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

Optimal. Leaf size=120 \[ \frac {8\ 2^{3/4} \sqrt {\frac {x^2}{\left (\sqrt {3 x^2-2}+\sqrt {2}\right )^2}} \left (\sqrt {3 x^2-2}+\sqrt {2}\right ) \operatorname {EllipticF}\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{3 x^2-2}}{\sqrt [4]{2}}\right ),\frac {1}{2}\right )}{63 \sqrt {3} x}+\frac {8}{63} \sqrt [4]{3 x^2-2} x+\frac {2}{21} \sqrt [4]{3 x^2-2} x^3 \]

[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 = 120, 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 \[ \frac {2 x \left (4 \sqrt [4]{2} \left (2-3 x^2\right )^{3/4} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {3}{2};\frac {3 x^2}{2}\right )+9 x^4+6 x^2-8\right )}{63 \left (3 x^2-2\right )^{3/4}} \]

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 = 0.62, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {x^{4}}{{\left (3 \, x^{2} - 2\right )}^{\frac {3}{4}}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x^4/(3*x^2 - 2)^(3/4), x)

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

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 [C]  time = 0.32, size = 60, normalized size = 0.50 \[ \frac {8 \,2^{\frac {1}{4}} \left (-\mathrm {signum}\left (\frac {3 x^{2}}{2}-1\right )\right )^{\frac {3}{4}} x \hypergeom \left (\left [\frac {1}{2}, \frac {3}{4}\right ], \left [\frac {3}{2}\right ], \frac {3 x^{2}}{2}\right )}{63 \mathrm {signum}\left (\frac {3 x^{2}}{2}-1\right )^{\frac {3}{4}}}+\frac {2 \left (3 x^{2}+4\right ) \left (3 x^{2}-2\right )^{\frac {1}{4}} x}{63} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/63*x*(3*x^2+4)*(3*x^2-2)^(1/4)+8/63*2^(1/4)/signum(3/2*x^2-1)^(3/4)*(-signum(3/2*x^2-1))^(3/4)*x*hypergeom([
1/2,3/4],[3/2],3/2*x^2)

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

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 = 0.77, size = 31, normalized size = 0.26 \[ \frac {\sqrt [4]{2} x^{5} e^{- \frac {3 i \pi }{4}} {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {5}{2} \\ \frac {7}{2} \end {matrix}\middle | {\frac {3 x^{2}}{2}} \right )}}{10} \]

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**2/2)/10

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