3.899 \(\int \frac {x^4}{\sqrt [4]{-2-3 x^2}} \, dx\)

Optimal. Leaf size=242 \[ -\frac {16 \sqrt [4]{2} \sqrt {-\frac {x^2}{\left (\sqrt {-3 x^2-2}+\sqrt {2}\right )^2}} \left (\sqrt {-3 x^2-2}+\sqrt {2}\right ) \operatorname {EllipticF}\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{-3 x^2-2}}{\sqrt [4]{2}}\right ),\frac {1}{2}\right )}{135 \sqrt {3} x}+\frac {8}{135} \left (-3 x^2-2\right )^{3/4} x+\frac {32 \sqrt [4]{-3 x^2-2} x}{135 \left (\sqrt {-3 x^2-2}+\sqrt {2}\right )}+\frac {32 \sqrt [4]{2} \sqrt {-\frac {x^2}{\left (\sqrt {-3 x^2-2}+\sqrt {2}\right )^2}} \left (\sqrt {-3 x^2-2}+\sqrt {2}\right ) E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{-3 x^2-2}}{\sqrt [4]{2}}\right )|\frac {1}{2}\right )}{135 \sqrt {3} x}-\frac {2}{27} \left (-3 x^2-2\right )^{3/4} x^3 \]

[Out]

8/135*x*(-3*x^2-2)^(3/4)-2/27*x^3*(-3*x^2-2)^(3/4)+32/135*x*(-3*x^2-2)^(1/4)/(2^(1/2)+(-3*x^2-2)^(1/2))+32/405
*2^(1/4)*(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)))*Ell
ipticE(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)-16/405*2^(1/4)*(cos(2*arctan(1/2*(-3*x^2-2)^(1/4)*2^(3/4)))^2)^(1/2)/cos(2*ar
ctan(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 = 0.11, antiderivative size = 242, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {321, 230, 305, 220, 1196} \[ -\frac {2}{27} \left (-3 x^2-2\right )^{3/4} x^3+\frac {8}{135} \left (-3 x^2-2\right )^{3/4} x+\frac {32 \sqrt [4]{-3 x^2-2} x}{135 \left (\sqrt {-3 x^2-2}+\sqrt {2}\right )}-\frac {16 \sqrt [4]{2} \sqrt {-\frac {x^2}{\left (\sqrt {-3 x^2-2}+\sqrt {2}\right )^2}} \left (\sqrt {-3 x^2-2}+\sqrt {2}\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{-3 x^2-2}}{\sqrt [4]{2}}\right )|\frac {1}{2}\right )}{135 \sqrt {3} x}+\frac {32 \sqrt [4]{2} \sqrt {-\frac {x^2}{\left (\sqrt {-3 x^2-2}+\sqrt {2}\right )^2}} \left (\sqrt {-3 x^2-2}+\sqrt {2}\right ) E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{-3 x^2-2}}{\sqrt [4]{2}}\right )|\frac {1}{2}\right )}{135 \sqrt {3} x} \]

Antiderivative was successfully verified.

[In]

Int[x^4/(-2 - 3*x^2)^(1/4),x]

[Out]

(8*x*(-2 - 3*x^2)^(3/4))/135 - (2*x^3*(-2 - 3*x^2)^(3/4))/27 + (32*x*(-2 - 3*x^2)^(1/4))/(135*(Sqrt[2] + Sqrt[
-2 - 3*x^2])) + (32*2^(1/4)*Sqrt[-(x^2/(Sqrt[2] + Sqrt[-2 - 3*x^2])^2)]*(Sqrt[2] + Sqrt[-2 - 3*x^2])*EllipticE
[2*ArcTan[(-2 - 3*x^2)^(1/4)/2^(1/4)], 1/2])/(135*Sqrt[3]*x) - (16*2^(1/4)*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])/(135*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 230

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

Rule 305

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

Rule 1196

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, -Simp[(d*x*Sqrt[a + c
*x^4])/(a*(1 + q^2*x^2)), x] + Simp[(d*(1 + q^2*x^2)*Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]*EllipticE[2*ArcTan[
q*x], 1/2])/(q*Sqrt[a + c*x^4]), x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

Rubi steps

\begin {align*} \int \frac {x^4}{\sqrt [4]{-2-3 x^2}} \, dx &=-\frac {2}{27} x^3 \left (-2-3 x^2\right )^{3/4}-\frac {4}{9} \int \frac {x^2}{\sqrt [4]{-2-3 x^2}} \, dx\\ &=\frac {8}{135} x \left (-2-3 x^2\right )^{3/4}-\frac {2}{27} x^3 \left (-2-3 x^2\right )^{3/4}+\frac {16}{135} \int \frac {1}{\sqrt [4]{-2-3 x^2}} \, dx\\ &=\frac {8}{135} x \left (-2-3 x^2\right )^{3/4}-\frac {2}{27} x^3 \left (-2-3 x^2\right )^{3/4}-\frac {\left (16 \sqrt {\frac {2}{3}} \sqrt {-x^2}\right ) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {1+\frac {x^4}{2}}} \, dx,x,\sqrt [4]{-2-3 x^2}\right )}{135 x}\\ &=\frac {8}{135} x \left (-2-3 x^2\right )^{3/4}-\frac {2}{27} x^3 \left (-2-3 x^2\right )^{3/4}-\frac {\left (32 \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 )}{135 \sqrt {3} x}+\frac {\left (32 \sqrt {-x^2}\right ) \operatorname {Subst}\left (\int \frac {1-\frac {x^2}{\sqrt {2}}}{\sqrt {1+\frac {x^4}{2}}} \, dx,x,\sqrt [4]{-2-3 x^2}\right )}{135 \sqrt {3} x}\\ &=\frac {8}{135} x \left (-2-3 x^2\right )^{3/4}-\frac {2}{27} x^3 \left (-2-3 x^2\right )^{3/4}+\frac {32 x \sqrt [4]{-2-3 x^2}}{135 \left (\sqrt {2}+\sqrt {-2-3 x^2}\right )}+\frac {32 \sqrt [4]{2} \sqrt {-\frac {x^2}{\left (\sqrt {2}+\sqrt {-2-3 x^2}\right )^2}} \left (\sqrt {2}+\sqrt {-2-3 x^2}\right ) E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{-2-3 x^2}}{\sqrt [4]{2}}\right )|\frac {1}{2}\right )}{135 \sqrt {3} x}-\frac {16 \sqrt [4]{2} \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 )}{135 \sqrt {3} x}\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 63, normalized size = 0.26 \[ \frac {2 x \left (4\ 2^{3/4} \sqrt [4]{3 x^2+2} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {3}{2};-\frac {3 x^2}{2}\right )+15 x^4-2 x^2-8\right )}{135 \sqrt [4]{-3 x^2-2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^4/(-2 - 3*x^2)^(1/4),x]

[Out]

(2*x*(-8 - 2*x^2 + 15*x^4 + 4*2^(3/4)*(2 + 3*x^2)^(1/4)*Hypergeometric2F1[1/4, 1/2, 3/2, (-3*x^2)/2]))/(135*(-
2 - 3*x^2)^(1/4))

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fricas [F]  time = 0.67, size = 0, normalized size = 0.00 \[ \frac {405 \, x {\rm integral}\left (-\frac {64 \, {\left (-3 \, x^{2} - 2\right )}^{\frac {3}{4}}}{405 \, {\left (3 \, x^{4} + 2 \, x^{2}\right )}}, x\right ) - 2 \, {\left (15 \, x^{4} - 12 \, x^{2} + 16\right )} {\left (-3 \, x^{2} - 2\right )}^{\frac {3}{4}}}{405 \, x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/405*(405*x*integral(-64/405*(-3*x^2 - 2)^(3/4)/(3*x^4 + 2*x^2), x) - 2*(15*x^4 - 12*x^2 + 16)*(-3*x^2 - 2)^(
3/4))/x

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^4/(-3*x^2 - 2)^(1/4), x)

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maple [C]  time = 0.27, size = 48, normalized size = 0.20 \[ -\frac {8 \left (-1\right )^{\frac {3}{4}} 2^{\frac {3}{4}} x \hypergeom \left (\left [\frac {1}{4}, \frac {1}{2}\right ], \left [\frac {3}{2}\right ], -\frac {3 x^{2}}{2}\right )}{135}+\frac {2 \left (5 x^{2}-4\right ) \left (3 x^{2}+2\right ) x}{135 \left (-3 x^{2}-2\right )^{\frac {1}{4}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/135*x*(5*x^2-4)*(3*x^2+2)/(-3*x^2-2)^(1/4)-8/135*(-1)^(3/4)*2^(3/4)*x*hypergeom([1/4,1/2],[3/2],-3/2*x^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^4/(-3*x^2 - 2)^(1/4), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/(- 3*x^2 - 2)^(1/4),x)

[Out]

int(x^4/(- 3*x^2 - 2)^(1/4), x)

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sympy [C]  time = 0.77, size = 34, normalized size = 0.14 \[ \frac {2^{\frac {3}{4}} x^{5} e^{- \frac {i \pi }{4}} {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {5}{2} \\ \frac {7}{2} \end {matrix}\middle | {\frac {3 x^{2} e^{i \pi }}{2}} \right )}}{10} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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