3.1072 \(\int \frac{1}{x^4 (2-3 x^2)^{3/4} (4-3 x^2)} \, dx\)

Optimal. Leaf size=184 \[ \frac{11 \sqrt{3} \text{EllipticF}\left (\frac{1}{2} \sin ^{-1}\left (\sqrt{\frac{3}{2}} x\right ),2\right )}{32 \sqrt [4]{2}}-\frac{\sqrt [4]{2-3 x^2}}{4 x}-\frac{\sqrt [4]{2-3 x^2}}{24 x^3}+\frac{3 \sqrt{3} \tan ^{-1}\left (\frac{2^{3/4}-\sqrt [4]{2} \sqrt{2-3 x^2}}{\sqrt{3} x \sqrt [4]{2-3 x^2}}\right )}{64 \sqrt [4]{2}}-\frac{3 \sqrt{3} \tanh ^{-1}\left (\frac{\sqrt [4]{2} \sqrt{2-3 x^2}+2^{3/4}}{\sqrt{3} x \sqrt [4]{2-3 x^2}}\right )}{64 \sqrt [4]{2}} \]

[Out]

-(2 - 3*x^2)^(1/4)/(24*x^3) - (2 - 3*x^2)^(1/4)/(4*x) + (3*Sqrt[3]*ArcTan[(2^(3/4) - 2^(1/4)*Sqrt[2 - 3*x^2])/
(Sqrt[3]*x*(2 - 3*x^2)^(1/4))])/(64*2^(1/4)) - (3*Sqrt[3]*ArcTanh[(2^(3/4) + 2^(1/4)*Sqrt[2 - 3*x^2])/(Sqrt[3]
*x*(2 - 3*x^2)^(1/4))])/(64*2^(1/4)) + (11*Sqrt[3]*EllipticF[ArcSin[Sqrt[3/2]*x]/2, 2])/(32*2^(1/4))

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Rubi [A]  time = 0.11609, antiderivative size = 184, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 5, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.208, Rules used = {443, 325, 232, 400, 441} \[ -\frac{\sqrt [4]{2-3 x^2}}{4 x}-\frac{\sqrt [4]{2-3 x^2}}{24 x^3}+\frac{3 \sqrt{3} \tan ^{-1}\left (\frac{2^{3/4}-\sqrt [4]{2} \sqrt{2-3 x^2}}{\sqrt{3} x \sqrt [4]{2-3 x^2}}\right )}{64 \sqrt [4]{2}}-\frac{3 \sqrt{3} \tanh ^{-1}\left (\frac{\sqrt [4]{2} \sqrt{2-3 x^2}+2^{3/4}}{\sqrt{3} x \sqrt [4]{2-3 x^2}}\right )}{64 \sqrt [4]{2}}+\frac{11 \sqrt{3} F\left (\left .\frac{1}{2} \sin ^{-1}\left (\sqrt{\frac{3}{2}} x\right )\right |2\right )}{32 \sqrt [4]{2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(2 - 3*x^2)^(1/4)/(24*x^3) - (2 - 3*x^2)^(1/4)/(4*x) + (3*Sqrt[3]*ArcTan[(2^(3/4) - 2^(1/4)*Sqrt[2 - 3*x^2])/
(Sqrt[3]*x*(2 - 3*x^2)^(1/4))])/(64*2^(1/4)) - (3*Sqrt[3]*ArcTanh[(2^(3/4) + 2^(1/4)*Sqrt[2 - 3*x^2])/(Sqrt[3]
*x*(2 - 3*x^2)^(1/4))])/(64*2^(1/4)) + (11*Sqrt[3]*EllipticF[ArcSin[Sqrt[3/2]*x]/2, 2])/(32*2^(1/4))

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 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

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 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{1}{x^4 \left (2-3 x^2\right )^{3/4} \left (4-3 x^2\right )} \, dx &=\int \left (\frac{1}{4 x^4 \left (2-3 x^2\right )^{3/4}}+\frac{3}{16 x^2 \left (2-3 x^2\right )^{3/4}}-\frac{9}{16 \left (2-3 x^2\right )^{3/4} \left (-4+3 x^2\right )}\right ) \, dx\\ &=\frac{3}{16} \int \frac{1}{x^2 \left (2-3 x^2\right )^{3/4}} \, dx+\frac{1}{4} \int \frac{1}{x^4 \left (2-3 x^2\right )^{3/4}} \, dx-\frac{9}{16} \int \frac{1}{\left (2-3 x^2\right )^{3/4} \left (-4+3 x^2\right )} \, dx\\ &=-\frac{\sqrt [4]{2-3 x^2}}{24 x^3}-\frac{3 \sqrt [4]{2-3 x^2}}{32 x}+2 \left (\frac{9}{64} \int \frac{1}{\left (2-3 x^2\right )^{3/4}} \, dx\right )+\frac{5}{16} \int \frac{1}{x^2 \left (2-3 x^2\right )^{3/4}} \, dx-\frac{27}{64} \int \frac{x^2}{\left (2-3 x^2\right )^{3/4} \left (-4+3 x^2\right )} \, dx\\ &=-\frac{\sqrt [4]{2-3 x^2}}{24 x^3}-\frac{\sqrt [4]{2-3 x^2}}{4 x}+\frac{3 \sqrt{3} \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 )}{64 \sqrt [4]{2}}-\frac{3 \sqrt{3} \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 )}{64 \sqrt [4]{2}}+\frac{3 \sqrt{3} F\left (\left .\frac{1}{2} \sin ^{-1}\left (\sqrt{\frac{3}{2}} x\right )\right |2\right )}{16 \sqrt [4]{2}}+\frac{15}{64} \int \frac{1}{\left (2-3 x^2\right )^{3/4}} \, dx\\ &=-\frac{\sqrt [4]{2-3 x^2}}{24 x^3}-\frac{\sqrt [4]{2-3 x^2}}{4 x}+\frac{3 \sqrt{3} \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 )}{64 \sqrt [4]{2}}-\frac{3 \sqrt{3} \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 )}{64 \sqrt [4]{2}}+\frac{11 \sqrt{3} F\left (\left .\frac{1}{2} \sin ^{-1}\left (\sqrt{\frac{3}{2}} x\right )\right |2\right )}{32 \sqrt [4]{2}}\\ \end{align*}

Mathematica [C]  time = 0.0507399, size = 37, normalized size = 0.2 \[ -\frac{F_1\left (-\frac{3}{2};\frac{3}{4},1;-\frac{1}{2};\frac{3 x^2}{2},\frac{3 x^2}{4}\right )}{12\ 2^{3/4} x^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [F]  time = 0.056, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{x}^{4} \left ( -3\,{x}^{2}+4 \right ) } \left ( -3\,{x}^{2}+2 \right ) ^{-{\frac{3}{4}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-integrate(1/((3*x^2 - 4)*(-3*x^2 + 2)^(3/4)*x^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}}}{9 \, x^{8} - 18 \, x^{6} + 8 \, x^{4}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((-3*x^2 + 2)^(1/4)/(9*x^8 - 18*x^6 + 8*x^4), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \int \frac{1}{3 x^{6} \left (2 - 3 x^{2}\right )^{\frac{3}{4}} - 4 x^{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(1/x**4/(-3*x**2+2)**(3/4)/(-3*x**2+4),x)

[Out]

-Integral(1/(3*x**6*(2 - 3*x**2)**(3/4) - 4*x**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{1}{{\left (3 \, x^{2} - 4\right )}{\left (-3 \, x^{2} + 2\right )}^{\frac{3}{4}} x^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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