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

Optimal. Leaf size=197 \[ \frac {\log \left (\sqrt {2-3 x^2}-2^{3/4} \sqrt [4]{2-3 x^2}+\sqrt {2}\right )}{16 \sqrt [4]{2}}-\frac {\log \left (\sqrt {2-3 x^2}+2^{3/4} \sqrt [4]{2-3 x^2}+\sqrt {2}\right )}{16 \sqrt [4]{2}}-\frac {\tan ^{-1}\left (\frac {\sqrt [4]{2-3 x^2}}{\sqrt [4]{2}}\right )}{4\ 2^{3/4}}-\frac {\tan ^{-1}\left (\sqrt [4]{4-6 x^2}+1\right )}{8 \sqrt [4]{2}}+\frac {\tan ^{-1}\left (1-\sqrt [4]{2} \sqrt [4]{2-3 x^2}\right )}{8 \sqrt [4]{2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{2-3 x^2}}{\sqrt [4]{2}}\right )}{4\ 2^{3/4}} \]

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Rubi [A]  time = 0.19, antiderivative size = 197, normalized size of antiderivative = 1.00, number of steps used = 18, number of rules used = 13, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.542, Rules used = {443, 266, 63, 212, 206, 203, 444, 211, 1165, 628, 1162, 617, 204} \begin {gather*} \frac {\log \left (\sqrt {2-3 x^2}-2^{3/4} \sqrt [4]{2-3 x^2}+\sqrt {2}\right )}{16 \sqrt [4]{2}}-\frac {\log \left (\sqrt {2-3 x^2}+2^{3/4} \sqrt [4]{2-3 x^2}+\sqrt {2}\right )}{16 \sqrt [4]{2}}-\frac {\tan ^{-1}\left (\frac {\sqrt [4]{2-3 x^2}}{\sqrt [4]{2}}\right )}{4\ 2^{3/4}}-\frac {\tan ^{-1}\left (\sqrt [4]{4-6 x^2}+1\right )}{8 \sqrt [4]{2}}+\frac {\tan ^{-1}\left (1-\sqrt [4]{2} \sqrt [4]{2-3 x^2}\right )}{8 \sqrt [4]{2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{2-3 x^2}}{\sqrt [4]{2}}\right )}{4\ 2^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-ArcTan[(2 - 3*x^2)^(1/4)/2^(1/4)]/(4*2^(3/4)) - ArcTan[1 + (4 - 6*x^2)^(1/4)]/(8*2^(1/4)) + ArcTan[1 - 2^(1/4
)*(2 - 3*x^2)^(1/4)]/(8*2^(1/4)) - ArcTanh[(2 - 3*x^2)^(1/4)/2^(1/4)]/(4*2^(3/4)) + Log[Sqrt[2] - 2^(3/4)*(2 -
3*x^2)^(1/4) + Sqrt[2 - 3*x^2]]/(16*2^(1/4)) - Log[Sqrt[2] + 2^(3/4)*(2 - 3*x^2)^(1/4) + Sqrt[2 - 3*x^2]]/(16
*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 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 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 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 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 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 266

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

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 {1}{x \left (2-3 x^2\right )^{3/4} \left (4-3 x^2\right )} \, dx &=\int \left (\frac {1}{4 x \left (2-3 x^2\right )^{3/4}}-\frac {3 x}{4 \left (2-3 x^2\right )^{3/4} \left (-4+3 x^2\right )}\right ) \, dx\\ &=\frac {1}{4} \int \frac {1}{x \left (2-3 x^2\right )^{3/4}} \, dx-\frac {3}{4} \int \frac {x}{\left (2-3 x^2\right )^{3/4} \left (-4+3 x^2\right )} \, dx\\ &=\frac {1}{8} \operatorname {Subst}\left (\int \frac {1}{(2-3 x)^{3/4} x} \, dx,x,x^2\right )-\frac {3}{8} \operatorname {Subst}\left (\int \frac {1}{(2-3 x)^{3/4} (-4+3 x)} \, dx,x,x^2\right )\\ &=-\left (\frac {1}{6} \operatorname {Subst}\left (\int \frac {1}{\frac {2}{3}-\frac {x^4}{3}} \, dx,x,\sqrt [4]{2-3 x^2}\right )\right )+\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{-2-x^4} \, dx,x,\sqrt [4]{2-3 x^2}\right )\\ &=-\frac {\operatorname {Subst}\left (\int \frac {1}{\sqrt {2}-x^2} \, dx,x,\sqrt [4]{2-3 x^2}\right )}{4 \sqrt {2}}-\frac {\operatorname {Subst}\left (\int \frac {1}{\sqrt {2}+x^2} \, dx,x,\sqrt [4]{2-3 x^2}\right )}{4 \sqrt {2}}+\frac {\operatorname {Subst}\left (\int \frac {\sqrt {2}-x^2}{-2-x^4} \, dx,x,\sqrt [4]{2-3 x^2}\right )}{4 \sqrt {2}}+\frac {\operatorname {Subst}\left (\int \frac {\sqrt {2}+x^2}{-2-x^4} \, dx,x,\sqrt [4]{2-3 x^2}\right )}{4 \sqrt {2}}\\ &=-\frac {\tan ^{-1}\left (\frac {\sqrt [4]{2-3 x^2}}{\sqrt [4]{2}}\right )}{4\ 2^{3/4}}-\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{2-3 x^2}}{\sqrt [4]{2}}\right )}{4\ 2^{3/4}}-\frac {\operatorname {Subst}\left (\int \frac {1}{\sqrt {2}-2^{3/4} x+x^2} \, dx,x,\sqrt [4]{2-3 x^2}\right )}{8 \sqrt {2}}-\frac {\operatorname {Subst}\left (\int \frac {1}{\sqrt {2}+2^{3/4} x+x^2} \, dx,x,\sqrt [4]{2-3 x^2}\right )}{8 \sqrt {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 )}{16 \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 )}{16 \sqrt [4]{2}}\\ &=-\frac {\tan ^{-1}\left (\frac {\sqrt [4]{2-3 x^2}}{\sqrt [4]{2}}\right )}{4\ 2^{3/4}}-\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{2-3 x^2}}{\sqrt [4]{2}}\right )}{4\ 2^{3/4}}+\frac {\log \left (\sqrt {2}-2^{3/4} \sqrt [4]{2-3 x^2}+\sqrt {2-3 x^2}\right )}{16 \sqrt [4]{2}}-\frac {\log \left (\sqrt {2}+2^{3/4} \sqrt [4]{2-3 x^2}+\sqrt {2-3 x^2}\right )}{16 \sqrt [4]{2}}-\frac {\operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt [4]{4-6 x^2}\right )}{8 \sqrt [4]{2}}+\frac {\operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt [4]{4-6 x^2}\right )}{8 \sqrt [4]{2}}\\ &=-\frac {\tan ^{-1}\left (\frac {\sqrt [4]{2-3 x^2}}{\sqrt [4]{2}}\right )}{4\ 2^{3/4}}-\frac {\tan ^{-1}\left (1+\sqrt [4]{4-6 x^2}\right )}{8 \sqrt [4]{2}}+\frac {\tan ^{-1}\left (1-\sqrt [4]{2} \sqrt [4]{2-3 x^2}\right )}{8 \sqrt [4]{2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{2-3 x^2}}{\sqrt [4]{2}}\right )}{4\ 2^{3/4}}+\frac {\log \left (\sqrt {2}-2^{3/4} \sqrt [4]{2-3 x^2}+\sqrt {2-3 x^2}\right )}{16 \sqrt [4]{2}}-\frac {\log \left (\sqrt {2}+2^{3/4} \sqrt [4]{2-3 x^2}+\sqrt {2-3 x^2}\right )}{16 \sqrt [4]{2}}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 156, normalized size = 0.79 \begin {gather*} \frac {-4 \tan ^{-1}\left (\sqrt [4]{1-\frac {3 x^2}{2}}\right )-4 \tanh ^{-1}\left (\sqrt [4]{1-\frac {3 x^2}{2}}\right )+\sqrt {2} \left (\log \left (\sqrt {2-3 x^2}-2^{3/4} \sqrt [4]{2-3 x^2}+\sqrt {2}\right )-\log \left (\sqrt {2-3 x^2}+2^{3/4} \sqrt [4]{2-3 x^2}+\sqrt {2}\right )+2 \tan ^{-1}\left (1-\sqrt [4]{4-6 x^2}\right )-2 \tan ^{-1}\left (\sqrt [4]{4-6 x^2}+1\right )\right )}{16\ 2^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*ArcTan[(1 - (3*x^2)/2)^(1/4)] - 4*ArcTanh[(1 - (3*x^2)/2)^(1/4)] + Sqrt[2]*(2*ArcTan[1 - (4 - 6*x^2)^(1/4)
] - 2*ArcTan[1 + (4 - 6*x^2)^(1/4)] + Log[Sqrt[2] - 2^(3/4)*(2 - 3*x^2)^(1/4) + Sqrt[2 - 3*x^2]] - Log[Sqrt[2]
+ 2^(3/4)*(2 - 3*x^2)^(1/4) + Sqrt[2 - 3*x^2]]))/(16*2^(3/4))

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IntegrateAlgebraic [A]  time = 0.19, size = 151, normalized size = 0.77 \begin {gather*} -\frac {\tan ^{-1}\left (\frac {\sqrt [4]{2-3 x^2}}{\sqrt [4]{2}}\right )}{4\ 2^{3/4}}-\frac {\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 )}{8 \sqrt [4]{2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{2-3 x^2}}{\sqrt [4]{2}}\right )}{4\ 2^{3/4}}-\frac {\tanh ^{-1}\left (\frac {2 \sqrt [4]{2} \sqrt [4]{2-3 x^2}}{\sqrt {2} \sqrt {2-3 x^2}+2}\right )}{8 \sqrt [4]{2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4*ArcTan[(2 - 3*x^2)^(1/4)/2^(1/4)]/2^(3/4) - ArcTan[(-2^(-1/4) + Sqrt[2 - 3*x^2]/2^(3/4))/(2 - 3*x^2)^(1/4
)]/(8*2^(1/4)) - ArcTanh[(2 - 3*x^2)^(1/4)/2^(1/4)]/(4*2^(3/4)) - ArcTanh[(2*2^(1/4)*(2 - 3*x^2)^(1/4))/(2 + S
qrt[2]*Sqrt[2 - 3*x^2])]/(8*2^(1/4))

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fricas [B]  time = 1.01, size = 315, normalized size = 1.60 \begin {gather*} \frac {1}{32} \cdot 8^{\frac {3}{4}} \sqrt {2} \arctan \left (\frac {1}{4} \cdot 8^{\frac {1}{4}} \sqrt {2} \sqrt {8^{\frac {3}{4}} \sqrt {2} {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}} + 4 \, \sqrt {2} + 4 \, \sqrt {-3 \, x^{2} + 2}} - \frac {1}{2} \cdot 8^{\frac {1}{4}} \sqrt {2} {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}} - 1\right ) + \frac {1}{32} \cdot 8^{\frac {3}{4}} \sqrt {2} \arctan \left (\frac {1}{16} \cdot 8^{\frac {1}{4}} \sqrt {2} \sqrt {-16 \cdot 8^{\frac {3}{4}} \sqrt {2} {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}} + 64 \, \sqrt {2} + 64 \, \sqrt {-3 \, x^{2} + 2}} - \frac {1}{2} \cdot 8^{\frac {1}{4}} \sqrt {2} {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}} + 1\right ) - \frac {1}{128} \cdot 8^{\frac {3}{4}} \sqrt {2} \log \left (16 \cdot 8^{\frac {3}{4}} \sqrt {2} {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}} + 64 \, \sqrt {2} + 64 \, \sqrt {-3 \, x^{2} + 2}\right ) + \frac {1}{128} \cdot 8^{\frac {3}{4}} \sqrt {2} \log \left (-16 \cdot 8^{\frac {3}{4}} \sqrt {2} {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}} + 64 \, \sqrt {2} + 64 \, \sqrt {-3 \, x^{2} + 2}\right ) + \frac {1}{16} \cdot 8^{\frac {3}{4}} \arctan \left (\frac {1}{2} \cdot 8^{\frac {1}{4}} \sqrt {\sqrt {2} + \sqrt {-3 \, x^{2} + 2}} - \frac {1}{2} \cdot 8^{\frac {1}{4}} {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}}\right ) - \frac {1}{64} \cdot 8^{\frac {3}{4}} \log \left (8^{\frac {3}{4}} + 4 \, {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}}\right ) + \frac {1}{64} \cdot 8^{\frac {3}{4}} \log \left (-8^{\frac {3}{4}} + 4 \, {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32*8^(3/4)*sqrt(2)*arctan(1/4*8^(1/4)*sqrt(2)*sqrt(8^(3/4)*sqrt(2)*(-3*x^2 + 2)^(1/4) + 4*sqrt(2) + 4*sqrt(-
3*x^2 + 2)) - 1/2*8^(1/4)*sqrt(2)*(-3*x^2 + 2)^(1/4) - 1) + 1/32*8^(3/4)*sqrt(2)*arctan(1/16*8^(1/4)*sqrt(2)*s
qrt(-16*8^(3/4)*sqrt(2)*(-3*x^2 + 2)^(1/4) + 64*sqrt(2) + 64*sqrt(-3*x^2 + 2)) - 1/2*8^(1/4)*sqrt(2)*(-3*x^2 +
2)^(1/4) + 1) - 1/128*8^(3/4)*sqrt(2)*log(16*8^(3/4)*sqrt(2)*(-3*x^2 + 2)^(1/4) + 64*sqrt(2) + 64*sqrt(-3*x^2
+ 2)) + 1/128*8^(3/4)*sqrt(2)*log(-16*8^(3/4)*sqrt(2)*(-3*x^2 + 2)^(1/4) + 64*sqrt(2) + 64*sqrt(-3*x^2 + 2))
+ 1/16*8^(3/4)*arctan(1/2*8^(1/4)*sqrt(sqrt(2) + sqrt(-3*x^2 + 2)) - 1/2*8^(1/4)*(-3*x^2 + 2)^(1/4)) - 1/64*8^
(3/4)*log(8^(3/4) + 4*(-3*x^2 + 2)^(1/4)) + 1/64*8^(3/4)*log(-8^(3/4) + 4*(-3*x^2 + 2)^(1/4))

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giac [A]  time = 0.50, size = 210, normalized size = 1.07 \begin {gather*} -\frac {1}{16} \cdot 4^{\frac {1}{8}} \sqrt {2} \arctan \left (\frac {1}{8} \cdot 4^{\frac {7}{8}} \sqrt {2} {\left (4^{\frac {1}{8}} \sqrt {2} + 2 \, {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}}\right )}\right ) - \frac {1}{16} \cdot 4^{\frac {1}{8}} \sqrt {2} \arctan \left (-\frac {1}{8} \cdot 4^{\frac {7}{8}} \sqrt {2} {\left (4^{\frac {1}{8}} \sqrt {2} - 2 \, {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}}\right )}\right ) - \frac {1}{32} \cdot 4^{\frac {1}{8}} \sqrt {2} \log \left (4^{\frac {1}{8}} \sqrt {2} {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}} + \sqrt {-3 \, x^{2} + 2} + 4^{\frac {1}{4}}\right ) + \frac {1}{32} \cdot 4^{\frac {1}{8}} \sqrt {2} \log \left (-4^{\frac {1}{8}} \sqrt {2} {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}} + \sqrt {-3 \, x^{2} + 2} + 4^{\frac {1}{4}}\right ) - \frac {1}{8} \cdot 4^{\frac {1}{8}} \arctan \left (\frac {1}{4} \cdot 4^{\frac {7}{8}} {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}}\right ) - \frac {1}{16} \cdot 4^{\frac {1}{8}} \log \left ({\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}} + 4^{\frac {1}{8}}\right ) + \frac {1}{16} \cdot 4^{\frac {1}{8}} \log \left (-{\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}} + 4^{\frac {1}{8}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*4^(1/8)*sqrt(2)*arctan(1/8*4^(7/8)*sqrt(2)*(4^(1/8)*sqrt(2) + 2*(-3*x^2 + 2)^(1/4))) - 1/16*4^(1/8)*sqrt
(2)*arctan(-1/8*4^(7/8)*sqrt(2)*(4^(1/8)*sqrt(2) - 2*(-3*x^2 + 2)^(1/4))) - 1/32*4^(1/8)*sqrt(2)*log(4^(1/8)*s
qrt(2)*(-3*x^2 + 2)^(1/4) + sqrt(-3*x^2 + 2) + 4^(1/4)) + 1/32*4^(1/8)*sqrt(2)*log(-4^(1/8)*sqrt(2)*(-3*x^2 +
2)^(1/4) + sqrt(-3*x^2 + 2) + 4^(1/4)) - 1/8*4^(1/8)*arctan(1/4*4^(7/8)*(-3*x^2 + 2)^(1/4)) - 1/16*4^(1/8)*log
((-3*x^2 + 2)^(1/4) + 4^(1/8)) + 1/16*4^(1/8)*log(-(-3*x^2 + 2)^(1/4) + 4^(1/8))

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maple [C]  time = 15.08, size = 560, normalized size = 2.84 \begin {gather*} \frac {\RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-2\right )^{2}\right ) \RootOf \left (\textit {\_Z}^{4}-2\right )^{2} \ln \left (\frac {3 x^{2} \RootOf \left (\textit {\_Z}^{4}-2\right )-4 \sqrt {-3 x^{2}+2}\, \RootOf \left (\textit {\_Z}^{4}-2\right )^{3}+4 \left (-3 x^{2}+2\right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{4}-2\right )^{2}-4 \RootOf \left (\textit {\_Z}^{4}-2\right )+4 \left (-3 x^{2}+2\right )^{\frac {3}{4}}}{3 x^{2}-4}\right )}{32}-\frac {\RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-2\right )^{2}\right ) \RootOf \left (\textit {\_Z}^{4}-2\right )^{2} \ln \left (\frac {3 x^{2} \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-2\right )^{2}\right )+3 x^{2} \RootOf \left (\textit {\_Z}^{4}-2\right )+2 \sqrt {-3 x^{2}+2}\, \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-2\right )^{2}\right ) \RootOf \left (\textit {\_Z}^{4}-2\right )^{2}-2 \sqrt {-3 x^{2}+2}\, \RootOf \left (\textit {\_Z}^{4}-2\right )^{3}-4 \left (-3 x^{2}+2\right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-2\right )^{2}\right ) \RootOf \left (\textit {\_Z}^{4}-2\right )+4 \left (-3 x^{2}+2\right )^{\frac {3}{4}}}{3 x^{2}-4}\right )}{16}+\frac {\RootOf \left (\textit {\_Z}^{4}-2\right )^{3} \ln \left (\frac {3 x^{2} \RootOf \left (\textit {\_Z}^{4}-2\right )-4 \sqrt {-3 x^{2}+2}\, \RootOf \left (\textit {\_Z}^{4}-2\right )^{3}+4 \left (-3 x^{2}+2\right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{4}-2\right )^{2}-4 \RootOf \left (\textit {\_Z}^{4}-2\right )+4 \left (-3 x^{2}+2\right )^{\frac {3}{4}}}{3 x^{2}-4}\right )}{32}+\frac {\RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-2\right )^{2}\right ) \ln \left (\frac {3 x^{2} \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-2\right )^{2}\right ) \RootOf \left (\textit {\_Z}^{4}-2\right )^{2}-4 \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-2\right )^{2}\right ) \RootOf \left (\textit {\_Z}^{4}-2\right )^{2}+4 \left (-3 x^{2}+2\right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{4}-2\right )^{2}+4 \sqrt {-3 x^{2}+2}\, \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-2\right )^{2}\right )-4 \left (-3 x^{2}+2\right )^{\frac {3}{4}}}{x^{2}}\right )}{16}+\frac {\RootOf \left (\textit {\_Z}^{4}-2\right ) \ln \left (-\frac {3 x^{2} \RootOf \left (\textit {\_Z}^{4}-2\right )^{3}-4 \RootOf \left (\textit {\_Z}^{4}-2\right )^{3}+4 \left (-3 x^{2}+2\right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{4}-2\right )^{2}-4 \sqrt {-3 x^{2}+2}\, \RootOf \left (\textit {\_Z}^{4}-2\right )+4 \left (-3 x^{2}+2\right )^{\frac {3}{4}}}{x^{2}}\right )}{16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*RootOf(_Z^2+RootOf(_Z^4-2)^2)*ln((3*RootOf(_Z^4-2)^2*RootOf(_Z^2+RootOf(_Z^4-2)^2)*x^2+4*RootOf(_Z^2+Root
Of(_Z^4-2)^2)*(-3*x^2+2)^(1/2)-4*RootOf(_Z^4-2)^2*RootOf(_Z^2+RootOf(_Z^4-2)^2)-4*(-3*x^2+2)^(3/4)+4*RootOf(_Z
^4-2)^2*(-3*x^2+2)^(1/4))/x^2)+1/16*RootOf(_Z^4-2)*ln(-(3*RootOf(_Z^4-2)^3*x^2+4*(-3*x^2+2)^(3/4)-4*(-3*x^2+2)
^(1/2)*RootOf(_Z^4-2)+4*RootOf(_Z^4-2)^2*(-3*x^2+2)^(1/4)-4*RootOf(_Z^4-2)^3)/x^2)+1/32*ln((-4*RootOf(_Z^4-2)^
3*(-3*x^2+2)^(1/2)+4*(-3*x^2+2)^(3/4)+4*RootOf(_Z^4-2)^2*(-3*x^2+2)^(1/4)+3*RootOf(_Z^4-2)*x^2-4*RootOf(_Z^4-2
))/(3*x^2-4))*RootOf(_Z^4-2)^3+1/32*ln((-4*RootOf(_Z^4-2)^3*(-3*x^2+2)^(1/2)+4*(-3*x^2+2)^(3/4)+4*RootOf(_Z^4-
2)^2*(-3*x^2+2)^(1/4)+3*RootOf(_Z^4-2)*x^2-4*RootOf(_Z^4-2))/(3*x^2-4))*RootOf(_Z^4-2)^2*RootOf(_Z^2+RootOf(_Z
^4-2)^2)-1/16*RootOf(_Z^4-2)^2*RootOf(_Z^2+RootOf(_Z^4-2)^2)*ln((2*(-3*x^2+2)^(1/2)*RootOf(_Z^2+RootOf(_Z^4-2)
^2)*RootOf(_Z^4-2)^2-2*RootOf(_Z^4-2)^3*(-3*x^2+2)^(1/2)-4*(-3*x^2+2)^(1/4)*RootOf(_Z^2+RootOf(_Z^4-2)^2)*Root
Of(_Z^4-2)+3*RootOf(_Z^2+RootOf(_Z^4-2)^2)*x^2+4*(-3*x^2+2)^(3/4)+3*RootOf(_Z^4-2)*x^2)/(3*x^2-4))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -\int \frac {1}{{\left (3 \, x^{2} - 4\right )} {\left (-3 \, x^{2} + 2\right )}^{\frac {3}{4}} x}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.24, size = 91, normalized size = 0.46 \begin {gather*} -\frac {2^{1/4}\,\mathrm {atan}\left (\frac {2^{3/4}\,{\left (2-3\,x^2\right )}^{1/4}}{2}\right )}{8}+\frac {2^{1/4}\,\mathrm {atan}\left (\frac {2^{3/4}\,{\left (2-3\,x^2\right )}^{1/4}\,1{}\mathrm {i}}{2}\right )\,1{}\mathrm {i}}{8}+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}{16}-\frac {1}{16}{}\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}{16}+\frac {1}{16}{}\mathrm {i}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2^(1/4)*atan((2^(3/4)*(2 - 3*x^2)^(1/4)*1i)/2)*1i)/8 - (2^(1/4)*atan((2^(3/4)*(2 - 3*x^2)^(1/4))/2))/8 - 2^(3
/4)*atan(2^(1/4)*(2 - 3*x^2)^(1/4)*(1/2 - 1i/2))*(1/16 + 1i/16) - 2^(3/4)*atan(2^(1/4)*(2 - 3*x^2)^(1/4)*(1/2
+ 1i/2))*(1/16 - 1i/16)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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