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

Optimal. Leaf size=83 \[ -\frac {9 x}{8 \sqrt [4]{3 x^2+2}}+\frac {3 \left (3 x^2+2\right )^{3/4}}{8 x}-\frac {\left (3 x^2+2\right )^{3/4}}{6 x^3}+\frac {3 \sqrt {3} E\left (\left .\frac {1}{2} \tan ^{-1}\left (\sqrt {\frac {3}{2}} x\right )\right |2\right )}{4\ 2^{3/4}} \]

[Out]

-9/8*x/(3*x^2+2)^(1/4)-1/6*(3*x^2+2)^(3/4)/x^3+3/8*(3*x^2+2)^(3/4)/x+3/8*2^(1/4)*(cos(1/2*arctan(1/2*x*6^(1/2)
))^2)^(1/2)/cos(1/2*arctan(1/2*x*6^(1/2)))*EllipticE(sin(1/2*arctan(1/2*x*6^(1/2))),2^(1/2))*3^(1/2)

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Rubi [A]  time = 0.02, antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {325, 227, 196} \[ -\frac {9 x}{8 \sqrt [4]{3 x^2+2}}+\frac {3 \left (3 x^2+2\right )^{3/4}}{8 x}-\frac {\left (3 x^2+2\right )^{3/4}}{6 x^3}+\frac {3 \sqrt {3} E\left (\left .\frac {1}{2} \tan ^{-1}\left (\sqrt {\frac {3}{2}} x\right )\right |2\right )}{4\ 2^{3/4}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^4*(2 + 3*x^2)^(1/4)),x]

[Out]

(-9*x)/(8*(2 + 3*x^2)^(1/4)) - (2 + 3*x^2)^(3/4)/(6*x^3) + (3*(2 + 3*x^2)^(3/4))/(8*x) + (3*Sqrt[3]*EllipticE[
ArcTan[Sqrt[3/2]*x]/2, 2])/(4*2^(3/4))

Rule 196

Int[((a_) + (b_.)*(x_)^2)^(-5/4), x_Symbol] :> Simp[(2*EllipticE[(1*ArcTan[Rt[b/a, 2]*x])/2, 2])/(a^(5/4)*Rt[b
/a, 2]), x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b/a]

Rule 227

Int[((a_) + (b_.)*(x_)^2)^(-1/4), x_Symbol] :> Simp[(2*x)/(a + b*x^2)^(1/4), x] - Dist[a, Int[1/(a + b*x^2)^(5
/4), x], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[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]

Rubi steps

\begin {align*} \int \frac {1}{x^4 \sqrt [4]{2+3 x^2}} \, dx &=-\frac {\left (2+3 x^2\right )^{3/4}}{6 x^3}-\frac {3}{4} \int \frac {1}{x^2 \sqrt [4]{2+3 x^2}} \, dx\\ &=-\frac {\left (2+3 x^2\right )^{3/4}}{6 x^3}+\frac {3 \left (2+3 x^2\right )^{3/4}}{8 x}-\frac {9}{16} \int \frac {1}{\sqrt [4]{2+3 x^2}} \, dx\\ &=-\frac {9 x}{8 \sqrt [4]{2+3 x^2}}-\frac {\left (2+3 x^2\right )^{3/4}}{6 x^3}+\frac {3 \left (2+3 x^2\right )^{3/4}}{8 x}+\frac {9}{8} \int \frac {1}{\left (2+3 x^2\right )^{5/4}} \, dx\\ &=-\frac {9 x}{8 \sqrt [4]{2+3 x^2}}-\frac {\left (2+3 x^2\right )^{3/4}}{6 x^3}+\frac {3 \left (2+3 x^2\right )^{3/4}}{8 x}+\frac {3 \sqrt {3} E\left (\left .\frac {1}{2} \tan ^{-1}\left (\sqrt {\frac {3}{2}} x\right )\right |2\right )}{4\ 2^{3/4}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 29, normalized size = 0.35 \[ -\frac {\, _2F_1\left (-\frac {3}{2},\frac {1}{4};-\frac {1}{2};-\frac {3 x^2}{2}\right )}{3 \sqrt [4]{2} x^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*Hypergeometric2F1[-3/2, 1/4, -1/2, (-3*x^2)/2]/(2^(1/4)*x^3)

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fricas [F]  time = 0.83, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (3 \, x^{2} + 2\right )}^{\frac {3}{4}}}{3 \, x^{6} + 2 \, x^{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((3*x^2 + 2)^(3/4)/(3*x^6 + 2*x^4), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.27, size = 45, normalized size = 0.54 \[ -\frac {9 \,2^{\frac {3}{4}} x \hypergeom \left (\left [\frac {1}{4}, \frac {1}{2}\right ], \left [\frac {3}{2}\right ], -\frac {3 x^{2}}{2}\right )}{32}+\frac {27 x^{4}+6 x^{2}-8}{24 \left (3 x^{2}+2\right )^{\frac {1}{4}} x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(27*x^4+6*x^2-8)/x^3/(3*x^2+2)^(1/4)-9/32*2^(3/4)*x*hypergeom([1/4,1/2],[3/2],-3/2*x^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^4*(3*x^2 + 2)^(1/4)),x)

[Out]

int(1/(x^4*(3*x^2 + 2)^(1/4)), x)

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sympy [C]  time = 0.84, size = 32, normalized size = 0.39 \[ - \frac {2^{\frac {3}{4}} {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{2}, \frac {1}{4} \\ - \frac {1}{2} \end {matrix}\middle | {\frac {3 x^{2} e^{i \pi }}{2}} \right )}}{6 x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2**(3/4)*hyper((-3/2, 1/4), (-1/2,), 3*x**2*exp_polar(I*pi)/2)/(6*x**3)

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