}^{-\frac {1}{15}} \sqrt {15}\, \left (i {\mathrm e}^{\frac {\left (-1+x \right )^{2}}{15}} \sqrt {15}-\operatorname {erf}\left (\frac {i \sqrt {15}\, \left (-1+x \right )}{15}\right ) x \sqrt {\pi }+\operatorname {erf}\left (\frac {i \sqrt {15}\, \left (-1+x \right )}{15}\right ) \sqrt {\pi }\right )}{15}\)	\(108\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.49 (sec) , antiderivative size = 10, normalized size of antiderivative = 1.00 \[ \int \frac {1}{15} e^{\frac {1}{15} \left (-2 x+x^2\right )} (-2+2 x) \, dx=e^{\left (\frac {1}{15} \, x^{2} - \frac {2}{15} \, x\right )} \]

[In]

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

[Out]

e^(1/15*x^2 - 2/15*x)

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 10, normalized size of antiderivative = 1.00 \[ \int \frac {1}{15} e^{\frac {1}{15} \left (-2 x+x^2\right )} (-2+2 x) \, dx=e^{\frac {x^{2}}{15} - \frac {2 x}{15}} \]

[In]

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

[Out]

exp(x**2/15 - 2*x/15)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 10, normalized size of antiderivative = 1.00 \[ \int \frac {1}{15} e^{\frac {1}{15} \left (-2 x+x^2\right )} (-2+2 x) \, dx=e^{\left (\frac {1}{15} \, x^{2} - \frac {2}{15} \, x\right )} \]

[In]

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

[Out]

e^(1/15*x^2 - 2/15*x)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 10, normalized size of antiderivative = 1.00 \[ \int \frac {1}{15} e^{\frac {1}{15} \left (-2 x+x^2\right )} (-2+2 x) \, dx=e^{\left (\frac {1}{15} \, x^{2} - \frac {2}{15} \, x\right )} \]

[In]

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

[Out]

e^(1/15*x^2 - 2/15*x)

Mupad [B] (verification not implemented)

Time = 13.11 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.70 \[ \int \frac {1}{15} e^{\frac {1}{15} \left (-2 x+x^2\right )} (-2+2 x) \, dx={\mathrm {e}}^{\frac {x\,\left (x-2\right )}{15}} \]

[In]

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

[Out]

exp((x*(x - 2))/15)