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

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

[Out]

1/6*(-3*x^2-2)^(1/4)/x^3-5/8*(-3*x^2-2)^(1/4)/x-5/32*2^(3/4)*(cos(2*arctan(1/2*(-3*x^2-2)^(1/4)*2^(3/4)))^2)^(
1/2)/cos(2*arctan(1/2*(-3*x^2-2)^(1/4)*2^(3/4)))*EllipticF(sin(2*arctan(1/2*(-3*x^2-2)^(1/4)*2^(3/4))),1/2*2^(
1/2))*(2^(1/2)+(-3*x^2-2)^(1/2))*(-x^2/(2^(1/2)+(-3*x^2-2)^(1/2))^2)^(1/2)/x*3^(1/2)

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Rubi [A]  time = 0.05, antiderivative size = 123, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, 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 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]

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 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^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*}

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Mathematica [C]  time = 0.01, 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*((1 + (3*x^2)/2)^(3/4)*Hypergeometric2F1[-3/2, 3/4, -1/2, (-3*x^2)/2])/(x^3*(-2 - 3*x^2)^(3/4))

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fricas [F]  time = 0.56, size = 0, normalized size = 0.00 \[ \frac {24 \, x^{3} {\rm integral}\left (-\frac {15 \, {\left (-3 \, x^{2} - 2\right )}^{\frac {1}{4}}}{16 \, {\left (3 \, x^{2} + 2\right )}}, x\right ) - {\left (15 \, x^{2} - 4\right )} {\left (-3 \, x^{2} - 2\right )}^{\frac {1}{4}}}{24 \, x^{3}} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (-3 \, x^{2} - 2\right )}^{\frac {3}{4}} x^{4}}\,{d x} \]

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)

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maple [F]  time = 0.04, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (-3 x^{2}-2\right )^{\frac {3}{4}} x^{4}}\, dx \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (-3 \, x^{2} - 2\right )}^{\frac {3}{4}} x^{4}}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{x^4\,{\left (-3\,x^2-2\right )}^{3/4}} \,d x \]

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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sympy [C]  time = 0.94, size = 37, normalized size = 0.30 \[ \frac {\sqrt [4]{2} e^{\frac {i \pi }{4}} {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{2}, \frac {3}{4} \\ - \frac {1}{2} \end {matrix}\middle | {\frac {3 x^{2} e^{i \pi }}{2}} \right )}}{6 x^{3}} \]

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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