3.1092 \(\int \frac{1}{x^4 (-2+3 x^2) (-1+3 x^2)^{3/4}} \, dx\)

Optimal. Leaf size=165 \[ -\frac{11 \sqrt{3} \sqrt{\frac{x^2}{\left (\sqrt{3 x^2-1}+1\right )^2}} \left (\sqrt{3 x^2-1}+1\right ) \text{EllipticF}\left (2 \tan ^{-1}\left (\sqrt [4]{3 x^2-1}\right ),\frac{1}{2}\right )}{8 x}-\frac{2 \sqrt [4]{3 x^2-1}}{x}-\frac{\sqrt [4]{3 x^2-1}}{6 x^3}+\frac{3}{8} \sqrt{\frac{3}{2}} \tan ^{-1}\left (\frac{\sqrt{\frac{3}{2}} x}{\sqrt [4]{3 x^2-1}}\right )-\frac{3}{8} \sqrt{\frac{3}{2}} \tanh ^{-1}\left (\frac{\sqrt{\frac{3}{2}} x}{\sqrt [4]{3 x^2-1}}\right ) \]

[Out]

-(-1 + 3*x^2)^(1/4)/(6*x^3) - (2*(-1 + 3*x^2)^(1/4))/x + (3*Sqrt[3/2]*ArcTan[(Sqrt[3/2]*x)/(-1 + 3*x^2)^(1/4)]
)/8 - (3*Sqrt[3/2]*ArcTanh[(Sqrt[3/2]*x)/(-1 + 3*x^2)^(1/4)])/8 - (11*Sqrt[3]*Sqrt[x^2/(1 + Sqrt[-1 + 3*x^2])^
2]*(1 + Sqrt[-1 + 3*x^2])*EllipticF[2*ArcTan[(-1 + 3*x^2)^(1/4)], 1/2])/(8*x)

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Rubi [A]  time = 0.166966, antiderivative size = 165, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 6, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {443, 325, 234, 220, 400, 442} \[ -\frac{2 \sqrt [4]{3 x^2-1}}{x}-\frac{\sqrt [4]{3 x^2-1}}{6 x^3}+\frac{3}{8} \sqrt{\frac{3}{2}} \tan ^{-1}\left (\frac{\sqrt{\frac{3}{2}} x}{\sqrt [4]{3 x^2-1}}\right )-\frac{3}{8} \sqrt{\frac{3}{2}} \tanh ^{-1}\left (\frac{\sqrt{\frac{3}{2}} x}{\sqrt [4]{3 x^2-1}}\right )-\frac{11 \sqrt{3} \sqrt{\frac{x^2}{\left (\sqrt{3 x^2-1}+1\right )^2}} \left (\sqrt{3 x^2-1}+1\right ) F\left (2 \tan ^{-1}\left (\sqrt [4]{3 x^2-1}\right )|\frac{1}{2}\right )}{8 x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(-1 + 3*x^2)^(1/4)/(6*x^3) - (2*(-1 + 3*x^2)^(1/4))/x + (3*Sqrt[3/2]*ArcTan[(Sqrt[3/2]*x)/(-1 + 3*x^2)^(1/4)]
)/8 - (3*Sqrt[3/2]*ArcTanh[(Sqrt[3/2]*x)/(-1 + 3*x^2)^(1/4)])/8 - (11*Sqrt[3]*Sqrt[x^2/(1 + Sqrt[-1 + 3*x^2])^
2]*(1 + Sqrt[-1 + 3*x^2])*EllipticF[2*ArcTan[(-1 + 3*x^2)^(1/4)], 1/2])/(8*x)

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 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 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 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 442

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

Rubi steps

\begin{align*} \int \frac{1}{x^4 \left (-2+3 x^2\right ) \left (-1+3 x^2\right )^{3/4}} \, dx &=\int \left (-\frac{1}{2 x^4 \left (-1+3 x^2\right )^{3/4}}-\frac{3}{4 x^2 \left (-1+3 x^2\right )^{3/4}}+\frac{9}{4 \left (-2+3 x^2\right ) \left (-1+3 x^2\right )^{3/4}}\right ) \, dx\\ &=-\left (\frac{1}{2} \int \frac{1}{x^4 \left (-1+3 x^2\right )^{3/4}} \, dx\right )-\frac{3}{4} \int \frac{1}{x^2 \left (-1+3 x^2\right )^{3/4}} \, dx+\frac{9}{4} \int \frac{1}{\left (-2+3 x^2\right ) \left (-1+3 x^2\right )^{3/4}} \, dx\\ &=-\frac{\sqrt [4]{-1+3 x^2}}{6 x^3}-\frac{3 \sqrt [4]{-1+3 x^2}}{4 x}-2 \left (\frac{9}{8} \int \frac{1}{\left (-1+3 x^2\right )^{3/4}} \, dx\right )-\frac{5}{4} \int \frac{1}{x^2 \left (-1+3 x^2\right )^{3/4}} \, dx+\frac{27}{8} \int \frac{x^2}{\left (-2+3 x^2\right ) \left (-1+3 x^2\right )^{3/4}} \, dx\\ &=-\frac{\sqrt [4]{-1+3 x^2}}{6 x^3}-\frac{2 \sqrt [4]{-1+3 x^2}}{x}+\frac{3}{8} \sqrt{\frac{3}{2}} \tan ^{-1}\left (\frac{\sqrt{\frac{3}{2}} x}{\sqrt [4]{-1+3 x^2}}\right )-\frac{3}{8} \sqrt{\frac{3}{2}} \tanh ^{-1}\left (\frac{\sqrt{\frac{3}{2}} x}{\sqrt [4]{-1+3 x^2}}\right )-\frac{15}{8} \int \frac{1}{\left (-1+3 x^2\right )^{3/4}} \, dx-2 \frac{\left (3 \sqrt{3} \sqrt{x^2}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1+x^4}} \, dx,x,\sqrt [4]{-1+3 x^2}\right )}{4 x}\\ &=-\frac{\sqrt [4]{-1+3 x^2}}{6 x^3}-\frac{2 \sqrt [4]{-1+3 x^2}}{x}+\frac{3}{8} \sqrt{\frac{3}{2}} \tan ^{-1}\left (\frac{\sqrt{\frac{3}{2}} x}{\sqrt [4]{-1+3 x^2}}\right )-\frac{3}{8} \sqrt{\frac{3}{2}} \tanh ^{-1}\left (\frac{\sqrt{\frac{3}{2}} x}{\sqrt [4]{-1+3 x^2}}\right )-\frac{3 \sqrt{3} \sqrt{\frac{x^2}{\left (1+\sqrt{-1+3 x^2}\right )^2}} \left (1+\sqrt{-1+3 x^2}\right ) F\left (2 \tan ^{-1}\left (\sqrt [4]{-1+3 x^2}\right )|\frac{1}{2}\right )}{4 x}-\frac{\left (5 \sqrt{3} \sqrt{x^2}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1+x^4}} \, dx,x,\sqrt [4]{-1+3 x^2}\right )}{4 x}\\ &=-\frac{\sqrt [4]{-1+3 x^2}}{6 x^3}-\frac{2 \sqrt [4]{-1+3 x^2}}{x}+\frac{3}{8} \sqrt{\frac{3}{2}} \tan ^{-1}\left (\frac{\sqrt{\frac{3}{2}} x}{\sqrt [4]{-1+3 x^2}}\right )-\frac{3}{8} \sqrt{\frac{3}{2}} \tanh ^{-1}\left (\frac{\sqrt{\frac{3}{2}} x}{\sqrt [4]{-1+3 x^2}}\right )-\frac{11 \sqrt{3} \sqrt{\frac{x^2}{\left (1+\sqrt{-1+3 x^2}\right )^2}} \left (1+\sqrt{-1+3 x^2}\right ) F\left (2 \tan ^{-1}\left (\sqrt [4]{-1+3 x^2}\right )|\frac{1}{2}\right )}{8 x}\\ \end{align*}

Mathematica [C]  time = 0.045282, size = 52, normalized size = 0.32 \[ \frac{\left (1-3 x^2\right )^{3/4} F_1\left (-\frac{3}{2};\frac{3}{4},1;-\frac{1}{2};3 x^2,\frac{3 x^2}{2}\right )}{6 x^3 \left (3 x^2-1\right )^{3/4}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((1 - 3*x^2)^(3/4)*AppellF1[-3/2, 3/4, 1, -1/2, 3*x^2, (3*x^2)/2])/(6*x^3*(-1 + 3*x^2)^(3/4))

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Maple [F]  time = 0.074, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{x}^{4} \left ( 3\,{x}^{2}-2 \right ) } \left ( 3\,{x}^{2}-1 \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*x^2-1)^(3/4),x)

[Out]

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

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

[Out]

integrate(1/((3*x^2 - 1)^(3/4)*(3*x^2 - 2)*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} - 1\right )}^{\frac{1}{4}}}{9 \, x^{8} - 9 \, x^{6} + 2 \, 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*x^2-1)^(3/4),x, algorithm="fricas")

[Out]

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

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

[Out]

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

[Out]

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