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

Optimal. Leaf size=30 \[ \frac{2 \tan ^{-1}\left (\frac{2 x+1}{\sqrt{3}}\right )}{\sqrt{3}}-\log \left (1-x^3\right ) \]

[Out]

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

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Rubi [A]  time = 0.0302431, antiderivative size = 30, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {1871, 1586, 618, 204, 260} \[ \frac{2 \tan ^{-1}\left (\frac{2 x+1}{\sqrt{3}}\right )}{\sqrt{3}}-\log \left (1-x^3\right ) \]

Antiderivative was successfully verified.

[In]

Int[(1 - x + 3*x^2)/(1 - x^3),x]

[Out]

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

Rule 1871

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

Rule 1586

Int[(u_.)*(Px_)^(p_.)*(Qx_)^(q_.), x_Symbol] :> Int[u*PolynomialQuotient[Px, Qx, x]^p*Qx^(p + q), x] /; FreeQ[
q, x] && PolyQ[Px, x] && PolyQ[Qx, x] && EqQ[PolynomialRemainder[Px, Qx, x], 0] && IntegerQ[p] && LtQ[p*q, 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 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 260

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

Rubi steps

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

Mathematica [A]  time = 0.008519, size = 30, normalized size = 1. \[ \frac{2 \tan ^{-1}\left (\frac{2 x+1}{\sqrt{3}}\right )}{\sqrt{3}}-\log \left (1-x^3\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.005, size = 33, normalized size = 1.1 \begin{align*} -\ln \left ( -1+x \right ) -\ln \left ({x}^{2}+x+1 \right ) +{\frac{2\,\sqrt{3}}{3}\arctan \left ({\frac{ \left ( 1+2\,x \right ) \sqrt{3}}{3}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.44615, size = 43, normalized size = 1.43 \begin{align*} \frac{2}{3} \, \sqrt{3} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x + 1\right )}\right ) - \log \left (x^{2} + x + 1\right ) - \log \left (x - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.23767, size = 101, normalized size = 3.37 \begin{align*} \frac{2}{3} \, \sqrt{3} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x + 1\right )}\right ) - \log \left (x^{2} + x + 1\right ) - \log \left (x - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.121207, size = 5, normalized size = 0.17 \begin{align*} - \log{\left (x - 1 \right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-log(x - 1)

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Giac [A]  time = 1.06748, size = 45, normalized size = 1.5 \begin{align*} \frac{2}{3} \, \sqrt{3} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x + 1\right )}\right ) - \log \left (x^{2} + x + 1\right ) - \log \left ({\left | x - 1 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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