3.2.19 \(\int \frac {x}{a^3+x^3} \, dx\) [119]

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

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 56, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.546, Rules used = {298, 31, 648, 631, 210, 642} \begin {gather*} \frac {\log \left (a^2-a x+x^2\right )}{6 a}-\frac {\text {ArcTan}\left (\frac {a-2 x}{\sqrt {3} a}\right )}{\sqrt {3} a}-\frac {\log (a+x)}{3 a} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x/(a^3 + x^3),x]

[Out]

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, 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 298

Int[(x_)/((a_) + (b_.)*(x_)^3), x_Symbol] :> Dist[-(3*Rt[a, 3]*Rt[b, 3])^(-1), Int[1/(Rt[a, 3] + Rt[b, 3]*x),
x], x] + Dist[1/(3*Rt[a, 3]*Rt[b, 3]), Int[(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 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 {x}{a^3+x^3} \, dx &=-\frac {\int \frac {1}{a+x} \, dx}{3 a}+\frac {\int \frac {a+x}{a^2-a x+x^2} \, dx}{3 a}\\ &=-\frac {\log (a+x)}{3 a}+\frac {1}{2} \int \frac {1}{a^2-a x+x^2} \, dx+\frac {\int \frac {-a+2 x}{a^2-a x+x^2} \, dx}{6 a}\\ &=-\frac {\log (a+x)}{3 a}+\frac {\log \left (a^2-a x+x^2\right )}{6 a}+\frac {\text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1-\frac {2 x}{a}\right )}{a}\\ &=-\frac {\tan ^{-1}\left (\frac {a-2 x}{\sqrt {3} a}\right )}{\sqrt {3} a}-\frac {\log (a+x)}{3 a}+\frac {\log \left (a^2-a x+x^2\right )}{6 a}\\ \end {align*}

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Mathematica [A]
time = 0.00, size = 50, normalized size = 0.89 \begin {gather*} \frac {2 \sqrt {3} \tan ^{-1}\left (\frac {-a+2 x}{\sqrt {3} a}\right )-2 \log (a+x)+\log \left (a^2-a x+x^2\right )}{6 a} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x/(a^3 + x^3),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)

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Maple [A]
time = 0.05, size = 51, normalized size = 0.91

method	result	size
risch	\(-\frac {\ln \left (a +x \right )}{3 a}+\frac {\left (\munderset {\textit {\_R} =\RootOf \left (\textit {\_Z}^{2} a^{2}-a \textit {\_Z} +1\right )}{\sum }\textit {\_R} \ln \left (a^{2} \textit {\_R} -a +x \right )\right )}{3}\)	\(43\)
default	\(\frac {\frac {\ln \left (a^{2}-a x +x^{2}\right )}{2}+\sqrt {3}\, \arctan \left (\frac {\left (2 x -a \right ) \sqrt {3}}{3 a}\right )}{3 a}-\frac {\ln \left (a +x \right )}{3 a}\)	\(51\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 2.54, size = 49, normalized size = 0.88 \begin {gather*} \frac {\sqrt {3} \arctan \left (-\frac {\sqrt {3} {\left (a - 2 \, x\right )}}{3 \, a}\right )}{3 \, a} + \frac {\log \left (a^{2} - a x + x^{2}\right )}{6 \, a} - \frac {\log \left (a + x\right )}{3 \, a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.36, size = 43, normalized size = 0.77 \begin {gather*} \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(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

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Sympy [C] Result contains complex when optimal does not.
time = 0.04, size = 71, normalized size = 1.27 \begin {gather*} \frac {- \frac {\log {\left (a + x \right )}}{3} + \left (\frac {1}{6} - \frac {\sqrt {3} i}{6}\right ) \log {\left (9 a \left (\frac {1}{6} - \frac {\sqrt {3} i}{6}\right )^{2} + x \right )} + \left (\frac {1}{6} + \frac {\sqrt {3} i}{6}\right ) \log {\left (9 a \left (\frac {1}{6} + \frac {\sqrt {3} i}{6}\right )^{2} + x \right )}}{a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-log(a + x)/3 + (1/6 - sqrt(3)*I/6)*log(9*a*(1/6 - sqrt(3)*I/6)**2 + x) + (1/6 + sqrt(3)*I/6)*log(9*a*(1/6 +
sqrt(3)*I/6)**2 + x))/a

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Giac [A]
time = 0.80, size = 50, normalized size = 0.89 \begin {gather*} \frac {\sqrt {3} \arctan \left (-\frac {\sqrt {3} {\left (a - 2 \, x\right )}}{3 \, a}\right )}{3 \, a} + \frac {\log \left (a^{2} - a x + x^{2}\right )}{6 \, a} - \frac {\log \left ({\left | a + x \right |}\right )}{3 \, a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.12, size = 68, normalized size = 1.21 \begin {gather*} -\frac {\ln \left (a+x\right )}{3\,a}-\frac {\ln \left (x+\frac {a\,{\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}^2}{4}\right )\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{6\,a}+\frac {\ln \left (x+\frac {a\,{\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}^2}{4}\right )\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{6\,a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(a^3 + x^3),x)

[Out]

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

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