3.1.33 \(\int \frac {1}{-1+2 x^3} \, dx\) [33]

Optimal. Leaf size=78 \[ -\frac {\tan ^{-1}\left (\frac {1+2 \sqrt [3]{2} x}{\sqrt {3}}\right )}{\sqrt [3]{2} \sqrt {3}}+\frac {\log \left (1-\sqrt [3]{2} x\right )}{3 \sqrt [3]{2}}-\frac {\log \left (1+\sqrt [3]{2} x+2^{2/3} x^2\right )}{6 \sqrt [3]{2}} \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {206, 31, 648, 631, 210, 642} \begin {gather*} -\frac {\log \left (2^{2/3} x^2+\sqrt [3]{2} x+1\right )}{6 \sqrt [3]{2}}+\frac {\log \left (1-\sqrt [3]{2} x\right )}{3 \sqrt [3]{2}}-\frac {\tan ^{-1}\left (\frac {2 \sqrt [3]{2} x+1}{\sqrt {3}}\right )}{\sqrt [3]{2} \sqrt {3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-1 + 2*x^3)^(-1),x]

[Out]

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

Rule 31

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

Rule 206

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 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 631

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 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]

Rubi steps

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

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Mathematica [A]
time = 0.01, size = 66, normalized size = 0.85 \begin {gather*} -\frac {2 \sqrt {3} \tan ^{-1}\left (\frac {1+2 \sqrt [3]{2} x}{\sqrt {3}}\right )-2 \log \left (1-\sqrt [3]{2} x\right )+\log \left (1+\sqrt [3]{2} x+2^{2/3} x^2\right )}{6 \sqrt [3]{2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-1 + 2*x^3)^(-1),x]

[Out]

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

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Mathics [A]
time = 2.20, size = 53, normalized size = 0.68 \begin {gather*} \frac {2^{\frac {2}{3}} \left (-2 \sqrt {3} \text {ArcTan}\left [\frac {\sqrt {3} \left (1+2 2^{\frac {1}{3}} x\right )}{3}\right ]-\text {Log}\left [\frac {2^{\frac {1}{3}}}{2}+\frac {2^{\frac {2}{3}} x}{2}+x^2\right ]+2 \text {Log}\left [-\frac {2^{\frac {2}{3}}}{2}+x\right ]\right )}{12} \end {gather*}

Antiderivative was successfully verified.

[In]

mathics('Integrate[1/(2*x^3-1),x]')

[Out]

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

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Maple [A]
time = 0.03, size = 58, normalized size = 0.74

method	result	size
risch	\(\frac {\left (\munderset {\textit {\_R} =\RootOf \left (2 \textit {\_Z}^{3}-1\right )}{\sum }\frac {\ln \left (-\textit {\_R} +x \right )}{\textit {\_R}^{2}}\right )}{6}\)	\(24\)
default	\(\frac {2^{\frac {2}{3}} \ln \left (x -\frac {2^{\frac {2}{3}}}{2}\right )}{6}-\frac {2^{\frac {2}{3}} \ln \left (x^{2}+\frac {2^{\frac {2}{3}} x}{2}+\frac {2^{\frac {1}{3}}}{2}\right )}{12}-\frac {\arctan \left (\frac {\left (1+2 \,2^{\frac {1}{3}} x \right ) \sqrt {3}}{3}\right ) 2^{\frac {2}{3}} \sqrt {3}}{6}\)	\(58\)
meijerg	\(\frac {2^{\frac {2}{3}} x \left (\ln \left (1-2^{\frac {1}{3}} \left (x^{3}\right )^{\frac {1}{3}}\right )-\frac {\ln \left (1+2^{\frac {1}{3}} \left (x^{3}\right )^{\frac {1}{3}}+2^{\frac {2}{3}} \left (x^{3}\right )^{\frac {2}{3}}\right )}{2}-\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, 2^{\frac {1}{3}} \left (x^{3}\right )^{\frac {1}{3}}}{2+2^{\frac {1}{3}} \left (x^{3}\right )^{\frac {1}{3}}}\right )\right )}{6 \left (x^{3}\right )^{\frac {1}{3}}}\)	\(82\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.37, size = 66, normalized size = 0.85 \begin {gather*} -\frac {1}{6} \, \sqrt {3} 2^{\frac {2}{3}} \arctan \left (\frac {1}{6} \, \sqrt {3} 2^{\frac {2}{3}} {\left (2 \cdot 2^{\frac {2}{3}} x + 2^{\frac {1}{3}}\right )}\right ) - \frac {1}{12} \cdot 2^{\frac {2}{3}} \log \left (2^{\frac {2}{3}} x^{2} + 2^{\frac {1}{3}} x + 1\right ) + \frac {1}{6} \cdot 2^{\frac {2}{3}} \log \left (\frac {1}{2} \cdot 2^{\frac {2}{3}} {\left (2^{\frac {1}{3}} x - 1\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.33, size = 63, normalized size = 0.81 \begin {gather*} -\frac {1}{6} \, \sqrt {6} 2^{\frac {1}{6}} \arctan \left (\frac {1}{6} \, \sqrt {6} 2^{\frac {1}{6}} {\left (2 \cdot 2^{\frac {2}{3}} x + 2^{\frac {1}{3}}\right )}\right ) - \frac {1}{12} \cdot 2^{\frac {2}{3}} \log \left (2 \, x^{2} + 2^{\frac {2}{3}} x + 2^{\frac {1}{3}}\right ) + \frac {1}{6} \cdot 2^{\frac {2}{3}} \log \left (2 \, x - 2^{\frac {2}{3}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.16, size = 78, normalized size = 1.00 \begin {gather*} \frac {2^{\frac {2}{3}} \log {\left (x - \frac {2^{\frac {2}{3}}}{2} \right )}}{6} - \frac {2^{\frac {2}{3}} \log {\left (x^{2} + \frac {2^{\frac {2}{3}} x}{2} + \frac {\sqrt [3]{2}}{2} \right )}}{12} - \frac {2^{\frac {2}{3}} \sqrt {3} \operatorname {atan}{\left (\frac {2 \cdot \sqrt [3]{2} \sqrt {3} x}{3} + \frac {\sqrt {3}}{3} \right )}}{6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.00, size = 107, normalized size = 1.37 \begin {gather*} -\frac {1}{12}\cdot 4^{\frac {1}{3}} \ln \left (x^{2}+\left (\frac {1}{2}\right )^{\frac {1}{3}} x+\left (\frac {1}{2}\right )^{\frac {1}{3}} \left (\frac {1}{2}\right )^{\frac {1}{3}}\right )-\frac {1}{3} \left (\frac {1}{2}\right )^{\frac {1}{3}} \sqrt {3} \arctan \left (\frac {x+\frac {\left (\frac {1}{2}\right )^{\frac {1}{3}}}{2}}{\frac {1}{2} \sqrt {3} \left (\frac {1}{2}\right )^{\frac {1}{3}}}\right )+\frac {1}{3} \left (\frac {1}{2}\right )^{\frac {1}{3}} \ln \left |x-\left (\frac {1}{2}\right )^{\frac {1}{3}}\right | \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.28, size = 72, normalized size = 0.92 \begin {gather*} \frac {2^{2/3}\,\ln \left (x-\frac {2^{2/3}}{2}\right )}{6}+\frac {2^{2/3}\,\ln \left (x-\frac {2^{2/3}\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{4}\right )\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{12}-\frac {2^{2/3}\,\ln \left (x+\frac {2^{2/3}\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{4}\right )\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{12} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(2*x^3 - 1),x)

[Out]

(2^(2/3)*log(x - 2^(2/3)/2))/6 + (2^(2/3)*log(x - (2^(2/3)*(3^(1/2)*1i - 1))/4)*(3^(1/2)*1i - 1))/12 - (2^(2/3
)*log(x + (2^(2/3)*(3^(1/2)*1i + 1))/4)*(3^(1/2)*1i + 1))/12

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