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

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

[Out]

-(b*x^2+a)^(1/3)/a/x+1/3*(a^(1/3)-(b*x^2+a)^(1/3))*EllipticF((-(b*x^2+a)^(1/3)+a^(1/3)*(1+3^(1/2)))/(-(b*x^2+a
)^(1/3)+a^(1/3)*(1-3^(1/2))),2*I-I*3^(1/2))*((a^(2/3)+a^(1/3)*(b*x^2+a)^(1/3)+(b*x^2+a)^(2/3))/(-(b*x^2+a)^(1/
3)+a^(1/3)*(1-3^(1/2)))^2)^(1/2)*(1/2*6^(1/2)-1/2*2^(1/2))*3^(3/4)/a/x/(-a^(1/3)*(a^(1/3)-(b*x^2+a)^(1/3))/(-(
b*x^2+a)^(1/3)+a^(1/3)*(1-3^(1/2)))^2)^(1/2)

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Rubi [A]  time = 0.12, antiderivative size = 265, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, 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 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]

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 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]

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*}

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Mathematica [C]  time = 0.01, 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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fricas [F]  time = 0.79, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b x^{2} + a\right )}^{\frac {1}{3}}}{b x^{4} + a x^{2}}, x\right ) \]

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

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)

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maple [F]  time = 0.28, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (b \,x^{2}+a \right )^{\frac {2}{3}} x^{2}}\, dx \]

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

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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mupad [B]  time = 5.46, size = 40, normalized size = 0.15 \[ -\frac {3\,{\left (\frac {a}{b\,x^2}+1\right )}^{2/3}\,{{}}_2{\mathrm {F}}_1\left (\frac {2}{3},\frac {7}{6};\ \frac {13}{6};\ -\frac {a}{b\,x^2}\right )}{7\,x\,{\left (b\,x^2+a\right )}^{2/3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^2*(a + b*x^2)^(2/3)),x)

[Out]

-(3*(a/(b*x^2) + 1)^(2/3)*hypergeom([2/3, 7/6], 13/6, -a/(b*x^2)))/(7*x*(a + b*x^2)^(2/3))

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sympy [A]  time = 0.90, size = 27, normalized size = 0.10 \[ - \frac {{{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {2}{3} \\ \frac {1}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{a^{\frac {2}{3}} x} \]

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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