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

Optimal. Leaf size=65 \[ -\frac {\log (a+x)}{3 a^5}+\frac {\tan ^{-1}\left (\frac {a-2 x}{\sqrt {3} a}\right )}{\sqrt {3} a^5}-\frac {1}{2 a^3 x^2}+\frac {\log \left (a^2-a x+x^2\right )}{6 a^5} \]

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Rubi [A]  time = 0.04, antiderivative size = 65, 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, 200, 31, 634, 617, 204, 628} \[ -\frac {1}{2 a^3 x^2}+\frac {\log \left (a^2-a x+x^2\right )}{6 a^5}-\frac {\log (a+x)}{3 a^5}+\frac {\tan ^{-1}\left (\frac {a-2 x}{\sqrt {3} a}\right )}{\sqrt {3} a^5} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^3*(a^3 + x^3)),x]

[Out]

-1/(2*a^3*x^2) + ArcTan[(a - 2*x)/(Sqrt[3]*a)]/(Sqrt[3]*a^5) - Log[a + x]/(3*a^5) + Log[a^2 - a*x + x^2]/(6*a^
5)

Rule 31

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

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 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 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^3 \left (a^3+x^3\right )} \, dx &=-\frac {1}{2 a^3 x^2}-\frac {\int \frac {1}{a^3+x^3} \, dx}{a^3}\\ &=-\frac {1}{2 a^3 x^2}-\frac {\int \frac {1}{a+x} \, dx}{3 a^5}-\frac {\int \frac {2 a-x}{a^2-a x+x^2} \, dx}{3 a^5}\\ &=-\frac {1}{2 a^3 x^2}-\frac {\log (a+x)}{3 a^5}+\frac {\int \frac {-a+2 x}{a^2-a x+x^2} \, dx}{6 a^5}-\frac {\int \frac {1}{a^2-a x+x^2} \, dx}{2 a^4}\\ &=-\frac {1}{2 a^3 x^2}-\frac {\log (a+x)}{3 a^5}+\frac {\log \left (a^2-a x+x^2\right )}{6 a^5}-\frac {\operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1-\frac {2 x}{a}\right )}{a^5}\\ &=-\frac {1}{2 a^3 x^2}+\frac {\tan ^{-1}\left (\frac {a-2 x}{\sqrt {3} a}\right )}{\sqrt {3} a^5}-\frac {\log (a+x)}{3 a^5}+\frac {\log \left (a^2-a x+x^2\right )}{6 a^5}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 68, normalized size = 1.05 \[ -\frac {\log (a+x)}{3 a^5}-\frac {\tan ^{-1}\left (\frac {2 x-a}{\sqrt {3} a}\right )}{\sqrt {3} a^5}-\frac {1}{2 a^3 x^2}+\frac {\log \left (a^2-a x+x^2\right )}{6 a^5} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^3*(a^3 + x^3)),x]

[Out]

-1/2*1/(a^3*x^2) - ArcTan[(-a + 2*x)/(Sqrt[3]*a)]/(Sqrt[3]*a^5) - Log[a + x]/(3*a^5) + Log[a^2 - a*x + x^2]/(6
*a^5)

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IntegrateAlgebraic [A]  time = 0.03, size = 68, normalized size = 1.05 \[ -\frac {\log (a+x)}{3 a^5}+\frac {\tan ^{-1}\left (\frac {1}{\sqrt {3}}-\frac {2 x}{\sqrt {3} a}\right )}{\sqrt {3} a^5}-\frac {1}{2 a^3 x^2}+\frac {\log \left (a^2-a x+x^2\right )}{6 a^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*1/(a^3*x^2) + ArcTan[1/Sqrt[3] - (2*x)/(Sqrt[3]*a)]/(Sqrt[3]*a^5) - Log[a + x]/(3*a^5) + Log[a^2 - a*x +
x^2]/(6*a^5)

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fricas [A]  time = 0.79, size = 62, normalized size = 0.95 \[ -\frac {2 \, \sqrt {3} x^{2} \arctan \left (-\frac {\sqrt {3} {\left (a - 2 \, x\right )}}{3 \, a}\right ) - x^{2} \log \left (a^{2} - a x + x^{2}\right ) + 2 \, x^{2} \log \left (a + x\right ) + 3 \, a^{2}}{6 \, a^{5} x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(2*sqrt(3)*x^2*arctan(-1/3*sqrt(3)*(a - 2*x)/a) - x^2*log(a^2 - a*x + x^2) + 2*x^2*log(a + x) + 3*a^2)/(a
^5*x^2)

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giac [A]  time = 0.99, size = 58, normalized size = 0.89 \[ -\frac {\sqrt {3} \arctan \left (-\frac {\sqrt {3} {\left (a - 2 \, x\right )}}{3 \, a}\right )}{3 \, a^{5}} + \frac {\log \left (a^{2} - a x + x^{2}\right )}{6 \, a^{5}} - \frac {\log \left ({\left | a + x \right |}\right )}{3 \, a^{5}} - \frac {1}{2 \, a^{3} x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*sqrt(3)*arctan(-1/3*sqrt(3)*(a - 2*x)/a)/a^5 + 1/6*log(a^2 - a*x + x^2)/a^5 - 1/3*log(abs(a + x))/a^5 - 1
/2/(a^3*x^2)

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maple [A]  time = 0.30, size = 60, normalized size = 0.92

method	result	size
default	\(\frac {\frac {\ln \left (a^{2}-a x +x^{2}\right )}{2}-\sqrt {3}\, \arctan \left (\frac {\left (2 x -a \right ) \sqrt {3}}{3 a}\right )}{3 a^{5}}-\frac {\ln \left (a +x \right )}{3 a^{5}}-\frac {1}{2 a^{3} x^{2}}\)	\(60\)
risch	\(-\frac {1}{2 a^{3} x^{2}}+\frac {\ln \left (4 a^{2}-4 a x +4 x^{2}\right )}{6 a^{5}}-\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 x -a \right ) \sqrt {3}}{3 a}\right )}{3 a^{5}}-\frac {\ln \left (a +x \right )}{3 a^{5}}\)	\(64\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3/a^5*(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^5-1/2/a^3/x^2

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maxima [A]  time = 0.96, size = 57, normalized size = 0.88 \[ -\frac {\sqrt {3} \arctan \left (-\frac {\sqrt {3} {\left (a - 2 \, x\right )}}{3 \, a}\right )}{3 \, a^{5}} + \frac {\log \left (a^{2} - a x + x^{2}\right )}{6 \, a^{5}} - \frac {\log \left (a + x\right )}{3 \, a^{5}} - \frac {1}{2 \, a^{3} x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*sqrt(3)*arctan(-1/3*sqrt(3)*(a - 2*x)/a)/a^5 + 1/6*log(a^2 - a*x + x^2)/a^5 - 1/3*log(a + x)/a^5 - 1/2/(a
^3*x^2)

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mupad [B]  time = 0.25, size = 86, normalized size = 1.32 \[ -\frac {\ln \left (a+x\right )}{3\,a^5}-\frac {1}{2\,a^3\,x^2}-\frac {\ln \left (\frac {3\,a^7\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{2}+3\,a^6\,x\right )\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{6\,a^5}+\frac {\ln \left (\frac {3\,a^7\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{2}-3\,a^6\,x\right )\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{6\,a^5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^3*(a^3 + x^3)),x)

[Out]

(log((3*a^7*(3^(1/2)*1i + 1))/2 - 3*a^6*x)*(3^(1/2)*1i + 1))/(6*a^5) - 1/(2*a^3*x^2) - (log((3*a^7*(3^(1/2)*1i
- 1))/2 + 3*a^6*x)*(3^(1/2)*1i - 1))/(6*a^5) - log(a + x)/(3*a^5)

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sympy [C]  time = 0.20, size = 80, normalized size = 1.23 \[ - \frac {1}{2 a^{3} x^{2}} + \frac {- \frac {\log {\left (a + x \right )}}{3} + \left (\frac {1}{6} - \frac {\sqrt {3} i}{6}\right ) \log {\left (- 3 a \left (\frac {1}{6} - \frac {\sqrt {3} i}{6}\right ) + x \right )} + \left (\frac {1}{6} + \frac {\sqrt {3} i}{6}\right ) \log {\left (- 3 a \left (\frac {1}{6} + \frac {\sqrt {3} i}{6}\right ) + x \right )}}{a^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/(2*a**3*x**2) + (-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**5

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