3.661 \(\int \frac{1}{x^2 \sqrt [3]{a^2+2 a b x^2+b^2 x^4}} \, dx\)

Optimal. Leaf size=289 \[ \frac{\sqrt{2-\sqrt{3}} \left (\frac{b x^2}{a}+1\right )^{2/3} \left (1-\sqrt [3]{\frac{b x^2}{a}+1}\right ) \sqrt{\frac{\left (\frac{b x^2}{a}+1\right )^{2/3}+\sqrt [3]{\frac{b x^2}{a}+1}+1}{\left (-\sqrt [3]{\frac{b x^2}{a}+1}-\sqrt{3}+1\right )^2}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{-\sqrt [3]{\frac{b x^2}{a}+1}+\sqrt{3}+1}{-\sqrt [3]{\frac{b x^2}{a}+1}-\sqrt{3}+1}\right ),4 \sqrt{3}-7\right )}{\sqrt [4]{3} x \sqrt [3]{a^2+2 a b x^2+b^2 x^4} \sqrt{-\frac{1-\sqrt [3]{\frac{b x^2}{a}+1}}{\left (-\sqrt [3]{\frac{b x^2}{a}+1}-\sqrt{3}+1\right )^2}}}-\frac{a+b x^2}{a x \sqrt [3]{a^2+2 a b x^2+b^2 x^4}} \]

[Out]

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

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Rubi [A]  time = 0.157868, antiderivative size = 289, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {1113, 325, 236, 219} \[ \frac{\sqrt{2-\sqrt{3}} \left (\frac{b x^2}{a}+1\right )^{2/3} \left (1-\sqrt [3]{\frac{b x^2}{a}+1}\right ) \sqrt{\frac{\left (\frac{b x^2}{a}+1\right )^{2/3}+\sqrt [3]{\frac{b x^2}{a}+1}+1}{\left (-\sqrt [3]{\frac{b x^2}{a}+1}-\sqrt{3}+1\right )^2}} F\left (\sin ^{-1}\left (\frac{-\sqrt [3]{\frac{b x^2}{a}+1}+\sqrt{3}+1}{-\sqrt [3]{\frac{b x^2}{a}+1}-\sqrt{3}+1}\right )|-7+4 \sqrt{3}\right )}{\sqrt [4]{3} x \sqrt [3]{a^2+2 a b x^2+b^2 x^4} \sqrt{-\frac{1-\sqrt [3]{\frac{b x^2}{a}+1}}{\left (-\sqrt [3]{\frac{b x^2}{a}+1}-\sqrt{3}+1\right )^2}}}-\frac{a+b x^2}{a x \sqrt [3]{a^2+2 a b x^2+b^2 x^4}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^2*(a^2 + 2*a*b*x^2 + b^2*x^4)^(1/3)),x]

[Out]

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

Rule 1113

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^2 +
c*x^4)^FracPart[p])/(1 + (2*c*x^2)/b)^(2*FracPart[p]), Int[(d*x)^m*(1 + (2*c*x^2)/b)^(2*p), x], x] /; FreeQ[{
a, b, c, d, m, p}, x] && EqQ[b^2 - 4*a*c, 0] &&  !IntegerQ[2*p]

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

Mathematica [C]  time = 0.0100523, size = 51, normalized size = 0.18 \[ -\frac{\left (\frac{b x^2}{a}+1\right )^{2/3} \, _2F_1\left (-\frac{1}{2},\frac{2}{3};\frac{1}{2};-\frac{b x^2}{a}\right )}{x \sqrt [3]{\left (a+b x^2\right )^2}} \]

Antiderivative was successfully verified.

[In]

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

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Maple [F]  time = 0.178, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{x}^{2}}{\frac{1}{\sqrt [3]{{b}^{2}{x}^{4}+2\,ab{x}^{2}+{a}^{2}}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(b^2*x^4+2*a*b*x^2+a^2)^(1/3),x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{\frac{2}{3}}}{b^{2} x^{6} + 2 \, a b x^{4} + a^{2} x^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(b^2*x^4+2*a*b*x^2+a^2)^(1/3),x, algorithm="fricas")

[Out]

integral((b^2*x^4 + 2*a*b*x^2 + a^2)^(2/3)/(b^2*x^6 + 2*a*b*x^4 + a^2*x^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(b^2*x^4+2*a*b*x^2+a^2)^(1/3),x, algorithm="giac")

[Out]

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