\(\int \frac {1-x+3 x^2}{1-x^3} \, dx\) [51]

Optimal result

Integrand size = 20, antiderivative size = 30 \[ \int \frac {1-x+3 x^2}{1-x^3} \, dx=\frac {2 \arctan \left (\frac {1+2 x}{\sqrt {3}}\right )}{\sqrt {3}}-\log \left (1-x^3\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1885, 1600, 632, 210, 266} \[ \int \frac {1-x+3 x^2}{1-x^3} \, dx=\frac {2 \arctan \left (\frac {2 x+1}{\sqrt {3}}\right )}{\sqrt {3}}-\log \left (1-x^3\right ) \]

[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 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 266

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]

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 1600

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 1885

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]

Rubi steps \begin{align*} \text {integral}& = 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 \text {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] (verified)

Time = 0.01 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00 \[ \int \frac {1-x+3 x^2}{1-x^3} \, dx=\frac {2 \arctan \left (\frac {1+2 x}{\sqrt {3}}\right )}{\sqrt {3}}-\log \left (1-x^3\right ) \]

[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]

Maple [A] (verified)

Time = 1.48 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.10

method	result	size
default	\(-\ln \left (-1+x \right )-\ln \left (x^{2}+x +1\right )+\frac {2 \arctan \left (\frac {\left (1+2 x \right ) \sqrt {3}}{3}\right ) \sqrt {3}}{3}\)	\(33\)
risch	\(\frac {2 \arctan \left (\frac {\left (1+2 x \right ) \sqrt {3}}{3}\right ) \sqrt {3}}{3}-\ln \left (4 x^{2}+4 x +4\right )-\ln \left (-1+x \right )\)	\(37\)
meijerg	\(-\ln \left (-x^{3}+1\right )+\frac {x^{2} \left (\ln \left (1-\left (x^{3}\right )^{\frac {1}{3}}\right )-\frac {\ln \left (1+\left (x^{3}\right )^{\frac {1}{3}}+\left (x^{3}\right )^{\frac {2}{3}}\right )}{2}+\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (x^{3}\right )^{\frac {1}{3}}}{2+\left (x^{3}\right )^{\frac {1}{3}}}\right )\right )}{3 \left (x^{3}\right )^{\frac {2}{3}}}-\frac {x \left (\ln \left (1-\left (x^{3}\right )^{\frac {1}{3}}\right )-\frac {\ln \left (1+\left (x^{3}\right )^{\frac {1}{3}}+\left (x^{3}\right )^{\frac {2}{3}}\right )}{2}-\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (x^{3}\right )^{\frac {1}{3}}}{2+\left (x^{3}\right )^{\frac {1}{3}}}\right )\right )}{3 \left (x^{3}\right )^{\frac {1}{3}}}\)	\(135\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.07 \[ \int \frac {1-x+3 x^2}{1-x^3} \, dx=\frac {2}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) - \log \left (x^{2} + x + 1\right ) - \log \left (x - 1\right ) \]

[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)

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 5, normalized size of antiderivative = 0.17 \[ \int \frac {1-x+3 x^2}{1-x^3} \, dx=- \log {\left (x - 1 \right )} \]

[In]

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

[Out]

-log(x - 1)

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.07 \[ \int \frac {1-x+3 x^2}{1-x^3} \, dx=\frac {2}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) - \log \left (x^{2} + x + 1\right ) - \log \left (x - 1\right ) \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.10 \[ \int \frac {1-x+3 x^2}{1-x^3} \, dx=\frac {2}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) - \log \left (x^{2} + x + 1\right ) - \log \left ({\left | x - 1 \right |}\right ) \]

[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))

Mupad [B] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 63, normalized size of antiderivative = 2.10 \[ \int \frac {1-x+3 x^2}{1-x^3} \, dx=-\ln \left (x+\frac {1}{2}-\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )-\ln \left (x+\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )-\ln \left (x-1\right )-\frac {\sqrt {3}\,\ln \left (x+\frac {1}{2}-\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,1{}\mathrm {i}}{3}+\frac {\sqrt {3}\,\ln \left (x+\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,1{}\mathrm {i}}{3} \]

[In]

int(-(3*x^2 - x + 1)/(x^3 - 1),x)

[Out]

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