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

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

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Rubi [A]  time = 0.02, antiderivative size = 33, normalized size of antiderivative = 1.00, 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, 298, 203, 206} \begin {gather*} \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 ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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])

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 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]

Rubi steps

\begin {align*} \int \frac {x}{\left (-2+3 x^2\right ) \sqrt [4]{-1+3 x^2}} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{(-2+3 x) \sqrt [4]{-1+3 x}} \, dx,x,x^2\right )\\ &=\frac {2}{3} \operatorname {Subst}\left (\int \frac {x^2}{-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*}

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Mathematica [A]  time = 0.01, size = 33, normalized size = 1.00 \begin {gather*} \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 ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.03, size = 33, normalized size = 1.00 \begin {gather*} \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 ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.90, size = 41, normalized size = 1.24 \begin {gather*} \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 {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(3*x^2-2)/(3*x^2-1)^(1/4),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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giac [A]  time = 0.36, size = 42, normalized size = 1.27 \begin {gather*} \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 {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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maple [C]  time = 0.93, size = 125, normalized size = 3.79 \begin {gather*} \frac {\RootOf \left (\textit {\_Z}^{2}+1\right ) \ln \left (-\frac {3 x^{2}+2 \left (3 x^{2}-1\right )^{\frac {3}{4}} \RootOf \left (\textit {\_Z}^{2}+1\right )-2 \left (3 x^{2}-1\right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{2}+1\right )-2 \sqrt {3 x^{2}-1}}{3 x^{2}-2}\right )}{6}+\frac {\ln \left (\frac {-3 x^{2}+2 \left (3 x^{2}-1\right )^{\frac {3}{4}}-2 \sqrt {3 x^{2}-1}+2 \left (3 x^{2}-1\right )^{\frac {1}{4}}}{3 x^{2}-2}\right )}{6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*ln((-3*x^2+2*(3*x^2-1)^(3/4)-2*(3*x^2-1)^(1/2)+2*(3*x^2-1)^(1/4))/(3*x^2-2))+1/6*RootOf(_Z^2+1)*ln(-(2*Roo
tOf(_Z^2+1)*(3*x^2-1)^(3/4)-2*RootOf(_Z^2+1)*(3*x^2-1)^(1/4)-2*(3*x^2-1)^(1/2)+3*x^2)/(3*x^2-2))

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maxima [A]  time = 2.02, size = 41, normalized size = 1.24 \begin {gather*} \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 {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(3*x^2-2)/(3*x^2-1)^(1/4),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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mupad [B]  time = 0.16, size = 25, normalized size = 0.76 \begin {gather*} \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 8.99, size = 42, normalized size = 1.27 \begin {gather*} \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 {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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