\(\int \frac {1-x}{1-x^3} \, dx\) [14]

Optimal result

Integrand size = 15, antiderivative size = 19 \[ \int \frac {1-x}{1-x^3} \, dx=\frac {2 \arctan \left (\frac {1+2 x}{\sqrt {3}}\right )}{\sqrt {3}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {1600, 632, 210} \[ \int \frac {1-x}{1-x^3} \, dx=\frac {2 \arctan \left (\frac {2 x+1}{\sqrt {3}}\right )}{\sqrt {3}} \]

[In]

Int[(1 - x)/(1 - x^3),x]

[Out]

(2*ArcTan[(1 + 2*x)/Sqrt[3]])/Sqrt[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 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]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \frac {1-x}{1-x^3} \, dx=\frac {2 \arctan \left (\frac {1+2 x}{\sqrt {3}}\right )}{\sqrt {3}} \]

[In]

Integrate[(1 - x)/(1 - x^3),x]

[Out]

(2*ArcTan[(1 + 2*x)/Sqrt[3]])/Sqrt[3]

Maple [A] (verified)

Time = 1.47 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89

method	result	size
default	\(\frac {2 \arctan \left (\frac {\left (1+2 x \right ) \sqrt {3}}{3}\right ) \sqrt {3}}{3}\)	\(17\)
risch	\(\frac {2 \arctan \left (\frac {\left (1+2 x \right ) \sqrt {3}}{3}\right ) \sqrt {3}}{3}\)	\(17\)
meijerg	\(-\frac {x \left (\ln \left (1-\left (x^{3}\right )^{\frac {1}{3}}\right )-\frac {\ln \left (1+\left (x^{3}\right )^{\frac {1}{3}}+\left (x^{3}\right )^{\frac {2}{3}}\right )}{2}-\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (x^{3}\right )^{\frac {1}{3}}}{2+\left (x^{3}\right )^{\frac {1}{3}}}\right )\right )}{3 \left (x^{3}\right )^{\frac {1}{3}}}+\frac {x^{2} \left (\ln \left (1-\left (x^{3}\right )^{\frac {1}{3}}\right )-\frac {\ln \left (1+\left (x^{3}\right )^{\frac {1}{3}}+\left (x^{3}\right )^{\frac {2}{3}}\right )}{2}+\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (x^{3}\right )^{\frac {1}{3}}}{2+\left (x^{3}\right )^{\frac {1}{3}}}\right )\right )}{3 \left (x^{3}\right )^{\frac {2}{3}}}\)	\(125\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.84 \[ \int \frac {1-x}{1-x^3} \, dx=\frac {2}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) \]

[In]

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

[Out]

2/3*sqrt(3)*arctan(1/3*sqrt(3)*(2*x + 1))

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.37 \[ \int \frac {1-x}{1-x^3} \, dx=\frac {2 \sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} x}{3} + \frac {\sqrt {3}}{3} \right )}}{3} \]

[In]

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

[Out]

2*sqrt(3)*atan(2*sqrt(3)*x/3 + sqrt(3)/3)/3

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.84 \[ \int \frac {1-x}{1-x^3} \, dx=\frac {2}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) \]

[In]

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

[Out]

2/3*sqrt(3)*arctan(1/3*sqrt(3)*(2*x + 1))

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.84 \[ \int \frac {1-x}{1-x^3} \, dx=\frac {2}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) \]

[In]

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

[Out]

2/3*sqrt(3)*arctan(1/3*sqrt(3)*(2*x + 1))

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.84 \[ \int \frac {1-x}{1-x^3} \, dx=\frac {2\,\sqrt {3}\,\mathrm {atan}\left (\frac {\sqrt {3}\,\left (2\,x+1\right )}{3}\right )}{3} \]

[In]

int((x - 1)/(x^3 - 1),x)

[Out]

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