\(\int \frac {x}{(-2+3 x^2) (-1+3 x^2)^{3/4}} \, dx\) [1084]

Optimal result

Integrand size = 22, antiderivative size = 33 \[ \int \frac {x}{\left (-2+3 x^2\right ) \left (-1+3 x^2\right )^{3/4}} \, dx=-\frac {1}{3} \arctan \left (\sqrt [4]{-1+3 x^2}\right )-\frac {1}{3} \text {arctanh}\left (\sqrt [4]{-1+3 x^2}\right ) \]

[Out]

-1/3*arctan((3*x^2-1)^(1/4))-1/3*arctanh((3*x^2-1)^(1/4))

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {455, 65, 218, 212, 209} \[ \int \frac {x}{\left (-2+3 x^2\right ) \left (-1+3 x^2\right )^{3/4}} \, dx=-\frac {1}{3} \arctan \left (\sqrt [4]{3 x^2-1}\right )-\frac {1}{3} \text {arctanh}\left (\sqrt [4]{3 x^2-1}\right ) \]

[In]

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

[Out]

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

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 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 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 218

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

Rule 455

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(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] && EqQ[m
- n + 1, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {1}{(-2+3 x) (-1+3 x)^{3/4}} \, dx,x,x^2\right ) \\ & = \frac {2}{3} \text {Subst}\left (\int \frac {1}{-1+x^4} \, dx,x,\sqrt [4]{-1+3 x^2}\right ) \\ & = -\left (\frac {1}{3} \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\sqrt [4]{-1+3 x^2}\right )\right )-\frac {1}{3} \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\sqrt [4]{-1+3 x^2}\right ) \\ & = -\frac {1}{3} \tan ^{-1}\left (\sqrt [4]{-1+3 x^2}\right )-\frac {1}{3} \tanh ^{-1}\left (\sqrt [4]{-1+3 x^2}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.00 \[ \int \frac {x}{\left (-2+3 x^2\right ) \left (-1+3 x^2\right )^{3/4}} \, dx=-\frac {1}{3} \arctan \left (\sqrt [4]{-1+3 x^2}\right )-\frac {1}{3} \text {arctanh}\left (\sqrt [4]{-1+3 x^2}\right ) \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 3.21 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.79

method	result	size
pseudoelliptic	\(-\frac {\arctan \left (\left (3 x^{2}-1\right )^{\frac {1}{4}}\right )}{3}-\frac {\operatorname {arctanh}\left (\left (3 x^{2}-1\right )^{\frac {1}{4}}\right )}{3}\)	\(26\)
trager	\(\frac {\ln \left (\frac {2 \left (3 x^{2}-1\right )^{\frac {3}{4}}-2 \sqrt {3 x^{2}-1}-3 x^{2}+2 \left (3 x^{2}-1\right )^{\frac {1}{4}}}{3 x^{2}-2}\right )}{6}-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (-\frac {2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \left (3 x^{2}-1\right )^{\frac {3}{4}}-2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \left (3 x^{2}-1\right )^{\frac {1}{4}}-2 \sqrt {3 x^{2}-1}+3 x^{2}}{3 x^{2}-2}\right )}{6}\)	\(125\)

[In]

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

[Out]

-1/3*arctan((3*x^2-1)^(1/4))-1/3*arctanh((3*x^2-1)^(1/4))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.24 \[ \int \frac {x}{\left (-2+3 x^2\right ) \left (-1+3 x^2\right )^{3/4}} \, dx=-\frac {1}{3} \, \arctan \left ({\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}}\right ) - \frac {1}{6} \, \log \left ({\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}} + 1\right ) + \frac {1}{6} \, \log \left ({\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}} - 1\right ) \]

[In]

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

[Out]

-1/3*arctan((3*x^2 - 1)^(1/4)) - 1/6*log((3*x^2 - 1)^(1/4) + 1) + 1/6*log((3*x^2 - 1)^(1/4) - 1)

Sympy [A] (verification not implemented)

Time = 3.14 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.27 \[ \int \frac {x}{\left (-2+3 x^2\right ) \left (-1+3 x^2\right )^{3/4}} \, dx=\frac {\log {\left (\sqrt [4]{3 x^{2} - 1} - 1 \right )}}{6} - \frac {\log {\left (\sqrt [4]{3 x^{2} - 1} + 1 \right )}}{6} - \frac {\operatorname {atan}{\left (\sqrt [4]{3 x^{2} - 1} \right )}}{3} \]

[In]

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

[Out]

log((3*x**2 - 1)**(1/4) - 1)/6 - log((3*x**2 - 1)**(1/4) + 1)/6 - atan((3*x**2 - 1)**(1/4))/3

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.24 \[ \int \frac {x}{\left (-2+3 x^2\right ) \left (-1+3 x^2\right )^{3/4}} \, dx=-\frac {1}{3} \, \arctan \left ({\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}}\right ) - \frac {1}{6} \, \log \left ({\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}} + 1\right ) + \frac {1}{6} \, \log \left ({\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}} - 1\right ) \]

[In]

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

[Out]

-1/3*arctan((3*x^2 - 1)^(1/4)) - 1/6*log((3*x^2 - 1)^(1/4) + 1) + 1/6*log((3*x^2 - 1)^(1/4) - 1)

Giac [A] (verification not implemented)

none

Time = 0.37 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.27 \[ \int \frac {x}{\left (-2+3 x^2\right ) \left (-1+3 x^2\right )^{3/4}} \, dx=-\frac {1}{3} \, \arctan \left ({\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}}\right ) - \frac {1}{6} \, \log \left ({\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}} + 1\right ) + \frac {1}{6} \, \log \left ({\left | {\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}} - 1 \right |}\right ) \]

[In]

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

[Out]

-1/3*arctan((3*x^2 - 1)^(1/4)) - 1/6*log((3*x^2 - 1)^(1/4) + 1) + 1/6*log(abs((3*x^2 - 1)^(1/4) - 1))

Mupad [B] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.76 \[ \int \frac {x}{\left (-2+3 x^2\right ) \left (-1+3 x^2\right )^{3/4}} \, dx=-\frac {\mathrm {atan}\left ({\left (3\,x^2-1\right )}^{1/4}\right )}{3}-\frac {\mathrm {atanh}\left ({\left (3\,x^2-1\right )}^{1/4}\right )}{3} \]

[In]

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

[Out]

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