3.20 $$\int \frac{x^8 (1-x^3)}{1-x^3+x^6} \, dx$$

Optimal. Leaf size=46 $-\frac{x^6}{6}+\frac{1}{6} \log \left (x^6-x^3+1\right )-\frac{\tan ^{-1}\left (\frac{1-2 x^3}{\sqrt{3}}\right )}{3 \sqrt{3}}$

[Out]

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

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Rubi [A]  time = 0.0578566, antiderivative size = 46, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 23, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.261, Rules used = {1474, 800, 634, 618, 204, 628} $-\frac{x^6}{6}+\frac{1}{6} \log \left (x^6-x^3+1\right )-\frac{\tan ^{-1}\left (\frac{1-2 x^3}{\sqrt{3}}\right )}{3 \sqrt{3}}$

Antiderivative was successfully veriﬁed.

[In]

Int[(x^8*(1 - x^3))/(1 - x^3 + x^6),x]

[Out]

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

Rule 1474

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.)*((d_) + (e_.)*(x_)^(n_))^(q_.), x_Symbol] :>
Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(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] && IntegerQ[Simplify[(m + 1)/n]]

Rule 800

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Int[Exp
andIntegrand[((d + e*x)^m*(f + g*x))/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b^2 -
4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[m]

Rule 634

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 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 628

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]

Rubi steps

\begin{align*} \int \frac{x^8 \left (1-x^3\right )}{1-x^3+x^6} \, dx &=\frac{1}{3} \operatorname{Subst}\left (\int \frac{(1-x) x^2}{1-x+x^2} \, dx,x,x^3\right )\\ &=\frac{1}{3} \operatorname{Subst}\left (\int \left (-x+\frac{x}{1-x+x^2}\right ) \, dx,x,x^3\right )\\ &=-\frac{x^6}{6}+\frac{1}{3} \operatorname{Subst}\left (\int \frac{x}{1-x+x^2} \, dx,x,x^3\right )\\ &=-\frac{x^6}{6}+\frac{1}{6} \operatorname{Subst}\left (\int \frac{1}{1-x+x^2} \, dx,x,x^3\right )+\frac{1}{6} \operatorname{Subst}\left (\int \frac{-1+2 x}{1-x+x^2} \, dx,x,x^3\right )\\ &=-\frac{x^6}{6}+\frac{1}{6} \log \left (1-x^3+x^6\right )-\frac{1}{3} \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,-1+2 x^3\right )\\ &=-\frac{x^6}{6}-\frac{\tan ^{-1}\left (\frac{1-2 x^3}{\sqrt{3}}\right )}{3 \sqrt{3}}+\frac{1}{6} \log \left (1-x^3+x^6\right )\\ \end{align*}

Mathematica [A]  time = 0.0158543, size = 46, normalized size = 1. $-\frac{x^6}{6}+\frac{1}{6} \log \left (x^6-x^3+1\right )+\frac{\tan ^{-1}\left (\frac{2 x^3-1}{\sqrt{3}}\right )}{3 \sqrt{3}}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(x^8*(1 - x^3))/(1 - x^3 + x^6),x]

[Out]

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

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Maple [A]  time = 0.003, size = 38, normalized size = 0.8 \begin{align*} -{\frac{{x}^{6}}{6}}+{\frac{\ln \left ({x}^{6}-{x}^{3}+1 \right ) }{6}}+{\frac{\sqrt{3}}{9}\arctan \left ({\frac{ \left ( 2\,{x}^{3}-1 \right ) \sqrt{3}}{3}} \right ) } \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(x^8*(-x^3+1)/(x^6-x^3+1),x)

[Out]

-1/6*x^6+1/6*ln(x^6-x^3+1)+1/9*3^(1/2)*arctan(1/3*(2*x^3-1)*3^(1/2))

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Maxima [A]  time = 1.49105, size = 50, normalized size = 1.09 \begin{align*} -\frac{1}{6} \, x^{6} + \frac{1}{9} \, \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 ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8*(-x^3+1)/(x^6-x^3+1),x, algorithm="maxima")

[Out]

-1/6*x^6 + 1/9*sqrt(3)*arctan(1/3*sqrt(3)*(2*x^3 - 1)) + 1/6*log(x^6 - x^3 + 1)

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Fricas [A]  time = 1.77774, size = 109, normalized size = 2.37 \begin{align*} -\frac{1}{6} \, x^{6} + \frac{1}{9} \, \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 ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8*(-x^3+1)/(x^6-x^3+1),x, algorithm="fricas")

[Out]

-1/6*x^6 + 1/9*sqrt(3)*arctan(1/3*sqrt(3)*(2*x^3 - 1)) + 1/6*log(x^6 - x^3 + 1)

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Sympy [A]  time = 0.149908, size = 42, normalized size = 0.91 \begin{align*} - \frac{x^{6}}{6} + \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 )}}{9} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**8*(-x**3+1)/(x**6-x**3+1),x)

[Out]

-x**6/6 + log(x**6 - x**3 + 1)/6 + sqrt(3)*atan(2*sqrt(3)*x**3/3 - sqrt(3)/3)/9

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Giac [A]  time = 1.11026, size = 50, normalized size = 1.09 \begin{align*} -\frac{1}{6} \, x^{6} + \frac{1}{9} \, \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 ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8*(-x^3+1)/(x^6-x^3+1),x, algorithm="giac")

[Out]

-1/6*x^6 + 1/9*sqrt(3)*arctan(1/3*sqrt(3)*(2*x^3 - 1)) + 1/6*log(x^6 - x^3 + 1)