\(\int \frac {1}{x^3 \sqrt [4]{2-3 x^2} (4-3 x^2)} \, dx\) [1036]

Optimal result

Integrand size = 24, antiderivative size = 163 \[ \int \frac {1}{x^3 \sqrt [4]{2-3 x^2} \left (4-3 x^2\right )} \, dx=-\frac {\left (2-3 x^2\right )^{3/4}}{16 x^2}+\frac {9 \arctan \left (\frac {\sqrt [4]{2-3 x^2}}{\sqrt [4]{2}}\right )}{32 \sqrt [4]{2}}+\frac {3 \arctan \left (\frac {\sqrt {2}-\sqrt {2-3 x^2}}{2^{3/4} \sqrt [4]{2-3 x^2}}\right )}{16\ 2^{3/4}}-\frac {9 \text {arctanh}\left (\frac {\sqrt [4]{2-3 x^2}}{\sqrt [4]{2}}\right )}{32 \sqrt [4]{2}}+\frac {3 \text {arctanh}\left (\frac {\sqrt {2}+\sqrt {2-3 x^2}}{2^{3/4} \sqrt [4]{2-3 x^2}}\right )}{16\ 2^{3/4}} \]

[Out]

-1/16*(-3*x^2+2)^(3/4)/x^2+9/64*arctan(1/2*2^(3/4)*(-3*x^2+2)^(1/4))*2^(3/4)+3/32*2^(1/4)*arctan(1/2*(2^(1/2)-
(-3*x^2+2)^(1/2))*2^(1/4)/(-3*x^2+2)^(1/4))-9/64*arctanh(1/2*2^(3/4)*(-3*x^2+2)^(1/4))*2^(3/4)+3/32*2^(1/4)*ar
ctanh(1/2*(2^(1/2)+(-3*x^2+2)^(1/2))*2^(1/4)/(-3*x^2+2)^(1/4))

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {451, 272, 44, 65, 304, 209, 212, 450} \[ \int \frac {1}{x^3 \sqrt [4]{2-3 x^2} \left (4-3 x^2\right )} \, dx=\frac {9 \arctan \left (\frac {\sqrt [4]{2-3 x^2}}{\sqrt [4]{2}}\right )}{32 \sqrt [4]{2}}+\frac {3 \arctan \left (\frac {\sqrt {2}-\sqrt {2-3 x^2}}{2^{3/4} \sqrt [4]{2-3 x^2}}\right )}{16\ 2^{3/4}}-\frac {9 \text {arctanh}\left (\frac {\sqrt [4]{2-3 x^2}}{\sqrt [4]{2}}\right )}{32 \sqrt [4]{2}}+\frac {3 \text {arctanh}\left (\frac {\sqrt {2-3 x^2}+\sqrt {2}}{2^{3/4} \sqrt [4]{2-3 x^2}}\right )}{16\ 2^{3/4}}-\frac {\left (2-3 x^2\right )^{3/4}}{16 x^2} \]

[In]

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

[Out]

-1/16*(2 - 3*x^2)^(3/4)/x^2 + (9*ArcTan[(2 - 3*x^2)^(1/4)/2^(1/4)])/(32*2^(1/4)) + (3*ArcTan[(Sqrt[2] - Sqrt[2
- 3*x^2])/(2^(3/4)*(2 - 3*x^2)^(1/4))])/(16*2^(3/4)) - (9*ArcTanh[(2 - 3*x^2)^(1/4)/2^(1/4)])/(32*2^(1/4)) +
(3*ArcTanh[(Sqrt[2] + Sqrt[2 - 3*x^2])/(2^(3/4)*(2 - 3*x^2)^(1/4))])/(16*2^(3/4))

Rule 44

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n + 1
)/((b*c - a*d)*(m + 1))), x] - Dist[d*((m + n + 2)/((b*c - a*d)*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && LtQ[n, 0]

Rule 65

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 209

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

Rule 212

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

Rule 272

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 304

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 450

Int[(x_)/(((a_) + (b_.)*(x_)^2)^(1/4)*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Simp[(-(Sqrt[2]*Rt[a, 4]*d)^(-1))*A
rcTan[(Rt[a, 4]^2 - Sqrt[a + b*x^2])/(Sqrt[2]*Rt[a, 4]*(a + b*x^2)^(1/4))], x] - Simp[(1/(Sqrt[2]*Rt[a, 4]*d))
*ArcTanh[(Rt[a, 4]^2 + Sqrt[a + b*x^2])/(Sqrt[2]*Rt[a, 4]*(a + b*x^2)^(1/4))], x] /; FreeQ[{a, b, c, d}, x] &&
EqQ[b*c - 2*a*d, 0] && PosQ[a]

Rule 451

Int[(x_)^(m_)/(((a_) + (b_.)*(x_)^2)^(1/4)*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Int[ExpandIntegrand[x^m/((a +
b*x^2)^(1/4)*(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])

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {1}{4 x^3 \sqrt [4]{2-3 x^2}}+\frac {3}{16 x \sqrt [4]{2-3 x^2}}-\frac {9 x}{16 \sqrt [4]{2-3 x^2} \left (-4+3 x^2\right )}\right ) \, dx \\ & = \frac {3}{16} \int \frac {1}{x \sqrt [4]{2-3 x^2}} \, dx+\frac {1}{4} \int \frac {1}{x^3 \sqrt [4]{2-3 x^2}} \, dx-\frac {9}{16} \int \frac {x}{\sqrt [4]{2-3 x^2} \left (-4+3 x^2\right )} \, dx \\ & = \frac {3 \tan ^{-1}\left (\frac {\sqrt {2}-\sqrt {2-3 x^2}}{2^{3/4} \sqrt [4]{2-3 x^2}}\right )}{16\ 2^{3/4}}+\frac {3 \tanh ^{-1}\left (\frac {\sqrt {2}+\sqrt {2-3 x^2}}{2^{3/4} \sqrt [4]{2-3 x^2}}\right )}{16\ 2^{3/4}}+\frac {3}{32} \text {Subst}\left (\int \frac {1}{\sqrt [4]{2-3 x} x} \, dx,x,x^2\right )+\frac {1}{8} \text {Subst}\left (\int \frac {1}{\sqrt [4]{2-3 x} x^2} \, dx,x,x^2\right ) \\ & = -\frac {\left (2-3 x^2\right )^{3/4}}{16 x^2}+\frac {3 \tan ^{-1}\left (\frac {\sqrt {2}-\sqrt {2-3 x^2}}{2^{3/4} \sqrt [4]{2-3 x^2}}\right )}{16\ 2^{3/4}}+\frac {3 \tanh ^{-1}\left (\frac {\sqrt {2}+\sqrt {2-3 x^2}}{2^{3/4} \sqrt [4]{2-3 x^2}}\right )}{16\ 2^{3/4}}+\frac {3}{64} \text {Subst}\left (\int \frac {1}{\sqrt [4]{2-3 x} x} \, dx,x,x^2\right )-\frac {1}{8} \text {Subst}\left (\int \frac {x^2}{\frac {2}{3}-\frac {x^4}{3}} \, dx,x,\sqrt [4]{2-3 x^2}\right ) \\ & = -\frac {\left (2-3 x^2\right )^{3/4}}{16 x^2}+\frac {3 \tan ^{-1}\left (\frac {\sqrt {2}-\sqrt {2-3 x^2}}{2^{3/4} \sqrt [4]{2-3 x^2}}\right )}{16\ 2^{3/4}}+\frac {3 \tanh ^{-1}\left (\frac {\sqrt {2}+\sqrt {2-3 x^2}}{2^{3/4} \sqrt [4]{2-3 x^2}}\right )}{16\ 2^{3/4}}-\frac {1}{16} \text {Subst}\left (\int \frac {x^2}{\frac {2}{3}-\frac {x^4}{3}} \, dx,x,\sqrt [4]{2-3 x^2}\right )-\frac {3}{16} \text {Subst}\left (\int \frac {1}{\sqrt {2}-x^2} \, dx,x,\sqrt [4]{2-3 x^2}\right )+\frac {3}{16} \text {Subst}\left (\int \frac {1}{\sqrt {2}+x^2} \, dx,x,\sqrt [4]{2-3 x^2}\right ) \\ & = -\frac {\left (2-3 x^2\right )^{3/4}}{16 x^2}+\frac {3 \tan ^{-1}\left (\frac {\sqrt [4]{2-3 x^2}}{\sqrt [4]{2}}\right )}{16 \sqrt [4]{2}}+\frac {3 \tan ^{-1}\left (\frac {\sqrt {2}-\sqrt {2-3 x^2}}{2^{3/4} \sqrt [4]{2-3 x^2}}\right )}{16\ 2^{3/4}}-\frac {3 \tanh ^{-1}\left (\frac {\sqrt [4]{2-3 x^2}}{\sqrt [4]{2}}\right )}{16 \sqrt [4]{2}}+\frac {3 \tanh ^{-1}\left (\frac {\sqrt {2}+\sqrt {2-3 x^2}}{2^{3/4} \sqrt [4]{2-3 x^2}}\right )}{16\ 2^{3/4}}-\frac {3}{32} \text {Subst}\left (\int \frac {1}{\sqrt {2}-x^2} \, dx,x,\sqrt [4]{2-3 x^2}\right )+\frac {3}{32} \text {Subst}\left (\int \frac {1}{\sqrt {2}+x^2} \, dx,x,\sqrt [4]{2-3 x^2}\right ) \\ & = -\frac {\left (2-3 x^2\right )^{3/4}}{16 x^2}+\frac {9 \tan ^{-1}\left (\frac {\sqrt [4]{2-3 x^2}}{\sqrt [4]{2}}\right )}{32 \sqrt [4]{2}}+\frac {3 \tan ^{-1}\left (\frac {\sqrt {2}-\sqrt {2-3 x^2}}{2^{3/4} \sqrt [4]{2-3 x^2}}\right )}{16\ 2^{3/4}}-\frac {9 \tanh ^{-1}\left (\frac {\sqrt [4]{2-3 x^2}}{\sqrt [4]{2}}\right )}{32 \sqrt [4]{2}}+\frac {3 \tanh ^{-1}\left (\frac {\sqrt {2}+\sqrt {2-3 x^2}}{2^{3/4} \sqrt [4]{2-3 x^2}}\right )}{16\ 2^{3/4}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.26 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.95 \[ \int \frac {1}{x^3 \sqrt [4]{2-3 x^2} \left (4-3 x^2\right )} \, dx=\frac {-4 \left (2-3 x^2\right )^{3/4}+9\ 2^{3/4} x^2 \arctan \left (\sqrt [4]{1-\frac {3 x^2}{2}}\right )+6 \sqrt [4]{2} x^2 \arctan \left (\frac {\sqrt {2}-\sqrt {2-3 x^2}}{2^{3/4} \sqrt [4]{2-3 x^2}}\right )-9\ 2^{3/4} x^2 \text {arctanh}\left (\sqrt [4]{1-\frac {3 x^2}{2}}\right )+6 \sqrt [4]{2} x^2 \text {arctanh}\left (\frac {2 \sqrt [4]{4-6 x^2}}{2+\sqrt {4-6 x^2}}\right )}{64 x^2} \]

[In]

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

[Out]

(-4*(2 - 3*x^2)^(3/4) + 9*2^(3/4)*x^2*ArcTan[(1 - (3*x^2)/2)^(1/4)] + 6*2^(1/4)*x^2*ArcTan[(Sqrt[2] - Sqrt[2 -
3*x^2])/(2^(3/4)*(2 - 3*x^2)^(1/4))] - 9*2^(3/4)*x^2*ArcTanh[(1 - (3*x^2)/2)^(1/4)] + 6*2^(1/4)*x^2*ArcTanh[(
2*(4 - 6*x^2)^(1/4))/(2 + Sqrt[4 - 6*x^2])])/(64*x^2)

Maple [A] (verified)

Time = 16.48 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.19

method	result	size
pseudoelliptic	\(\frac {18 \arctan \left (\frac {2^{\frac {3}{4}} \left (-3 x^{2}+2\right )^{\frac {1}{4}}}{2}\right ) 2^{\frac {3}{4}} x^{2}-9 \ln \left (\frac {\left (-3 x^{2}+2\right )^{\frac {1}{4}}+2^{\frac {1}{4}}}{\left (-3 x^{2}+2\right )^{\frac {1}{4}}-2^{\frac {1}{4}}}\right ) 2^{\frac {3}{4}} x^{2}-6 \ln \left (\frac {-2^{\frac {3}{4}} \left (-3 x^{2}+2\right )^{\frac {1}{4}}+\sqrt {2}+\sqrt {-3 x^{2}+2}}{2^{\frac {3}{4}} \left (-3 x^{2}+2\right )^{\frac {1}{4}}+\sqrt {2}+\sqrt {-3 x^{2}+2}}\right ) 2^{\frac {1}{4}} x^{2}-12 \arctan \left (2^{\frac {1}{4}} \left (-3 x^{2}+2\right )^{\frac {1}{4}}+1\right ) 2^{\frac {1}{4}} x^{2}-12 \arctan \left (-1+2^{\frac {1}{4}} \left (-3 x^{2}+2\right )^{\frac {1}{4}}\right ) 2^{\frac {1}{4}} x^{2}-8 \left (-3 x^{2}+2\right )^{\frac {3}{4}}}{128 x^{2}}\)	\(194\)

[In]

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

[Out]

1/128*(18*arctan(1/2*2^(3/4)*(-3*x^2+2)^(1/4))*2^(3/4)*x^2-9*ln(((-3*x^2+2)^(1/4)+2^(1/4))/((-3*x^2+2)^(1/4)-2
^(1/4)))*2^(3/4)*x^2-6*ln((-2^(3/4)*(-3*x^2+2)^(1/4)+2^(1/2)+(-3*x^2+2)^(1/2))/(2^(3/4)*(-3*x^2+2)^(1/4)+2^(1/
2)+(-3*x^2+2)^(1/2)))*2^(1/4)*x^2-12*arctan(2^(1/4)*(-3*x^2+2)^(1/4)+1)*2^(1/4)*x^2-12*arctan(-1+2^(1/4)*(-3*x
^2+2)^(1/4))*2^(1/4)*x^2-8*(-3*x^2+2)^(3/4))/x^2

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.29 (sec) , antiderivative size = 215, normalized size of antiderivative = 1.32 \[ \int \frac {1}{x^3 \sqrt [4]{2-3 x^2} \left (4-3 x^2\right )} \, dx=-\frac {9 \cdot 2^{\frac {3}{4}} x^{2} \log \left (2^{\frac {1}{4}} + {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}}\right ) - 9 i \cdot 2^{\frac {3}{4}} x^{2} \log \left (i \cdot 2^{\frac {1}{4}} + {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}}\right ) + 9 i \cdot 2^{\frac {3}{4}} x^{2} \log \left (-i \cdot 2^{\frac {1}{4}} + {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}}\right ) - 9 \cdot 2^{\frac {3}{4}} x^{2} \log \left (-2^{\frac {1}{4}} + {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}}\right ) + \left (6 i - 6\right ) \cdot 2^{\frac {1}{4}} x^{2} \log \left (\left (i + 1\right ) \cdot 2^{\frac {3}{4}} + 2 \, {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}}\right ) - \left (6 i + 6\right ) \cdot 2^{\frac {1}{4}} x^{2} \log \left (-\left (i - 1\right ) \cdot 2^{\frac {3}{4}} + 2 \, {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}}\right ) + \left (6 i + 6\right ) \cdot 2^{\frac {1}{4}} x^{2} \log \left (\left (i - 1\right ) \cdot 2^{\frac {3}{4}} + 2 \, {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}}\right ) - \left (6 i - 6\right ) \cdot 2^{\frac {1}{4}} x^{2} \log \left (-\left (i + 1\right ) \cdot 2^{\frac {3}{4}} + 2 \, {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}}\right ) + 8 \, {\left (-3 \, x^{2} + 2\right )}^{\frac {3}{4}}}{128 \, x^{2}} \]

[In]

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

[Out]

-1/128*(9*2^(3/4)*x^2*log(2^(1/4) + (-3*x^2 + 2)^(1/4)) - 9*I*2^(3/4)*x^2*log(I*2^(1/4) + (-3*x^2 + 2)^(1/4))
+ 9*I*2^(3/4)*x^2*log(-I*2^(1/4) + (-3*x^2 + 2)^(1/4)) - 9*2^(3/4)*x^2*log(-2^(1/4) + (-3*x^2 + 2)^(1/4)) + (6
*I - 6)*2^(1/4)*x^2*log((I + 1)*2^(3/4) + 2*(-3*x^2 + 2)^(1/4)) - (6*I + 6)*2^(1/4)*x^2*log(-(I - 1)*2^(3/4) +
2*(-3*x^2 + 2)^(1/4)) + (6*I + 6)*2^(1/4)*x^2*log((I - 1)*2^(3/4) + 2*(-3*x^2 + 2)^(1/4)) - (6*I - 6)*2^(1/4)
*x^2*log(-(I + 1)*2^(3/4) + 2*(-3*x^2 + 2)^(1/4)) + 8*(-3*x^2 + 2)^(3/4))/x^2

Sympy [F]

\[ \int \frac {1}{x^3 \sqrt [4]{2-3 x^2} \left (4-3 x^2\right )} \, dx=- \int \frac {1}{3 x^{5} \sqrt [4]{2 - 3 x^{2}} - 4 x^{3} \sqrt [4]{2 - 3 x^{2}}}\, dx \]

[In]

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

[Out]

-Integral(1/(3*x**5*(2 - 3*x**2)**(1/4) - 4*x**3*(2 - 3*x**2)**(1/4)), x)

Maxima [F]

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

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.18 \[ \int \frac {1}{x^3 \sqrt [4]{2-3 x^2} \left (4-3 x^2\right )} \, dx=\frac {9}{64} \cdot 2^{\frac {3}{4}} \arctan \left (\frac {1}{2} \cdot 2^{\frac {3}{4}} {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}}\right ) - \frac {9}{128} \cdot 2^{\frac {3}{4}} \log \left (2^{\frac {1}{4}} + {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}}\right ) + \frac {9}{128} \cdot 2^{\frac {3}{4}} \log \left (2^{\frac {1}{4}} - {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}}\right ) - \frac {3}{32} \cdot 2^{\frac {1}{4}} \arctan \left (\frac {1}{2} \cdot 2^{\frac {1}{4}} {\left (2^{\frac {3}{4}} + 2 \, {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}}\right )}\right ) - \frac {3}{32} \cdot 2^{\frac {1}{4}} \arctan \left (-\frac {1}{2} \cdot 2^{\frac {1}{4}} {\left (2^{\frac {3}{4}} - 2 \, {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}}\right )}\right ) + \frac {3}{64} \cdot 2^{\frac {1}{4}} \log \left (2^{\frac {3}{4}} {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}} + \sqrt {2} + \sqrt {-3 \, x^{2} + 2}\right ) - \frac {3}{64} \cdot 2^{\frac {1}{4}} \log \left (-2^{\frac {3}{4}} {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}} + \sqrt {2} + \sqrt {-3 \, x^{2} + 2}\right ) - \frac {{\left (-3 \, x^{2} + 2\right )}^{\frac {3}{4}}}{16 \, x^{2}} \]

[In]

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

[Out]

9/64*2^(3/4)*arctan(1/2*2^(3/4)*(-3*x^2 + 2)^(1/4)) - 9/128*2^(3/4)*log(2^(1/4) + (-3*x^2 + 2)^(1/4)) + 9/128*
2^(3/4)*log(2^(1/4) - (-3*x^2 + 2)^(1/4)) - 3/32*2^(1/4)*arctan(1/2*2^(1/4)*(2^(3/4) + 2*(-3*x^2 + 2)^(1/4)))
- 3/32*2^(1/4)*arctan(-1/2*2^(1/4)*(2^(3/4) - 2*(-3*x^2 + 2)^(1/4))) + 3/64*2^(1/4)*log(2^(3/4)*(-3*x^2 + 2)^(
1/4) + sqrt(2) + sqrt(-3*x^2 + 2)) - 3/64*2^(1/4)*log(-2^(3/4)*(-3*x^2 + 2)^(1/4) + sqrt(2) + sqrt(-3*x^2 + 2)
) - 1/16*(-3*x^2 + 2)^(3/4)/x^2

Mupad [B] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.67 \[ \int \frac {1}{x^3 \sqrt [4]{2-3 x^2} \left (4-3 x^2\right )} \, dx=\frac {9\,2^{3/4}\,\mathrm {atan}\left (\frac {2^{3/4}\,{\left (2-3\,x^2\right )}^{1/4}}{2}\right )}{64}-\frac {{\left (2-3\,x^2\right )}^{3/4}}{16\,x^2}+\frac {2^{3/4}\,\mathrm {atan}\left (\frac {2^{3/4}\,{\left (2-3\,x^2\right )}^{1/4}\,1{}\mathrm {i}}{2}\right )\,9{}\mathrm {i}}{64}-\frac {{\left (-1\right )}^{1/4}\,2^{3/4}\,\mathrm {atan}\left (\frac {{\left (-1\right )}^{1/4}\,2^{3/4}\,{\left (2-3\,x^2\right )}^{1/4}\,1{}\mathrm {i}}{2}\right )\,3{}\mathrm {i}}{32}-\frac {{\left (-1\right )}^{3/4}\,2^{3/4}\,\mathrm {atan}\left (\frac {{\left (-1\right )}^{3/4}\,2^{3/4}\,{\left (2-3\,x^2\right )}^{1/4}\,1{}\mathrm {i}}{2}\right )\,3{}\mathrm {i}}{32} \]

[In]

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

[Out]

(9*2^(3/4)*atan((2^(3/4)*(2 - 3*x^2)^(1/4))/2))/64 - (2 - 3*x^2)^(3/4)/(16*x^2) + (2^(3/4)*atan((2^(3/4)*(2 -
3*x^2)^(1/4)*1i)/2)*9i)/64 - ((-1)^(1/4)*2^(3/4)*atan(((-1)^(1/4)*2^(3/4)*(2 - 3*x^2)^(1/4)*1i)/2)*3i)/32 - ((
-1)^(3/4)*2^(3/4)*atan(((-1)^(3/4)*2^(3/4)*(2 - 3*x^2)^(1/4)*1i)/2)*3i)/32