\(\int \frac {x}{6-5 x+x^2} \, dx\) [2256]

Optimal result

Integrand size = 12, antiderivative size = 17 \[ \int \frac {x}{6-5 x+x^2} \, dx=-2 \log (2-x)+3 \log (3-x) \]

[Out]

-2*ln(2-x)+3*ln(3-x)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {646, 31} \[ \int \frac {x}{6-5 x+x^2} \, dx=3 \log (3-x)-2 \log (2-x) \]

[In]

Int[x/(6 - 5*x + x^2),x]

[Out]

-2*Log[2 - x] + 3*Log[3 - x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 646

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Dist[
(c*d - e*(b/2 - q/2))/q, Int[1/(b/2 - q/2 + c*x), x], x] - Dist[(c*d - e*(b/2 + q/2))/q, Int[1/(b/2 + q/2 + c*
x), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] && NiceSqrtQ[b^2 - 4*a*
c]

Rubi steps \begin{align*} \text {integral}& = -\left (2 \int \frac {1}{-2+x} \, dx\right )+3 \int \frac {1}{-3+x} \, dx \\ & = -2 \log (2-x)+3 \log (3-x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int \frac {x}{6-5 x+x^2} \, dx=-2 \log (2-x)+3 \log (3-x) \]

[In]

Integrate[x/(6 - 5*x + x^2),x]

[Out]

-2*Log[2 - x] + 3*Log[3 - x]

Maple [A] (verified)

Time = 0.17 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82

method	result	size
default	\(-2 \ln \left (-2+x \right )+3 \ln \left (-3+x \right )\)	\(14\)
norman	\(-2 \ln \left (-2+x \right )+3 \ln \left (-3+x \right )\)	\(14\)
risch	\(-2 \ln \left (-2+x \right )+3 \ln \left (-3+x \right )\)	\(14\)
parallelrisch	\(-2 \ln \left (-2+x \right )+3 \ln \left (-3+x \right )\)	\(14\)

[In]

int(x/(x^2-5*x+6),x,method=_RETURNVERBOSE)

[Out]

-2*ln(-2+x)+3*ln(-3+x)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.76 \[ \int \frac {x}{6-5 x+x^2} \, dx=-2 \, \log \left (x - 2\right ) + 3 \, \log \left (x - 3\right ) \]

[In]

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

[Out]

-2*log(x - 2) + 3*log(x - 3)

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.71 \[ \int \frac {x}{6-5 x+x^2} \, dx=3 \log {\left (x - 3 \right )} - 2 \log {\left (x - 2 \right )} \]

[In]

integrate(x/(x**2-5*x+6),x)

[Out]

3*log(x - 3) - 2*log(x - 2)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.76 \[ \int \frac {x}{6-5 x+x^2} \, dx=-2 \, \log \left (x - 2\right ) + 3 \, \log \left (x - 3\right ) \]

[In]

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

[Out]

-2*log(x - 2) + 3*log(x - 3)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88 \[ \int \frac {x}{6-5 x+x^2} \, dx=-2 \, \log \left ({\left | x - 2 \right |}\right ) + 3 \, \log \left ({\left | x - 3 \right |}\right ) \]

[In]

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

[Out]

-2*log(abs(x - 2)) + 3*log(abs(x - 3))

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.76 \[ \int \frac {x}{6-5 x+x^2} \, dx=3\,\ln \left (x-3\right )-2\,\ln \left (x-2\right ) \]

[In]

int(x/(x^2 - 5*x + 6),x)

[Out]

3*log(x - 3) - 2*log(x - 2)