3.118 \(\int \frac{1}{a^3+x^3} \, dx\)

Optimal. Leaf size=56 \[ -\frac{\log \left (a^2-a x+x^2\right )}{6 a^2}+\frac{\log (a+x)}{3 a^2}-\frac{\tan ^{-1}\left (\frac{a-2 x}{\sqrt{3} a}\right )}{\sqrt{3} a^2} \]

[Out]

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

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Rubi [A]  time = 0.0303337, antiderivative size = 56, normalized size of antiderivative = 1., 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 = {200, 31, 634, 617, 204, 628} \[ -\frac{\log \left (a^2-a x+x^2\right )}{6 a^2}+\frac{\log (a+x)}{3 a^2}-\frac{\tan ^{-1}\left (\frac{a-2 x}{\sqrt{3} a}\right )}{\sqrt{3} a^2} \]

Antiderivative was successfully verified.

[In]

Int[(a^3 + x^3)^(-1),x]

[Out]

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

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

Mathematica [A]  time = 0.0123992, size = 52, normalized size = 0.93 \[ \frac{-\log \left (a^2-a x+x^2\right )+2 \log (a+x)+2 \sqrt{3} \tan ^{-1}\left (\frac{2 x-a}{\sqrt{3} a}\right )}{6 a^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(a^3 + x^3)^(-1),x]

[Out]

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

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Maple [A]  time = 0.008, size = 52, normalized size = 0.9 \begin{align*} -{\frac{\ln \left ({a}^{2}-ax+{x}^{2} \right ) }{6\,{a}^{2}}}+{\frac{\sqrt{3}}{3\,{a}^{2}}\arctan \left ({\frac{ \left ( 2\,x-a \right ) \sqrt{3}}{3\,a}} \right ) }+{\frac{\ln \left ( a+x \right ) }{3\,{a}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a^3+x^3),x)

[Out]

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

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Maxima [A]  time = 1.40532, size = 66, normalized size = 1.18 \begin{align*} \frac{\sqrt{3} \arctan \left (-\frac{\sqrt{3}{\left (a - 2 \, x\right )}}{3 \, a}\right )}{3 \, a^{2}} - \frac{\log \left (a^{2} - a x + x^{2}\right )}{6 \, a^{2}} + \frac{\log \left (a + x\right )}{3 \, a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.05402, size = 124, normalized size = 2.21 \begin{align*} \frac{2 \, \sqrt{3} \arctan \left (-\frac{\sqrt{3}{\left (a - 2 \, x\right )}}{3 \, a}\right ) - \log \left (a^{2} - a x + x^{2}\right ) + 2 \, \log \left (a + x\right )}{6 \, a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [C]  time = 0.119672, size = 73, normalized size = 1.3 \begin{align*} \frac{\frac{\log{\left (a + x \right )}}{3} + \left (- \frac{1}{6} - \frac{\sqrt{3} i}{6}\right ) \log{\left (3 a \left (- \frac{1}{6} - \frac{\sqrt{3} i}{6}\right ) + x \right )} + \left (- \frac{1}{6} + \frac{\sqrt{3} i}{6}\right ) \log{\left (3 a \left (- \frac{1}{6} + \frac{\sqrt{3} i}{6}\right ) + x \right )}}{a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.05349, size = 68, normalized size = 1.21 \begin{align*} \frac{\sqrt{3} \arctan \left (-\frac{\sqrt{3}{\left (a - 2 \, x\right )}}{3 \, a}\right )}{3 \, a^{2}} - \frac{\log \left (a^{2} - a x + x^{2}\right )}{6 \, a^{2}} + \frac{\log \left ({\left | a + x \right |}\right )}{3 \, a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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