3.1044 \(\int \frac{x^3}{(-2+3 x^2) \sqrt [4]{-1+3 x^2}} \, dx\)

Optimal. Leaf size=48 \[ \frac{2}{27} \left (3 x^2-1\right )^{3/4}+\frac{2}{9} \tan ^{-1}\left (\sqrt [4]{3 x^2-1}\right )-\frac{2}{9} \tanh ^{-1}\left (\sqrt [4]{3 x^2-1}\right ) \]

[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

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Rubi [A]  time = 0.032934, antiderivative size = 48, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {446, 80, 63, 298, 203, 206} \[ \frac{2}{27} \left (3 x^2-1\right )^{3/4}+\frac{2}{9} \tan ^{-1}\left (\sqrt [4]{3 x^2-1}\right )-\frac{2}{9} \tanh ^{-1}\left (\sqrt [4]{3 x^2-1}\right ) \]

Antiderivative was successfully verified.

[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 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 80

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 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 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 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 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])

Rubi steps

\begin{align*} \int \frac{x^3}{\left (-2+3 x^2\right ) \sqrt [4]{-1+3 x^2}} \, dx &=\frac{1}{2} \operatorname{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} \operatorname{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} \operatorname{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} \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\sqrt [4]{-1+3 x^2}\right )+\frac{2}{9} \operatorname{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]  time = 0.0101469, size = 44, normalized size = 0.92 \[ \frac{2}{27} \left (\left (3 x^2-1\right )^{3/4}+3 \tan ^{-1}\left (\sqrt [4]{3 x^2-1}\right )-3 \tanh ^{-1}\left (\sqrt [4]{3 x^2-1}\right )\right ) \]

Antiderivative was successfully verified.

[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

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Maple [F]  time = 0.065, size = 0, normalized size = 0. \begin{align*} \int{\frac{{x}^{3}}{3\,{x}^{2}-2}{\frac{1}{\sqrt [4]{3\,{x}^{2}-1}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.51309, size = 70, normalized size = 1.46 \begin{align*} \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 ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [A]  time = 1.32912, size = 165, normalized size = 3.44 \begin{align*} \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 ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [A]  time = 10.2749, size = 58, normalized size = 1.21 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Giac [A]  time = 1.23998, size = 72, normalized size = 1.5 \begin{align*} \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 ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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))