3.6.65 \(\int \frac {1}{x (a+b x^3)^{2/3}} \, dx\) [565]

Optimal. Leaf size=84 \[ -\frac {\tan ^{-1}\left (\frac {\sqrt [3]{a}+2 \sqrt [3]{a+b x^3}}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} a^{2/3}}-\frac {\log (x)}{2 a^{2/3}}+\frac {\log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^3}\right )}{2 a^{2/3}} \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {272, 59, 631, 210, 31} \begin {gather*} -\frac {\text {ArcTan}\left (\frac {2 \sqrt [3]{a+b x^3}+\sqrt [3]{a}}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} a^{2/3}}+\frac {\log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^3}\right )}{2 a^{2/3}}-\frac {\log (x)}{2 a^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x*(a + b*x^3)^(2/3)),x]

[Out]

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

Rule 31

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

Rule 59

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(2/3)), x_Symbol] :> With[{q = Rt[(b*c - a*d)/b, 3]}, Simp[-L
og[RemoveContent[a + b*x, x]]/(2*b*q^2), x] + (-Dist[3/(2*b*q), Subst[Int[1/(q^2 + q*x + x^2), x], x, (c + d*x
)^(1/3)], x] - Dist[3/(2*b*q^2), Subst[Int[1/(q - x), x], x, (c + d*x)^(1/3)], x])] /; FreeQ[{a, b, c, d}, x]
&& PosQ[(b*c - a*d)/b]

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 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

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]

Rubi steps

\begin {align*} \int \frac {1}{x \left (a+b x^3\right )^{2/3}} \, dx &=\frac {1}{3} \text {Subst}\left (\int \frac {1}{x (a+b x)^{2/3}} \, dx,x,x^3\right )\\ &=-\frac {\log (x)}{2 a^{2/3}}-\frac {\text {Subst}\left (\int \frac {1}{\sqrt [3]{a}-x} \, dx,x,\sqrt [3]{a+b x^3}\right )}{2 a^{2/3}}-\frac {\text {Subst}\left (\int \frac {1}{a^{2/3}+\sqrt [3]{a} x+x^2} \, dx,x,\sqrt [3]{a+b x^3}\right )}{2 \sqrt [3]{a}}\\ &=-\frac {\log (x)}{2 a^{2/3}}+\frac {\log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^3}\right )}{2 a^{2/3}}+\frac {\text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+\frac {2 \sqrt [3]{a+b x^3}}{\sqrt [3]{a}}\right )}{a^{2/3}}\\ &=-\frac {\tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{a+b x^3}}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{\sqrt {3} a^{2/3}}-\frac {\log (x)}{2 a^{2/3}}+\frac {\log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^3}\right )}{2 a^{2/3}}\\ \end {align*}

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Mathematica [A]
time = 0.08, size = 101, normalized size = 1.20 \begin {gather*} -\frac {2 \sqrt {3} \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{a+b x^3}}{\sqrt [3]{a}}}{\sqrt {3}}\right )-2 \log \left (-\sqrt [3]{a}+\sqrt [3]{a+b x^3}\right )+\log \left (a^{2/3}+\sqrt [3]{a} \sqrt [3]{a+b x^3}+\left (a+b x^3\right )^{2/3}\right )}{6 a^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x*(a + b*x^3)^(2/3)),x]

[Out]

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

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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {1}{x \left (b \,x^{3}+a \right )^{\frac {2}{3}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x/(b*x^3+a)^(2/3),x)

[Out]

int(1/x/(b*x^3+a)^(2/3),x)

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Maxima [A]
time = 0.50, size = 86, normalized size = 1.02 \begin {gather*} -\frac {\sqrt {3} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} + a^{\frac {1}{3}}\right )}}{3 \, a^{\frac {1}{3}}}\right )}{3 \, a^{\frac {2}{3}}} - \frac {\log \left ({\left (b x^{3} + a\right )}^{\frac {2}{3}} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + a^{\frac {2}{3}}\right )}{6 \, a^{\frac {2}{3}}} + \frac {\log \left ({\left (b x^{3} + a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}}\right )}{3 \, a^{\frac {2}{3}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 126 vs. \(2 (61) = 122\).
time = 0.36, size = 126, normalized size = 1.50 \begin {gather*} -\frac {2 \, \sqrt {3} {\left (a^{2}\right )}^{\frac {1}{6}} a \arctan \left (\frac {{\left (a^{2}\right )}^{\frac {1}{6}} {\left (\sqrt {3} {\left (a^{2}\right )}^{\frac {1}{3}} a + 2 \, \sqrt {3} {\left (b x^{3} + a\right )}^{\frac {1}{3}} {\left (a^{2}\right )}^{\frac {2}{3}}\right )}}{3 \, a^{2}}\right ) + {\left (a^{2}\right )}^{\frac {2}{3}} \log \left ({\left (b x^{3} + a\right )}^{\frac {2}{3}} a + {\left (a^{2}\right )}^{\frac {1}{3}} a + {\left (b x^{3} + a\right )}^{\frac {1}{3}} {\left (a^{2}\right )}^{\frac {2}{3}}\right ) - 2 \, {\left (a^{2}\right )}^{\frac {2}{3}} \log \left ({\left (b x^{3} + a\right )}^{\frac {1}{3}} a - {\left (a^{2}\right )}^{\frac {2}{3}}\right )}{6 \, a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [C] Result contains complex when optimal does not.
time = 0.46, size = 39, normalized size = 0.46 \begin {gather*} - \frac {\Gamma \left (\frac {2}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {2}{3}, \frac {2}{3} \\ \frac {5}{3} \end {matrix}\middle | {\frac {a e^{i \pi }}{b x^{3}}} \right )}}{3 b^{\frac {2}{3}} x^{2} \Gamma \left (\frac {5}{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-gamma(2/3)*hyper((2/3, 2/3), (5/3,), a*exp_polar(I*pi)/(b*x**3))/(3*b**(2/3)*x**2*gamma(5/3))

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Giac [A]
time = 2.39, size = 87, normalized size = 1.04 \begin {gather*} -\frac {\sqrt {3} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} + a^{\frac {1}{3}}\right )}}{3 \, a^{\frac {1}{3}}}\right )}{3 \, a^{\frac {2}{3}}} - \frac {\log \left ({\left (b x^{3} + a\right )}^{\frac {2}{3}} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + a^{\frac {2}{3}}\right )}{6 \, a^{\frac {2}{3}}} + \frac {\log \left ({\left | {\left (b x^{3} + a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}} \right |}\right )}{3 \, a^{\frac {2}{3}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 1.23, size = 102, normalized size = 1.21 \begin {gather*} \frac {\ln \left (3\,{\left (b\,x^3+a\right )}^{1/3}-3\,a^{1/3}\right )}{3\,a^{2/3}}+\frac {\ln \left (\frac {3\,a^{1/3}\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{2}-3\,{\left (b\,x^3+a\right )}^{1/3}\right )\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{6\,a^{2/3}}-\frac {\ln \left (\frac {3\,a^{1/3}\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{2}+3\,{\left (b\,x^3+a\right )}^{1/3}\right )\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{6\,a^{2/3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x*(a + b*x^3)^(2/3)),x)

[Out]

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

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