\(\int \frac {x^2 (-2+x^3)}{1-x^3+x^6} \, dx\) [32]

Optimal result

Integrand size = 21, antiderivative size = 36 \[ \int \frac {x^2 \left (-2+x^3\right )}{1-x^3+x^6} \, dx=\frac {\arctan \left (\frac {1-2 x^3}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {1}{6} \log \left (1-x^3+x^6\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {1482, 648, 632, 210, 642} \[ \int \frac {x^2 \left (-2+x^3\right )}{1-x^3+x^6} \, dx=\frac {\arctan \left (\frac {1-2 x^3}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {1}{6} \log \left (x^6-x^3+1\right ) \]

[In]

Int[(x^2*(-2 + x^3))/(1 - x^3 + x^6),x]

[Out]

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

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 1482

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.)*((d_) + (e_.)*(x_)^(n_))^(q_.), x_Symbol] :>
Dist[1/n, Subst[Int[(d + e*x)^q*(a + b*x + c*x^2)^p, x], x, x^n], x] /; FreeQ[{a, b, c, d, e, m, n, p, q}, x]
&& EqQ[n2, 2*n] && EqQ[Simplify[m - n + 1], 0]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.03 \[ \int \frac {x^2 \left (-2+x^3\right )}{1-x^3+x^6} \, dx=-\frac {\arctan \left (\frac {-1+2 x^3}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {1}{6} \log \left (1-x^3+x^6\right ) \]

[In]

Integrate[(x^2*(-2 + x^3))/(1 - x^3 + x^6),x]

[Out]

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

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.92

method	result	size
default	\(\frac {\ln \left (x^{6}-x^{3}+1\right )}{6}-\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 x^{3}-1\right ) \sqrt {3}}{3}\right )}{3}\)	\(33\)
risch	\(\frac {\ln \left (4 x^{6}-4 x^{3}+4\right )}{6}-\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 x^{3}-1\right ) \sqrt {3}}{3}\right )}{3}\)	\(35\)

[In]

int(x^2*(x^3-2)/(x^6-x^3+1),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.89 \[ \int \frac {x^2 \left (-2+x^3\right )}{1-x^3+x^6} \, dx=-\frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{3} - 1\right )}\right ) + \frac {1}{6} \, \log \left (x^{6} - x^{3} + 1\right ) \]

[In]

integrate(x^2*(x^3-2)/(x^6-x^3+1),x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.03 \[ \int \frac {x^2 \left (-2+x^3\right )}{1-x^3+x^6} \, dx=\frac {\log {\left (x^{6} - x^{3} + 1 \right )}}{6} - \frac {\sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} x^{3}}{3} - \frac {\sqrt {3}}{3} \right )}}{3} \]

[In]

integrate(x**2*(x**3-2)/(x**6-x**3+1),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.89 \[ \int \frac {x^2 \left (-2+x^3\right )}{1-x^3+x^6} \, dx=-\frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{3} - 1\right )}\right ) + \frac {1}{6} \, \log \left (x^{6} - x^{3} + 1\right ) \]

[In]

integrate(x^2*(x^3-2)/(x^6-x^3+1),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.89 \[ \int \frac {x^2 \left (-2+x^3\right )}{1-x^3+x^6} \, dx=-\frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{3} - 1\right )}\right ) + \frac {1}{6} \, \log \left (x^{6} - x^{3} + 1\right ) \]

[In]

integrate(x^2*(x^3-2)/(x^6-x^3+1),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.94 \[ \int \frac {x^2 \left (-2+x^3\right )}{1-x^3+x^6} \, dx=\frac {\ln \left (x^6-x^3+1\right )}{6}+\frac {\sqrt {3}\,\mathrm {atan}\left (\frac {\sqrt {3}}{3}-\frac {2\,\sqrt {3}\,x^3}{3}\right )}{3} \]

[In]

int((x^2*(x^3 - 2))/(x^6 - x^3 + 1),x)

[Out]

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