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

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

[Out]

8/63*x*(-3*x^2-2)^(1/4)-2/21*x^3*(-3*x^2-2)^(1/4)-8/189*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 = 121, 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 = {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 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 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]

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*}

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Mathematica [C]  time = 0.02, 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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fricas [F]  time = 0.59, size = 0, normalized size = 0.00 \[ -\frac {2}{63} \, {\left (3 \, x^{3} - 4 \, x\right )} {\left (-3 \, x^{2} - 2\right )}^{\frac {1}{4}} + {\rm integral}\left (-\frac {16 \, {\left (-3 \, x^{2} - 2\right )}^{\frac {1}{4}}}{63 \, {\left (3 \, x^{2} + 2\right )}}, x\right ) \]

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

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)

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maple [F]  time = 0.03, size = 0, normalized size = 0.00 \[ \int \frac {x^{4}}{\left (-3 x^{2}-2\right )^{\frac {3}{4}}}\, dx \]

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

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

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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sympy [C]  time = 0.76, size = 36, normalized size = 0.30 \[ \frac {\sqrt [4]{2} x^{5} e^{- \frac {3 i \pi }{4}} {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {5}{2} \\ \frac {7}{2} \end {matrix}\middle | {\frac {3 x^{2} e^{i \pi }}{2}} \right )}}{10} \]

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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