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

Optimal. Leaf size=63 \[ \frac{2}{15} \left (3 x^2+2\right )^{3/4} x-\frac{8 x}{15 \sqrt [4]{3 x^2+2}}+\frac{8 \sqrt [4]{2} E\left (\left .\frac{1}{2} \tan ^{-1}\left (\sqrt{\frac{3}{2}} x\right )\right |2\right )}{15 \sqrt{3}} \]

[Out]

(-8*x)/(15*(2 + 3*x^2)^(1/4)) + (2*x*(2 + 3*x^2)^(3/4))/15 + (8*2^(1/4)*EllipticE[ArcTan[Sqrt[3/2]*x]/2, 2])/(
15*Sqrt[3])

________________________________________________________________________________________

Rubi [A]  time = 0.012527, antiderivative size = 63, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {321, 227, 196} \[ \frac{2}{15} \left (3 x^2+2\right )^{3/4} x-\frac{8 x}{15 \sqrt [4]{3 x^2+2}}+\frac{8 \sqrt [4]{2} E\left (\left .\frac{1}{2} \tan ^{-1}\left (\sqrt{\frac{3}{2}} x\right )\right |2\right )}{15 \sqrt{3}} \]

Antiderivative was successfully verified.

[In]

Int[x^2/(2 + 3*x^2)^(1/4),x]

[Out]

(-8*x)/(15*(2 + 3*x^2)^(1/4)) + (2*x*(2 + 3*x^2)^(3/4))/15 + (8*2^(1/4)*EllipticE[ArcTan[Sqrt[3/2]*x]/2, 2])/(
15*Sqrt[3])

Rule 321

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

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 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]

Rubi steps

\begin{align*} \int \frac{x^2}{\sqrt [4]{2+3 x^2}} \, dx &=\frac{2}{15} x \left (2+3 x^2\right )^{3/4}-\frac{4}{15} \int \frac{1}{\sqrt [4]{2+3 x^2}} \, dx\\ &=-\frac{8 x}{15 \sqrt [4]{2+3 x^2}}+\frac{2}{15} x \left (2+3 x^2\right )^{3/4}+\frac{8}{15} \int \frac{1}{\left (2+3 x^2\right )^{5/4}} \, dx\\ &=-\frac{8 x}{15 \sqrt [4]{2+3 x^2}}+\frac{2}{15} x \left (2+3 x^2\right )^{3/4}+\frac{8 \sqrt [4]{2} E\left (\left .\frac{1}{2} \tan ^{-1}\left (\sqrt{\frac{3}{2}} x\right )\right |2\right )}{15 \sqrt{3}}\\ \end{align*}

Mathematica [C]  time = 0.0091322, size = 41, normalized size = 0.65 \[ \frac{2}{15} x \left (\left (3 x^2+2\right )^{3/4}-2^{3/4} \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{3}{2};-\frac{3 x^2}{2}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^2/(2 + 3*x^2)^(1/4),x]

[Out]

(2*x*((2 + 3*x^2)^(3/4) - 2^(3/4)*Hypergeometric2F1[1/4, 1/2, 3/2, (-3*x^2)/2]))/15

________________________________________________________________________________________

Maple [C]  time = 0.02, size = 31, normalized size = 0.5 \begin{align*}{\frac{2\,x}{15} \left ( 3\,{x}^{2}+2 \right ) ^{{\frac{3}{4}}}}-{\frac{2\,{2}^{3/4}x}{15}{\mbox{$_2$F$_1$}({\frac{1}{4}},{\frac{1}{2}};\,{\frac{3}{2}};\,-{\frac{3\,{x}^{2}}{2}})}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*x*(3*x^2+2)^(3/4)-2/15*2^(3/4)*x*hypergeom([1/4,1/2],[3/2],-3/2*x^2)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2/(3*x^2 + 2)^(1/4), x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x^2/(3*x^2 + 2)^(1/4), x)

________________________________________________________________________________________

Sympy [C]  time = 0.597907, size = 27, normalized size = 0.43 \begin{align*} \frac{2^{\frac{3}{4}} x^{3}{{}_{2}F_{1}\left (\begin{matrix} \frac{1}{4}, \frac{3}{2} \\ \frac{5}{2} \end{matrix}\middle |{\frac{3 x^{2} e^{i \pi }}{2}} \right )}}{6} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2**(3/4)*x**3*hyper((1/4, 3/2), (5/2,), 3*x**2*exp_polar(I*pi)/2)/6

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2/(3*x^2 + 2)^(1/4), x)