3.1090 \(\int \frac{1}{(-2+3 x^2) (-1+3 x^2)^{3/4}} \, dx\)

Optimal. Leaf size=127 \[ -\frac{\sqrt{\frac{x^2}{\left (\sqrt{3 x^2-1}+1\right )^2}} \left (\sqrt{3 x^2-1}+1\right ) \text{EllipticF}\left (2 \tan ^{-1}\left (\sqrt [4]{3 x^2-1}\right ),\frac{1}{2}\right )}{2 \sqrt{3} x}+\frac{\tan ^{-1}\left (\frac{\sqrt{\frac{3}{2}} x}{\sqrt [4]{3 x^2-1}}\right )}{2 \sqrt{6}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{\frac{3}{2}} x}{\sqrt [4]{3 x^2-1}}\right )}{2 \sqrt{6}} \]

[Out]

ArcTan[(Sqrt[3/2]*x)/(-1 + 3*x^2)^(1/4)]/(2*Sqrt[6]) - ArcTanh[(Sqrt[3/2]*x)/(-1 + 3*x^2)^(1/4)]/(2*Sqrt[6]) -
(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])/(2*S
qrt[3]*x)

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Rubi [A]  time = 0.0481679, antiderivative size = 127, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {400, 234, 220, 442} \[ \frac{\tan ^{-1}\left (\frac{\sqrt{\frac{3}{2}} x}{\sqrt [4]{3 x^2-1}}\right )}{2 \sqrt{6}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{\frac{3}{2}} x}{\sqrt [4]{3 x^2-1}}\right )}{2 \sqrt{6}}-\frac{\sqrt{\frac{x^2}{\left (\sqrt{3 x^2-1}+1\right )^2}} \left (\sqrt{3 x^2-1}+1\right ) F\left (2 \tan ^{-1}\left (\sqrt [4]{3 x^2-1}\right )|\frac{1}{2}\right )}{2 \sqrt{3} x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

ArcTan[(Sqrt[3/2]*x)/(-1 + 3*x^2)^(1/4)]/(2*Sqrt[6]) - ArcTanh[(Sqrt[3/2]*x)/(-1 + 3*x^2)^(1/4)]/(2*Sqrt[6]) -
(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])/(2*S
qrt[3]*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 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 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}{\left (-2+3 x^2\right ) \left (-1+3 x^2\right )^{3/4}} \, dx &=-\left (\frac{1}{2} \int \frac{1}{\left (-1+3 x^2\right )^{3/4}} \, dx\right )+\frac{3}{2} \int \frac{x^2}{\left (-2+3 x^2\right ) \left (-1+3 x^2\right )^{3/4}} \, dx\\ &=\frac{\tan ^{-1}\left (\frac{\sqrt{\frac{3}{2}} x}{\sqrt [4]{-1+3 x^2}}\right )}{2 \sqrt{6}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{\frac{3}{2}} x}{\sqrt [4]{-1+3 x^2}}\right )}{2 \sqrt{6}}-\frac{\sqrt{x^2} \operatorname{Subst}\left (\int \frac{1}{\sqrt{1+x^4}} \, dx,x,\sqrt [4]{-1+3 x^2}\right )}{\sqrt{3} x}\\ &=\frac{\tan ^{-1}\left (\frac{\sqrt{\frac{3}{2}} x}{\sqrt [4]{-1+3 x^2}}\right )}{2 \sqrt{6}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{\frac{3}{2}} x}{\sqrt [4]{-1+3 x^2}}\right )}{2 \sqrt{6}}-\frac{\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 )}{2 \sqrt{3} x}\\ \end{align*}

Mathematica [C]  time = 0.0153465, size = 68, normalized size = 0.54 \[ \frac{\sqrt [4]{-1} \sqrt{x^2} \left (\Pi \left (-i;\left .\sin ^{-1}\left ((-1)^{3/4} \sqrt [4]{3 x^2-1}\right )\right |-1\right )+\Pi \left (i;\left .\sin ^{-1}\left ((-1)^{3/4} \sqrt [4]{3 x^2-1}\right )\right |-1\right )\right )}{\sqrt{3} x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-1)^(1/4)*Sqrt[x^2]*(EllipticPi[-I, ArcSin[(-1)^(3/4)*(-1 + 3*x^2)^(1/4)], -1] + EllipticPi[I, ArcSin[(-1)^(
3/4)*(-1 + 3*x^2)^(1/4)], -1]))/(Sqrt[3]*x)

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Maple [F]  time = 0.057, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{3\,{x}^{2}-2} \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/(3*x^2-2)/(3*x^2-1)^(3/4),x)

[Out]

int(1/(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 )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(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)

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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^{4} - 9 \, x^{2} + 2}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Integral(1/((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 )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(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)