3.93 \(\int \frac {-1+3 x}{1-x+x^2} \, dx\)

Optimal. Leaf size=33 \[ \frac {3}{2} \log \left (x^2-x+1\right )-\frac {\tan ^{-1}\left (\frac {1-2 x}{\sqrt {3}}\right )}{\sqrt {3}} \]

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Rubi [A]  time = 0.02, antiderivative size = 33, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {634, 618, 204, 628} \[ \frac {3}{2} \log \left (x^2-x+1\right )-\frac {\tan ^{-1}\left (\frac {1-2 x}{\sqrt {3}}\right )}{\sqrt {3}} \]

Antiderivative was successfully verified.

[In]

Int[(-1 + 3*x)/(1 - x + x^2),x]

[Out]

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

Rule 204

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

Rule 618

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 628

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 634

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*} \int \frac {-1+3 x}{1-x+x^2} \, dx &=\frac {1}{2} \int \frac {1}{1-x+x^2} \, dx+\frac {3}{2} \int \frac {-1+2 x}{1-x+x^2} \, dx\\ &=\frac {3}{2} \log \left (1-x+x^2\right )-\operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,-1+2 x\right )\\ &=\frac {\tan ^{-1}\left (\frac {-1+2 x}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {3}{2} \log \left (1-x+x^2\right )\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 32, normalized size = 0.97 \[ \frac {3}{2} \log \left (x^2-x+1\right )+\frac {\tan ^{-1}\left (\frac {2 x-1}{\sqrt {3}}\right )}{\sqrt {3}} \]

Antiderivative was successfully verified.

[In]

Integrate[(-1 + 3*x)/(1 - x + x^2),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.02, size = 36, normalized size = 1.09 \[ \frac {3}{2} \log \left (x^2-x+1\right )-\frac {\tan ^{-1}\left (\frac {1}{\sqrt {3}}-\frac {2 x}{\sqrt {3}}\right )}{\sqrt {3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.84, size = 28, normalized size = 0.85 \[ \frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) + \frac {3}{2} \, \log \left (x^{2} - x + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*sqrt(3)*arctan(1/3*sqrt(3)*(2*x - 1)) + 3/2*log(x^2 - x + 1)

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giac [A]  time = 0.86, size = 28, normalized size = 0.85 \[ \frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) + \frac {3}{2} \, \log \left (x^{2} - x + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*sqrt(3)*arctan(1/3*sqrt(3)*(2*x - 1)) + 3/2*log(x^2 - x + 1)

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maple [A]  time = 0.42, size = 29, normalized size = 0.88

method	result	size
default	\(\frac {3 \ln \left (x^{2}-x +1\right )}{2}+\frac {\sqrt {3}\, \arctan \left (\frac {\left (-1+2 x \right ) \sqrt {3}}{3}\right )}{3}\)	\(29\)
risch	\(\frac {3 \ln \left (4 x^{2}-4 x +4\right )}{2}+\frac {\sqrt {3}\, \arctan \left (\frac {\left (-1+2 x \right ) \sqrt {3}}{3}\right )}{3}\)	\(31\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/2*ln(x^2-x+1)+1/3*3^(1/2)*arctan(1/3*(-1+2*x)*3^(1/2))

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maxima [A]  time = 1.00, size = 28, normalized size = 0.85 \[ \frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) + \frac {3}{2} \, \log \left (x^{2} - x + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*sqrt(3)*arctan(1/3*sqrt(3)*(2*x - 1)) + 3/2*log(x^2 - x + 1)

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mupad [B]  time = 0.04, size = 30, normalized size = 0.91 \[ \frac {3\,\ln \left (x^2-x+1\right )}{2}+\frac {\sqrt {3}\,\mathrm {atan}\left (\frac {2\,\sqrt {3}\,x}{3}-\frac {\sqrt {3}}{3}\right )}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((3*x - 1)/(x^2 - x + 1),x)

[Out]

(3*log(x^2 - x + 1))/2 + (3^(1/2)*atan((2*3^(1/2)*x)/3 - 3^(1/2)/3))/3

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sympy [A]  time = 0.12, size = 36, normalized size = 1.09 \[ \frac {3 \log {\left (x^{2} - x + 1 \right )}}{2} + \frac {\sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} x}{3} - \frac {\sqrt {3}}{3} \right )}}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*log(x**2 - x + 1)/2 + sqrt(3)*atan(2*sqrt(3)*x/3 - sqrt(3)/3)/3

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