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

Optimal. Leaf size=158 \[ \frac {2}{9} \sqrt [4]{2-3 x^2}+\frac {\log \left (\sqrt {2-3 x^2}-2^{3/4} \sqrt [4]{2-3 x^2}+\sqrt {2}\right )}{9 \sqrt [4]{2}}-\frac {\log \left (\sqrt {2-3 x^2}+2^{3/4} \sqrt [4]{2-3 x^2}+\sqrt {2}\right )}{9 \sqrt [4]{2}}-\frac {1}{9} 2^{3/4} \tan ^{-1}\left (\sqrt [4]{4-6 x^2}+1\right )+\frac {1}{9} 2^{3/4} \tan ^{-1}\left (1-\sqrt [4]{2} \sqrt [4]{2-3 x^2}\right ) \]

________________________________________________________________________________________

Rubi [A]  time = 0.15, antiderivative size = 158, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 10, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {443, 261, 444, 63, 211, 1165, 628, 1162, 617, 204} \begin {gather*} \frac {2}{9} \sqrt [4]{2-3 x^2}+\frac {\log \left (\sqrt {2-3 x^2}-2^{3/4} \sqrt [4]{2-3 x^2}+\sqrt {2}\right )}{9 \sqrt [4]{2}}-\frac {\log \left (\sqrt {2-3 x^2}+2^{3/4} \sqrt [4]{2-3 x^2}+\sqrt {2}\right )}{9 \sqrt [4]{2}}-\frac {1}{9} 2^{3/4} \tan ^{-1}\left (\sqrt [4]{4-6 x^2}+1\right )+\frac {1}{9} 2^{3/4} \tan ^{-1}\left (1-\sqrt [4]{2} \sqrt [4]{2-3 x^2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(2 - 3*x^2)^(1/4))/9 - (2^(3/4)*ArcTan[1 + (4 - 6*x^2)^(1/4)])/9 + (2^(3/4)*ArcTan[1 - 2^(1/4)*(2 - 3*x^2)^
(1/4)])/9 + Log[Sqrt[2] - 2^(3/4)*(2 - 3*x^2)^(1/4) + Sqrt[2 - 3*x^2]]/(9*2^(1/4)) - Log[Sqrt[2] + 2^(3/4)*(2
- 3*x^2)^(1/4) + Sqrt[2 - 3*x^2]]/(9*2^(1/4))

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 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 211

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]}, Di
st[1/(2*r), Int[(r - s*x^2)/(a + b*x^4), x], x] + Dist[1/(2*r), Int[(r + s*x^2)/(a + b*x^4), x], x]] /; FreeQ[
{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ, b
]]))

Rule 261

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

Rule 443

Int[(x_)^(m_)/(((a_) + (b_.)*(x_)^2)^(3/4)*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Int[ExpandIntegrand[x^m/((a +
b*x^2)^(3/4)*(c + d*x^2)), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c - 2*a*d, 0] && IntegerQ[m] && (PosQ[a]
|| IntegerQ[m/2])

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 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1162

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e, 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 1165

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(-2*d)/e, 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rubi steps

\begin {align*} \int \frac {x^3}{\left (2-3 x^2\right )^{3/4} \left (4-3 x^2\right )} \, dx &=\int \left (-\frac {x}{3 \left (2-3 x^2\right )^{3/4}}+\frac {4 x}{3 \left (2-3 x^2\right )^{3/4} \left (4-3 x^2\right )}\right ) \, dx\\ &=-\left (\frac {1}{3} \int \frac {x}{\left (2-3 x^2\right )^{3/4}} \, dx\right )+\frac {4}{3} \int \frac {x}{\left (2-3 x^2\right )^{3/4} \left (4-3 x^2\right )} \, dx\\ &=\frac {2}{9} \sqrt [4]{2-3 x^2}+\frac {2}{3} \operatorname {Subst}\left (\int \frac {1}{(2-3 x)^{3/4} (4-3 x)} \, dx,x,x^2\right )\\ &=\frac {2}{9} \sqrt [4]{2-3 x^2}-\frac {8}{9} \operatorname {Subst}\left (\int \frac {1}{2+x^4} \, dx,x,\sqrt [4]{2-3 x^2}\right )\\ &=\frac {2}{9} \sqrt [4]{2-3 x^2}-\frac {1}{9} \left (2 \sqrt {2}\right ) \operatorname {Subst}\left (\int \frac {\sqrt {2}-x^2}{2+x^4} \, dx,x,\sqrt [4]{2-3 x^2}\right )-\frac {1}{9} \left (2 \sqrt {2}\right ) \operatorname {Subst}\left (\int \frac {\sqrt {2}+x^2}{2+x^4} \, dx,x,\sqrt [4]{2-3 x^2}\right )\\ &=\frac {2}{9} \sqrt [4]{2-3 x^2}+\frac {\operatorname {Subst}\left (\int \frac {2^{3/4}+2 x}{-\sqrt {2}-2^{3/4} x-x^2} \, dx,x,\sqrt [4]{2-3 x^2}\right )}{9 \sqrt [4]{2}}+\frac {\operatorname {Subst}\left (\int \frac {2^{3/4}-2 x}{-\sqrt {2}+2^{3/4} x-x^2} \, dx,x,\sqrt [4]{2-3 x^2}\right )}{9 \sqrt [4]{2}}-\frac {1}{9} \sqrt {2} \operatorname {Subst}\left (\int \frac {1}{\sqrt {2}-2^{3/4} x+x^2} \, dx,x,\sqrt [4]{2-3 x^2}\right )-\frac {1}{9} \sqrt {2} \operatorname {Subst}\left (\int \frac {1}{\sqrt {2}+2^{3/4} x+x^2} \, dx,x,\sqrt [4]{2-3 x^2}\right )\\ &=\frac {2}{9} \sqrt [4]{2-3 x^2}+\frac {\log \left (\sqrt {2}-2^{3/4} \sqrt [4]{2-3 x^2}+\sqrt {2-3 x^2}\right )}{9 \sqrt [4]{2}}-\frac {\log \left (\sqrt {2}+2^{3/4} \sqrt [4]{2-3 x^2}+\sqrt {2-3 x^2}\right )}{9 \sqrt [4]{2}}-\frac {1}{9} 2^{3/4} \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt [4]{4-6 x^2}\right )+\frac {1}{9} 2^{3/4} \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt [4]{4-6 x^2}\right )\\ &=\frac {2}{9} \sqrt [4]{2-3 x^2}-\frac {1}{9} 2^{3/4} \tan ^{-1}\left (1+\sqrt [4]{4-6 x^2}\right )+\frac {1}{9} 2^{3/4} \tan ^{-1}\left (1-\sqrt [4]{2} \sqrt [4]{2-3 x^2}\right )+\frac {\log \left (\sqrt {2}-2^{3/4} \sqrt [4]{2-3 x^2}+\sqrt {2-3 x^2}\right )}{9 \sqrt [4]{2}}-\frac {\log \left (\sqrt {2}+2^{3/4} \sqrt [4]{2-3 x^2}+\sqrt {2-3 x^2}\right )}{9 \sqrt [4]{2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.04, size = 146, normalized size = 0.92 \begin {gather*} \frac {1}{18} \left (4 \sqrt [4]{2-3 x^2}+2^{3/4} \log \left (\sqrt {2-3 x^2}-2^{3/4} \sqrt [4]{2-3 x^2}+\sqrt {2}\right )-2^{3/4} \log \left (\sqrt {2-3 x^2}+2^{3/4} \sqrt [4]{2-3 x^2}+\sqrt {2}\right )+2\ 2^{3/4} \tan ^{-1}\left (1-\sqrt [4]{4-6 x^2}\right )-2\ 2^{3/4} \tan ^{-1}\left (\sqrt [4]{4-6 x^2}+1\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(4*(2 - 3*x^2)^(1/4) + 2*2^(3/4)*ArcTan[1 - (4 - 6*x^2)^(1/4)] - 2*2^(3/4)*ArcTan[1 + (4 - 6*x^2)^(1/4)] + 2^(
3/4)*Log[Sqrt[2] - 2^(3/4)*(2 - 3*x^2)^(1/4) + Sqrt[2 - 3*x^2]] - 2^(3/4)*Log[Sqrt[2] + 2^(3/4)*(2 - 3*x^2)^(1
/4) + Sqrt[2 - 3*x^2]])/18

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.13, size = 112, normalized size = 0.71 \begin {gather*} \frac {2}{9} \sqrt [4]{2-3 x^2}-\frac {1}{9} 2^{3/4} \tan ^{-1}\left (\frac {\frac {\sqrt {2-3 x^2}}{2^{3/4}}-\frac {1}{\sqrt [4]{2}}}{\sqrt [4]{2-3 x^2}}\right )-\frac {1}{9} 2^{3/4} \tanh ^{-1}\left (\frac {2 \sqrt [4]{2} \sqrt [4]{2-3 x^2}}{\sqrt {2} \sqrt {2-3 x^2}+2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.97, size = 186, normalized size = 1.18 \begin {gather*} \frac {2}{9} \cdot 2^{\frac {3}{4}} \arctan \left (2^{\frac {1}{4}} \sqrt {2^{\frac {3}{4}} {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}} + \sqrt {2} + \sqrt {-3 \, x^{2} + 2}} - 2^{\frac {1}{4}} {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}} - 1\right ) + \frac {2}{9} \cdot 2^{\frac {3}{4}} \arctan \left (2^{\frac {1}{4}} \sqrt {-2^{\frac {3}{4}} {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}} + \sqrt {2} + \sqrt {-3 \, x^{2} + 2}} - 2^{\frac {1}{4}} {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}} + 1\right ) - \frac {1}{18} \cdot 2^{\frac {3}{4}} \log \left (2^{\frac {3}{4}} {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}} + \sqrt {2} + \sqrt {-3 \, x^{2} + 2}\right ) + \frac {1}{18} \cdot 2^{\frac {3}{4}} \log \left (-2^{\frac {3}{4}} {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}} + \sqrt {2} + \sqrt {-3 \, x^{2} + 2}\right ) + \frac {2}{9} \, {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/9*2^(3/4)*arctan(2^(1/4)*sqrt(2^(3/4)*(-3*x^2 + 2)^(1/4) + sqrt(2) + sqrt(-3*x^2 + 2)) - 2^(1/4)*(-3*x^2 + 2
)^(1/4) - 1) + 2/9*2^(3/4)*arctan(2^(1/4)*sqrt(-2^(3/4)*(-3*x^2 + 2)^(1/4) + sqrt(2) + sqrt(-3*x^2 + 2)) - 2^(
1/4)*(-3*x^2 + 2)^(1/4) + 1) - 1/18*2^(3/4)*log(2^(3/4)*(-3*x^2 + 2)^(1/4) + sqrt(2) + sqrt(-3*x^2 + 2)) + 1/1
8*2^(3/4)*log(-2^(3/4)*(-3*x^2 + 2)^(1/4) + sqrt(2) + sqrt(-3*x^2 + 2)) + 2/9*(-3*x^2 + 2)^(1/4)

________________________________________________________________________________________

giac [A]  time = 0.46, size = 129, normalized size = 0.82 \begin {gather*} -\frac {1}{9} \cdot 2^{\frac {3}{4}} \arctan \left (\frac {1}{2} \cdot 2^{\frac {1}{4}} {\left (2^{\frac {3}{4}} + 2 \, {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}}\right )}\right ) - \frac {1}{9} \cdot 2^{\frac {3}{4}} \arctan \left (-\frac {1}{2} \cdot 2^{\frac {1}{4}} {\left (2^{\frac {3}{4}} - 2 \, {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}}\right )}\right ) - \frac {1}{18} \cdot 2^{\frac {3}{4}} \log \left (2^{\frac {3}{4}} {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}} + \sqrt {2} + \sqrt {-3 \, x^{2} + 2}\right ) + \frac {1}{18} \cdot 2^{\frac {3}{4}} \log \left (-2^{\frac {3}{4}} {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}} + \sqrt {2} + \sqrt {-3 \, x^{2} + 2}\right ) + \frac {2}{9} \, {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/9*2^(3/4)*arctan(1/2*2^(1/4)*(2^(3/4) + 2*(-3*x^2 + 2)^(1/4))) - 1/9*2^(3/4)*arctan(-1/2*2^(1/4)*(2^(3/4) -
2*(-3*x^2 + 2)^(1/4))) - 1/18*2^(3/4)*log(2^(3/4)*(-3*x^2 + 2)^(1/4) + sqrt(2) + sqrt(-3*x^2 + 2)) + 1/18*2^(
3/4)*log(-2^(3/4)*(-3*x^2 + 2)^(1/4) + sqrt(2) + sqrt(-3*x^2 + 2)) + 2/9*(-3*x^2 + 2)^(1/4)

________________________________________________________________________________________

maple [C]  time = 2.39, size = 528, normalized size = 3.34 \begin {gather*} -\frac {2 \left (3 x^{2}-2\right )}{9 \left (-3 x^{2}+2\right )^{\frac {3}{4}}}-\frac {\left (-\frac {\RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}+2\right )^{2}\right ) \ln \left (\frac {-27 x^{6}-18 \left (-27 x^{6}+54 x^{4}-36 x^{2}+8\right )^{\frac {1}{4}} x^{4} \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}+2\right )^{2}\right )+36 x^{4}+6 \sqrt {-27 x^{6}+54 x^{4}-36 x^{2}+8}\, x^{2} \RootOf \left (\textit {\_Z}^{4}+2\right )^{2}+24 \left (-27 x^{6}+54 x^{4}-36 x^{2}+8\right )^{\frac {1}{4}} x^{2} \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}+2\right )^{2}\right )+2 \left (-27 x^{6}+54 x^{4}-36 x^{2}+8\right )^{\frac {3}{4}} \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}+2\right )^{2}\right ) \RootOf \left (\textit {\_Z}^{4}+2\right )^{2}-12 x^{2}-4 \sqrt {-27 x^{6}+54 x^{4}-36 x^{2}+8}\, \RootOf \left (\textit {\_Z}^{4}+2\right )^{2}-8 \left (-27 x^{6}+54 x^{4}-36 x^{2}+8\right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}+2\right )^{2}\right )}{\left (3 x^{2}-4\right ) \left (3 x^{2}-2\right )^{2}}\right )}{9}+\frac {\RootOf \left (\textit {\_Z}^{4}+2\right ) \ln \left (\frac {-27 x^{6}+18 \left (-27 x^{6}+54 x^{4}-36 x^{2}+8\right )^{\frac {1}{4}} x^{4} \RootOf \left (\textit {\_Z}^{4}+2\right )+36 x^{4}-6 \sqrt {-27 x^{6}+54 x^{4}-36 x^{2}+8}\, x^{2} \RootOf \left (\textit {\_Z}^{4}+2\right )^{2}-24 \left (-27 x^{6}+54 x^{4}-36 x^{2}+8\right )^{\frac {1}{4}} x^{2} \RootOf \left (\textit {\_Z}^{4}+2\right )+2 \left (-27 x^{6}+54 x^{4}-36 x^{2}+8\right )^{\frac {3}{4}} \RootOf \left (\textit {\_Z}^{4}+2\right )^{3}-12 x^{2}+4 \sqrt {-27 x^{6}+54 x^{4}-36 x^{2}+8}\, \RootOf \left (\textit {\_Z}^{4}+2\right )^{2}+8 \left (-27 x^{6}+54 x^{4}-36 x^{2}+8\right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{4}+2\right )}{\left (3 x^{2}-4\right ) \left (3 x^{2}-2\right )^{2}}\right )}{9}\right ) \left (-\left (3 x^{2}-2\right )^{3}\right )^{\frac {1}{4}}}{\left (-3 x^{2}+2\right )^{\frac {3}{4}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/9*(3*x^2-2)/(-3*x^2+2)^(3/4)-(1/9*RootOf(_Z^4+2)*ln((2*RootOf(_Z^4+2)^3*(-27*x^6+54*x^4-36*x^2+8)^(3/4)-6*R
ootOf(_Z^4+2)^2*(-27*x^6+54*x^4-36*x^2+8)^(1/2)*x^2+18*RootOf(_Z^4+2)*(-27*x^6+54*x^4-36*x^2+8)^(1/4)*x^4-27*x
^6+4*RootOf(_Z^4+2)^2*(-27*x^6+54*x^4-36*x^2+8)^(1/2)-24*RootOf(_Z^4+2)*(-27*x^6+54*x^4-36*x^2+8)^(1/4)*x^2+36
*x^4+8*RootOf(_Z^4+2)*(-27*x^6+54*x^4-36*x^2+8)^(1/4)-12*x^2)/(3*x^2-4)/(3*x^2-2)^2)-1/9*RootOf(_Z^2+RootOf(_Z
^4+2)^2)*ln((2*RootOf(_Z^2+RootOf(_Z^4+2)^2)*RootOf(_Z^4+2)^2*(-27*x^6+54*x^4-36*x^2+8)^(3/4)-18*RootOf(_Z^2+R
ootOf(_Z^4+2)^2)*(-27*x^6+54*x^4-36*x^2+8)^(1/4)*x^4+6*RootOf(_Z^4+2)^2*(-27*x^6+54*x^4-36*x^2+8)^(1/2)*x^2-27
*x^6+24*RootOf(_Z^2+RootOf(_Z^4+2)^2)*(-27*x^6+54*x^4-36*x^2+8)^(1/4)*x^2-4*RootOf(_Z^4+2)^2*(-27*x^6+54*x^4-3
6*x^2+8)^(1/2)+36*x^4-8*RootOf(_Z^2+RootOf(_Z^4+2)^2)*(-27*x^6+54*x^4-36*x^2+8)^(1/4)-12*x^2)/(3*x^2-4)/(3*x^2
-2)^2))/(-3*x^2+2)^(3/4)*(-(3*x^2-2)^3)^(1/4)

________________________________________________________________________________________

maxima [A]  time = 2.01, size = 129, normalized size = 0.82 \begin {gather*} -\frac {1}{9} \cdot 2^{\frac {3}{4}} \arctan \left (\frac {1}{2} \cdot 2^{\frac {1}{4}} {\left (2^{\frac {3}{4}} + 2 \, {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}}\right )}\right ) - \frac {1}{9} \cdot 2^{\frac {3}{4}} \arctan \left (-\frac {1}{2} \cdot 2^{\frac {1}{4}} {\left (2^{\frac {3}{4}} - 2 \, {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}}\right )}\right ) - \frac {1}{18} \cdot 2^{\frac {3}{4}} \log \left (2^{\frac {3}{4}} {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}} + \sqrt {2} + \sqrt {-3 \, x^{2} + 2}\right ) + \frac {1}{18} \cdot 2^{\frac {3}{4}} \log \left (-2^{\frac {3}{4}} {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}} + \sqrt {2} + \sqrt {-3 \, x^{2} + 2}\right ) + \frac {2}{9} \, {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/9*2^(3/4)*arctan(1/2*2^(1/4)*(2^(3/4) + 2*(-3*x^2 + 2)^(1/4))) - 1/9*2^(3/4)*arctan(-1/2*2^(1/4)*(2^(3/4) -
2*(-3*x^2 + 2)^(1/4))) - 1/18*2^(3/4)*log(2^(3/4)*(-3*x^2 + 2)^(1/4) + sqrt(2) + sqrt(-3*x^2 + 2)) + 1/18*2^(
3/4)*log(-2^(3/4)*(-3*x^2 + 2)^(1/4) + sqrt(2) + sqrt(-3*x^2 + 2)) + 2/9*(-3*x^2 + 2)^(1/4)

________________________________________________________________________________________

mupad [B]  time = 0.17, size = 60, normalized size = 0.38 \begin {gather*} \frac {2\,{\left (2-3\,x^2\right )}^{1/4}}{9}+2^{3/4}\,\mathrm {atan}\left (2^{1/4}\,{\left (2-3\,x^2\right )}^{1/4}\,\left (\frac {1}{2}-\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (-\frac {1}{9}-\frac {1}{9}{}\mathrm {i}\right )+2^{3/4}\,\mathrm {atan}\left (2^{1/4}\,{\left (2-3\,x^2\right )}^{1/4}\,\left (\frac {1}{2}+\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (-\frac {1}{9}+\frac {1}{9}{}\mathrm {i}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________