3.1037 \(\int \frac{x^4}{\sqrt [4]{2-3 x^2} (4-3 x^2)} \, dx\)

Optimal. Leaf size=164 \[ \frac{2}{45} \left (2-3 x^2\right )^{3/4} x+\frac{4 \sqrt [4]{2} \tan ^{-1}\left (\frac{2^{3/4}-\sqrt [4]{2} \sqrt{2-3 x^2}}{\sqrt{3} x \sqrt [4]{2-3 x^2}}\right )}{9 \sqrt{3}}+\frac{4 \sqrt [4]{2} \tanh ^{-1}\left (\frac{\sqrt [4]{2} \sqrt{2-3 x^2}+2^{3/4}}{\sqrt{3} x \sqrt [4]{2-3 x^2}}\right )}{9 \sqrt{3}}-\frac{16 \sqrt [4]{2} E\left (\left .\frac{1}{2} \sin ^{-1}\left (\sqrt{\frac{3}{2}} x\right )\right |2\right )}{15 \sqrt{3}} \]

[Out]

(2*x*(2 - 3*x^2)^(3/4))/45 + (4*2^(1/4)*ArcTan[(2^(3/4) - 2^(1/4)*Sqrt[2 - 3*x^2])/(Sqrt[3]*x*(2 - 3*x^2)^(1/4
))])/(9*Sqrt[3]) + (4*2^(1/4)*ArcTanh[(2^(3/4) + 2^(1/4)*Sqrt[2 - 3*x^2])/(Sqrt[3]*x*(2 - 3*x^2)^(1/4))])/(9*S
qrt[3]) - (16*2^(1/4)*EllipticE[ArcSin[Sqrt[3/2]*x]/2, 2])/(15*Sqrt[3])

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Rubi [A]  time = 0.072696, antiderivative size = 164, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {440, 228, 321, 397} \[ \frac{2}{45} \left (2-3 x^2\right )^{3/4} x+\frac{4 \sqrt [4]{2} \tan ^{-1}\left (\frac{2^{3/4}-\sqrt [4]{2} \sqrt{2-3 x^2}}{\sqrt{3} x \sqrt [4]{2-3 x^2}}\right )}{9 \sqrt{3}}+\frac{4 \sqrt [4]{2} \tanh ^{-1}\left (\frac{\sqrt [4]{2} \sqrt{2-3 x^2}+2^{3/4}}{\sqrt{3} x \sqrt [4]{2-3 x^2}}\right )}{9 \sqrt{3}}-\frac{16 \sqrt [4]{2} E\left (\left .\frac{1}{2} \sin ^{-1}\left (\sqrt{\frac{3}{2}} x\right )\right |2\right )}{15 \sqrt{3}} \]

Antiderivative was successfully verified.

[In]

Int[x^4/((2 - 3*x^2)^(1/4)*(4 - 3*x^2)),x]

[Out]

(2*x*(2 - 3*x^2)^(3/4))/45 + (4*2^(1/4)*ArcTan[(2^(3/4) - 2^(1/4)*Sqrt[2 - 3*x^2])/(Sqrt[3]*x*(2 - 3*x^2)^(1/4
))])/(9*Sqrt[3]) + (4*2^(1/4)*ArcTanh[(2^(3/4) + 2^(1/4)*Sqrt[2 - 3*x^2])/(Sqrt[3]*x*(2 - 3*x^2)^(1/4))])/(9*S
qrt[3]) - (16*2^(1/4)*EllipticE[ArcSin[Sqrt[3/2]*x]/2, 2])/(15*Sqrt[3])

Rule 440

Int[(x_)^(m_)/(((a_) + (b_.)*(x_)^2)^(1/4)*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Int[ExpandIntegrand[x^m/((a +
b*x^2)^(1/4)*(c + d*x^2)), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c - 2*a*d, 0] && IntegerQ[m] && (PosQ[a]
|| IntegerQ[m/2])

Rule 228

Int[((a_) + (b_.)*(x_)^2)^(-1/4), x_Symbol] :> Simp[(2*EllipticE[(1*ArcSin[Rt[-(b/a), 2]*x])/2, 2])/(a^(1/4)*R
t[-(b/a), 2]), x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && NegQ[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 397

Int[1/(((a_) + (b_.)*(x_)^2)^(1/4)*((c_) + (d_.)*(x_)^2)), x_Symbol] :> With[{q = Rt[b^2/a, 4]}, -Simp[(b*ArcT
an[(b + q^2*Sqrt[a + b*x^2])/(q^3*x*(a + b*x^2)^(1/4))])/(2*a*d*q), x] - Simp[(b*ArcTanh[(b - q^2*Sqrt[a + b*x
^2])/(q^3*x*(a + b*x^2)^(1/4))])/(2*a*d*q), x]] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c - 2*a*d, 0] && PosQ[b^2/a
]

Rubi steps

\begin{align*} \int \frac{x^4}{\sqrt [4]{2-3 x^2} \left (4-3 x^2\right )} \, dx &=\int \left (-\frac{4}{9 \sqrt [4]{2-3 x^2}}-\frac{x^2}{3 \sqrt [4]{2-3 x^2}}+\frac{16}{9 \sqrt [4]{2-3 x^2} \left (4-3 x^2\right )}\right ) \, dx\\ &=-\left (\frac{1}{3} \int \frac{x^2}{\sqrt [4]{2-3 x^2}} \, dx\right )-\frac{4}{9} \int \frac{1}{\sqrt [4]{2-3 x^2}} \, dx+\frac{16}{9} \int \frac{1}{\sqrt [4]{2-3 x^2} \left (4-3 x^2\right )} \, dx\\ &=\frac{2}{45} x \left (2-3 x^2\right )^{3/4}+\frac{4 \sqrt [4]{2} \tan ^{-1}\left (\frac{2^{3/4}-\sqrt [4]{2} \sqrt{2-3 x^2}}{\sqrt{3} x \sqrt [4]{2-3 x^2}}\right )}{9 \sqrt{3}}+\frac{4 \sqrt [4]{2} \tanh ^{-1}\left (\frac{2^{3/4}+\sqrt [4]{2} \sqrt{2-3 x^2}}{\sqrt{3} x \sqrt [4]{2-3 x^2}}\right )}{9 \sqrt{3}}-\frac{8 \sqrt [4]{2} E\left (\left .\frac{1}{2} \sin ^{-1}\left (\sqrt{\frac{3}{2}} x\right )\right |2\right )}{9 \sqrt{3}}-\frac{4}{45} \int \frac{1}{\sqrt [4]{2-3 x^2}} \, dx\\ &=\frac{2}{45} x \left (2-3 x^2\right )^{3/4}+\frac{4 \sqrt [4]{2} \tan ^{-1}\left (\frac{2^{3/4}-\sqrt [4]{2} \sqrt{2-3 x^2}}{\sqrt{3} x \sqrt [4]{2-3 x^2}}\right )}{9 \sqrt{3}}+\frac{4 \sqrt [4]{2} \tanh ^{-1}\left (\frac{2^{3/4}+\sqrt [4]{2} \sqrt{2-3 x^2}}{\sqrt{3} x \sqrt [4]{2-3 x^2}}\right )}{9 \sqrt{3}}-\frac{16 \sqrt [4]{2} E\left (\left .\frac{1}{2} \sin ^{-1}\left (\sqrt{\frac{3}{2}} x\right )\right |2\right )}{15 \sqrt{3}}\\ \end{align*}

Mathematica [C]  time = 0.128451, size = 184, normalized size = 1.12 \[ \frac{1}{45} x \left (3\ 2^{3/4} x^2 F_1\left (\frac{3}{2};\frac{1}{4},1;\frac{5}{2};\frac{3 x^2}{2},\frac{3 x^2}{4}\right )+\frac{2 \left (\frac{32 F_1\left (\frac{1}{2};\frac{1}{4},1;\frac{3}{2};\frac{3 x^2}{2},\frac{3 x^2}{4}\right )}{\left (3 x^2-4\right ) \left (x^2 \left (2 F_1\left (\frac{3}{2};\frac{1}{4},2;\frac{5}{2};\frac{3 x^2}{2},\frac{3 x^2}{4}\right )+F_1\left (\frac{3}{2};\frac{5}{4},1;\frac{5}{2};\frac{3 x^2}{2},\frac{3 x^2}{4}\right )\right )+4 F_1\left (\frac{1}{2};\frac{1}{4},1;\frac{3}{2};\frac{3 x^2}{2},\frac{3 x^2}{4}\right )\right )}-3 x^2+2\right )}{\sqrt [4]{2-3 x^2}}\right ) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[x^4/((2 - 3*x^2)^(1/4)*(4 - 3*x^2)),x]

[Out]

(x*(3*2^(3/4)*x^2*AppellF1[3/2, 1/4, 1, 5/2, (3*x^2)/2, (3*x^2)/4] + (2*(2 - 3*x^2 + (32*AppellF1[1/2, 1/4, 1,
3/2, (3*x^2)/2, (3*x^2)/4])/((-4 + 3*x^2)*(4*AppellF1[1/2, 1/4, 1, 3/2, (3*x^2)/2, (3*x^2)/4] + x^2*(2*Appell
F1[3/2, 1/4, 2, 5/2, (3*x^2)/2, (3*x^2)/4] + AppellF1[3/2, 5/4, 1, 5/2, (3*x^2)/2, (3*x^2)/4])))))/(2 - 3*x^2)
^(1/4)))/45

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Maple [F]  time = 0.066, size = 0, normalized size = 0. \begin{align*} \int{\frac{{x}^{4}}{-3\,{x}^{2}+4}{\frac{1}{\sqrt [4]{-3\,{x}^{2}+2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(x^4/(-3*x^2+2)^(1/4)/(-3*x^2+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} - 4\right )}{\left (-3 \, x^{2} + 2\right )}^{\frac{1}{4}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (-3 \, x^{2} + 2\right )}^{\frac{3}{4}} x^{4}}{9 \, x^{4} - 18 \, x^{2} + 8}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((-3*x^2 + 2)^(3/4)*x^4/(9*x^4 - 18*x^2 + 8), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \int \frac{x^{4}}{3 x^{2} \sqrt [4]{2 - 3 x^{2}} - 4 \sqrt [4]{2 - 3 x^{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int -\frac{x^{4}}{{\left (3 \, x^{2} - 4\right )}{\left (-3 \, x^{2} + 2\right )}^{\frac{1}{4}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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