3.1068 \(\int \frac{x^4}{(2-3 x^2)^{3/4} (4-3 x^2)} \, dx\)

Optimal. Leaf size=164 \[ -\frac{4\ 2^{3/4} \text{EllipticF}\left (\frac{1}{2} \sin ^{-1}\left (\sqrt{\frac{3}{2}} x\right ),2\right )}{27 \sqrt{3}}+\frac{2}{27} \sqrt [4]{2-3 x^2} x+\frac{2\ 2^{3/4} \tan ^{-1}\left (\frac{2^{3/4}-\sqrt [4]{2} \sqrt{2-3 x^2}}{\sqrt{3} x \sqrt [4]{2-3 x^2}}\right )}{9 \sqrt{3}}-\frac{2\ 2^{3/4} \tanh ^{-1}\left (\frac{\sqrt [4]{2} \sqrt{2-3 x^2}+2^{3/4}}{\sqrt{3} x \sqrt [4]{2-3 x^2}}\right )}{9 \sqrt{3}} \]

[Out]

(2*x*(2 - 3*x^2)^(1/4))/27 + (2*2^(3/4)*ArcTan[(2^(3/4) - 2^(1/4)*Sqrt[2 - 3*x^2])/(Sqrt[3]*x*(2 - 3*x^2)^(1/4
))])/(9*Sqrt[3]) - (2*2^(3/4)*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]) - (4*2^(3/4)*EllipticF[ArcSin[Sqrt[3/2]*x]/2, 2])/(27*Sqrt[3])

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Rubi [A]  time = 0.0984093, antiderivative size = 164, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 5, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.208, Rules used = {443, 232, 321, 400, 441} \[ \frac{2}{27} \sqrt [4]{2-3 x^2} x+\frac{2\ 2^{3/4} \tan ^{-1}\left (\frac{2^{3/4}-\sqrt [4]{2} \sqrt{2-3 x^2}}{\sqrt{3} x \sqrt [4]{2-3 x^2}}\right )}{9 \sqrt{3}}-\frac{2\ 2^{3/4} \tanh ^{-1}\left (\frac{\sqrt [4]{2} \sqrt{2-3 x^2}+2^{3/4}}{\sqrt{3} x \sqrt [4]{2-3 x^2}}\right )}{9 \sqrt{3}}-\frac{4\ 2^{3/4} F\left (\left .\frac{1}{2} \sin ^{-1}\left (\sqrt{\frac{3}{2}} x\right )\right |2\right )}{27 \sqrt{3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x*(2 - 3*x^2)^(1/4))/27 + (2*2^(3/4)*ArcTan[(2^(3/4) - 2^(1/4)*Sqrt[2 - 3*x^2])/(Sqrt[3]*x*(2 - 3*x^2)^(1/4
))])/(9*Sqrt[3]) - (2*2^(3/4)*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]) - (4*2^(3/4)*EllipticF[ArcSin[Sqrt[3/2]*x]/2, 2])/(27*Sqrt[3])

Rule 443

Int[(x_)^(m_)/(((a_) + (b_.)*(x_)^2)^(3/4)*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Int[ExpandIntegrand[x^m/((a +
b*x^2)^(3/4)*(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 232

Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Simp[(2*EllipticF[(1*ArcSin[Rt[-(b/a), 2]*x])/2, 2])/(a^(3/4)*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 400

Int[1/(((a_) + (b_.)*(x_)^2)^(3/4)*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Dist[1/c, Int[1/(a + b*x^2)^(3/4), x],
x] - Dist[d/c, Int[x^2/((a + b*x^2)^(3/4)*(c + d*x^2)), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c - 2*a*d,
0]

Rule 441

Int[(x_)^2/(((a_) + (b_.)*(x_)^2)^(3/4)*((c_) + (d_.)*(x_)^2)), x_Symbol] :> -Simp[(b*ArcTan[(b + Rt[b^2/a, 4]
^2*Sqrt[a + b*x^2])/(Rt[b^2/a, 4]^3*x*(a + b*x^2)^(1/4))])/(a*d*Rt[b^2/a, 4]^3), x] + Simp[(b*ArcTanh[(b - Rt[
b^2/a, 4]^2*Sqrt[a + b*x^2])/(Rt[b^2/a, 4]^3*x*(a + b*x^2)^(1/4))])/(a*d*Rt[b^2/a, 4]^3), 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}{\left (2-3 x^2\right )^{3/4} \left (4-3 x^2\right )} \, dx &=\int \left (-\frac{4}{9 \left (2-3 x^2\right )^{3/4}}-\frac{x^2}{3 \left (2-3 x^2\right )^{3/4}}+\frac{16}{9 \left (2-3 x^2\right )^{3/4} \left (4-3 x^2\right )}\right ) \, dx\\ &=-\left (\frac{1}{3} \int \frac{x^2}{\left (2-3 x^2\right )^{3/4}} \, dx\right )-\frac{4}{9} \int \frac{1}{\left (2-3 x^2\right )^{3/4}} \, dx+\frac{16}{9} \int \frac{1}{\left (2-3 x^2\right )^{3/4} \left (4-3 x^2\right )} \, dx\\ &=\frac{2}{27} x \sqrt [4]{2-3 x^2}-\frac{4\ 2^{3/4} F\left (\left .\frac{1}{2} \sin ^{-1}\left (\sqrt{\frac{3}{2}} x\right )\right |2\right )}{9 \sqrt{3}}-\frac{4}{27} \int \frac{1}{\left (2-3 x^2\right )^{3/4}} \, dx+\frac{4}{9} \int \frac{1}{\left (2-3 x^2\right )^{3/4}} \, dx+\frac{4}{3} \int \frac{x^2}{\left (2-3 x^2\right )^{3/4} \left (4-3 x^2\right )} \, dx\\ &=\frac{2}{27} x \sqrt [4]{2-3 x^2}+\frac{2\ 2^{3/4} \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{2\ 2^{3/4} \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{4\ 2^{3/4} F\left (\left .\frac{1}{2} \sin ^{-1}\left (\sqrt{\frac{3}{2}} x\right )\right |2\right )}{27 \sqrt{3}}\\ \end{align*}

Mathematica [C]  time = 0.120838, size = 184, normalized size = 1.12 \[ \frac{2}{27} x \left (\sqrt [4]{2} x^2 F_1\left (\frac{3}{2};\frac{3}{4},1;\frac{5}{2};\frac{3 x^2}{2},\frac{3 x^2}{4}\right )+\frac{\frac{32 F_1\left (\frac{1}{2};\frac{3}{4},1;\frac{3}{2};\frac{3 x^2}{2},\frac{3 x^2}{4}\right )}{\left (3 x^2-4\right ) \left (x^2 \left (2 F_1\left (\frac{3}{2};\frac{3}{4},2;\frac{5}{2};\frac{3 x^2}{2},\frac{3 x^2}{4}\right )+3 F_1\left (\frac{3}{2};\frac{7}{4},1;\frac{5}{2};\frac{3 x^2}{2},\frac{3 x^2}{4}\right )\right )+4 F_1\left (\frac{1}{2};\frac{3}{4},1;\frac{3}{2};\frac{3 x^2}{2},\frac{3 x^2}{4}\right )\right )}-3 x^2+2}{\left (2-3 x^2\right )^{3/4}}\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(2*x*(2^(1/4)*x^2*AppellF1[3/2, 3/4, 1, 5/2, (3*x^2)/2, (3*x^2)/4] + (2 - 3*x^2 + (32*AppellF1[1/2, 3/4, 1, 3/
2, (3*x^2)/2, (3*x^2)/4])/((-4 + 3*x^2)*(4*AppellF1[1/2, 3/4, 1, 3/2, (3*x^2)/2, (3*x^2)/4] + x^2*(2*AppellF1[
3/2, 3/4, 2, 5/2, (3*x^2)/2, (3*x^2)/4] + 3*AppellF1[3/2, 7/4, 1, 5/2, (3*x^2)/2, (3*x^2)/4]))))/(2 - 3*x^2)^(
3/4)))/27

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Maple [F]  time = 0.068, size = 0, normalized size = 0. \begin{align*} \int{\frac{{x}^{4}}{-3\,{x}^{2}+4} \left ( -3\,{x}^{2}+2 \right ) ^{-{\frac{3}{4}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-Integral(x**4/(3*x**2*(2 - 3*x**2)**(3/4) - 4*(2 - 3*x**2)**(3/4)), 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{3}{4}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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