\(\int \frac {1}{x (-2+3 x^2) \sqrt [4]{-1+3 x^2}} \, dx\) [1046]

Optimal result

Integrand size = 24, antiderivative size = 173 \[ \int \frac {1}{x \left (-2+3 x^2\right ) \sqrt [4]{-1+3 x^2}} \, dx=\frac {1}{2} \arctan \left (\sqrt [4]{-1+3 x^2}\right )+\frac {\arctan \left (1-\sqrt {2} \sqrt [4]{-1+3 x^2}\right )}{2 \sqrt {2}}-\frac {\arctan \left (1+\sqrt {2} \sqrt [4]{-1+3 x^2}\right )}{2 \sqrt {2}}-\frac {1}{2} \text {arctanh}\left (\sqrt [4]{-1+3 x^2}\right )-\frac {\log \left (1-\sqrt {2} \sqrt [4]{-1+3 x^2}+\sqrt {-1+3 x^2}\right )}{4 \sqrt {2}}+\frac {\log \left (1+\sqrt {2} \sqrt [4]{-1+3 x^2}+\sqrt {-1+3 x^2}\right )}{4 \sqrt {2}} \]

[Out]

1/2*arctan((3*x^2-1)^(1/4))-1/2*arctanh((3*x^2-1)^(1/4))-1/4*arctan(-1+(3*x^2-1)^(1/4)*2^(1/2))*2^(1/2)-1/4*ar
ctan(1+(3*x^2-1)^(1/4)*2^(1/2))*2^(1/2)-1/8*ln(1-(3*x^2-1)^(1/4)*2^(1/2)+(3*x^2-1)^(1/2))*2^(1/2)+1/8*ln(1+(3*
x^2-1)^(1/4)*2^(1/2)+(3*x^2-1)^(1/2))*2^(1/2)

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {457, 88, 65, 303, 1176, 631, 210, 1179, 642, 304, 209, 212} \[ \int \frac {1}{x \left (-2+3 x^2\right ) \sqrt [4]{-1+3 x^2}} \, dx=\frac {1}{2} \arctan \left (\sqrt [4]{3 x^2-1}\right )+\frac {\arctan \left (1-\sqrt {2} \sqrt [4]{3 x^2-1}\right )}{2 \sqrt {2}}-\frac {\arctan \left (\sqrt {2} \sqrt [4]{3 x^2-1}+1\right )}{2 \sqrt {2}}-\frac {1}{2} \text {arctanh}\left (\sqrt [4]{3 x^2-1}\right )-\frac {\log \left (\sqrt {3 x^2-1}-\sqrt {2} \sqrt [4]{3 x^2-1}+1\right )}{4 \sqrt {2}}+\frac {\log \left (\sqrt {3 x^2-1}+\sqrt {2} \sqrt [4]{3 x^2-1}+1\right )}{4 \sqrt {2}} \]

[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 65

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 88

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 209

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

Rule 210

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

Rule 212

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

Rule 303

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 304

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 457

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 631

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 642

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 1176

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 1179

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*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {1}{x (-2+3 x) \sqrt [4]{-1+3 x}} \, dx,x,x^2\right ) \\ & = -\left (\frac {1}{4} \text {Subst}\left (\int \frac {1}{x \sqrt [4]{-1+3 x}} \, dx,x,x^2\right )\right )+\frac {3}{4} \text {Subst}\left (\int \frac {1}{(-2+3 x) \sqrt [4]{-1+3 x}} \, dx,x,x^2\right ) \\ & = -\left (\frac {1}{3} \text {Subst}\left (\int \frac {x^2}{\frac {1}{3}+\frac {x^4}{3}} \, dx,x,\sqrt [4]{-1+3 x^2}\right )\right )+\text {Subst}\left (\int \frac {x^2}{-1+x^4} \, dx,x,\sqrt [4]{-1+3 x^2}\right ) \\ & = \frac {1}{6} \text {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} \text {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} \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\sqrt [4]{-1+3 x^2}\right )+\frac {1}{2} \text {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} \text {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,\sqrt [4]{-1+3 x^2}\right )-\frac {1}{4} \text {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,\sqrt [4]{-1+3 x^2}\right )-\frac {\text {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 {\text {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 {\text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} \sqrt [4]{-1+3 x^2}\right )}{2 \sqrt {2}}+\frac {\text {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*}

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.64 \[ \int \frac {1}{x \left (-2+3 x^2\right ) \sqrt [4]{-1+3 x^2}} \, dx=\frac {1}{4} \left (2 \arctan \left (\sqrt [4]{-1+3 x^2}\right )-\sqrt {2} \arctan \left (\frac {-1+\sqrt {-1+3 x^2}}{\sqrt {2} \sqrt [4]{-1+3 x^2}}\right )-2 \text {arctanh}\left (\sqrt [4]{-1+3 x^2}\right )+\sqrt {2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{-1+3 x^2}}{1+\sqrt {-1+3 x^2}}\right )\right ) \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 4.52 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.82

method	result	size
pseudoelliptic	\(\frac {\ln \left (-1+\left (3 x^{2}-1\right )^{\frac {1}{4}}\right )}{4}-\frac {\ln \left (\frac {1-\left (3 x^{2}-1\right )^{\frac {1}{4}} \sqrt {2}+\sqrt {3 x^{2}-1}}{1+\left (3 x^{2}-1\right )^{\frac {1}{4}} \sqrt {2}+\sqrt {3 x^{2}-1}}\right ) \sqrt {2}}{8}-\frac {\arctan \left (1+\left (3 x^{2}-1\right )^{\frac {1}{4}} \sqrt {2}\right ) \sqrt {2}}{4}-\frac {\arctan \left (-1+\left (3 x^{2}-1\right )^{\frac {1}{4}} \sqrt {2}\right ) \sqrt {2}}{4}-\frac {\ln \left (1+\left (3 x^{2}-1\right )^{\frac {1}{4}}\right )}{4}+\frac {\arctan \left (\left (3 x^{2}-1\right )^{\frac {1}{4}}\right )}{2}\)	\(142\)
trager	\(-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right ) \ln \left (-\frac {2 \sqrt {3 x^{2}-1}\, \operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{3}-2 \operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{2} \left (3 x^{2}-1\right )^{\frac {1}{4}}-3 \operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right ) x^{2}+2 \left (3 x^{2}-1\right )^{\frac {3}{4}}+2 \operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )}{x^{2}}\right )}{4}+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{3} \ln \left (-\frac {3 \operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{3} x^{2}-2 \operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{3}+2 \operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{2} \left (3 x^{2}-1\right )^{\frac {1}{4}}-2 \sqrt {3 x^{2}-1}\, \operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )+2 \left (3 x^{2}-1\right )^{\frac {3}{4}}}{x^{2}}\right )}{4}+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{2} \ln \left (-\frac {2 \operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{2} \sqrt {3 x^{2}-1}-3 \operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{2} x^{2}+2 \left (3 x^{2}-1\right )^{\frac {3}{4}}-2 \left (3 x^{2}-1\right )^{\frac {1}{4}}}{3 x^{2}-2}\right )}{4}-\frac {\ln \left (-\frac {2 \left (3 x^{2}-1\right )^{\frac {3}{4}}+2 \sqrt {3 x^{2}-1}+3 x^{2}+2 \left (3 x^{2}-1\right )^{\frac {1}{4}}}{3 x^{2}-2}\right )}{4}\)	\(302\)

[In]

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

[Out]

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

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.27 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.77 \[ \int \frac {1}{x \left (-2+3 x^2\right ) \sqrt [4]{-1+3 x^2}} \, dx=-\left (\frac {1}{8} i - \frac {1}{8}\right ) \, \sqrt {2} \log \left (\left (i + 1\right ) \, \sqrt {2} + 2 \, {\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}}\right ) + \left (\frac {1}{8} i + \frac {1}{8}\right ) \, \sqrt {2} \log \left (-\left (i - 1\right ) \, \sqrt {2} + 2 \, {\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}}\right ) - \left (\frac {1}{8} i + \frac {1}{8}\right ) \, \sqrt {2} \log \left (\left (i - 1\right ) \, \sqrt {2} + 2 \, {\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}}\right ) + \left (\frac {1}{8} i - \frac {1}{8}\right ) \, \sqrt {2} \log \left (-\left (i + 1\right ) \, \sqrt {2} + 2 \, {\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}}\right ) + \frac {1}{2} \, \arctan \left ({\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}}\right ) - \frac {1}{4} \, \log \left ({\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}} + 1\right ) + \frac {1}{4} \, \log \left ({\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}} - 1\right ) \]

[In]

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

[Out]

-(1/8*I - 1/8)*sqrt(2)*log((I + 1)*sqrt(2) + 2*(3*x^2 - 1)^(1/4)) + (1/8*I + 1/8)*sqrt(2)*log(-(I - 1)*sqrt(2)
+ 2*(3*x^2 - 1)^(1/4)) - (1/8*I + 1/8)*sqrt(2)*log((I - 1)*sqrt(2) + 2*(3*x^2 - 1)^(1/4)) + (1/8*I - 1/8)*sqr
t(2)*log(-(I + 1)*sqrt(2) + 2*(3*x^2 - 1)^(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)

Sympy [F]

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

[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)

Maxima [F]

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

[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)

Giac [A] (verification not implemented)

none

Time = 0.37 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.90 \[ \int \frac {1}{x \left (-2+3 x^2\right ) \sqrt [4]{-1+3 x^2}} \, dx=-\frac {1}{4} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, {\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}}\right )}\right ) - \frac {1}{4} \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, {\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}}\right )}\right ) + \frac {1}{8} \, \sqrt {2} \log \left (\sqrt {2} {\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}} + \sqrt {3 \, x^{2} - 1} + 1\right ) - \frac {1}{8} \, \sqrt {2} \log \left (-\sqrt {2} {\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}} + \sqrt {3 \, x^{2} - 1} + 1\right ) + \frac {1}{2} \, \arctan \left ({\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}}\right ) - \frac {1}{4} \, \log \left ({\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}} + 1\right ) + \frac {1}{4} \, \log \left ({\left | {\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}} - 1 \right |}\right ) \]

[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))

Mupad [B] (verification not implemented)

Time = 5.49 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.45 \[ \int \frac {1}{x \left (-2+3 x^2\right ) \sqrt [4]{-1+3 x^2}} \, dx=\frac {\mathrm {atan}\left ({\left (3\,x^2-1\right )}^{1/4}\right )}{2}+\frac {\mathrm {atan}\left ({\left (3\,x^2-1\right )}^{1/4}\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2}+\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,{\left (3\,x^2-1\right )}^{1/4}\,\left (\frac {1}{2}-\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (-\frac {1}{4}+\frac {1}{4}{}\mathrm {i}\right )+\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,{\left (3\,x^2-1\right )}^{1/4}\,\left (\frac {1}{2}+\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (-\frac {1}{4}-\frac {1}{4}{}\mathrm {i}\right ) \]

[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)