3.1.89 \(\int \frac {1}{1+x^3} \, dx\) [89]

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

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 41, normalized size of antiderivative = 0.95, 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 = {206, 31, 648, 632, 210, 642} \begin {gather*} -\frac {\arctan \left (\frac {1-2 x}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {1}{6} \log \left (x^2-x+1\right )+\frac {1}{3} \log (x+1) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 + x^3)^(-1),x]

[Out]

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

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 632

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 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 {gather*} \begin {aligned} \text {Integral} &=\frac {1}{3} \int \frac {1}{1+x} \, dx+\frac {1}{3} \int \frac {2-x}{1-x+x^2} \, dx\\ &=\frac {1}{3} \log (1+x)-\frac {1}{6} \int \frac {-1+2 x}{1-x+x^2} \, dx+\frac {1}{2} \int \frac {1}{1-x+x^2} \, dx\\ &=\frac {1}{3} \log (1+x)-\frac {1}{6} \log \left (1-x+x^2\right )-\text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,-1+2 x\right )\\ &=\frac {\arctan \left (\frac {-1+2 x}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {1}{3} \log (1+x)-\frac {1}{6} \log \left (1-x+x^2\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]
time = 0.01, size = 40, normalized size = 0.93 \begin {gather*} \frac {\arctan \left (\frac {-1+2 x}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {1}{3} \log (1+x)-\frac {1}{6} \log \left (1-x+x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 + x^3)^(-1),x]

[Out]

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

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Maple [A]
time = 0.03, size = 35, normalized size = 0.81

method	result	size
default	\(\frac {\ln \left (1+x \right )}{3}-\frac {\ln \left (x^{2}-x +1\right )}{6}+\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 x -1\right ) \sqrt {3}}{3}\right )}{3}\)	\(35\)
risch	\(-\frac {\ln \left (4 x^{2}-4 x +4\right )}{6}+\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 x -1\right ) \sqrt {3}}{3}\right )}{3}+\frac {\ln \left (1+x \right )}{3}\)	\(37\)
meijerg	\(\frac {x \ln \left (1+\left (x^{3}\right )^{\frac {1}{3}}\right )}{3 \left (x^{3}\right )^{\frac {1}{3}}}-\frac {x \ln \left (1-\left (x^{3}\right )^{\frac {1}{3}}+\left (x^{3}\right )^{\frac {2}{3}}\right )}{6 \left (x^{3}\right )^{\frac {1}{3}}}+\frac {x \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (x^{3}\right )^{\frac {1}{3}}}{2-\left (x^{3}\right )^{\frac {1}{3}}}\right )}{3 \left (x^{3}\right )^{\frac {1}{3}}}\)	\(74\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^3+1),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.49, size = 34, normalized size = 0.79 \begin {gather*} \frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) - \frac {1}{6} \, \log \left (x^{2} - x + 1\right ) + \frac {1}{3} \, \log \left (x + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.64, size = 34, normalized size = 0.79 \begin {gather*} \frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) - \frac {1}{6} \, \log \left (x^{2} - x + 1\right ) + \frac {1}{3} \, \log \left (x + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.04, size = 41, normalized size = 0.95 \begin {gather*} \frac {\log {\left (x + 1 \right )}}{3} - \frac {\log {\left (x^{2} - x + 1 \right )}}{6} + \frac {\sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} x}{3} - \frac {\sqrt {3}}{3} \right )}}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(x**3+1),x)

[Out]

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

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Giac [A]
time = 0.45, size = 35, normalized size = 0.81 \begin {gather*} \frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) - \frac {1}{6} \, \log \left (x^{2} - x + 1\right ) + \frac {1}{3} \, \log \left ({\left | x + 1 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.09, size = 31, normalized size = 0.72 \begin {gather*} \frac {\ln \left (x+1\right )}{3}-\frac {\ln \left ({\left (x-\frac {1}{2}\right )}^2+\frac {3}{4}\right )}{6}+\frac {\sqrt {3}\,\mathrm {atan}\left (\frac {2\,\sqrt {3}\,\left (x-\frac {1}{2}\right )}{3}\right )}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^3 + 1),x)

[Out]

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

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