3.917 \(\int \frac{1}{x^4 (-2-3 x^2)^{3/4}} \, dx\)

Optimal. Leaf size=123 \[ -\frac{5 \sqrt{3} \sqrt{-\frac{x^2}{\left (\sqrt{-3 x^2-2}+\sqrt{2}\right )^2}} \left (\sqrt{-3 x^2-2}+\sqrt{2}\right ) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{-3 x^2-2}}{\sqrt [4]{2}}\right ),\frac{1}{2}\right )}{16 \sqrt [4]{2} x}-\frac{5 \sqrt [4]{-3 x^2-2}}{8 x}+\frac{\sqrt [4]{-3 x^2-2}}{6 x^3} \]

[Out]

(-2 - 3*x^2)^(1/4)/(6*x^3) - (5*(-2 - 3*x^2)^(1/4))/(8*x) - (5*Sqrt[3]*Sqrt[-(x^2/(Sqrt[2] + Sqrt[-2 - 3*x^2])
^2)]*(Sqrt[2] + Sqrt[-2 - 3*x^2])*EllipticF[2*ArcTan[(-2 - 3*x^2)^(1/4)/2^(1/4)], 1/2])/(16*2^(1/4)*x)

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Rubi [A]  time = 0.0485651, antiderivative size = 123, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {325, 234, 220} \[ -\frac{5 \sqrt [4]{-3 x^2-2}}{8 x}+\frac{\sqrt [4]{-3 x^2-2}}{6 x^3}-\frac{5 \sqrt{3} \sqrt{-\frac{x^2}{\left (\sqrt{-3 x^2-2}+\sqrt{2}\right )^2}} \left (\sqrt{-3 x^2-2}+\sqrt{2}\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{-3 x^2-2}}{\sqrt [4]{2}}\right )|\frac{1}{2}\right )}{16 \sqrt [4]{2} x} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^4*(-2 - 3*x^2)^(3/4)),x]

[Out]

(-2 - 3*x^2)^(1/4)/(6*x^3) - (5*(-2 - 3*x^2)^(1/4))/(8*x) - (5*Sqrt[3]*Sqrt[-(x^2/(Sqrt[2] + Sqrt[-2 - 3*x^2])
^2)]*(Sqrt[2] + Sqrt[-2 - 3*x^2])*EllipticF[2*ArcTan[(-2 - 3*x^2)^(1/4)/2^(1/4)], 1/2])/(16*2^(1/4)*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]

Rule 234

Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Dist[(2*Sqrt[-((b*x^2)/a)])/(b*x), Subst[Int[1/Sqrt[1 - x^4/a],
x], x, (a + b*x^2)^(1/4)], x] /; FreeQ[{a, b}, x] && NegQ[a]

Rule 220

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[((1 + q^2*x^2)*Sqrt[(a + b*x^4)/(a*(
1 + q^2*x^2)^2)]*EllipticF[2*ArcTan[q*x], 1/2])/(2*q*Sqrt[a + b*x^4]), x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rubi steps

\begin{align*} \int \frac{1}{x^4 \left (-2-3 x^2\right )^{3/4}} \, dx &=\frac{\sqrt [4]{-2-3 x^2}}{6 x^3}-\frac{5}{4} \int \frac{1}{x^2 \left (-2-3 x^2\right )^{3/4}} \, dx\\ &=\frac{\sqrt [4]{-2-3 x^2}}{6 x^3}-\frac{5 \sqrt [4]{-2-3 x^2}}{8 x}+\frac{15}{16} \int \frac{1}{\left (-2-3 x^2\right )^{3/4}} \, dx\\ &=\frac{\sqrt [4]{-2-3 x^2}}{6 x^3}-\frac{5 \sqrt [4]{-2-3 x^2}}{8 x}-\frac{\left (5 \sqrt{\frac{3}{2}} \sqrt{-x^2}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1+\frac{x^4}{2}}} \, dx,x,\sqrt [4]{-2-3 x^2}\right )}{8 x}\\ &=\frac{\sqrt [4]{-2-3 x^2}}{6 x^3}-\frac{5 \sqrt [4]{-2-3 x^2}}{8 x}-\frac{5 \sqrt{3} \sqrt{-\frac{x^2}{\left (\sqrt{2}+\sqrt{-2-3 x^2}\right )^2}} \left (\sqrt{2}+\sqrt{-2-3 x^2}\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{-2-3 x^2}}{\sqrt [4]{2}}\right )|\frac{1}{2}\right )}{16 \sqrt [4]{2} x}\\ \end{align*}

Mathematica [C]  time = 0.0079847, size = 48, normalized size = 0.39 \[ -\frac{\left (\frac{3 x^2}{2}+1\right )^{3/4} \, _2F_1\left (-\frac{3}{2},\frac{3}{4};-\frac{1}{2};-\frac{3 x^2}{2}\right )}{3 x^3 \left (-3 x^2-2\right )^{3/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^4*(-2 - 3*x^2)^(3/4)),x]

[Out]

-((1 + (3*x^2)/2)^(3/4)*Hypergeometric2F1[-3/2, 3/4, -1/2, (-3*x^2)/2])/(3*x^3*(-2 - 3*x^2)^(3/4))

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Maple [F]  time = 0.013, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{x}^{4}} \left ( -3\,{x}^{2}-2 \right ) ^{-{\frac{3}{4}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^4/(-3*x^2-2)^(3/4),x)

[Out]

int(1/x^4/(-3*x^2-2)^(3/4),x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/((-3*x^2 - 2)^(3/4)*x^4), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{24 \, x^{3}{\rm integral}\left (-\frac{15 \,{\left (-3 \, x^{2} - 2\right )}^{\frac{1}{4}}}{16 \,{\left (3 \, x^{2} + 2\right )}}, x\right ) -{\left (15 \, x^{2} - 4\right )}{\left (-3 \, x^{2} - 2\right )}^{\frac{1}{4}}}{24 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(24*x^3*integral(-15/16*(-3*x^2 - 2)^(1/4)/(3*x^2 + 2), x) - (15*x^2 - 4)*(-3*x^2 - 2)^(1/4))/x^3

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Sympy [C]  time = 0.955024, size = 37, normalized size = 0.3 \begin{align*} \frac{\sqrt [4]{2} e^{\frac{i \pi }{4}}{{}_{2}F_{1}\left (\begin{matrix} - \frac{3}{2}, \frac{3}{4} \\ - \frac{1}{2} \end{matrix}\middle |{\frac{3 x^{2} e^{i \pi }}{2}} \right )}}{6 x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**4/(-3*x**2-2)**(3/4),x)

[Out]

2**(1/4)*exp(I*pi/4)*hyper((-3/2, 3/4), (-1/2,), 3*x**2*exp_polar(I*pi)/2)/(6*x**3)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/((-3*x^2 - 2)^(3/4)*x^4), x)