3.1038 \(\int \frac{x^2}{\sqrt [4]{2-3 x^2} (4-3 x^2)} \, dx\)

Optimal. Leaf size=148 \[ \frac{\sqrt [4]{2} \tan ^{-1}\left (\frac{2^{3/4}-\sqrt [4]{2} \sqrt{2-3 x^2}}{\sqrt{3} x \sqrt [4]{2-3 x^2}}\right )}{3 \sqrt{3}}+\frac{\sqrt [4]{2} \tanh ^{-1}\left (\frac{\sqrt [4]{2} \sqrt{2-3 x^2}+2^{3/4}}{\sqrt{3} x \sqrt [4]{2-3 x^2}}\right )}{3 \sqrt{3}}-\frac{2 \sqrt [4]{2} E\left (\left .\frac{1}{2} \sin ^{-1}\left (\sqrt{\frac{3}{2}} x\right )\right |2\right )}{3 \sqrt{3}} \]

[Out]

(2^(1/4)*ArcTan[(2^(3/4) - 2^(1/4)*Sqrt[2 - 3*x^2])/(Sqrt[3]*x*(2 - 3*x^2)^(1/4))])/(3*Sqrt[3]) + (2^(1/4)*Arc
Tanh[(2^(3/4) + 2^(1/4)*Sqrt[2 - 3*x^2])/(Sqrt[3]*x*(2 - 3*x^2)^(1/4))])/(3*Sqrt[3]) - (2*2^(1/4)*EllipticE[Ar
cSin[Sqrt[3/2]*x]/2, 2])/(3*Sqrt[3])

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Rubi [A]  time = 0.0555915, antiderivative size = 148, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {440, 228, 397} \[ \frac{\sqrt [4]{2} \tan ^{-1}\left (\frac{2^{3/4}-\sqrt [4]{2} \sqrt{2-3 x^2}}{\sqrt{3} x \sqrt [4]{2-3 x^2}}\right )}{3 \sqrt{3}}+\frac{\sqrt [4]{2} \tanh ^{-1}\left (\frac{\sqrt [4]{2} \sqrt{2-3 x^2}+2^{3/4}}{\sqrt{3} x \sqrt [4]{2-3 x^2}}\right )}{3 \sqrt{3}}-\frac{2 \sqrt [4]{2} E\left (\left .\frac{1}{2} \sin ^{-1}\left (\sqrt{\frac{3}{2}} x\right )\right |2\right )}{3 \sqrt{3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2^(1/4)*ArcTan[(2^(3/4) - 2^(1/4)*Sqrt[2 - 3*x^2])/(Sqrt[3]*x*(2 - 3*x^2)^(1/4))])/(3*Sqrt[3]) + (2^(1/4)*Arc
Tanh[(2^(3/4) + 2^(1/4)*Sqrt[2 - 3*x^2])/(Sqrt[3]*x*(2 - 3*x^2)^(1/4))])/(3*Sqrt[3]) - (2*2^(1/4)*EllipticE[Ar
cSin[Sqrt[3/2]*x]/2, 2])/(3*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 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^2}{\sqrt [4]{2-3 x^2} \left (4-3 x^2\right )} \, dx &=\int \left (-\frac{1}{3 \sqrt [4]{2-3 x^2}}+\frac{4}{3 \sqrt [4]{2-3 x^2} \left (4-3 x^2\right )}\right ) \, dx\\ &=-\left (\frac{1}{3} \int \frac{1}{\sqrt [4]{2-3 x^2}} \, dx\right )+\frac{4}{3} \int \frac{1}{\sqrt [4]{2-3 x^2} \left (4-3 x^2\right )} \, dx\\ &=\frac{\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 )}{3 \sqrt{3}}+\frac{\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 )}{3 \sqrt{3}}-\frac{2 \sqrt [4]{2} E\left (\left .\frac{1}{2} \sin ^{-1}\left (\sqrt{\frac{3}{2}} x\right )\right |2\right )}{3 \sqrt{3}}\\ \end{align*}

Mathematica [C]  time = 0.0175562, size = 37, normalized size = 0.25 \[ \frac{x^3 F_1\left (\frac{3}{2};\frac{1}{4},1;\frac{5}{2};\frac{3 x^2}{2},\frac{3 x^2}{4}\right )}{12 \sqrt [4]{2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(x^3*AppellF1[3/2, 1/4, 1, 5/2, (3*x^2)/2, (3*x^2)/4])/(12*2^(1/4))

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Maple [F]  time = 0.045, size = 0, normalized size = 0. \begin{align*} \int{\frac{{x}^{2}}{-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^2/(-3*x^2+2)^(1/4)/(-3*x^2+4),x)

[Out]

int(x^2/(-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^{2}}{{\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^2/(-3*x^2+2)^(1/4)/(-3*x^2+4),x, algorithm="maxima")

[Out]

-integrate(x^2/((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^{2}}{9 \, x^{4} - 18 \, x^{2} + 8}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((-3*x^2 + 2)^(3/4)*x^2/(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^{2}}{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**2/(-3*x**2+2)**(1/4)/(-3*x**2+4),x)

[Out]

-Integral(x**2/(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^{2}}{{\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^2/(-3*x^2+2)^(1/4)/(-3*x^2+4),x, algorithm="giac")

[Out]

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