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

Optimal. Leaf size=86 \[ \frac{3 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{4 a^{2/3}}-\frac{\sqrt{3} \tan ^{-1}\left (\frac{2 \sqrt [3]{a+b x^2}+\sqrt [3]{a}}{\sqrt{3} \sqrt [3]{a}}\right )}{2 a^{2/3}}-\frac{\log (x)}{2 a^{2/3}} \]

[Out]

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

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Rubi [A]  time = 0.0518777, antiderivative size = 86, normalized size of antiderivative = 1., 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 = {266, 57, 617, 204, 31} \[ \frac{3 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{4 a^{2/3}}-\frac{\sqrt{3} \tan ^{-1}\left (\frac{2 \sqrt [3]{a+b x^2}+\sqrt [3]{a}}{\sqrt{3} \sqrt [3]{a}}\right )}{2 a^{2/3}}-\frac{\log (x)}{2 a^{2/3}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x*(a + b*x^2)^(2/3)),x]

[Out]

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

Rule 266

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 57

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 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 31

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

Rubi steps

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

Mathematica [A]  time = 0.0310799, size = 101, normalized size = 1.17 \[ -\frac{\log \left (a^{2/3}+\sqrt [3]{a} \sqrt [3]{a+b x^2}+\left (a+b x^2\right )^{2/3}\right )-2 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )+2 \sqrt{3} \tan ^{-1}\left (\frac{\frac{2 \sqrt [3]{a+b x^2}}{\sqrt [3]{a}}+1}{\sqrt{3}}\right )}{4 a^{2/3}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x*(a + b*x^2)^(2/3)),x]

[Out]

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

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Maple [F]  time = 0.023, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{x} \left ( b{x}^{2}+a \right ) ^{-{\frac{2}{3}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.78634, size = 356, normalized size = 4.14 \begin{align*} -\frac{2 \, \sqrt{3}{\left (a^{2}\right )}^{\frac{1}{6}} a \arctan \left (\frac{\sqrt{3}{\left (a^{2}\right )}^{\frac{1}{6}}{\left ({\left (a^{2}\right )}^{\frac{1}{3}} a + 2 \,{\left (b x^{2} + 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^{2} + a\right )}^{\frac{2}{3}} a +{\left (a^{2}\right )}^{\frac{1}{3}} a +{\left (b x^{2} + 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^{2} + a\right )}^{\frac{1}{3}} a -{\left (a^{2}\right )}^{\frac{2}{3}}\right )}{4 \, a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [C]  time = 1.06003, size = 41, normalized size = 0.48 \begin{align*} - \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^{2}}} \right )}}{2 b^{\frac{2}{3}} x^{\frac{4}{3}} \Gamma \left (\frac{5}{3}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-gamma(2/3)*hyper((2/3, 2/3), (5/3,), a*exp_polar(I*pi)/(b*x**2))/(2*b**(2/3)*x**(4/3)*gamma(5/3))

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Giac [A]  time = 4.04531, size = 117, normalized size = 1.36 \begin{align*} -\frac{\sqrt{3} \arctan \left (\frac{\sqrt{3}{\left (2 \,{\left (b x^{2} + a\right )}^{\frac{1}{3}} + a^{\frac{1}{3}}\right )}}{3 \, a^{\frac{1}{3}}}\right )}{2 \, a^{\frac{2}{3}}} - \frac{\log \left ({\left (b x^{2} + a\right )}^{\frac{2}{3}} +{\left (b x^{2} + a\right )}^{\frac{1}{3}} a^{\frac{1}{3}} + a^{\frac{2}{3}}\right )}{4 \, a^{\frac{2}{3}}} + \frac{\log \left ({\left |{\left (b x^{2} + a\right )}^{\frac{1}{3}} - a^{\frac{1}{3}} \right |}\right )}{2 \, a^{\frac{2}{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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