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

Optimal. Leaf size=265 \[ \frac{\sqrt{2-\sqrt{3}} \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right ) \sqrt{\frac{a^{2/3}+\sqrt [3]{a} \sqrt [3]{a+b x^2}+\left (a+b x^2\right )^{2/3}}{\left (\left (1-\sqrt{3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )^2}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\left (1+\sqrt{3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}}{\left (1-\sqrt{3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}}\right ),4 \sqrt{3}-7\right )}{\sqrt [4]{3} a x \sqrt{-\frac{\sqrt [3]{a} \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{\left (\left (1-\sqrt{3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )^2}}}-\frac{\sqrt [3]{a+b x^2}}{a x} \]

[Out]

-((a + b*x^2)^(1/3)/(a*x)) + (Sqrt[2 - Sqrt[3]]*(a^(1/3) - (a + b*x^2)^(1/3))*Sqrt[(a^(2/3) + a^(1/3)*(a + b*x
^2)^(1/3) + (a + b*x^2)^(2/3))/((1 - Sqrt[3])*a^(1/3) - (a + b*x^2)^(1/3))^2]*EllipticF[ArcSin[((1 + Sqrt[3])*
a^(1/3) - (a + b*x^2)^(1/3))/((1 - Sqrt[3])*a^(1/3) - (a + b*x^2)^(1/3))], -7 + 4*Sqrt[3]])/(3^(1/4)*a*x*Sqrt[
-((a^(1/3)*(a^(1/3) - (a + b*x^2)^(1/3)))/((1 - Sqrt[3])*a^(1/3) - (a + b*x^2)^(1/3))^2)])

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Rubi [A]  time = 0.120547, antiderivative size = 265, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {325, 236, 219} \[ \frac{\sqrt{2-\sqrt{3}} \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right ) \sqrt{\frac{a^{2/3}+\sqrt [3]{a} \sqrt [3]{a+b x^2}+\left (a+b x^2\right )^{2/3}}{\left (\left (1-\sqrt{3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )^2}} F\left (\sin ^{-1}\left (\frac{\left (1+\sqrt{3}\right ) \sqrt [3]{a}-\sqrt [3]{b x^2+a}}{\left (1-\sqrt{3}\right ) \sqrt [3]{a}-\sqrt [3]{b x^2+a}}\right )|-7+4 \sqrt{3}\right )}{\sqrt [4]{3} a x \sqrt{-\frac{\sqrt [3]{a} \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{\left (\left (1-\sqrt{3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )^2}}}-\frac{\sqrt [3]{a+b x^2}}{a x} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^2*(a + b*x^2)^(2/3)),x]

[Out]

-((a + b*x^2)^(1/3)/(a*x)) + (Sqrt[2 - Sqrt[3]]*(a^(1/3) - (a + b*x^2)^(1/3))*Sqrt[(a^(2/3) + a^(1/3)*(a + b*x
^2)^(1/3) + (a + b*x^2)^(2/3))/((1 - Sqrt[3])*a^(1/3) - (a + b*x^2)^(1/3))^2]*EllipticF[ArcSin[((1 + Sqrt[3])*
a^(1/3) - (a + b*x^2)^(1/3))/((1 - Sqrt[3])*a^(1/3) - (a + b*x^2)^(1/3))], -7 + 4*Sqrt[3]])/(3^(1/4)*a*x*Sqrt[
-((a^(1/3)*(a^(1/3) - (a + b*x^2)^(1/3)))/((1 - Sqrt[3])*a^(1/3) - (a + b*x^2)^(1/3))^2)])

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 236

Int[((a_) + (b_.)*(x_)^2)^(-2/3), x_Symbol] :> Dist[(3*Sqrt[b*x^2])/(2*b*x), Subst[Int[1/Sqrt[-a + x^3], x], x
, (a + b*x^2)^(1/3)], x] /; FreeQ[{a, b}, x]

Rule 219

Int[1/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Simp[(2*Sqr
t[2 - Sqrt[3]]*(s + r*x)*Sqrt[(s^2 - r*s*x + r^2*x^2)/((1 - Sqrt[3])*s + r*x)^2]*EllipticF[ArcSin[((1 + Sqrt[3
])*s + r*x)/((1 - Sqrt[3])*s + r*x)], -7 + 4*Sqrt[3]])/(3^(1/4)*r*Sqrt[a + b*x^3]*Sqrt[-((s*(s + r*x))/((1 - S
qrt[3])*s + r*x)^2)]), x]] /; FreeQ[{a, b}, x] && NegQ[a]

Rubi steps

\begin{align*} \int \frac{1}{x^2 \left (a+b x^2\right )^{2/3}} \, dx &=-\frac{\sqrt [3]{a+b x^2}}{a x}-\frac{b \int \frac{1}{\left (a+b x^2\right )^{2/3}} \, dx}{3 a}\\ &=-\frac{\sqrt [3]{a+b x^2}}{a x}-\frac{\sqrt{b x^2} \operatorname{Subst}\left (\int \frac{1}{\sqrt{-a+x^3}} \, dx,x,\sqrt [3]{a+b x^2}\right )}{2 a x}\\ &=-\frac{\sqrt [3]{a+b x^2}}{a x}+\frac{\sqrt{2-\sqrt{3}} \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right ) \sqrt{\frac{a^{2/3}+\sqrt [3]{a} \sqrt [3]{a+b x^2}+\left (a+b x^2\right )^{2/3}}{\left (\left (1-\sqrt{3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )^2}} F\left (\sin ^{-1}\left (\frac{\left (1+\sqrt{3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}}{\left (1-\sqrt{3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}}\right )|-7+4 \sqrt{3}\right )}{\sqrt [4]{3} a x \sqrt{-\frac{\sqrt [3]{a} \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{\left (\left (1-\sqrt{3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )^2}}}\\ \end{align*}

Mathematica [C]  time = 0.0086949, size = 49, normalized size = 0.18 \[ -\frac{\left (\frac{b x^2}{a}+1\right )^{2/3} \, _2F_1\left (-\frac{1}{2},\frac{2}{3};\frac{1}{2};-\frac{b x^2}{a}\right )}{x \left (a+b x^2\right )^{2/3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((1 + (b*x^2)/a)^(2/3)*Hypergeometric2F1[-1/2, 2/3, 1/2, -((b*x^2)/a)])/(x*(a + b*x^2)^(2/3)))

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Maple [F]  time = 0.025, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{x}^{2}} \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^2/(b*x^2+a)^(2/3),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (b x^{2} + a\right )}^{\frac{1}{3}}}{b x^{4} + a x^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*x^2 + a)^(1/3)/(b*x^4 + a*x^2), x)

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Sympy [A]  time = 0.772984, size = 27, normalized size = 0.1 \begin{align*} - \frac{{{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, \frac{2}{3} \\ \frac{1}{2} \end{matrix}\middle |{\frac{b x^{2} e^{i \pi }}{a}} \right )}}{a^{\frac{2}{3}} x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-hyper((-1/2, 2/3), (1/2,), b*x**2*exp_polar(I*pi)/a)/(a**(2/3)*x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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