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

Optimal. Leaf size=40 \[ \frac {5}{6} \log \left (x^2+x+1\right )-\frac {2}{3} \log (1-x)+\sqrt {3} \tan ^{-1}\left (\frac {2 x+1}{\sqrt {3}}\right ) \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 40, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {1875, 31, 634, 618, 204, 628} \[ \frac {5}{6} \log \left (x^2+x+1\right )-\frac {2}{3} \log (1-x)+\sqrt {3} \tan ^{-1}\left (\frac {2 x+1}{\sqrt {3}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[(-3 + x^2)/(-1 + x^3),x]

[Out]

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

Rule 31

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

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]

Rule 1875

Int[(P2_)/((a_) + (b_.)*(x_)^3), x_Symbol] :> With[{A = Coeff[P2, x, 0], B = Coeff[P2, x, 1], C = Coeff[P2, x,
2], q = (-(a/b))^(1/3)}, Dist[(q*(A + B*q + C*q^2))/(3*a), Int[1/(q - x), x], x] + Dist[q/(3*a), Int[(q*(2*A
- B*q - C*q^2) + (A + B*q - 2*C*q^2)*x)/(q^2 + q*x + x^2), x], x] /; NeQ[a*B^3 - b*A^3, 0] && NeQ[A + B*q + C*
q^2, 0]] /; FreeQ[{a, b}, x] && PolyQ[P2, x, 2] && LtQ[a/b, 0]

Rubi steps

\begin {align*} \int \frac {-3+x^2}{-1+x^3} \, dx &=-\left (\frac {1}{3} \int \frac {-7-5 x}{1+x+x^2} \, dx\right )+\frac {2}{3} \int \frac {1}{1-x} \, dx\\ &=-\frac {2}{3} \log (1-x)+\frac {5}{6} \int \frac {1+2 x}{1+x+x^2} \, dx+\frac {3}{2} \int \frac {1}{1+x+x^2} \, dx\\ &=-\frac {2}{3} \log (1-x)+\frac {5}{6} \log \left (1+x+x^2\right )-3 \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+2 x\right )\\ &=\sqrt {3} \tan ^{-1}\left (\frac {1+2 x}{\sqrt {3}}\right )-\frac {2}{3} \log (1-x)+\frac {5}{6} \log \left (1+x+x^2\right )\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 50, normalized size = 1.25 \[ \frac {1}{3} \log \left (1-x^3\right )+\frac {1}{2} \log \left (x^2+x+1\right )-\log (1-x)+\sqrt {3} \tan ^{-1}\left (\frac {2 x+1}{\sqrt {3}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(-3 + x^2)/(-1 + x^3),x]

[Out]

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

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fricas [A]  time = 0.87, size = 31, normalized size = 0.78 \[ \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) + \frac {5}{6} \, \log \left (x^{2} + x + 1\right ) - \frac {2}{3} \, \log \left (x - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.32, size = 32, normalized size = 0.80 \[ \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) + \frac {5}{6} \, \log \left (x^{2} + x + 1\right ) - \frac {2}{3} \, \log \left ({\left | x - 1 \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.05, size = 32, normalized size = 0.80 \[ \sqrt {3}\, \arctan \left (\frac {\left (2 x +1\right ) \sqrt {3}}{3}\right )-\frac {2 \ln \left (x -1\right )}{3}+\frac {5 \ln \left (x^{2}+x +1\right )}{6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2-3)/(x^3-1),x)

[Out]

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

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maxima [A]  time = 2.99, size = 31, normalized size = 0.78 \[ \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) + \frac {5}{6} \, \log \left (x^{2} + x + 1\right ) - \frac {2}{3} \, \log \left (x - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.16, size = 46, normalized size = 1.15 \[ -\frac {2\,\ln \left (x-1\right )}{3}-\ln \left (x+\frac {1}{2}-\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (-\frac {5}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )+\ln \left (x+\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {5}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2 - 3)/(x^3 - 1),x)

[Out]

log(x + (3^(1/2)*1i)/2 + 1/2)*((3^(1/2)*1i)/2 + 5/6) - log(x - (3^(1/2)*1i)/2 + 1/2)*((3^(1/2)*1i)/2 - 5/6) -
(2*log(x - 1))/3

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sympy [A]  time = 0.29, size = 42, normalized size = 1.05 \[ - \frac {2 \log {\left (x - 1 \right )}}{3} + \frac {5 \log {\left (x^{2} + x + 1 \right )}}{6} + \sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} x}{3} + \frac {\sqrt {3}}{3} \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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