3.1036 \(\int \frac{1}{x^3 \sqrt [4]{2-3 x^2} (4-3 x^2)} \, dx\)

Optimal. Leaf size=163 \[ -\frac{\left (2-3 x^2\right )^{3/4}}{16 x^2}+\frac{9 \tan ^{-1}\left (\frac{\sqrt [4]{2-3 x^2}}{\sqrt [4]{2}}\right )}{32 \sqrt [4]{2}}+\frac{3 \tan ^{-1}\left (\frac{\sqrt{2}-\sqrt{2-3 x^2}}{2^{3/4} \sqrt [4]{2-3 x^2}}\right )}{16\ 2^{3/4}}-\frac{9 \tanh ^{-1}\left (\frac{\sqrt [4]{2-3 x^2}}{\sqrt [4]{2}}\right )}{32 \sqrt [4]{2}}+\frac{3 \tanh ^{-1}\left (\frac{\sqrt{2-3 x^2}+\sqrt{2}}{2^{3/4} \sqrt [4]{2-3 x^2}}\right )}{16\ 2^{3/4}} \]

[Out]

-(2 - 3*x^2)^(3/4)/(16*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))

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Rubi [A]  time = 0.132789, antiderivative size = 163, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 8, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {440, 266, 51, 63, 298, 203, 206, 439} \[ -\frac{\left (2-3 x^2\right )^{3/4}}{16 x^2}+\frac{9 \tan ^{-1}\left (\frac{\sqrt [4]{2-3 x^2}}{\sqrt [4]{2}}\right )}{32 \sqrt [4]{2}}+\frac{3 \tan ^{-1}\left (\frac{\sqrt{2}-\sqrt{2-3 x^2}}{2^{3/4} \sqrt [4]{2-3 x^2}}\right )}{16\ 2^{3/4}}-\frac{9 \tanh ^{-1}\left (\frac{\sqrt [4]{2-3 x^2}}{\sqrt [4]{2}}\right )}{32 \sqrt [4]{2}}+\frac{3 \tanh ^{-1}\left (\frac{\sqrt{2-3 x^2}+\sqrt{2}}{2^{3/4} \sqrt [4]{2-3 x^2}}\right )}{16\ 2^{3/4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(2 - 3*x^2)^(3/4)/(16*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 440

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])

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 51

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] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

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 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 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 439

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

Rubi steps

\begin{align*} \int \frac{1}{x^3 \sqrt [4]{2-3 x^2} \left (4-3 x^2\right )} \, dx &=\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} \operatorname{Subst}\left (\int \frac{1}{\sqrt [4]{2-3 x} x} \, dx,x,x^2\right )+\frac{1}{8} \operatorname{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} \operatorname{Subst}\left (\int \frac{1}{\sqrt [4]{2-3 x} x} \, dx,x,x^2\right )-\frac{1}{8} \operatorname{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} \operatorname{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} \operatorname{Subst}\left (\int \frac{1}{\sqrt{2}-x^2} \, dx,x,\sqrt [4]{2-3 x^2}\right )+\frac{3}{16} \operatorname{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} \operatorname{Subst}\left (\int \frac{1}{\sqrt{2}-x^2} \, dx,x,\sqrt [4]{2-3 x^2}\right )+\frac{3}{32} \operatorname{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 [C]  time = 0.0506956, size = 102, normalized size = 0.63 \[ -\frac{4 \left (2-3 x^2\right )^{3/4} x^2 \, _2F_1\left (\frac{3}{4},1;\frac{7}{4};\frac{3 x^2}{2}-1\right )+4 \left (2-3 x^2\right )^{3/4}-9\ 2^{3/4} x^2 \tan ^{-1}\left (\sqrt [4]{1-\frac{3 x^2}{2}}\right )+9\ 2^{3/4} x^2 \tanh ^{-1}\left (\sqrt [4]{1-\frac{3 x^2}{2}}\right )}{64 x^2} \]

Antiderivative was successfully verified.

[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)] + 9*2^(3/4)*x^2*ArcTanh[(1 - (3*x^2)/2)^(1
/4)] + 4*x^2*(2 - 3*x^2)^(3/4)*Hypergeometric2F1[3/4, 1, 7/4, -1 + (3*x^2)/2])/(64*x^2)

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Maple [F]  time = 0.077, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{x}^{3} \left ( -3\,{x}^{2}+4 \right ) }{\frac{1}{\sqrt [4]{-3\,{x}^{2}+2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [B]  time = 1.45525, size = 909, normalized size = 5.58 \begin{align*} -\frac{36 \cdot 2^{\frac{3}{4}} x^{2} \arctan \left (\frac{1}{2} \cdot 2^{\frac{3}{4}} \sqrt{\sqrt{2} + \sqrt{-3 \, x^{2} + 2}} - \frac{1}{2} \cdot 2^{\frac{3}{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 ) - 9 \cdot 2^{\frac{3}{4}} x^{2} \log \left (-2^{\frac{1}{4}} +{\left (-3 \, x^{2} + 2\right )}^{\frac{1}{4}}\right ) - 24 \cdot 2^{\frac{1}{4}} x^{2} \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 ) - 24 \cdot 2^{\frac{1}{4}} x^{2} \arctan \left (\frac{1}{2} \cdot 2^{\frac{1}{4}} \sqrt{-4 \cdot 2^{\frac{3}{4}}{\left (-3 \, x^{2} + 2\right )}^{\frac{1}{4}} + 4 \, \sqrt{2} + 4 \, \sqrt{-3 \, x^{2} + 2}} - 2^{\frac{1}{4}}{\left (-3 \, x^{2} + 2\right )}^{\frac{1}{4}} + 1\right ) - 6 \cdot 2^{\frac{1}{4}} x^{2} \log \left (4 \cdot 2^{\frac{3}{4}}{\left (-3 \, x^{2} + 2\right )}^{\frac{1}{4}} + 4 \, \sqrt{2} + 4 \, \sqrt{-3 \, x^{2} + 2}\right ) + 6 \cdot 2^{\frac{1}{4}} x^{2} \log \left (-4 \cdot 2^{\frac{3}{4}}{\left (-3 \, x^{2} + 2\right )}^{\frac{1}{4}} + 4 \, \sqrt{2} + 4 \, \sqrt{-3 \, x^{2} + 2}\right ) + 8 \,{\left (-3 \, x^{2} + 2\right )}^{\frac{3}{4}}}{128 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/128*(36*2^(3/4)*x^2*arctan(1/2*2^(3/4)*sqrt(sqrt(2) + sqrt(-3*x^2 + 2)) - 1/2*2^(3/4)*(-3*x^2 + 2)^(1/4)) +
9*2^(3/4)*x^2*log(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)) - 24*2^(1/
4)*x^2*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) - 24*2^(1/4)*x^2*arctan(1/2*2^(1/4)*sqrt(-4*2^(3/4)*(-3*x^2 + 2)^(1/4) + 4*sqrt(2) + 4*sqrt(-3*x^2 + 2
)) - 2^(1/4)*(-3*x^2 + 2)^(1/4) + 1) - 6*2^(1/4)*x^2*log(4*2^(3/4)*(-3*x^2 + 2)^(1/4) + 4*sqrt(2) + 4*sqrt(-3*
x^2 + 2)) + 6*2^(1/4)*x^2*log(-4*2^(3/4)*(-3*x^2 + 2)^(1/4) + 4*sqrt(2) + 4*sqrt(-3*x^2 + 2)) + 8*(-3*x^2 + 2)
^(3/4))/x^2

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \int \frac{1}{3 x^{5} \sqrt [4]{2 - 3 x^{2}} - 4 x^{3} \sqrt [4]{2 - 3 x^{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Giac [A]  time = 1.33623, size = 259, normalized size = 1.59 \begin{align*} \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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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