\(\int \frac {-1+x}{3-4 x+3 x^2} \, dx\) [464]

Optimal result

Integrand size = 16, antiderivative size = 37 \[ \int \frac {-1+x}{3-4 x+3 x^2} \, dx=\frac {\arctan \left (\frac {2-3 x}{\sqrt {5}}\right )}{3 \sqrt {5}}+\frac {1}{6} \log \left (3-4 x+3 x^2\right ) \]

[Out]

1/6*ln(3*x^2-4*x+3)+1/15*arctan(1/5*(2-3*x)*5^(1/2))*5^(1/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {648, 632, 210, 642} \[ \int \frac {-1+x}{3-4 x+3 x^2} \, dx=\frac {\arctan \left (\frac {2-3 x}{\sqrt {5}}\right )}{3 \sqrt {5}}+\frac {1}{6} \log \left (3 x^2-4 x+3\right ) \]

[In]

Int[(-1 + x)/(3 - 4*x + 3*x^2),x]

[Out]

ArcTan[(2 - 3*x)/Sqrt[5]]/(3*Sqrt[5]) + Log[3 - 4*x + 3*x^2]/6

Rule 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 632

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[d*(Log[RemoveContent[a + b*x +
c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 648

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), 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}& = \frac {1}{6} \int \frac {-4+6 x}{3-4 x+3 x^2} \, dx-\frac {1}{3} \int \frac {1}{3-4 x+3 x^2} \, dx \\ & = \frac {1}{6} \log \left (3-4 x+3 x^2\right )+\frac {2}{3} \text {Subst}\left (\int \frac {1}{-20-x^2} \, dx,x,-4+6 x\right ) \\ & = \frac {\tan ^{-1}\left (\frac {2-3 x}{\sqrt {5}}\right )}{3 \sqrt {5}}+\frac {1}{6} \log \left (3-4 x+3 x^2\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.00 \[ \int \frac {-1+x}{3-4 x+3 x^2} \, dx=-\frac {\arctan \left (\frac {-2+3 x}{\sqrt {5}}\right )}{3 \sqrt {5}}+\frac {1}{6} \log \left (3-4 x+3 x^2\right ) \]

[In]

Integrate[(-1 + x)/(3 - 4*x + 3*x^2),x]

[Out]

-1/3*ArcTan[(-2 + 3*x)/Sqrt[5]]/Sqrt[5] + Log[3 - 4*x + 3*x^2]/6

Maple [A] (verified)

Time = 1.36 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.84

method	result	size
default	\(\frac {\ln \left (3 x^{2}-4 x +3\right )}{6}-\frac {\sqrt {5}\, \arctan \left (\frac {\left (6 x -4\right ) \sqrt {5}}{10}\right )}{15}\)	\(31\)
risch	\(\frac {\ln \left (9 x^{2}-12 x +9\right )}{6}-\frac {\sqrt {5}\, \arctan \left (\frac {\left (3 x -2\right ) \sqrt {5}}{5}\right )}{15}\)	\(31\)

[In]

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

[Out]

1/6*ln(3*x^2-4*x+3)-1/15*5^(1/2)*arctan(1/10*(6*x-4)*5^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.81 \[ \int \frac {-1+x}{3-4 x+3 x^2} \, dx=-\frac {1}{15} \, \sqrt {5} \arctan \left (\frac {1}{5} \, \sqrt {5} {\left (3 \, x - 2\right )}\right ) + \frac {1}{6} \, \log \left (3 \, x^{2} - 4 \, x + 3\right ) \]

[In]

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

[Out]

-1/15*sqrt(5)*arctan(1/5*sqrt(5)*(3*x - 2)) + 1/6*log(3*x^2 - 4*x + 3)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.05 \[ \int \frac {-1+x}{3-4 x+3 x^2} \, dx=\frac {\log {\left (x^{2} - \frac {4 x}{3} + 1 \right )}}{6} - \frac {\sqrt {5} \operatorname {atan}{\left (\frac {3 \sqrt {5} x}{5} - \frac {2 \sqrt {5}}{5} \right )}}{15} \]

[In]

integrate((-1+x)/(3*x**2-4*x+3),x)

[Out]

log(x**2 - 4*x/3 + 1)/6 - sqrt(5)*atan(3*sqrt(5)*x/5 - 2*sqrt(5)/5)/15

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.81 \[ \int \frac {-1+x}{3-4 x+3 x^2} \, dx=-\frac {1}{15} \, \sqrt {5} \arctan \left (\frac {1}{5} \, \sqrt {5} {\left (3 \, x - 2\right )}\right ) + \frac {1}{6} \, \log \left (3 \, x^{2} - 4 \, x + 3\right ) \]

[In]

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

[Out]

-1/15*sqrt(5)*arctan(1/5*sqrt(5)*(3*x - 2)) + 1/6*log(3*x^2 - 4*x + 3)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.81 \[ \int \frac {-1+x}{3-4 x+3 x^2} \, dx=-\frac {1}{15} \, \sqrt {5} \arctan \left (\frac {1}{5} \, \sqrt {5} {\left (3 \, x - 2\right )}\right ) + \frac {1}{6} \, \log \left (3 \, x^{2} - 4 \, x + 3\right ) \]

[In]

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

[Out]

-1/15*sqrt(5)*arctan(1/5*sqrt(5)*(3*x - 2)) + 1/6*log(3*x^2 - 4*x + 3)

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.81 \[ \int \frac {-1+x}{3-4 x+3 x^2} \, dx=\frac {\ln \left (x^2-\frac {4\,x}{3}+1\right )}{6}-\frac {\sqrt {5}\,\mathrm {atan}\left (\frac {3\,\sqrt {5}\,x}{5}-\frac {2\,\sqrt {5}}{5}\right )}{15} \]

[In]

int((x - 1)/(3*x^2 - 4*x + 3),x)

[Out]

log(x^2 - (4*x)/3 + 1)/6 - (5^(1/2)*atan((3*5^(1/2)*x)/5 - (2*5^(1/2))/5))/15