3.1071 \(\int \frac {1}{x^2 (2-3 x^2)^{3/4} (4-3 x^2)} \, dx\)

Optimal. Leaf size=166 \[ -\frac {\sqrt [4]{2-3 x^2}}{8 x}+\frac {\sqrt {3} \tan ^{-1}\left (\frac {2^{3/4}-\sqrt [4]{2} \sqrt {2-3 x^2}}{\sqrt {3} x \sqrt [4]{2-3 x^2}}\right )}{16 \sqrt [4]{2}}-\frac {\sqrt {3} \tanh ^{-1}\left (\frac {\sqrt [4]{2} \sqrt {2-3 x^2}+2^{3/4}}{\sqrt {3} x \sqrt [4]{2-3 x^2}}\right )}{16 \sqrt [4]{2}}+\frac {\sqrt {3} F\left (\left .\frac {1}{2} \sin ^{-1}\left (\sqrt {\frac {3}{2}} x\right )\right |2\right )}{4 \sqrt [4]{2}} \]

[Out]

-1/8*(-3*x^2+2)^(1/4)/x+1/32*2^(3/4)*arctan(1/3*(2^(3/4)-2^(1/4)*(-3*x^2+2)^(1/2))/x/(-3*x^2+2)^(1/4)*3^(1/2))
*3^(1/2)-1/32*2^(3/4)*arctanh(1/3*(2^(3/4)+2^(1/4)*(-3*x^2+2)^(1/2))/x/(-3*x^2+2)^(1/4)*3^(1/2))*3^(1/2)+1/8*2
^(3/4)*(cos(1/2*arcsin(1/2*x*6^(1/2)))^2)^(1/2)/cos(1/2*arcsin(1/2*x*6^(1/2)))*EllipticF(sin(1/2*arcsin(1/2*x*
6^(1/2))),2^(1/2))*3^(1/2)

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Rubi [A]  time = 0.09, antiderivative size = 166, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {443, 325, 232, 400, 441} \[ -\frac {\sqrt [4]{2-3 x^2}}{8 x}+\frac {\sqrt {3} \tan ^{-1}\left (\frac {2^{3/4}-\sqrt [4]{2} \sqrt {2-3 x^2}}{\sqrt {3} x \sqrt [4]{2-3 x^2}}\right )}{16 \sqrt [4]{2}}-\frac {\sqrt {3} \tanh ^{-1}\left (\frac {\sqrt [4]{2} \sqrt {2-3 x^2}+2^{3/4}}{\sqrt {3} x \sqrt [4]{2-3 x^2}}\right )}{16 \sqrt [4]{2}}+\frac {\sqrt {3} F\left (\left .\frac {1}{2} \sin ^{-1}\left (\sqrt {\frac {3}{2}} x\right )\right |2\right )}{4 \sqrt [4]{2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(2 - 3*x^2)^(1/4)/(8*x) + (Sqrt[3]*ArcTan[(2^(3/4) - 2^(1/4)*Sqrt[2 - 3*x^2])/(Sqrt[3]*x*(2 - 3*x^2)^(1/4))])
/(16*2^(1/4)) - (Sqrt[3]*ArcTanh[(2^(3/4) + 2^(1/4)*Sqrt[2 - 3*x^2])/(Sqrt[3]*x*(2 - 3*x^2)^(1/4))])/(16*2^(1/
4)) + (Sqrt[3]*EllipticF[ArcSin[Sqrt[3/2]*x]/2, 2])/(4*2^(1/4))

Rule 232

Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Simp[(2*EllipticF[(1*ArcSin[Rt[-(b/a), 2]*x])/2, 2])/(a^(3/4)*R
t[-(b/a), 2]), x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && NegQ[b/a]

Rule 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 400

Int[1/(((a_) + (b_.)*(x_)^2)^(3/4)*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Dist[1/c, Int[1/(a + b*x^2)^(3/4), x],
x] - Dist[d/c, Int[x^2/((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]

Rule 441

Int[(x_)^2/(((a_) + (b_.)*(x_)^2)^(3/4)*((c_) + (d_.)*(x_)^2)), x_Symbol] :> -Simp[(b*ArcTan[(b + Rt[b^2/a, 4]
^2*Sqrt[a + b*x^2])/(Rt[b^2/a, 4]^3*x*(a + b*x^2)^(1/4))])/(a*d*Rt[b^2/a, 4]^3), x] + Simp[(b*ArcTanh[(b - Rt[
b^2/a, 4]^2*Sqrt[a + b*x^2])/(Rt[b^2/a, 4]^3*x*(a + b*x^2)^(1/4))])/(a*d*Rt[b^2/a, 4]^3), x] /; FreeQ[{a, b, c
, d}, x] && EqQ[b*c - 2*a*d, 0] && PosQ[b^2/a]

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])

Rubi steps

\begin {align*} \int \frac {1}{x^2 \left (2-3 x^2\right )^{3/4} \left (4-3 x^2\right )} \, dx &=\int \left (\frac {1}{4 x^2 \left (2-3 x^2\right )^{3/4}}-\frac {3}{4 \left (2-3 x^2\right )^{3/4} \left (-4+3 x^2\right )}\right ) \, dx\\ &=\frac {1}{4} \int \frac {1}{x^2 \left (2-3 x^2\right )^{3/4}} \, dx-\frac {3}{4} \int \frac {1}{\left (2-3 x^2\right )^{3/4} \left (-4+3 x^2\right )} \, dx\\ &=-\frac {\sqrt [4]{2-3 x^2}}{8 x}+2 \left (\frac {3}{16} \int \frac {1}{\left (2-3 x^2\right )^{3/4}} \, dx\right )-\frac {9}{16} \int \frac {x^2}{\left (2-3 x^2\right )^{3/4} \left (-4+3 x^2\right )} \, dx\\ &=-\frac {\sqrt [4]{2-3 x^2}}{8 x}+\frac {\sqrt {3} \tan ^{-1}\left (\frac {2^{3/4}-\sqrt [4]{2} \sqrt {2-3 x^2}}{\sqrt {3} x \sqrt [4]{2-3 x^2}}\right )}{16 \sqrt [4]{2}}-\frac {\sqrt {3} \tanh ^{-1}\left (\frac {2^{3/4}+\sqrt [4]{2} \sqrt {2-3 x^2}}{\sqrt {3} x \sqrt [4]{2-3 x^2}}\right )}{16 \sqrt [4]{2}}+\frac {\sqrt {3} F\left (\left .\frac {1}{2} \sin ^{-1}\left (\sqrt {\frac {3}{2}} x\right )\right |2\right )}{4 \sqrt [4]{2}}\\ \end {align*}

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Mathematica [C]  time = 0.09, size = 37, normalized size = 0.22 \[ -\frac {F_1\left (-\frac {1}{2};\frac {3}{4},1;\frac {1}{2};\frac {3 x^2}{2},\frac {3 x^2}{4}\right )}{4\ 2^{3/4} x} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-1/4*AppellF1[-1/2, 3/4, 1, 1/2, (3*x^2)/2, (3*x^2)/4]/(2^(3/4)*x)

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fricas [F]  time = 5.53, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}}}{9 \, x^{6} - 18 \, x^{4} + 8 \, x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((-3*x^2 + 2)^(1/4)/(9*x^6 - 18*x^4 + 8*x^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 5.90, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (-3 x^{2}+2\right )^{\frac {3}{4}} \left (-3 x^{2}+4\right ) x^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(-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^2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ -\int \frac {1}{x^2\,{\left (2-3\,x^2\right )}^{3/4}\,\left (3\,x^2-4\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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