\(\int \frac {x^3}{(-2+3 x^2) \sqrt [4]{-1+3 x^2}} \, dx\) [1044]

Optimal result

Integrand size = 24, antiderivative size = 48 \[ \int \frac {x^3}{\left (-2+3 x^2\right ) \sqrt [4]{-1+3 x^2}} \, dx=\frac {2}{27} \left (-1+3 x^2\right )^{3/4}+\frac {2}{9} \arctan \left (\sqrt [4]{-1+3 x^2}\right )-\frac {2}{9} \text {arctanh}\left (\sqrt [4]{-1+3 x^2}\right ) \]

[Out]

2/27*(3*x^2-1)^(3/4)+2/9*arctan((3*x^2-1)^(1/4))-2/9*arctanh((3*x^2-1)^(1/4))

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {457, 81, 65, 304, 209, 212} \[ \int \frac {x^3}{\left (-2+3 x^2\right ) \sqrt [4]{-1+3 x^2}} \, dx=\frac {2}{9} \arctan \left (\sqrt [4]{3 x^2-1}\right )-\frac {2}{9} \text {arctanh}\left (\sqrt [4]{3 x^2-1}\right )+\frac {2}{27} \left (3 x^2-1\right )^{3/4} \]

[In]

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

[Out]

(2*(-1 + 3*x^2)^(3/4))/27 + (2*ArcTan[(-1 + 3*x^2)^(1/4)])/9 - (2*ArcTanh[(-1 + 3*x^2)^(1/4)])/9

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 81

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[b*(c + d*x)^
(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
0]

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 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]]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {x}{(-2+3 x) \sqrt [4]{-1+3 x}} \, dx,x,x^2\right ) \\ & = \frac {2}{27} \left (-1+3 x^2\right )^{3/4}+\frac {1}{3} \text {Subst}\left (\int \frac {1}{(-2+3 x) \sqrt [4]{-1+3 x}} \, dx,x,x^2\right ) \\ & = \frac {2}{27} \left (-1+3 x^2\right )^{3/4}+\frac {4}{9} \text {Subst}\left (\int \frac {x^2}{-1+x^4} \, dx,x,\sqrt [4]{-1+3 x^2}\right ) \\ & = \frac {2}{27} \left (-1+3 x^2\right )^{3/4}-\frac {2}{9} \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\sqrt [4]{-1+3 x^2}\right )+\frac {2}{9} \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\sqrt [4]{-1+3 x^2}\right ) \\ & = \frac {2}{27} \left (-1+3 x^2\right )^{3/4}+\frac {2}{9} \tan ^{-1}\left (\sqrt [4]{-1+3 x^2}\right )-\frac {2}{9} \tanh ^{-1}\left (\sqrt [4]{-1+3 x^2}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.92 \[ \int \frac {x^3}{\left (-2+3 x^2\right ) \sqrt [4]{-1+3 x^2}} \, dx=\frac {2}{27} \left (\left (-1+3 x^2\right )^{3/4}+3 \arctan \left (\sqrt [4]{-1+3 x^2}\right )-3 \text {arctanh}\left (\sqrt [4]{-1+3 x^2}\right )\right ) \]

[In]

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

[Out]

(2*((-1 + 3*x^2)^(3/4) + 3*ArcTan[(-1 + 3*x^2)^(1/4)] - 3*ArcTanh[(-1 + 3*x^2)^(1/4)]))/27

Maple [A] (verified)

Time = 3.88 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.10

method	result	size
pseudoelliptic	\(\frac {2 \left (3 x^{2}-1\right )^{\frac {3}{4}}}{27}+\frac {\ln \left (-1+\left (3 x^{2}-1\right )^{\frac {1}{4}}\right )}{9}-\frac {\ln \left (1+\left (3 x^{2}-1\right )^{\frac {1}{4}}\right )}{9}+\frac {2 \arctan \left (\left (3 x^{2}-1\right )^{\frac {1}{4}}\right )}{9}\)	\(53\)
trager	\(\frac {2 \left (3 x^{2}-1\right )^{\frac {3}{4}}}{27}-\frac {\ln \left (-\frac {2 \left (3 x^{2}-1\right )^{\frac {3}{4}}+2 \sqrt {3 x^{2}-1}+3 x^{2}+2 \left (3 x^{2}-1\right )^{\frac {1}{4}}}{3 x^{2}-2}\right )}{9}-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (\frac {2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \left (3 x^{2}-1\right )^{\frac {3}{4}}-2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \left (3 x^{2}-1\right )^{\frac {1}{4}}+2 \sqrt {3 x^{2}-1}-3 x^{2}}{3 x^{2}-2}\right )}{9}\)	\(136\)
risch	\(\frac {2 \left (3 x^{2}-1\right )^{\frac {3}{4}}}{27}+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (-\frac {2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \left (3 x^{2}-1\right )^{\frac {3}{4}}-2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \left (3 x^{2}-1\right )^{\frac {1}{4}}-2 \sqrt {3 x^{2}-1}+3 x^{2}}{3 x^{2}-2}\right )}{9}-\frac {\ln \left (-\frac {2 \left (3 x^{2}-1\right )^{\frac {3}{4}}+2 \sqrt {3 x^{2}-1}+3 x^{2}+2 \left (3 x^{2}-1\right )^{\frac {1}{4}}}{3 x^{2}-2}\right )}{9}\)	\(137\)

[In]

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

[Out]

2/27*(3*x^2-1)^(3/4)+1/9*ln(-1+(3*x^2-1)^(1/4))-1/9*ln(1+(3*x^2-1)^(1/4))+2/9*arctan((3*x^2-1)^(1/4))

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.08 \[ \int \frac {x^3}{\left (-2+3 x^2\right ) \sqrt [4]{-1+3 x^2}} \, dx=\frac {2}{27} \, {\left (3 \, x^{2} - 1\right )}^{\frac {3}{4}} + \frac {2}{9} \, \arctan \left ({\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}}\right ) - \frac {1}{9} \, \log \left ({\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}} + 1\right ) + \frac {1}{9} \, \log \left ({\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}} - 1\right ) \]

[In]

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

[Out]

2/27*(3*x^2 - 1)^(3/4) + 2/9*arctan((3*x^2 - 1)^(1/4)) - 1/9*log((3*x^2 - 1)^(1/4) + 1) + 1/9*log((3*x^2 - 1)^
(1/4) - 1)

Sympy [A] (verification not implemented)

Time = 4.05 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.21 \[ \int \frac {x^3}{\left (-2+3 x^2\right ) \sqrt [4]{-1+3 x^2}} \, dx=\frac {2 \left (3 x^{2} - 1\right )^{\frac {3}{4}}}{27} + \frac {\log {\left (\sqrt [4]{3 x^{2} - 1} - 1 \right )}}{9} - \frac {\log {\left (\sqrt [4]{3 x^{2} - 1} + 1 \right )}}{9} + \frac {2 \operatorname {atan}{\left (\sqrt [4]{3 x^{2} - 1} \right )}}{9} \]

[In]

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

[Out]

2*(3*x**2 - 1)**(3/4)/27 + log((3*x**2 - 1)**(1/4) - 1)/9 - log((3*x**2 - 1)**(1/4) + 1)/9 + 2*atan((3*x**2 -
1)**(1/4))/9

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.08 \[ \int \frac {x^3}{\left (-2+3 x^2\right ) \sqrt [4]{-1+3 x^2}} \, dx=\frac {2}{27} \, {\left (3 \, x^{2} - 1\right )}^{\frac {3}{4}} + \frac {2}{9} \, \arctan \left ({\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}}\right ) - \frac {1}{9} \, \log \left ({\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}} + 1\right ) + \frac {1}{9} \, \log \left ({\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}} - 1\right ) \]

[In]

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

[Out]

2/27*(3*x^2 - 1)^(3/4) + 2/9*arctan((3*x^2 - 1)^(1/4)) - 1/9*log((3*x^2 - 1)^(1/4) + 1) + 1/9*log((3*x^2 - 1)^
(1/4) - 1)

Giac [A] (verification not implemented)

none

Time = 0.38 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.10 \[ \int \frac {x^3}{\left (-2+3 x^2\right ) \sqrt [4]{-1+3 x^2}} \, dx=\frac {2}{27} \, {\left (3 \, x^{2} - 1\right )}^{\frac {3}{4}} + \frac {2}{9} \, \arctan \left ({\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}}\right ) - \frac {1}{9} \, \log \left ({\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}} + 1\right ) + \frac {1}{9} \, \log \left ({\left | {\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}} - 1 \right |}\right ) \]

[In]

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

[Out]

2/27*(3*x^2 - 1)^(3/4) + 2/9*arctan((3*x^2 - 1)^(1/4)) - 1/9*log((3*x^2 - 1)^(1/4) + 1) + 1/9*log(abs((3*x^2 -
1)^(1/4) - 1))

Mupad [B] (verification not implemented)

Time = 5.39 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.75 \[ \int \frac {x^3}{\left (-2+3 x^2\right ) \sqrt [4]{-1+3 x^2}} \, dx=\frac {2\,\mathrm {atan}\left ({\left (3\,x^2-1\right )}^{1/4}\right )}{9}-\frac {2\,\mathrm {atanh}\left ({\left (3\,x^2-1\right )}^{1/4}\right )}{9}+\frac {2\,{\left (3\,x^2-1\right )}^{3/4}}{27} \]

[In]

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

[Out]

(2*atan((3*x^2 - 1)^(1/4)))/9 - (2*atanh((3*x^2 - 1)^(1/4)))/9 + (2*(3*x^2 - 1)^(3/4))/27