3.1083 \(\int \frac {x^3}{(-2+3 x^2) (-1+3 x^2)^{3/4}} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {446, 80, 63, 212, 206, 203} \[ \frac {2}{9} \sqrt [4]{3 x^2-1}-\frac {2}{9} \tan ^{-1}\left (\sqrt [4]{3 x^2-1}\right )-\frac {2}{9} \tanh ^{-1}\left (\sqrt [4]{3 x^2-1}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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 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 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]]

Rubi steps

\begin {align*} \int \frac {x^3}{\left (-2+3 x^2\right ) \left (-1+3 x^2\right )^{3/4}} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {x}{(-2+3 x) (-1+3 x)^{3/4}} \, dx,x,x^2\right )\\ &=\frac {2}{9} \sqrt [4]{-1+3 x^2}+\frac {1}{3} \operatorname {Subst}\left (\int \frac {1}{(-2+3 x) (-1+3 x)^{3/4}} \, dx,x,x^2\right )\\ &=\frac {2}{9} \sqrt [4]{-1+3 x^2}+\frac {4}{9} \operatorname {Subst}\left (\int \frac {1}{-1+x^4} \, dx,x,\sqrt [4]{-1+3 x^2}\right )\\ &=\frac {2}{9} \sqrt [4]{-1+3 x^2}-\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}{9} \sqrt [4]{-1+3 x^2}-\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*}

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Mathematica [A]  time = 0.01, size = 44, normalized size = 0.92 \[ \frac {2}{9} \left (\sqrt [4]{3 x^2-1}-\tan ^{-1}\left (\sqrt [4]{3 x^2-1}\right )-\tanh ^{-1}\left (\sqrt [4]{3 x^2-1}\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.57, size = 52, normalized size = 1.08 \[ \frac {2}{9} \, {\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}} - \frac {2}{9} \, \arctan \left ({\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}}\right ) - \frac {1}{9} \, \log \left ({\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}} + 1\right ) + \frac {1}{9} \, \log \left ({\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}} - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/9*(3*x^2 - 1)^(1/4) - 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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giac [A]  time = 0.41, size = 53, normalized size = 1.10 \[ \frac {2}{9} \, {\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}} - \frac {2}{9} \, \arctan \left ({\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}}\right ) - \frac {1}{9} \, \log \left ({\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}} + 1\right ) + \frac {1}{9} \, \log \left ({\left | {\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}} - 1 \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/9*(3*x^2 - 1)^(1/4) - 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))

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maple [C]  time = 0.76, size = 412, normalized size = 8.58 \[ \frac {2 \left (3 x^{2}-1\right )^{\frac {1}{4}}}{9}+\frac {\left (\frac {\RootOf \left (\textit {\_Z}^{2}+1\right ) \ln \left (-\frac {27 x^{6}-18 \left (27 x^{6}-27 x^{4}+9 x^{2}-1\right )^{\frac {1}{4}} x^{4} \RootOf \left (\textit {\_Z}^{2}+1\right )-18 x^{4}+12 \left (27 x^{6}-27 x^{4}+9 x^{2}-1\right )^{\frac {1}{4}} x^{2} \RootOf \left (\textit {\_Z}^{2}+1\right )-6 \sqrt {27 x^{6}-27 x^{4}+9 x^{2}-1}\, x^{2}+3 x^{2}+2 \left (27 x^{6}-27 x^{4}+9 x^{2}-1\right )^{\frac {3}{4}} \RootOf \left (\textit {\_Z}^{2}+1\right )-2 \left (27 x^{6}-27 x^{4}+9 x^{2}-1\right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{2}+1\right )+2 \sqrt {27 x^{6}-27 x^{4}+9 x^{2}-1}}{\left (3 x^{2}-2\right ) \left (3 x^{2}-1\right )^{2}}\right )}{9}+\frac {\ln \left (\frac {-27 x^{6}+18 \left (27 x^{6}-27 x^{4}+9 x^{2}-1\right )^{\frac {1}{4}} x^{4}+18 x^{4}-6 \sqrt {27 x^{6}-27 x^{4}+9 x^{2}-1}\, x^{2}-12 \left (27 x^{6}-27 x^{4}+9 x^{2}-1\right )^{\frac {1}{4}} x^{2}-3 x^{2}+2 \left (27 x^{6}-27 x^{4}+9 x^{2}-1\right )^{\frac {3}{4}}+2 \sqrt {27 x^{6}-27 x^{4}+9 x^{2}-1}+2 \left (27 x^{6}-27 x^{4}+9 x^{2}-1\right )^{\frac {1}{4}}}{\left (3 x^{2}-2\right ) \left (3 x^{2}-1\right )^{2}}\right )}{9}\right ) \left (\left (3 x^{2}-1\right )^{3}\right )^{\frac {1}{4}}}{\left (3 x^{2}-1\right )^{\frac {3}{4}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/9*(3*x^2-1)^(1/4)+(1/9*RootOf(_Z^2+1)*ln(-(-18*RootOf(_Z^2+1)*(27*x^6-27*x^4+9*x^2-1)^(1/4)*x^4+27*x^6+2*Roo
tOf(_Z^2+1)*(27*x^6-27*x^4+9*x^2-1)^(3/4)-6*(27*x^6-27*x^4+9*x^2-1)^(1/2)*x^2+12*RootOf(_Z^2+1)*(27*x^6-27*x^4
+9*x^2-1)^(1/4)*x^2-18*x^4+2*(27*x^6-27*x^4+9*x^2-1)^(1/2)-2*RootOf(_Z^2+1)*(27*x^6-27*x^4+9*x^2-1)^(1/4)+3*x^
2)/(3*x^2-2)/(3*x^2-1)^2)+1/9*ln((-27*x^6+18*(27*x^6-27*x^4+9*x^2-1)^(1/4)*x^4-6*(27*x^6-27*x^4+9*x^2-1)^(1/2)
*x^2+18*x^4+2*(27*x^6-27*x^4+9*x^2-1)^(3/4)-12*(27*x^6-27*x^4+9*x^2-1)^(1/4)*x^2+2*(27*x^6-27*x^4+9*x^2-1)^(1/
2)-3*x^2+2*(27*x^6-27*x^4+9*x^2-1)^(1/4))/(3*x^2-2)/(3*x^2-1)^2))/(3*x^2-1)^(3/4)*((3*x^2-1)^3)^(1/4)

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maxima [A]  time = 1.81, size = 52, normalized size = 1.08 \[ \frac {2}{9} \, {\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}} - \frac {2}{9} \, \arctan \left ({\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}}\right ) - \frac {1}{9} \, \log \left ({\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}} + 1\right ) + \frac {1}{9} \, \log \left ({\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}} - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/9*(3*x^2 - 1)^(1/4) - 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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mupad [B]  time = 0.12, size = 36, normalized size = 0.75 \[ \frac {2\,{\left (3\,x^2-1\right )}^{1/4}}{9}-\frac {2\,\mathrm {atanh}\left ({\left (3\,x^2-1\right )}^{1/4}\right )}{9}-\frac {2\,\mathrm {atan}\left ({\left (3\,x^2-1\right )}^{1/4}\right )}{9} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{3}}{\left (3 x^{2} - 2\right ) \left (3 x^{2} - 1\right )^{\frac {3}{4}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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