3.1047 \(\int \frac{1}{x^3 (-2+3 x^2) \sqrt [4]{-1+3 x^2}} \, dx\)

Optimal. Leaf size=191 \[ -\frac{\left (3 x^2-1\right )^{3/4}}{4 x^2}-\frac{9 \log \left (\sqrt{3 x^2-1}-\sqrt{2} \sqrt [4]{3 x^2-1}+1\right )}{16 \sqrt{2}}+\frac{9 \log \left (\sqrt{3 x^2-1}+\sqrt{2} \sqrt [4]{3 x^2-1}+1\right )}{16 \sqrt{2}}+\frac{3}{4} \tan ^{-1}\left (\sqrt [4]{3 x^2-1}\right )+\frac{9 \tan ^{-1}\left (1-\sqrt{2} \sqrt [4]{3 x^2-1}\right )}{8 \sqrt{2}}-\frac{9 \tan ^{-1}\left (\sqrt{2} \sqrt [4]{3 x^2-1}+1\right )}{8 \sqrt{2}}-\frac{3}{4} \tanh ^{-1}\left (\sqrt [4]{3 x^2-1}\right ) \]

[Out]

-(-1 + 3*x^2)^(3/4)/(4*x^2) + (3*ArcTan[(-1 + 3*x^2)^(1/4)])/4 + (9*ArcTan[1 - Sqrt[2]*(-1 + 3*x^2)^(1/4)])/(8
*Sqrt[2]) - (9*ArcTan[1 + Sqrt[2]*(-1 + 3*x^2)^(1/4)])/(8*Sqrt[2]) - (3*ArcTanh[(-1 + 3*x^2)^(1/4)])/4 - (9*Lo
g[1 - Sqrt[2]*(-1 + 3*x^2)^(1/4) + Sqrt[-1 + 3*x^2]])/(16*Sqrt[2]) + (9*Log[1 + Sqrt[2]*(-1 + 3*x^2)^(1/4) + S
qrt[-1 + 3*x^2]])/(16*Sqrt[2])

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Rubi [A]  time = 0.147334, antiderivative size = 191, normalized size of antiderivative = 1., number of steps used = 17, number of rules used = 13, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.542, Rules used = {446, 103, 156, 63, 297, 1162, 617, 204, 1165, 628, 298, 203, 206} \[ -\frac{\left (3 x^2-1\right )^{3/4}}{4 x^2}-\frac{9 \log \left (\sqrt{3 x^2-1}-\sqrt{2} \sqrt [4]{3 x^2-1}+1\right )}{16 \sqrt{2}}+\frac{9 \log \left (\sqrt{3 x^2-1}+\sqrt{2} \sqrt [4]{3 x^2-1}+1\right )}{16 \sqrt{2}}+\frac{3}{4} \tan ^{-1}\left (\sqrt [4]{3 x^2-1}\right )+\frac{9 \tan ^{-1}\left (1-\sqrt{2} \sqrt [4]{3 x^2-1}\right )}{8 \sqrt{2}}-\frac{9 \tan ^{-1}\left (\sqrt{2} \sqrt [4]{3 x^2-1}+1\right )}{8 \sqrt{2}}-\frac{3}{4} \tanh ^{-1}\left (\sqrt [4]{3 x^2-1}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(-1 + 3*x^2)^(3/4)/(4*x^2) + (3*ArcTan[(-1 + 3*x^2)^(1/4)])/4 + (9*ArcTan[1 - Sqrt[2]*(-1 + 3*x^2)^(1/4)])/(8
*Sqrt[2]) - (9*ArcTan[1 + Sqrt[2]*(-1 + 3*x^2)^(1/4)])/(8*Sqrt[2]) - (3*ArcTanh[(-1 + 3*x^2)^(1/4)])/4 - (9*Lo
g[1 - Sqrt[2]*(-1 + 3*x^2)^(1/4) + Sqrt[-1 + 3*x^2]])/(16*Sqrt[2]) + (9*Log[1 + Sqrt[2]*(-1 + 3*x^2)^(1/4) + S
qrt[-1 + 3*x^2]])/(16*Sqrt[2])

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 103

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a +
b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), x] + Dist[1/((m + 1)*(b*
c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*(m + 1) - b*(d*e*(m + n + 2) +
c*f*(m + p + 2)) - b*d*f*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && LtQ[m, -1] &&
IntegerQ[m] && (IntegerQ[n] || IntegersQ[2*n, 2*p])

Rule 156

Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :>
Dist[(b*g - a*h)/(b*c - a*d), Int[(e + f*x)^p/(a + b*x), x], x] - Dist[(d*g - c*h)/(b*c - a*d), Int[(e + f*x)
^p/(c + d*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 297

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]},
Dist[1/(2*s), Int[(r + s*x^2)/(a + b*x^4), x], x] - Dist[1/(2*s), Int[(r - s*x^2)/(a + b*x^4), x], x]] /; Free
Q[{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ,
b]]))

Rule 1162

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e, 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 1165

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(-2*d)/e, 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 298

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b),
2]]}, Dist[s/(2*b), Int[1/(r + s*x^2), x], x] - Dist[s/(2*b), Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &
&  !GtQ[a/b, 0]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{1}{x^3 \left (-2+3 x^2\right ) \sqrt [4]{-1+3 x^2}} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{x^2 (-2+3 x) \sqrt [4]{-1+3 x}} \, dx,x,x^2\right )\\ &=-\frac{\left (-1+3 x^2\right )^{3/4}}{4 x^2}-\frac{1}{4} \operatorname{Subst}\left (\int \frac{-\frac{9}{2}+\frac{9 x}{4}}{x (-2+3 x) \sqrt [4]{-1+3 x}} \, dx,x,x^2\right )\\ &=-\frac{\left (-1+3 x^2\right )^{3/4}}{4 x^2}-\frac{9}{16} \operatorname{Subst}\left (\int \frac{1}{x \sqrt [4]{-1+3 x}} \, dx,x,x^2\right )+\frac{9}{8} \operatorname{Subst}\left (\int \frac{1}{(-2+3 x) \sqrt [4]{-1+3 x}} \, dx,x,x^2\right )\\ &=-\frac{\left (-1+3 x^2\right )^{3/4}}{4 x^2}-\frac{3}{4} \operatorname{Subst}\left (\int \frac{x^2}{\frac{1}{3}+\frac{x^4}{3}} \, dx,x,\sqrt [4]{-1+3 x^2}\right )+\frac{3}{2} \operatorname{Subst}\left (\int \frac{x^2}{-1+x^4} \, dx,x,\sqrt [4]{-1+3 x^2}\right )\\ &=-\frac{\left (-1+3 x^2\right )^{3/4}}{4 x^2}+\frac{3}{8} \operatorname{Subst}\left (\int \frac{1-x^2}{\frac{1}{3}+\frac{x^4}{3}} \, dx,x,\sqrt [4]{-1+3 x^2}\right )-\frac{3}{8} \operatorname{Subst}\left (\int \frac{1+x^2}{\frac{1}{3}+\frac{x^4}{3}} \, dx,x,\sqrt [4]{-1+3 x^2}\right )-\frac{3}{4} \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\sqrt [4]{-1+3 x^2}\right )+\frac{3}{4} \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\sqrt [4]{-1+3 x^2}\right )\\ &=-\frac{\left (-1+3 x^2\right )^{3/4}}{4 x^2}+\frac{3}{4} \tan ^{-1}\left (\sqrt [4]{-1+3 x^2}\right )-\frac{3}{4} \tanh ^{-1}\left (\sqrt [4]{-1+3 x^2}\right )-\frac{9}{16} \operatorname{Subst}\left (\int \frac{1}{1-\sqrt{2} x+x^2} \, dx,x,\sqrt [4]{-1+3 x^2}\right )-\frac{9}{16} \operatorname{Subst}\left (\int \frac{1}{1+\sqrt{2} x+x^2} \, dx,x,\sqrt [4]{-1+3 x^2}\right )-\frac{9 \operatorname{Subst}\left (\int \frac{\sqrt{2}+2 x}{-1-\sqrt{2} x-x^2} \, dx,x,\sqrt [4]{-1+3 x^2}\right )}{16 \sqrt{2}}-\frac{9 \operatorname{Subst}\left (\int \frac{\sqrt{2}-2 x}{-1+\sqrt{2} x-x^2} \, dx,x,\sqrt [4]{-1+3 x^2}\right )}{16 \sqrt{2}}\\ &=-\frac{\left (-1+3 x^2\right )^{3/4}}{4 x^2}+\frac{3}{4} \tan ^{-1}\left (\sqrt [4]{-1+3 x^2}\right )-\frac{3}{4} \tanh ^{-1}\left (\sqrt [4]{-1+3 x^2}\right )-\frac{9 \log \left (1-\sqrt{2} \sqrt [4]{-1+3 x^2}+\sqrt{-1+3 x^2}\right )}{16 \sqrt{2}}+\frac{9 \log \left (1+\sqrt{2} \sqrt [4]{-1+3 x^2}+\sqrt{-1+3 x^2}\right )}{16 \sqrt{2}}-\frac{9 \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1-\sqrt{2} \sqrt [4]{-1+3 x^2}\right )}{8 \sqrt{2}}+\frac{9 \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1+\sqrt{2} \sqrt [4]{-1+3 x^2}\right )}{8 \sqrt{2}}\\ &=-\frac{\left (-1+3 x^2\right )^{3/4}}{4 x^2}+\frac{3}{4} \tan ^{-1}\left (\sqrt [4]{-1+3 x^2}\right )+\frac{9 \tan ^{-1}\left (1-\sqrt{2} \sqrt [4]{-1+3 x^2}\right )}{8 \sqrt{2}}-\frac{9 \tan ^{-1}\left (1+\sqrt{2} \sqrt [4]{-1+3 x^2}\right )}{8 \sqrt{2}}-\frac{3}{4} \tanh ^{-1}\left (\sqrt [4]{-1+3 x^2}\right )-\frac{9 \log \left (1-\sqrt{2} \sqrt [4]{-1+3 x^2}+\sqrt{-1+3 x^2}\right )}{16 \sqrt{2}}+\frac{9 \log \left (1+\sqrt{2} \sqrt [4]{-1+3 x^2}+\sqrt{-1+3 x^2}\right )}{16 \sqrt{2}}\\ \end{align*}

Mathematica [C]  time = 0.0308546, size = 77, normalized size = 0.4 \[ \frac{1}{4} \left (-3 \left (3 x^2-1\right )^{3/4} \, _2F_1\left (\frac{3}{4},1;\frac{7}{4};1-3 x^2\right )-\frac{\left (3 x^2-1\right )^{3/4}}{x^2}+3 \tan ^{-1}\left (\sqrt [4]{3 x^2-1}\right )-3 \tanh ^{-1}\left (\sqrt [4]{3 x^2-1}\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-((-1 + 3*x^2)^(3/4)/x^2) + 3*ArcTan[(-1 + 3*x^2)^(1/4)] - 3*ArcTanh[(-1 + 3*x^2)^(1/4)] - 3*(-1 + 3*x^2)^(3/
4)*Hypergeometric2F1[3/4, 1, 7/4, 1 - 3*x^2])/4

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Maple [F]  time = 0.087, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{x}^{3} \left ( 3\,{x}^{2}-2 \right ) }{\frac{1}{\sqrt [4]{3\,{x}^{2}-1}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.36684, size = 720, normalized size = 3.77 \begin{align*} \frac{36 \, \sqrt{2} x^{2} \arctan \left (\sqrt{2} \sqrt{\sqrt{2}{\left (3 \, x^{2} - 1\right )}^{\frac{1}{4}} + \sqrt{3 \, x^{2} - 1} + 1} - \sqrt{2}{\left (3 \, x^{2} - 1\right )}^{\frac{1}{4}} - 1\right ) + 36 \, \sqrt{2} x^{2} \arctan \left (\frac{1}{2} \, \sqrt{2} \sqrt{-4 \, \sqrt{2}{\left (3 \, x^{2} - 1\right )}^{\frac{1}{4}} + 4 \, \sqrt{3 \, x^{2} - 1} + 4} - \sqrt{2}{\left (3 \, x^{2} - 1\right )}^{\frac{1}{4}} + 1\right ) + 9 \, \sqrt{2} x^{2} \log \left (4 \, \sqrt{2}{\left (3 \, x^{2} - 1\right )}^{\frac{1}{4}} + 4 \, \sqrt{3 \, x^{2} - 1} + 4\right ) - 9 \, \sqrt{2} x^{2} \log \left (-4 \, \sqrt{2}{\left (3 \, x^{2} - 1\right )}^{\frac{1}{4}} + 4 \, \sqrt{3 \, x^{2} - 1} + 4\right ) + 24 \, x^{2} \arctan \left ({\left (3 \, x^{2} - 1\right )}^{\frac{1}{4}}\right ) - 12 \, x^{2} \log \left ({\left (3 \, x^{2} - 1\right )}^{\frac{1}{4}} + 1\right ) + 12 \, x^{2} \log \left ({\left (3 \, x^{2} - 1\right )}^{\frac{1}{4}} - 1\right ) - 8 \,{\left (3 \, x^{2} - 1\right )}^{\frac{3}{4}}}{32 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32*(36*sqrt(2)*x^2*arctan(sqrt(2)*sqrt(sqrt(2)*(3*x^2 - 1)^(1/4) + sqrt(3*x^2 - 1) + 1) - sqrt(2)*(3*x^2 - 1
)^(1/4) - 1) + 36*sqrt(2)*x^2*arctan(1/2*sqrt(2)*sqrt(-4*sqrt(2)*(3*x^2 - 1)^(1/4) + 4*sqrt(3*x^2 - 1) + 4) -
sqrt(2)*(3*x^2 - 1)^(1/4) + 1) + 9*sqrt(2)*x^2*log(4*sqrt(2)*(3*x^2 - 1)^(1/4) + 4*sqrt(3*x^2 - 1) + 4) - 9*sq
rt(2)*x^2*log(-4*sqrt(2)*(3*x^2 - 1)^(1/4) + 4*sqrt(3*x^2 - 1) + 4) + 24*x^2*arctan((3*x^2 - 1)^(1/4)) - 12*x^
2*log((3*x^2 - 1)^(1/4) + 1) + 12*x^2*log((3*x^2 - 1)^(1/4) - 1) - 8*(3*x^2 - 1)^(3/4))/x^2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/(x**3*(3*x**2 - 2)*(3*x**2 - 1)**(1/4)), x)

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Giac [A]  time = 1.23334, size = 228, normalized size = 1.19 \begin{align*} -\frac{9}{16} \, \sqrt{2} \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} + 2 \,{\left (3 \, x^{2} - 1\right )}^{\frac{1}{4}}\right )}\right ) - \frac{9}{16} \, \sqrt{2} \arctan \left (-\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} - 2 \,{\left (3 \, x^{2} - 1\right )}^{\frac{1}{4}}\right )}\right ) + \frac{9}{32} \, \sqrt{2} \log \left (\sqrt{2}{\left (3 \, x^{2} - 1\right )}^{\frac{1}{4}} + \sqrt{3 \, x^{2} - 1} + 1\right ) - \frac{9}{32} \, \sqrt{2} \log \left (-\sqrt{2}{\left (3 \, x^{2} - 1\right )}^{\frac{1}{4}} + \sqrt{3 \, x^{2} - 1} + 1\right ) - \frac{{\left (3 \, x^{2} - 1\right )}^{\frac{3}{4}}}{4 \, x^{2}} + \frac{3}{4} \, \arctan \left ({\left (3 \, x^{2} - 1\right )}^{\frac{1}{4}}\right ) - \frac{3}{8} \, \log \left ({\left (3 \, x^{2} - 1\right )}^{\frac{1}{4}} + 1\right ) + \frac{3}{8} \, \log \left ({\left |{\left (3 \, x^{2} - 1\right )}^{\frac{1}{4}} - 1 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-9/16*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*(3*x^2 - 1)^(1/4))) - 9/16*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(2)
- 2*(3*x^2 - 1)^(1/4))) + 9/32*sqrt(2)*log(sqrt(2)*(3*x^2 - 1)^(1/4) + sqrt(3*x^2 - 1) + 1) - 9/32*sqrt(2)*log
(-sqrt(2)*(3*x^2 - 1)^(1/4) + sqrt(3*x^2 - 1) + 1) - 1/4*(3*x^2 - 1)^(3/4)/x^2 + 3/4*arctan((3*x^2 - 1)^(1/4))
- 3/8*log((3*x^2 - 1)^(1/4) + 1) + 3/8*log(abs((3*x^2 - 1)^(1/4) - 1))