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

Optimal. Leaf size=121 \[ -\frac{8\ 2^{3/4} \sqrt{-\frac{x^2}{\left (\sqrt{-3 x^2-2}+\sqrt{2}\right )^2}} \left (\sqrt{-3 x^2-2}+\sqrt{2}\right ) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{-3 x^2-2}}{\sqrt [4]{2}}\right ),\frac{1}{2}\right )}{63 \sqrt{3} x}-\frac{2}{21} \sqrt [4]{-3 x^2-2} x^3+\frac{8}{63} \sqrt [4]{-3 x^2-2} x \]

[Out]

(8*x*(-2 - 3*x^2)^(1/4))/63 - (2*x^3*(-2 - 3*x^2)^(1/4))/21 - (8*2^(3/4)*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])/(63*Sqrt[3]*x)

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Rubi [A]  time = 0.0461336, antiderivative size = 121, 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 = {321, 234, 220} \[ -\frac{2}{21} \sqrt [4]{-3 x^2-2} x^3+\frac{8}{63} \sqrt [4]{-3 x^2-2} x-\frac{8\ 2^{3/4} \sqrt{-\frac{x^2}{\left (\sqrt{-3 x^2-2}+\sqrt{2}\right )^2}} \left (\sqrt{-3 x^2-2}+\sqrt{2}\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{-3 x^2-2}}{\sqrt [4]{2}}\right )|\frac{1}{2}\right )}{63 \sqrt{3} x} \]

Antiderivative was successfully verified.

[In]

Int[x^4/(-2 - 3*x^2)^(3/4),x]

[Out]

(8*x*(-2 - 3*x^2)^(1/4))/63 - (2*x^3*(-2 - 3*x^2)^(1/4))/21 - (8*2^(3/4)*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])/(63*Sqrt[3]*x)

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && 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{x^4}{\left (-2-3 x^2\right )^{3/4}} \, dx &=-\frac{2}{21} x^3 \sqrt [4]{-2-3 x^2}-\frac{4}{7} \int \frac{x^2}{\left (-2-3 x^2\right )^{3/4}} \, dx\\ &=\frac{8}{63} x \sqrt [4]{-2-3 x^2}-\frac{2}{21} x^3 \sqrt [4]{-2-3 x^2}+\frac{16}{63} \int \frac{1}{\left (-2-3 x^2\right )^{3/4}} \, dx\\ &=\frac{8}{63} x \sqrt [4]{-2-3 x^2}-\frac{2}{21} x^3 \sqrt [4]{-2-3 x^2}-\frac{\left (16 \sqrt{\frac{2}{3}} \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 )}{63 x}\\ &=\frac{8}{63} x \sqrt [4]{-2-3 x^2}-\frac{2}{21} x^3 \sqrt [4]{-2-3 x^2}-\frac{8\ 2^{3/4} \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 )}{63 \sqrt{3} x}\\ \end{align*}

Mathematica [C]  time = 0.0153001, size = 63, normalized size = 0.52 \[ \frac{2 x \left (4 \sqrt [4]{2} \left (3 x^2+2\right )^{3/4} \, _2F_1\left (\frac{1}{2},\frac{3}{4};\frac{3}{2};-\frac{3 x^2}{2}\right )+9 x^4-6 x^2-8\right )}{63 \left (-3 x^2-2\right )^{3/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^4/(-2 - 3*x^2)^(3/4),x]

[Out]

(2*x*(-8 - 6*x^2 + 9*x^4 + 4*2^(1/4)*(2 + 3*x^2)^(3/4)*Hypergeometric2F1[1/2, 3/4, 3/2, (-3*x^2)/2]))/(63*(-2
- 3*x^2)^(3/4))

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Maple [F]  time = 0.011, size = 0, normalized size = 0. \begin{align*} \int{{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(x^4/(-3*x^2-2)^(3/4),x)

[Out]

int(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{x^{4}}{{\left (-3 \, x^{2} - 2\right )}^{\frac{3}{4}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^4/(-3*x^2 - 2)^(3/4), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{2}{63} \,{\left (3 \, x^{3} - 4 \, x\right )}{\left (-3 \, x^{2} - 2\right )}^{\frac{1}{4}} +{\rm integral}\left (-\frac{16 \,{\left (-3 \, x^{2} - 2\right )}^{\frac{1}{4}}}{63 \,{\left (3 \, x^{2} + 2\right )}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/63*(3*x^3 - 4*x)*(-3*x^2 - 2)^(1/4) + integral(-16/63*(-3*x^2 - 2)^(1/4)/(3*x^2 + 2), x)

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Sympy [C]  time = 0.685824, size = 36, normalized size = 0.3 \begin{align*} \frac{\sqrt [4]{2} x^{5} e^{- \frac{3 i \pi }{4}}{{}_{2}F_{1}\left (\begin{matrix} \frac{3}{4}, \frac{5}{2} \\ \frac{7}{2} \end{matrix}\middle |{\frac{3 x^{2} e^{i \pi }}{2}} \right )}}{10} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2**(1/4)*x**5*exp(-3*I*pi/4)*hyper((3/4, 5/2), (7/2,), 3*x**2*exp_polar(I*pi)/2)/10

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^4/(-3*x^2 - 2)^(3/4), x)