3.122 \(\int \frac {1}{x^2 (a^3+x^3)} \, dx\)

Optimal. Leaf size=63 \[ \frac {\log (a+x)}{3 a^4}+\frac {\tan ^{-1}\left (\frac {a-2 x}{\sqrt {3} a}\right )}{\sqrt {3} a^4}-\frac {1}{a^3 x}-\frac {\log \left (a^2-a x+x^2\right )}{6 a^4} \]

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Rubi [A]  time = 0.04, antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.538, Rules used = {325, 292, 31, 634, 617, 204, 628} \[ -\frac {\log \left (a^2-a x+x^2\right )}{6 a^4}-\frac {1}{a^3 x}+\frac {\log (a+x)}{3 a^4}+\frac {\tan ^{-1}\left (\frac {a-2 x}{\sqrt {3} a}\right )}{\sqrt {3} a^4} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^2*(a^3 + x^3)),x]

[Out]

-(1/(a^3*x)) + ArcTan[(a - 2*x)/(Sqrt[3]*a)]/(Sqrt[3]*a^4) + Log[a + x]/(3*a^4) - Log[a^2 - a*x + x^2]/(6*a^4)

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, 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 292

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 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

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}{x^2 \left (a^3+x^3\right )} \, dx &=-\frac {1}{a^3 x}-\frac {\int \frac {x}{a^3+x^3} \, dx}{a^3}\\ &=-\frac {1}{a^3 x}+\frac {\int \frac {1}{a+x} \, dx}{3 a^4}-\frac {\int \frac {a+x}{a^2-a x+x^2} \, dx}{3 a^4}\\ &=-\frac {1}{a^3 x}+\frac {\log (a+x)}{3 a^4}-\frac {\int \frac {-a+2 x}{a^2-a x+x^2} \, dx}{6 a^4}-\frac {\int \frac {1}{a^2-a x+x^2} \, dx}{2 a^3}\\ &=-\frac {1}{a^3 x}+\frac {\log (a+x)}{3 a^4}-\frac {\log \left (a^2-a x+x^2\right )}{6 a^4}-\frac {\operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1-\frac {2 x}{a}\right )}{a^4}\\ &=-\frac {1}{a^3 x}+\frac {\tan ^{-1}\left (\frac {a-2 x}{\sqrt {3} a}\right )}{\sqrt {3} a^4}+\frac {\log (a+x)}{3 a^4}-\frac {\log \left (a^2-a x+x^2\right )}{6 a^4}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 60, normalized size = 0.95 \[ -\frac {x \log \left (a^2-a x+x^2\right )-2 x \log (a+x)+2 \sqrt {3} x \tan ^{-1}\left (\frac {2 x-a}{\sqrt {3} a}\right )+6 a}{6 a^4 x} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^2*(a^3 + x^3)),x]

[Out]

-1/6*(6*a + 2*Sqrt[3]*x*ArcTan[(-a + 2*x)/(Sqrt[3]*a)] - 2*x*Log[a + x] + x*Log[a^2 - a*x + x^2])/(a^4*x)

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IntegrateAlgebraic [A]  time = 0.03, size = 66, normalized size = 1.05 \[ \frac {\log (a+x)}{3 a^4}+\frac {\tan ^{-1}\left (\frac {1}{\sqrt {3}}-\frac {2 x}{\sqrt {3} a}\right )}{\sqrt {3} a^4}-\frac {1}{a^3 x}-\frac {\log \left (a^2-a x+x^2\right )}{6 a^4} \]

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/(x^2*(a^3 + x^3)),x]

[Out]

-(1/(a^3*x)) + ArcTan[1/Sqrt[3] - (2*x)/(Sqrt[3]*a)]/(Sqrt[3]*a^4) + Log[a + x]/(3*a^4) - Log[a^2 - a*x + x^2]
/(6*a^4)

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fricas [A]  time = 0.92, size = 53, normalized size = 0.84 \[ -\frac {2 \, \sqrt {3} x \arctan \left (-\frac {\sqrt {3} {\left (a - 2 \, x\right )}}{3 \, a}\right ) + x \log \left (a^{2} - a x + x^{2}\right ) - 2 \, x \log \left (a + x\right ) + 6 \, a}{6 \, a^{4} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.93, size = 58, normalized size = 0.92 \[ -\frac {\sqrt {3} \arctan \left (-\frac {\sqrt {3} {\left (a - 2 \, x\right )}}{3 \, a}\right )}{3 \, a^{4}} - \frac {\log \left (a^{2} - a x + x^{2}\right )}{6 \, a^{4}} + \frac {\log \left ({\left | a + x \right |}\right )}{3 \, a^{4}} - \frac {1}{a^{3} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.28, size = 60, normalized size = 0.95

method	result	size
default	\(\frac {-\frac {\ln \left (a^{2}-a x +x^{2}\right )}{2}-\sqrt {3}\, \arctan \left (\frac {\left (2 x -a \right ) \sqrt {3}}{3 a}\right )}{3 a^{4}}-\frac {1}{a^{3} x}+\frac {\ln \left (a +x \right )}{3 a^{4}}\)	\(60\)
risch	\(-\frac {1}{a^{3} x}+\frac {\ln \left (-a -x \right )}{3 a^{4}}+\frac {\left (\munderset {\textit {\_R} =\RootOf \left (a^{8} \textit {\_Z}^{2}+a^{4} \textit {\_Z} +1\right )}{\sum }\textit {\_R} \ln \left (\left (-4 \textit {\_R}^{3} a^{12}+3\right ) x -a^{9} \textit {\_R}^{2}\right )\right )}{3}\)	\(67\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^2/(a^3+x^3),x,method=_RETURNVERBOSE)

[Out]

1/3/a^4*(-1/2*ln(a^2-a*x+x^2)-3^(1/2)*arctan(1/3*(2*x-a)*3^(1/2)/a))-1/a^3/x+1/3*ln(a+x)/a^4

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maxima [A]  time = 0.96, size = 57, normalized size = 0.90 \[ -\frac {\sqrt {3} \arctan \left (-\frac {\sqrt {3} {\left (a - 2 \, x\right )}}{3 \, a}\right )}{3 \, a^{4}} - \frac {\log \left (a^{2} - a x + x^{2}\right )}{6 \, a^{4}} + \frac {\log \left (a + x\right )}{3 \, a^{4}} - \frac {1}{a^{3} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.25, size = 88, normalized size = 1.40 \[ \frac {\ln \left (a+x\right )}{3\,a^4}-\frac {1}{a^3\,x}+\frac {\ln \left (\frac {{\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}^2\,a^4}{4}+x\,a^3\right )\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{6\,a^4}-\frac {\ln \left (\frac {{\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}^2\,a^4}{4}+x\,a^3\right )\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{6\,a^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^2*(a^3 + x^3)),x)

[Out]

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

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sympy [C]  time = 0.17, size = 83, normalized size = 1.32 \[ - \frac {1}{a^{3} x} + \frac {\frac {\log {\left (a + x \right )}}{3} + \left (- \frac {1}{6} - \frac {\sqrt {3} i}{6}\right ) \log {\left (9 a \left (- \frac {1}{6} - \frac {\sqrt {3} i}{6}\right )^{2} + x \right )} + \left (- \frac {1}{6} + \frac {\sqrt {3} i}{6}\right ) \log {\left (9 a \left (- \frac {1}{6} + \frac {\sqrt {3} i}{6}\right )^{2} + x \right )}}{a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**2/(a**3+x**3),x)

[Out]

-1/(a**3*x) + (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**4

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