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]

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

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Rubi [A]  time = 0.02, antiderivative size = 43, normalized size of antiderivative = 1.00, 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 31

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

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 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 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 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]

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]

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*}

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Mathematica [A]  time = 0.01, size = 43, normalized size = 1.00 \[ -\frac {1}{24} \log \left (x^2+2 x+4\right )+\frac {1}{12} \log (2-x)-\frac {\tan ^{-1}\left (\frac {x+1}{\sqrt {3}}\right )}{4 \sqrt {3}} \]

Antiderivative was successfully verified.

[In]

Integrate[(-8 + x^3)^(-1),x]

[Out]

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

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fricas [A]  time = 0.41, size = 32, normalized size = 0.74 \[ -\frac {1}{12} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (x + 1\right )}\right ) - \frac {1}{24} \, \log \left (x^{2} + 2 \, x + 4\right ) + \frac {1}{12} \, \log \left (x - 2\right ) \]

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)

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giac [A]  time = 0.82, size = 33, normalized size = 0.77 \[ -\frac {1}{12} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (x + 1\right )}\right ) - \frac {1}{24} \, \log \left (x^{2} + 2 \, x + 4\right ) + \frac {1}{12} \, \log \left ({\left | x - 2 \right |}\right ) \]

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))

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maple [A]  time = 0.00, size = 35, normalized size = 0.81 \[ -\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 x +2\right ) \sqrt {3}}{6}\right )}{12}+\frac {\ln \left (x -2\right )}{12}-\frac {\ln \left (x^{2}+2 x +4\right )}{24} \]

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(x-2)

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maxima [A]  time = 1.11, size = 32, normalized size = 0.74 \[ -\frac {1}{12} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (x + 1\right )}\right ) - \frac {1}{24} \, \log \left (x^{2} + 2 \, x + 4\right ) + \frac {1}{12} \, \log \left (x - 2\right ) \]

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)

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mupad [B]  time = 0.09, size = 46, normalized size = 1.07 \[ \frac {\ln \left (x-2\right )}{12}+\ln \left (x+1-\sqrt {3}\,1{}\mathrm {i}\right )\,\left (-\frac {1}{24}+\frac {\sqrt {3}\,1{}\mathrm {i}}{24}\right )-\ln \left (x+1+\sqrt {3}\,1{}\mathrm {i}\right )\,\left (\frac {1}{24}+\frac {\sqrt {3}\,1{}\mathrm {i}}{24}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^3 - 8),x)

[Out]

log(x - 2)/12 + log(x - 3^(1/2)*1i + 1)*((3^(1/2)*1i)/24 - 1/24) - log(x + 3^(1/2)*1i + 1)*((3^(1/2)*1i)/24 +
1/24)

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sympy [A]  time = 0.14, size = 41, normalized size = 0.95 \[ \frac {\log {\left (x - 2 \right )}}{12} - \frac {\log {\left (x^{2} + 2 x + 4 \right )}}{24} - \frac {\sqrt {3} \operatorname {atan}{\left (\frac {\sqrt {3} x}{3} + \frac {\sqrt {3}}{3} \right )}}{12} \]

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

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