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

Optimal. Leaf size=173 \[ -\frac {\log \left (\sqrt {3 x^2-1}-\sqrt {2} \sqrt [4]{3 x^2-1}+1\right )}{4 \sqrt {2}}+\frac {\log \left (\sqrt {3 x^2-1}+\sqrt {2} \sqrt [4]{3 x^2-1}+1\right )}{4 \sqrt {2}}+\frac {1}{2} \tan ^{-1}\left (\sqrt [4]{3 x^2-1}\right )+\frac {\tan ^{-1}\left (1-\sqrt {2} \sqrt [4]{3 x^2-1}\right )}{2 \sqrt {2}}-\frac {\tan ^{-1}\left (\sqrt {2} \sqrt [4]{3 x^2-1}+1\right )}{2 \sqrt {2}}-\frac {1}{2} \tanh ^{-1}\left (\sqrt [4]{3 x^2-1}\right ) \]

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Rubi [A]  time = 0.13, antiderivative size = 173, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 12, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {446, 86, 63, 297, 1162, 617, 204, 1165, 628, 298, 203, 206} \begin {gather*} -\frac {\log \left (\sqrt {3 x^2-1}-\sqrt {2} \sqrt [4]{3 x^2-1}+1\right )}{4 \sqrt {2}}+\frac {\log \left (\sqrt {3 x^2-1}+\sqrt {2} \sqrt [4]{3 x^2-1}+1\right )}{4 \sqrt {2}}+\frac {1}{2} \tan ^{-1}\left (\sqrt [4]{3 x^2-1}\right )+\frac {\tan ^{-1}\left (1-\sqrt {2} \sqrt [4]{3 x^2-1}\right )}{2 \sqrt {2}}-\frac {\tan ^{-1}\left (\sqrt {2} \sqrt [4]{3 x^2-1}+1\right )}{2 \sqrt {2}}-\frac {1}{2} \tanh ^{-1}\left (\sqrt [4]{3 x^2-1}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 86

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

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 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 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])

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 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 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 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 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 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 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]

Rubi steps

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

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Mathematica [C]  time = 0.02, size = 63, normalized size = 0.36 \begin {gather*} -\frac {1}{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 {1}{2} \tan ^{-1}\left (\sqrt [4]{3 x^2-1}\right )-\frac {1}{2} \tanh ^{-1}\left (\sqrt [4]{3 x^2-1}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.13, size = 122, normalized size = 0.71 \begin {gather*} \frac {1}{2} \tan ^{-1}\left (\sqrt [4]{3 x^2-1}\right )-\frac {\tan ^{-1}\left (\frac {\frac {\sqrt {3 x^2-1}}{\sqrt {2}}-\frac {1}{\sqrt {2}}}{\sqrt [4]{3 x^2-1}}\right )}{2 \sqrt {2}}-\frac {1}{2} \tanh ^{-1}\left (\sqrt [4]{3 x^2-1}\right )+\frac {\tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{3 x^2-1}}{\sqrt {3 x^2-1}+1}\right )}{2 \sqrt {2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.77, size = 215, normalized size = 1.24 \begin {gather*} \frac {1}{2} \, \sqrt {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 ) + \frac {1}{2} \, \sqrt {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 ) + \frac {1}{8} \, \sqrt {2} \log \left (4 \, \sqrt {2} {\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}} + 4 \, \sqrt {3 \, x^{2} - 1} + 4\right ) - \frac {1}{8} \, \sqrt {2} \log \left (-4 \, \sqrt {2} {\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}} + 4 \, \sqrt {3 \, x^{2} - 1} + 4\right ) + \frac {1}{2} \, \arctan \left ({\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}}\right ) - \frac {1}{4} \, \log \left ({\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}} + 1\right ) + \frac {1}{4} \, \log \left ({\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}} - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.36, size = 155, normalized size = 0.90 \begin {gather*} -\frac {1}{4} \, \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 {1}{4} \, \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 {1}{8} \, \sqrt {2} \log \left (\sqrt {2} {\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}} + \sqrt {3 \, x^{2} - 1} + 1\right ) - \frac {1}{8} \, \sqrt {2} \log \left (-\sqrt {2} {\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}} + \sqrt {3 \, x^{2} - 1} + 1\right ) + \frac {1}{2} \, \arctan \left ({\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}}\right ) - \frac {1}{4} \, \log \left ({\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}} + 1\right ) + \frac {1}{4} \, \log \left ({\left | {\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}} - 1 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*(3*x^2 - 1)^(1/4))) - 1/4*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(2) -
2*(3*x^2 - 1)^(1/4))) + 1/8*sqrt(2)*log(sqrt(2)*(3*x^2 - 1)^(1/4) + sqrt(3*x^2 - 1) + 1) - 1/8*sqrt(2)*log(-sq
rt(2)*(3*x^2 - 1)^(1/4) + sqrt(3*x^2 - 1) + 1) + 1/2*arctan((3*x^2 - 1)^(1/4)) - 1/4*log((3*x^2 - 1)^(1/4) + 1
) + 1/4*log(abs((3*x^2 - 1)^(1/4) - 1))

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maple [C]  time = 5.78, size = 301, normalized size = 1.74 \begin {gather*} \frac {\RootOf \left (\textit {\_Z}^{4}+1\right )^{3} \ln \left (-\frac {3 x^{2} \RootOf \left (\textit {\_Z}^{4}+1\right )^{3}-2 \RootOf \left (\textit {\_Z}^{4}+1\right )^{3}+2 \left (3 x^{2}-1\right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{4}+1\right )^{2}-2 \sqrt {3 x^{2}-1}\, \RootOf \left (\textit {\_Z}^{4}+1\right )+2 \left (3 x^{2}-1\right )^{\frac {3}{4}}}{x^{2}}\right )}{4}+\frac {\RootOf \left (\textit {\_Z}^{4}+1\right )^{2} \ln \left (-\frac {-3 x^{2} \RootOf \left (\textit {\_Z}^{4}+1\right )^{2}+2 \sqrt {3 x^{2}-1}\, \RootOf \left (\textit {\_Z}^{4}+1\right )^{2}+2 \left (3 x^{2}-1\right )^{\frac {3}{4}}-2 \left (3 x^{2}-1\right )^{\frac {1}{4}}}{3 x^{2}-2}\right )}{4}+\frac {\RootOf \left (\textit {\_Z}^{4}+1\right ) \ln \left (\frac {-3 x^{2} \RootOf \left (\textit {\_Z}^{4}+1\right )+2 \sqrt {3 x^{2}-1}\, \RootOf \left (\textit {\_Z}^{4}+1\right )^{3}+2 \left (3 x^{2}-1\right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{4}+1\right )^{2}+2 \RootOf \left (\textit {\_Z}^{4}+1\right )-2 \left (3 x^{2}-1\right )^{\frac {3}{4}}}{x^{2}}\right )}{4}+\frac {\ln \left (-\frac {-3 x^{2}+2 \left (3 x^{2}-1\right )^{\frac {3}{4}}-2 \sqrt {3 x^{2}-1}+2 \left (3 x^{2}-1\right )^{\frac {1}{4}}}{3 x^{2}-2}\right )}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*RootOf(_Z^4+1)*ln((2*(3*x^2-1)^(1/2)*RootOf(_Z^4+1)^3+2*RootOf(_Z^4+1)^2*(3*x^2-1)^(1/4)-3*RootOf(_Z^4+1)*
x^2-2*(3*x^2-1)^(3/4)+2*RootOf(_Z^4+1))/x^2)+1/4*RootOf(_Z^4+1)^3*ln(-(3*RootOf(_Z^4+1)^3*x^2-2*RootOf(_Z^4+1)
^3+2*RootOf(_Z^4+1)^2*(3*x^2-1)^(1/4)-2*(3*x^2-1)^(1/2)*RootOf(_Z^4+1)+2*(3*x^2-1)^(3/4))/x^2)+1/4*ln(-(-3*x^2
+2*(3*x^2-1)^(3/4)-2*(3*x^2-1)^(1/2)+2*(3*x^2-1)^(1/4))/(3*x^2-2))+1/4*RootOf(_Z^4+1)^2*ln(-(2*(3*x^2-1)^(1/2)
*RootOf(_Z^4+1)^2-3*RootOf(_Z^4+1)^2*x^2+2*(3*x^2-1)^(3/4)-2*(3*x^2-1)^(1/4))/(3*x^2-2))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{{\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}} {\left (3 \, x^{2} - 2\right )} x}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(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), x)

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mupad [B]  time = 0.19, size = 77, normalized size = 0.45 \begin {gather*} \frac {\mathrm {atan}\left ({\left (3\,x^2-1\right )}^{1/4}\right )}{2}+\frac {\mathrm {atan}\left ({\left (3\,x^2-1\right )}^{1/4}\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2}+\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,{\left (3\,x^2-1\right )}^{1/4}\,\left (\frac {1}{2}-\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (-\frac {1}{4}+\frac {1}{4}{}\mathrm {i}\right )+\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,{\left (3\,x^2-1\right )}^{1/4}\,\left (\frac {1}{2}+\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (-\frac {1}{4}-\frac {1}{4}{}\mathrm {i}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

atan((3*x^2 - 1)^(1/4))/2 + (atan((3*x^2 - 1)^(1/4)*1i)*1i)/2 - 2^(1/2)*atan(2^(1/2)*(3*x^2 - 1)^(1/4)*(1/2 -
1i/2))*(1/4 - 1i/4) - 2^(1/2)*atan(2^(1/2)*(3*x^2 - 1)^(1/4)*(1/2 + 1i/2))*(1/4 + 1i/4)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x \left (3 x^{2} - 2\right ) \sqrt [4]{3 x^{2} - 1}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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