\(\int \frac {-1+x^2}{1+x^3} \, dx\) [416]

Optimal result

Integrand size = 13, antiderivative size = 32 \[ \int \frac {-1+x^2}{1+x^3} \, dx=\frac {\arctan \left (\frac {1-2 x}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {1}{2} \log \left (1-x+x^2\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {1886, 648, 632, 210, 642} \[ \int \frac {-1+x^2}{1+x^3} \, dx=\frac {\arctan \left (\frac {1-2 x}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {1}{2} \log \left (x^2-x+1\right ) \]

[In]

Int[(-1 + x^2)/(1 + x^3),x]

[Out]

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

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 642

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 648

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 1886

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

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.03 \[ \int \frac {-1+x^2}{1+x^3} \, dx=-\frac {\arctan \left (\frac {-1+2 x}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {1}{2} \log \left (1-x+x^2\right ) \]

[In]

Integrate[(-1 + x^2)/(1 + x^3),x]

[Out]

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

Maple [A] (verified)

Time = 0.74 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.91

method	result	size
default	\(\frac {\ln \left (x^{2}-x +1\right )}{2}-\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 x -1\right ) \sqrt {3}}{3}\right )}{3}\)	\(29\)
risch	\(\frac {\ln \left (4 x^{2}-4 x +4\right )}{2}-\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 x -1\right ) \sqrt {3}}{3}\right )}{3}\)	\(31\)
meijerg	\(\frac {\ln \left (x^{3}+1\right )}{3}-\frac {x \ln \left (1+\left (x^{3}\right )^{\frac {1}{3}}\right )}{3 \left (x^{3}\right )^{\frac {1}{3}}}+\frac {x \ln \left (1-\left (x^{3}\right )^{\frac {1}{3}}+\left (x^{3}\right )^{\frac {2}{3}}\right )}{6 \left (x^{3}\right )^{\frac {1}{3}}}-\frac {x \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (x^{3}\right )^{\frac {1}{3}}}{2-\left (x^{3}\right )^{\frac {1}{3}}}\right )}{3 \left (x^{3}\right )^{\frac {1}{3}}}\)	\(82\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.88 \[ \int \frac {-1+x^2}{1+x^3} \, dx=-\frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) + \frac {1}{2} \, \log \left (x^{2} - x + 1\right ) \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.06 \[ \int \frac {-1+x^2}{1+x^3} \, dx=\frac {\log {\left (x^{2} - x + 1 \right )}}{2} - \frac {\sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} x}{3} - \frac {\sqrt {3}}{3} \right )}}{3} \]

[In]

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

[Out]

log(x**2 - x + 1)/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 = 28, normalized size of antiderivative = 0.88 \[ \int \frac {-1+x^2}{1+x^3} \, dx=-\frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) + \frac {1}{2} \, \log \left (x^{2} - x + 1\right ) \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.88 \[ \int \frac {-1+x^2}{1+x^3} \, dx=-\frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) + \frac {1}{2} \, \log \left (x^{2} - x + 1\right ) \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.94 \[ \int \frac {-1+x^2}{1+x^3} \, dx=\frac {\ln \left (x^2-x+1\right )}{2}-\frac {\sqrt {3}\,\mathrm {atan}\left (\frac {2\,\sqrt {3}\,x}{3}-\frac {\sqrt {3}}{3}\right )}{3} \]

[In]

int((x^2 - 1)/(x^3 + 1),x)

[Out]

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