3.817 \(\int \frac{x^2}{\sqrt [4]{a+b x^2}} \, dx\)

Optimal. Leaf size=98 \[ \frac{4 a^{3/2} \sqrt [4]{\frac{b x^2}{a}+1} E\left (\left .\frac{1}{2} \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )\right |2\right )}{5 b^{3/2} \sqrt [4]{a+b x^2}}-\frac{4 a x}{5 b \sqrt [4]{a+b x^2}}+\frac{2 x \left (a+b x^2\right )^{3/4}}{5 b} \]

[Out]

(-4*a*x)/(5*b*(a + b*x^2)^(1/4)) + (2*x*(a + b*x^2)^(3/4))/(5*b) + (4*a^(3/2)*(1 + (b*x^2)/a)^(1/4)*EllipticE[
ArcTan[(Sqrt[b]*x)/Sqrt[a]]/2, 2])/(5*b^(3/2)*(a + b*x^2)^(1/4))

________________________________________________________________________________________

Rubi [A]  time = 0.0239423, antiderivative size = 98, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.267, Rules used = {321, 229, 227, 196} \[ \frac{4 a^{3/2} \sqrt [4]{\frac{b x^2}{a}+1} E\left (\left .\frac{1}{2} \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )\right |2\right )}{5 b^{3/2} \sqrt [4]{a+b x^2}}-\frac{4 a x}{5 b \sqrt [4]{a+b x^2}}+\frac{2 x \left (a+b x^2\right )^{3/4}}{5 b} \]

Antiderivative was successfully verified.

[In]

Int[x^2/(a + b*x^2)^(1/4),x]

[Out]

(-4*a*x)/(5*b*(a + b*x^2)^(1/4)) + (2*x*(a + b*x^2)^(3/4))/(5*b) + (4*a^(3/2)*(1 + (b*x^2)/a)^(1/4)*EllipticE[
ArcTan[(Sqrt[b]*x)/Sqrt[a]]/2, 2])/(5*b^(3/2)*(a + b*x^2)^(1/4))

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 229

Int[((a_) + (b_.)*(x_)^2)^(-1/4), x_Symbol] :> Dist[(1 + (b*x^2)/a)^(1/4)/(a + b*x^2)^(1/4), Int[1/(1 + (b*x^2
)/a)^(1/4), x], x] /; FreeQ[{a, b}, x] && PosQ[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 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]{a+b x^2}} \, dx &=\frac{2 x \left (a+b x^2\right )^{3/4}}{5 b}-\frac{(2 a) \int \frac{1}{\sqrt [4]{a+b x^2}} \, dx}{5 b}\\ &=\frac{2 x \left (a+b x^2\right )^{3/4}}{5 b}-\frac{\left (2 a \sqrt [4]{1+\frac{b x^2}{a}}\right ) \int \frac{1}{\sqrt [4]{1+\frac{b x^2}{a}}} \, dx}{5 b \sqrt [4]{a+b x^2}}\\ &=-\frac{4 a x}{5 b \sqrt [4]{a+b x^2}}+\frac{2 x \left (a+b x^2\right )^{3/4}}{5 b}+\frac{\left (2 a \sqrt [4]{1+\frac{b x^2}{a}}\right ) \int \frac{1}{\left (1+\frac{b x^2}{a}\right )^{5/4}} \, dx}{5 b \sqrt [4]{a+b x^2}}\\ &=-\frac{4 a x}{5 b \sqrt [4]{a+b x^2}}+\frac{2 x \left (a+b x^2\right )^{3/4}}{5 b}+\frac{4 a^{3/2} \sqrt [4]{1+\frac{b x^2}{a}} E\left (\left .\frac{1}{2} \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )\right |2\right )}{5 b^{3/2} \sqrt [4]{a+b x^2}}\\ \end{align*}

Mathematica [C]  time = 0.0155101, size = 62, normalized size = 0.63 \[ \frac{2 x \left (-a \sqrt [4]{\frac{b x^2}{a}+1} \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{3}{2};-\frac{b x^2}{a}\right )+a+b x^2\right )}{5 b \sqrt [4]{a+b x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2/(a + b*x^2)^(1/4),x]

[Out]

(2*x*(a + b*x^2 - a*(1 + (b*x^2)/a)^(1/4)*Hypergeometric2F1[1/4, 1/2, 3/2, -((b*x^2)/a)]))/(5*b*(a + b*x^2)^(1
/4))

________________________________________________________________________________________

Maple [F]  time = 0.023, size = 0, normalized size = 0. \begin{align*} \int{{x}^{2}{\frac{1}{\sqrt [4]{b{x}^{2}+a}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2/(b*x^2 + a)^(1/4), x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x^2/(b*x^2 + a)^(1/4), x)

________________________________________________________________________________________

Sympy [C]  time = 0.691345, size = 27, normalized size = 0.28 \begin{align*} \frac{x^{3}{{}_{2}F_{1}\left (\begin{matrix} \frac{1}{4}, \frac{3}{2} \\ \frac{5}{2} \end{matrix}\middle |{\frac{b x^{2} e^{i \pi }}{a}} \right )}}{3 \sqrt [4]{a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2/(b*x**2+a)**(1/4),x)

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2/(b*x^2 + a)^(1/4), x)