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

Optimal. Leaf size=74 \[ -\frac {\log \left (x^2+\sqrt [3]{2} x+2^{2/3}\right )}{6\ 2^{2/3}}+\frac {\log \left (\sqrt [3]{2}-x\right )}{3\ 2^{2/3}}-\frac {\tan ^{-1}\left (\frac {2^{2/3} x+1}{\sqrt {3}}\right )}{2^{2/3} \sqrt {3}} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 74, 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, 617, 204, 628} \[ -\frac {\log \left (x^2+\sqrt [3]{2} x+2^{2/3}\right )}{6\ 2^{2/3}}+\frac {\log \left (\sqrt [3]{2}-x\right )}{3\ 2^{2/3}}-\frac {\tan ^{-1}\left (\frac {2^{2/3} x+1}{\sqrt {3}}\right )}{2^{2/3} \sqrt {3}} \]

Antiderivative was successfully verified.

[In]

Int[(-2 + x^3)^(-1),x]

[Out]

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

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 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
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}{-2+x^3} \, dx &=\frac {\int \frac {1}{-\sqrt [3]{2}+x} \, dx}{3\ 2^{2/3}}+\frac {\int \frac {-2 \sqrt [3]{2}-x}{2^{2/3}+\sqrt [3]{2} x+x^2} \, dx}{3\ 2^{2/3}}\\ &=\frac {\log \left (\sqrt [3]{2}-x\right )}{3\ 2^{2/3}}-\frac {\int \frac {\sqrt [3]{2}+2 x}{2^{2/3}+\sqrt [3]{2} x+x^2} \, dx}{6\ 2^{2/3}}-\frac {\int \frac {1}{2^{2/3}+\sqrt [3]{2} x+x^2} \, dx}{2 \sqrt [3]{2}}\\ &=\frac {\log \left (\sqrt [3]{2}-x\right )}{3\ 2^{2/3}}-\frac {\log \left (2^{2/3}+\sqrt [3]{2} x+x^2\right )}{6\ 2^{2/3}}+\frac {\operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+2^{2/3} x\right )}{2^{2/3}}\\ &=-\frac {\tan ^{-1}\left (\frac {1+2^{2/3} x}{\sqrt {3}}\right )}{2^{2/3} \sqrt {3}}+\frac {\log \left (\sqrt [3]{2}-x\right )}{3\ 2^{2/3}}-\frac {\log \left (2^{2/3}+\sqrt [3]{2} x+x^2\right )}{6\ 2^{2/3}}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 65, normalized size = 0.88 \[ -\frac {\log \left (\sqrt [3]{2} x^2+2^{2/3} x+2\right )-2 \log \left (2-2^{2/3} x\right )+2 \sqrt {3} \tan ^{-1}\left (\frac {2^{2/3} x+1}{\sqrt {3}}\right )}{6\ 2^{2/3}} \]

Antiderivative was successfully verified.

[In]

Integrate[(-2 + x^3)^(-1),x]

[Out]

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

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fricas [A]  time = 0.41, size = 68, normalized size = 0.92 \[ -\frac {1}{6} \cdot 4^{\frac {1}{6}} \sqrt {3} \arctan \left (\frac {1}{6} \cdot 4^{\frac {1}{6}} {\left (4^{\frac {2}{3}} \sqrt {3} x + 4^{\frac {1}{3}} \sqrt {3}\right )}\right ) - \frac {1}{24} \cdot 4^{\frac {2}{3}} \log \left (2 \, x^{2} + 4^{\frac {2}{3}} x + 2 \cdot 4^{\frac {1}{3}}\right ) + \frac {1}{12} \cdot 4^{\frac {2}{3}} \log \left (2 \, x - 4^{\frac {2}{3}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 1.28, size = 57, normalized size = 0.77 \[ -\frac {1}{6} \, \sqrt {3} 2^{\frac {1}{3}} \arctan \left (\frac {1}{6} \, \sqrt {3} 2^{\frac {2}{3}} {\left (2 \, x + 2^{\frac {1}{3}}\right )}\right ) - \frac {1}{12} \cdot 2^{\frac {1}{3}} \log \left (x^{2} + 2^{\frac {1}{3}} x + 2^{\frac {2}{3}}\right ) + \frac {1}{6} \cdot 2^{\frac {1}{3}} \log \left ({\left | x - 2^{\frac {1}{3}} \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*sqrt(3)*2^(1/3)*arctan(1/6*sqrt(3)*2^(2/3)*(2*x + 2^(1/3))) - 1/12*2^(1/3)*log(x^2 + 2^(1/3)*x + 2^(2/3))
+ 1/6*2^(1/3)*log(abs(x - 2^(1/3)))

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maple [A]  time = 0.00, size = 54, normalized size = 0.73 \[ -\frac {2^{\frac {1}{3}} \sqrt {3}\, \arctan \left (\frac {\left (2^{\frac {2}{3}} x +1\right ) \sqrt {3}}{3}\right )}{6}+\frac {2^{\frac {1}{3}} \ln \left (x -2^{\frac {1}{3}}\right )}{6}-\frac {2^{\frac {1}{3}} \ln \left (x^{2}+2^{\frac {1}{3}} x +2^{\frac {2}{3}}\right )}{12} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^3-2),x)

[Out]

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

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maxima [A]  time = 0.96, size = 56, normalized size = 0.76 \[ -\frac {1}{6} \, \sqrt {3} 2^{\frac {1}{3}} \arctan \left (\frac {1}{6} \, \sqrt {3} 2^{\frac {2}{3}} {\left (2 \, x + 2^{\frac {1}{3}}\right )}\right ) - \frac {1}{12} \cdot 2^{\frac {1}{3}} \log \left (x^{2} + 2^{\frac {1}{3}} x + 2^{\frac {2}{3}}\right ) + \frac {1}{6} \cdot 2^{\frac {1}{3}} \log \left (x - 2^{\frac {1}{3}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*sqrt(3)*2^(1/3)*arctan(1/6*sqrt(3)*2^(2/3)*(2*x + 2^(1/3))) - 1/12*2^(1/3)*log(x^2 + 2^(1/3)*x + 2^(2/3))
+ 1/6*2^(1/3)*log(x - 2^(1/3))

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mupad [B]  time = 0.36, size = 72, normalized size = 0.97 \[ \frac {2^{1/3}\,\ln \left (x-2^{1/3}\right )}{6}+\frac {2^{1/3}\,\ln \left (x-\frac {2^{1/3}\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{2}\right )\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{12}-\frac {2^{1/3}\,\ln \left (x+\frac {2^{1/3}\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{2}\right )\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{12} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^3 - 2),x)

[Out]

(2^(1/3)*log(x - 2^(1/3)))/6 + (2^(1/3)*log(x - (2^(1/3)*(3^(1/2)*1i - 1))/2)*(3^(1/2)*1i - 1))/12 - (2^(1/3)*
log(x + (2^(1/3)*(3^(1/2)*1i + 1))/2)*(3^(1/2)*1i + 1))/12

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sympy [A]  time = 0.30, size = 71, normalized size = 0.96 \[ \frac {\sqrt [3]{2} \log {\left (x - \sqrt [3]{2} \right )}}{6} - \frac {\sqrt [3]{2} \log {\left (x^{2} + \sqrt [3]{2} x + 2^{\frac {2}{3}} \right )}}{12} - \frac {\sqrt [3]{2} \sqrt {3} \operatorname {atan}{\left (\frac {2^{\frac {2}{3}} \sqrt {3} x}{3} + \frac {\sqrt {3}}{3} \right )}}{6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2**(1/3)*log(x - 2**(1/3))/6 - 2**(1/3)*log(x**2 + 2**(1/3)*x + 2**(2/3))/12 - 2**(1/3)*sqrt(3)*atan(2**(2/3)*
sqrt(3)*x/3 + sqrt(3)/3)/6

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