3.11.85 \(\int \frac {1}{x (-2+3 x^2) (-1+3 x^2)^{3/4}} \, dx\) [1085]

3.11.85.1 Optimal result

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

output

-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*arctan(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)
 

3.11.85.2 Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.64 \[ \int \frac {1}{x \left (-2+3 x^2\right ) \left (-1+3 x^2\right )^{3/4}} \, dx=\frac {1}{4} \left (-2 \arctan \left (\sqrt [4]{-1+3 x^2}\right )-\sqrt {2} \arctan \left (\frac {-1+\sqrt {-1+3 x^2}}{\sqrt {2} \sqrt [4]{-1+3 x^2}}\right )-2 \text {arctanh}\left (\sqrt [4]{-1+3 x^2}\right )-\sqrt {2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{-1+3 x^2}}{1+\sqrt {-1+3 x^2}}\right )\right ) \]

input

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

output

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

3.11.85.3 Rubi [A] (warning: unable to verify)

Time = 0.35 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.01, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {354, 25, 97, 73, 755, 27, 756, 216, 219, 1476, 1082, 217, 1479, 25, 27, 1103}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {1}{x \left (3 x^2-2\right ) \left (3 x^2-1\right )^{3/4}} \, dx\)

\(\Big \downarrow \) 354

\(\displaystyle \frac {1}{2} \int -\frac {1}{x^2 \left (2-3 x^2\right ) \left (3 x^2-1\right )^{3/4}}dx^2\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {1}{2} \int \frac {1}{x^2 \left (2-3 x^2\right ) \left (3 x^2-1\right )^{3/4}}dx^2\)

\(\Big \downarrow \) 97

\(\displaystyle \frac {1}{2} \left (-\frac {1}{2} \int \frac {1}{x^2 \left (3 x^2-1\right )^{3/4}}dx^2-\frac {3}{2} \int \frac {1}{\left (2-3 x^2\right ) \left (3 x^2-1\right )^{3/4}}dx^2\right )\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {1}{2} \left (-2 \int \frac {1}{1-x^8}d\sqrt [4]{3 x^2-1}-\frac {2}{3} \int \frac {1}{\frac {x^8}{3}+\frac {1}{3}}d\sqrt [4]{3 x^2-1}\right )\)

\(\Big \downarrow \) 755

\(\displaystyle \frac {1}{2} \left (-2 \int \frac {1}{1-x^8}d\sqrt [4]{3 x^2-1}-\frac {2}{3} \left (\frac {1}{2} \int \frac {3 \left (1-x^4\right )}{x^8+1}d\sqrt [4]{3 x^2-1}+\frac {1}{2} \int \frac {3 \left (x^4+1\right )}{x^8+1}d\sqrt [4]{3 x^2-1}\right )\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{2} \left (-2 \int \frac {1}{1-x^8}d\sqrt [4]{3 x^2-1}-\frac {2}{3} \left (\frac {3}{2} \int \frac {1-x^4}{x^8+1}d\sqrt [4]{3 x^2-1}+\frac {3}{2} \int \frac {x^4+1}{x^8+1}d\sqrt [4]{3 x^2-1}\right )\right )\)

\(\Big \downarrow \) 756

\(\displaystyle \frac {1}{2} \left (-2 \left (\frac {1}{2} \int \frac {1}{1-x^4}d\sqrt [4]{3 x^2-1}+\frac {1}{2} \int \frac {1}{x^4+1}d\sqrt [4]{3 x^2-1}\right )-\frac {2}{3} \left (\frac {3}{2} \int \frac {1-x^4}{x^8+1}d\sqrt [4]{3 x^2-1}+\frac {3}{2} \int \frac {x^4+1}{x^8+1}d\sqrt [4]{3 x^2-1}\right )\right )\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {1}{2} \left (-2 \left (\frac {1}{2} \int \frac {1}{1-x^4}d\sqrt [4]{3 x^2-1}+\frac {1}{2} \arctan \left (\sqrt [4]{3 x^2-1}\right )\right )-\frac {2}{3} \left (\frac {3}{2} \int \frac {1-x^4}{x^8+1}d\sqrt [4]{3 x^2-1}+\frac {3}{2} \int \frac {x^4+1}{x^8+1}d\sqrt [4]{3 x^2-1}\right )\right )\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {1}{2} \left (-\frac {2}{3} \left (\frac {3}{2} \int \frac {1-x^4}{x^8+1}d\sqrt [4]{3 x^2-1}+\frac {3}{2} \int \frac {x^4+1}{x^8+1}d\sqrt [4]{3 x^2-1}\right )-2 \left (\frac {1}{2} \arctan \left (\sqrt [4]{3 x^2-1}\right )+\frac {1}{2} \text {arctanh}\left (\sqrt [4]{3 x^2-1}\right )\right )\right )\)

\(\Big \downarrow \) 1476

\(\displaystyle \frac {1}{2} \left (-\frac {2}{3} \left (\frac {3}{2} \left (\frac {1}{2} \int \frac {1}{x^4-\sqrt {2} \sqrt [4]{3 x^2-1}+1}d\sqrt [4]{3 x^2-1}+\frac {1}{2} \int \frac {1}{x^4+\sqrt {2} \sqrt [4]{3 x^2-1}+1}d\sqrt [4]{3 x^2-1}\right )+\frac {3}{2} \int \frac {1-x^4}{x^8+1}d\sqrt [4]{3 x^2-1}\right )-2 \left (\frac {1}{2} \arctan \left (\sqrt [4]{3 x^2-1}\right )+\frac {1}{2} \text {arctanh}\left (\sqrt [4]{3 x^2-1}\right )\right )\right )\)

\(\Big \downarrow \) 1082

\(\displaystyle \frac {1}{2} \left (-\frac {2}{3} \left (\frac {3}{2} \left (\frac {\int \frac {1}{-x^4-1}d\left (1-\sqrt {2} \sqrt [4]{3 x^2-1}\right )}{\sqrt {2}}-\frac {\int \frac {1}{-x^4-1}d\left (\sqrt {2} \sqrt [4]{3 x^2-1}+1\right )}{\sqrt {2}}\right )+\frac {3}{2} \int \frac {1-x^4}{x^8+1}d\sqrt [4]{3 x^2-1}\right )-2 \left (\frac {1}{2} \arctan \left (\sqrt [4]{3 x^2-1}\right )+\frac {1}{2} \text {arctanh}\left (\sqrt [4]{3 x^2-1}\right )\right )\right )\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {1}{2} \left (-\frac {2}{3} \left (\frac {3}{2} \int \frac {1-x^4}{x^8+1}d\sqrt [4]{3 x^2-1}+\frac {3}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt [4]{3 x^2-1}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt [4]{3 x^2-1}\right )}{\sqrt {2}}\right )\right )-2 \left (\frac {1}{2} \arctan \left (\sqrt [4]{3 x^2-1}\right )+\frac {1}{2} \text {arctanh}\left (\sqrt [4]{3 x^2-1}\right )\right )\right )\)

\(\Big \downarrow \) 1479

\(\displaystyle \frac {1}{2} \left (-\frac {2}{3} \left (\frac {3}{2} \left (-\frac {\int -\frac {\sqrt {2}-2 \sqrt [4]{3 x^2-1}}{x^4-\sqrt {2} \sqrt [4]{3 x^2-1}+1}d\sqrt [4]{3 x^2-1}}{2 \sqrt {2}}-\frac {\int -\frac {\sqrt {2} \left (\sqrt {2} \sqrt [4]{3 x^2-1}+1\right )}{x^4+\sqrt {2} \sqrt [4]{3 x^2-1}+1}d\sqrt [4]{3 x^2-1}}{2 \sqrt {2}}\right )+\frac {3}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt [4]{3 x^2-1}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt [4]{3 x^2-1}\right )}{\sqrt {2}}\right )\right )-2 \left (\frac {1}{2} \arctan \left (\sqrt [4]{3 x^2-1}\right )+\frac {1}{2} \text {arctanh}\left (\sqrt [4]{3 x^2-1}\right )\right )\right )\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {1}{2} \left (-\frac {2}{3} \left (\frac {3}{2} \left (\frac {\int \frac {\sqrt {2}-2 \sqrt [4]{3 x^2-1}}{x^4-\sqrt {2} \sqrt [4]{3 x^2-1}+1}d\sqrt [4]{3 x^2-1}}{2 \sqrt {2}}+\frac {\int \frac {\sqrt {2} \left (\sqrt {2} \sqrt [4]{3 x^2-1}+1\right )}{x^4+\sqrt {2} \sqrt [4]{3 x^2-1}+1}d\sqrt [4]{3 x^2-1}}{2 \sqrt {2}}\right )+\frac {3}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt [4]{3 x^2-1}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt [4]{3 x^2-1}\right )}{\sqrt {2}}\right )\right )-2 \left (\frac {1}{2} \arctan \left (\sqrt [4]{3 x^2-1}\right )+\frac {1}{2} \text {arctanh}\left (\sqrt [4]{3 x^2-1}\right )\right )\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{2} \left (-\frac {2}{3} \left (\frac {3}{2} \left (\frac {\int \frac {\sqrt {2}-2 \sqrt [4]{3 x^2-1}}{x^4-\sqrt {2} \sqrt [4]{3 x^2-1}+1}d\sqrt [4]{3 x^2-1}}{2 \sqrt {2}}+\frac {1}{2} \int \frac {\sqrt {2} \sqrt [4]{3 x^2-1}+1}{x^4+\sqrt {2} \sqrt [4]{3 x^2-1}+1}d\sqrt [4]{3 x^2-1}\right )+\frac {3}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt [4]{3 x^2-1}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt [4]{3 x^2-1}\right )}{\sqrt {2}}\right )\right )-2 \left (\frac {1}{2} \arctan \left (\sqrt [4]{3 x^2-1}\right )+\frac {1}{2} \text {arctanh}\left (\sqrt [4]{3 x^2-1}\right )\right )\right )\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {1}{2} \left (-2 \left (\frac {1}{2} \arctan \left (\sqrt [4]{3 x^2-1}\right )+\frac {1}{2} \text {arctanh}\left (\sqrt [4]{3 x^2-1}\right )\right )-\frac {2}{3} \left (\frac {3}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt [4]{3 x^2-1}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt [4]{3 x^2-1}\right )}{\sqrt {2}}\right )+\frac {3}{2} \left (\frac {\log \left (x^4+\sqrt {2} \sqrt [4]{3 x^2-1}+1\right )}{2 \sqrt {2}}-\frac {\log \left (x^4-\sqrt {2} \sqrt [4]{3 x^2-1}+1\right )}{2 \sqrt {2}}\right )\right )\right )\)

input

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

output

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

3.11.85.3.1 Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[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] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 97

Int[((e_.) + (f_.)*(x_))^(p_)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), 
x_] :> Simp[b/(b*c - a*d)   Int[(e + f*x)^p/(a + b*x), x], x] - Simp[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 216

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

rule 217

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 219

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] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 354

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_S 
ymbol] :> Simp[1/2   Subst[Int[x^((m - 1)/2)*(a + b*x)^p*(c + d*x)^q, x], x 
, x^2], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IntegerQ 
[(m - 1)/2]
 

rule 755

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

rule 756

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

rule 1082

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S 
implify[a*(c/b^2)]}, Simp[-2/b   Subst[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])] /; Fre 
eQ[{a, b, c}, x]
 

rule 1103

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[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 1476

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 
2*(d/e), 2]}, Simp[e/(2*c)   Int[1/Simp[d/e + q*x + x^2, x], x], x] + Simp[ 
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 1479

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

3.11.85.4 Maple [A] (verified)

Time = 3.88 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.84

method	result	size
pseudoelliptic	\(\frac {\ln \left (-1+\left (3 x^{2}-1\right )^{\frac {1}{4}}\right )}{4}-\frac {\ln \left (\frac {-\left (3 x^{2}-1\right )^{\frac {1}{4}} \sqrt {2}-\sqrt {3 x^{2}-1}-1}{\left (3 x^{2}-1\right )^{\frac {1}{4}} \sqrt {2}-\sqrt {3 x^{2}-1}-1}\right ) \sqrt {2}}{8}-\frac {\arctan \left (1+\left (3 x^{2}-1\right )^{\frac {1}{4}} \sqrt {2}\right ) \sqrt {2}}{4}-\frac {\arctan \left (-1+\left (3 x^{2}-1\right )^{\frac {1}{4}} \sqrt {2}\right ) \sqrt {2}}{4}-\frac {\ln \left (1+\left (3 x^{2}-1\right )^{\frac {1}{4}}\right )}{4}-\frac {\arctan \left (\left (3 x^{2}-1\right )^{\frac {1}{4}}\right )}{2}\)	\(146\)
trager	\(-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{3} \ln \left (-\frac {2 \sqrt {3 x^{2}-1}\, \operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{3}-2 \operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{2} \left (3 x^{2}-1\right )^{\frac {1}{4}}-3 \operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right ) x^{2}+2 \left (3 x^{2}-1\right )^{\frac {3}{4}}+2 \operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )}{x^{2}}\right )}{4}+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right ) \ln \left (-\frac {3 \operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{3} x^{2}-2 \operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{3}+2 \operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{2} \left (3 x^{2}-1\right )^{\frac {1}{4}}-2 \sqrt {3 x^{2}-1}\, \operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )+2 \left (3 x^{2}-1\right )^{\frac {3}{4}}}{x^{2}}\right )}{4}-\frac {\ln \left (-\frac {2 \left (3 x^{2}-1\right )^{\frac {3}{4}}+2 \sqrt {3 x^{2}-1}+3 x^{2}+2 \left (3 x^{2}-1\right )^{\frac {1}{4}}}{3 x^{2}-2}\right )}{4}+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{2} \ln \left (\frac {2 \operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{2} \sqrt {3 x^{2}-1}-3 \operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{2} x^{2}-2 \left (3 x^{2}-1\right )^{\frac {3}{4}}+2 \left (3 x^{2}-1\right )^{\frac {1}{4}}}{3 x^{2}-2}\right )}{4}\)	\(301\)

input

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

output

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

3.11.85.5 Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.28 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.77 \[ \int \frac {1}{x \left (-2+3 x^2\right ) \left (-1+3 x^2\right )^{3/4}} \, dx=-\left (\frac {1}{8} i + \frac {1}{8}\right ) \, \sqrt {2} \log \left (\left (i + 1\right ) \, \sqrt {2} + 2 \, {\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}}\right ) + \left (\frac {1}{8} i - \frac {1}{8}\right ) \, \sqrt {2} \log \left (-\left (i - 1\right ) \, \sqrt {2} + 2 \, {\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}}\right ) - \left (\frac {1}{8} i - \frac {1}{8}\right ) \, \sqrt {2} \log \left (\left (i - 1\right ) \, \sqrt {2} + 2 \, {\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}}\right ) + \left (\frac {1}{8} i + \frac {1}{8}\right ) \, \sqrt {2} \log \left (-\left (i + 1\right ) \, \sqrt {2} + 2 \, {\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}}\right ) - \frac {1}{2} \, \arctan \left ({\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}}\right ) - \frac {1}{4} \, \log \left ({\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}} + 1\right ) + \frac {1}{4} \, \log \left ({\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}} - 1\right ) \]

input

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

output

-(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)*sqrt( 
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)
 

3.11.85.6 Sympy [F]

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

input

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

output

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

3.11.85.7 Maxima [F]

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

input

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

output

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

3.11.85.8 Giac [A] (verification not implemented)

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

input

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

output

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

3.11.85.9 Mupad [B] (verification not implemented)

Time = 5.37 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.45 \[ \int \frac {1}{x \left (-2+3 x^2\right ) \left (-1+3 x^2\right )^{3/4}} \, dx=-\frac {\mathrm {atan}\left ({\left (3\,x^2-1\right )}^{1/4}\right )}{2}+\frac {\mathrm {atan}\left ({\left (3\,x^2-1\right )}^{1/4}\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2}+\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,{\left (3\,x^2-1\right )}^{1/4}\,\left (\frac {1}{2}-\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (-\frac {1}{4}-\frac {1}{4}{}\mathrm {i}\right )+\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,{\left (3\,x^2-1\right )}^{1/4}\,\left (\frac {1}{2}+\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (-\frac {1}{4}+\frac {1}{4}{}\mathrm {i}\right ) \]

input

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

output

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