\(\int \frac {12 e^{-\frac {12}{-2+x}}}{4-4 x+x^2} \, dx\) [144]

Optimal result

Integrand size = 21, antiderivative size = 17 \[ \int \frac {12 e^{-\frac {12}{-2+x}}}{4-4 x+x^2} \, dx=e^{\frac {6}{-x+\frac {2+x}{2}}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.65, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {12, 27, 2240} \[ \int \frac {12 e^{-\frac {12}{-2+x}}}{4-4 x+x^2} \, dx=e^{\frac {12}{2-x}} \]

[In]

Int[12/(E^(12/(-2 + x))*(4 - 4*x + x^2)),x]

[Out]

E^(12/(2 - x))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 2240

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[(e + f*x)^n*(
F^(a + b*(c + d*x)^n)/(b*f*n*(c + d*x)^n*Log[F])), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[m, n - 1] &
& EqQ[d*e - c*f, 0]

Rubi steps \begin{align*} \text {integral}& = 12 \int \frac {e^{-\frac {12}{-2+x}}}{4-4 x+x^2} \, dx \\ & = 12 \int \frac {e^{-\frac {12}{-2+x}}}{(-2+x)^2} \, dx \\ & = e^{\frac {12}{2-x}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.53 \[ \int \frac {12 e^{-\frac {12}{-2+x}}}{4-4 x+x^2} \, dx=e^{-\frac {12}{-2+x}} \]

[In]

Integrate[12/(E^(12/(-2 + x))*(4 - 4*x + x^2)),x]

[Out]

E^(-12/(-2 + x))

Maple [A] (verified)

Time = 0.22 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.53

method	result	size
risch	\({\mathrm e}^{-\frac {12}{-2+x}}\)	\(9\)
gosper	\({\mathrm e}^{-\frac {12}{-2+x}}\)	\(11\)
derivativedivides	\({\mathrm e}^{-\frac {12}{-2+x}}\)	\(11\)
default	\({\mathrm e}^{-\frac {12}{-2+x}}\)	\(11\)
norman	\(\frac {x \,{\mathrm e}^{-\frac {12}{-2+x}}-2 \,{\mathrm e}^{-\frac {12}{-2+x}}}{-2+x}\)	\(32\)

[In]

int(12*exp(-6/(-2+x))^2/(x^2-4*x+4),x,method=_RETURNVERBOSE)

[Out]

exp(-12/(-2+x))

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.47 \[ \int \frac {12 e^{-\frac {12}{-2+x}}}{4-4 x+x^2} \, dx=e^{\left (-\frac {12}{x - 2}\right )} \]

[In]

integrate(12*exp(-6/(-2+x))^2/(x^2-4*x+4),x, algorithm="fricas")

[Out]

e^(-12/(x - 2))

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.41 \[ \int \frac {12 e^{-\frac {12}{-2+x}}}{4-4 x+x^2} \, dx=e^{- \frac {12}{x - 2}} \]

[In]

integrate(12*exp(-6/(-2+x))**2/(x**2-4*x+4),x)

[Out]

exp(-12/(x - 2))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.47 \[ \int \frac {12 e^{-\frac {12}{-2+x}}}{4-4 x+x^2} \, dx=e^{\left (-\frac {12}{x - 2}\right )} \]

[In]

integrate(12*exp(-6/(-2+x))^2/(x^2-4*x+4),x, algorithm="maxima")

[Out]

e^(-12/(x - 2))

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.47 \[ \int \frac {12 e^{-\frac {12}{-2+x}}}{4-4 x+x^2} \, dx=e^{\left (-\frac {12}{x - 2}\right )} \]

[In]

integrate(12*exp(-6/(-2+x))^2/(x^2-4*x+4),x, algorithm="giac")

[Out]

e^(-12/(x - 2))

Mupad [B] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.47 \[ \int \frac {12 e^{-\frac {12}{-2+x}}}{4-4 x+x^2} \, dx={\mathrm {e}}^{-\frac {12}{x-2}} \]

[In]

int((12*exp(-12/(x - 2)))/(x^2 - 4*x + 4),x)

[Out]

exp(-12/(x - 2))