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

Optimal. Leaf size=43 \[ -\frac{1}{24} \log \left (x^2+2 x+4\right )+\frac{1}{12} \log (2-x)-\frac{\tan ^{-1}\left (\frac{x+1}{\sqrt{3}}\right )}{4 \sqrt{3}} \]

[Out]

-ArcTan[(1 + x)/Sqrt[3]]/(4*Sqrt[3]) + Log[2 - x]/12 - Log[4 + 2*x + x^2]/24

________________________________________________________________________________________

Rubi [A]  time = 0.0203541, antiderivative size = 43, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 7, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.857, Rules used = {200, 31, 634, 618, 204, 628} \[ -\frac{1}{24} \log \left (x^2+2 x+4\right )+\frac{1}{12} \log (2-x)-\frac{\tan ^{-1}\left (\frac{x+1}{\sqrt{3}}\right )}{4 \sqrt{3}} \]

Antiderivative was successfully verified.

[In]

Int[(-8 + x^3)^(-1),x]

[Out]

-ArcTan[(1 + x)/Sqrt[3]]/(4*Sqrt[3]) + Log[2 - x]/12 - Log[4 + 2*x + x^2]/24

Rule 200

Int[((a_) + (b_.)*(x_)^3)^(-1), x_Symbol] :> Dist[1/(3*Rt[a, 3]^2), Int[1/(Rt[a, 3] + Rt[b, 3]*x), x], x] + Di
st[1/(3*Rt[a, 3]^2), Int[(2*Rt[a, 3] - Rt[b, 3]*x)/(Rt[a, 3]^2 - Rt[a, 3]*Rt[b, 3]*x + Rt[b, 3]^2*x^2), x], x]
/; FreeQ[{a, b}, x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

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{1}{-8+x^3} \, dx &=\frac{1}{12} \int \frac{1}{-2+x} \, dx+\frac{1}{12} \int \frac{-4-x}{4+2 x+x^2} \, dx\\ &=\frac{1}{12} \log (2-x)-\frac{1}{24} \int \frac{2+2 x}{4+2 x+x^2} \, dx-\frac{1}{4} \int \frac{1}{4+2 x+x^2} \, dx\\ &=\frac{1}{12} \log (2-x)-\frac{1}{24} \log \left (4+2 x+x^2\right )+\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{-12-x^2} \, dx,x,2+2 x\right )\\ &=-\frac{\tan ^{-1}\left (\frac{1+x}{\sqrt{3}}\right )}{4 \sqrt{3}}+\frac{1}{12} \log (2-x)-\frac{1}{24} \log \left (4+2 x+x^2\right )\\ \end{align*}

Mathematica [A]  time = 0.0083778, size = 43, normalized size = 1. \[ -\frac{1}{24} \log \left (x^2+2 x+4\right )+\frac{1}{12} \log (2-x)-\frac{\tan ^{-1}\left (\frac{x+1}{\sqrt{3}}\right )}{4 \sqrt{3}} \]

Antiderivative was successfully verified.

[In]

Integrate[(-8 + x^3)^(-1),x]

[Out]

-ArcTan[(1 + x)/Sqrt[3]]/(4*Sqrt[3]) + Log[2 - x]/12 - Log[4 + 2*x + x^2]/24

________________________________________________________________________________________

Maple [A]  time = 0.005, size = 35, normalized size = 0.8 \begin{align*} -{\frac{\ln \left ({x}^{2}+2\,x+4 \right ) }{24}}-{\frac{\sqrt{3}}{12}\arctan \left ({\frac{ \left ( 2\,x+2 \right ) \sqrt{3}}{6}} \right ) }+{\frac{\ln \left ( -2+x \right ) }{12}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^3-8),x)

[Out]

-1/24*ln(x^2+2*x+4)-1/12*3^(1/2)*arctan(1/6*(2*x+2)*3^(1/2))+1/12*ln(-2+x)

________________________________________________________________________________________

Maxima [A]  time = 1.4085, size = 43, normalized size = 1. \begin{align*} -\frac{1}{12} \, \sqrt{3} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (x + 1\right )}\right ) - \frac{1}{24} \, \log \left (x^{2} + 2 \, x + 4\right ) + \frac{1}{12} \, \log \left (x - 2\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*sqrt(3)*arctan(1/3*sqrt(3)*(x + 1)) - 1/24*log(x^2 + 2*x + 4) + 1/12*log(x - 2)

________________________________________________________________________________________

Fricas [A]  time = 1.987, size = 117, normalized size = 2.72 \begin{align*} -\frac{1}{12} \, \sqrt{3} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (x + 1\right )}\right ) - \frac{1}{24} \, \log \left (x^{2} + 2 \, x + 4\right ) + \frac{1}{12} \, \log \left (x - 2\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*sqrt(3)*arctan(1/3*sqrt(3)*(x + 1)) - 1/24*log(x^2 + 2*x + 4) + 1/12*log(x - 2)

________________________________________________________________________________________

Sympy [A]  time = 0.122397, size = 41, normalized size = 0.95 \begin{align*} \frac{\log{\left (x - 2 \right )}}{12} - \frac{\log{\left (x^{2} + 2 x + 4 \right )}}{24} - \frac{\sqrt{3} \operatorname{atan}{\left (\frac{\sqrt{3} x}{3} + \frac{\sqrt{3}}{3} \right )}}{12} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x - 2)/12 - log(x**2 + 2*x + 4)/24 - sqrt(3)*atan(sqrt(3)*x/3 + sqrt(3)/3)/12

________________________________________________________________________________________

Giac [A]  time = 1.04449, size = 45, normalized size = 1.05 \begin{align*} -\frac{1}{12} \, \sqrt{3} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (x + 1\right )}\right ) - \frac{1}{24} \, \log \left (x^{2} + 2 \, x + 4\right ) + \frac{1}{12} \, \log \left ({\left | x - 2 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*sqrt(3)*arctan(1/3*sqrt(3)*(x + 1)) - 1/24*log(x^2 + 2*x + 4) + 1/12*log(abs(x - 2))