3.1084 \(\int \frac{x}{(-2+3 x^2) (-1+3 x^2)^{3/4}} \, dx\)

Optimal. Leaf size=33 \[ -\frac{1}{3} \tan ^{-1}\left (\sqrt [4]{3 x^2-1}\right )-\frac{1}{3} \tanh ^{-1}\left (\sqrt [4]{3 x^2-1}\right ) \]

[Out]

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

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Rubi [A]  time = 0.0212163, antiderivative size = 33, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.227, Rules used = {444, 63, 212, 206, 203} \[ -\frac{1}{3} \tan ^{-1}\left (\sqrt [4]{3 x^2-1}\right )-\frac{1}{3} \tanh ^{-1}\left (\sqrt [4]{3 x^2-1}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 444

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]

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 212

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] &&
 !GtQ[a/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])

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])

Rubi steps

\begin{align*} \int \frac{x}{\left (-2+3 x^2\right ) \left (-1+3 x^2\right )^{3/4}} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{(-2+3 x) (-1+3 x)^{3/4}} \, dx,x,x^2\right )\\ &=\frac{2}{3} \operatorname{Subst}\left (\int \frac{1}{-1+x^4} \, dx,x,\sqrt [4]{-1+3 x^2}\right )\\ &=-\left (\frac{1}{3} \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\sqrt [4]{-1+3 x^2}\right )\right )-\frac{1}{3} \operatorname{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]  time = 0.0057361, size = 33, normalized size = 1. \[ -\frac{1}{3} \tan ^{-1}\left (\sqrt [4]{3 x^2-1}\right )-\frac{1}{3} \tanh ^{-1}\left (\sqrt [4]{3 x^2-1}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.057, size = 0, normalized size = 0. \begin{align*} \int{\frac{x}{3\,{x}^{2}-2} \left ( 3\,{x}^{2}-1 \right ) ^{-{\frac{3}{4}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.57283, size = 55, normalized size = 1.67 \begin{align*} -\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 ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [A]  time = 1.57698, size = 132, normalized size = 4. \begin{align*} -\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 ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [A]  time = 4.23762, size = 42, normalized size = 1.27 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Giac [A]  time = 1.18501, size = 57, normalized size = 1.73 \begin{align*} -\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 ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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))