3.815 \(\int \frac{x^6}{\sqrt [4]{a+b x^2}} \, dx\)

Optimal. Leaf size=146 \[ \frac{8 a^2 x \left (a+b x^2\right )^{3/4}}{39 b^3}-\frac{16 a^3 x}{39 b^3 \sqrt [4]{a+b x^2}}+\frac{16 a^{7/2} \sqrt [4]{\frac{b x^2}{a}+1} E\left (\left .\frac{1}{2} \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )\right |2\right )}{39 b^{7/2} \sqrt [4]{a+b x^2}}-\frac{20 a x^3 \left (a+b x^2\right )^{3/4}}{117 b^2}+\frac{2 x^5 \left (a+b x^2\right )^{3/4}}{13 b} \]

[Out]

(-16*a^3*x)/(39*b^3*(a + b*x^2)^(1/4)) + (8*a^2*x*(a + b*x^2)^(3/4))/(39*b^3) - (20*a*x^3*(a + b*x^2)^(3/4))/(
117*b^2) + (2*x^5*(a + b*x^2)^(3/4))/(13*b) + (16*a^(7/2)*(1 + (b*x^2)/a)^(1/4)*EllipticE[ArcTan[(Sqrt[b]*x)/S
qrt[a]]/2, 2])/(39*b^(7/2)*(a + b*x^2)^(1/4))

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Rubi [A]  time = 0.0518924, antiderivative size = 146, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.267, Rules used = {321, 229, 227, 196} \[ \frac{8 a^2 x \left (a+b x^2\right )^{3/4}}{39 b^3}-\frac{16 a^3 x}{39 b^3 \sqrt [4]{a+b x^2}}+\frac{16 a^{7/2} \sqrt [4]{\frac{b x^2}{a}+1} E\left (\left .\frac{1}{2} \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )\right |2\right )}{39 b^{7/2} \sqrt [4]{a+b x^2}}-\frac{20 a x^3 \left (a+b x^2\right )^{3/4}}{117 b^2}+\frac{2 x^5 \left (a+b x^2\right )^{3/4}}{13 b} \]

Antiderivative was successfully verified.

[In]

Int[x^6/(a + b*x^2)^(1/4),x]

[Out]

(-16*a^3*x)/(39*b^3*(a + b*x^2)^(1/4)) + (8*a^2*x*(a + b*x^2)^(3/4))/(39*b^3) - (20*a*x^3*(a + b*x^2)^(3/4))/(
117*b^2) + (2*x^5*(a + b*x^2)^(3/4))/(13*b) + (16*a^(7/2)*(1 + (b*x^2)/a)^(1/4)*EllipticE[ArcTan[(Sqrt[b]*x)/S
qrt[a]]/2, 2])/(39*b^(7/2)*(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^6}{\sqrt [4]{a+b x^2}} \, dx &=\frac{2 x^5 \left (a+b x^2\right )^{3/4}}{13 b}-\frac{(10 a) \int \frac{x^4}{\sqrt [4]{a+b x^2}} \, dx}{13 b}\\ &=-\frac{20 a x^3 \left (a+b x^2\right )^{3/4}}{117 b^2}+\frac{2 x^5 \left (a+b x^2\right )^{3/4}}{13 b}+\frac{\left (20 a^2\right ) \int \frac{x^2}{\sqrt [4]{a+b x^2}} \, dx}{39 b^2}\\ &=\frac{8 a^2 x \left (a+b x^2\right )^{3/4}}{39 b^3}-\frac{20 a x^3 \left (a+b x^2\right )^{3/4}}{117 b^2}+\frac{2 x^5 \left (a+b x^2\right )^{3/4}}{13 b}-\frac{\left (8 a^3\right ) \int \frac{1}{\sqrt [4]{a+b x^2}} \, dx}{39 b^3}\\ &=\frac{8 a^2 x \left (a+b x^2\right )^{3/4}}{39 b^3}-\frac{20 a x^3 \left (a+b x^2\right )^{3/4}}{117 b^2}+\frac{2 x^5 \left (a+b x^2\right )^{3/4}}{13 b}-\frac{\left (8 a^3 \sqrt [4]{1+\frac{b x^2}{a}}\right ) \int \frac{1}{\sqrt [4]{1+\frac{b x^2}{a}}} \, dx}{39 b^3 \sqrt [4]{a+b x^2}}\\ &=-\frac{16 a^3 x}{39 b^3 \sqrt [4]{a+b x^2}}+\frac{8 a^2 x \left (a+b x^2\right )^{3/4}}{39 b^3}-\frac{20 a x^3 \left (a+b x^2\right )^{3/4}}{117 b^2}+\frac{2 x^5 \left (a+b x^2\right )^{3/4}}{13 b}+\frac{\left (8 a^3 \sqrt [4]{1+\frac{b x^2}{a}}\right ) \int \frac{1}{\left (1+\frac{b x^2}{a}\right )^{5/4}} \, dx}{39 b^3 \sqrt [4]{a+b x^2}}\\ &=-\frac{16 a^3 x}{39 b^3 \sqrt [4]{a+b x^2}}+\frac{8 a^2 x \left (a+b x^2\right )^{3/4}}{39 b^3}-\frac{20 a x^3 \left (a+b x^2\right )^{3/4}}{117 b^2}+\frac{2 x^5 \left (a+b x^2\right )^{3/4}}{13 b}+\frac{16 a^{7/2} \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 )}{39 b^{7/2} \sqrt [4]{a+b x^2}}\\ \end{align*}

Mathematica [C]  time = 0.0328298, size = 90, normalized size = 0.62 \[ \frac{2 \left (-12 a^3 x \sqrt [4]{\frac{b x^2}{a}+1} \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{3}{2};-\frac{b x^2}{a}\right )+2 a^2 b x^3+12 a^3 x-a b^2 x^5+9 b^3 x^7\right )}{117 b^3 \sqrt [4]{a+b x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^6/(a + b*x^2)^(1/4),x]

[Out]

(2*(12*a^3*x + 2*a^2*b*x^3 - a*b^2*x^5 + 9*b^3*x^7 - 12*a^3*x*(1 + (b*x^2)/a)^(1/4)*Hypergeometric2F1[1/4, 1/2
, 3/2, -((b*x^2)/a)]))/(117*b^3*(a + b*x^2)^(1/4))

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Maple [F]  time = 0.025, size = 0, normalized size = 0. \begin{align*} \int{{x}^{6}{\frac{1}{\sqrt [4]{b{x}^{2}+a}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^6/(b*x^2+a)^(1/4),x)

[Out]

int(x^6/(b*x^2+a)^(1/4),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{6}}{{\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^6/(b*x^2+a)^(1/4),x, algorithm="maxima")

[Out]

integrate(x^6/(b*x^2 + a)^(1/4), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{x^{6}}{{\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^6/(b*x^2+a)^(1/4),x, algorithm="fricas")

[Out]

integral(x^6/(b*x^2 + a)^(1/4), x)

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Sympy [C]  time = 0.917057, size = 27, normalized size = 0.18 \begin{align*} \frac{x^{7}{{}_{2}F_{1}\left (\begin{matrix} \frac{1}{4}, \frac{7}{2} \\ \frac{9}{2} \end{matrix}\middle |{\frac{b x^{2} e^{i \pi }}{a}} \right )}}{7 \sqrt [4]{a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**6/(b*x**2+a)**(1/4),x)

[Out]

x**7*hyper((1/4, 7/2), (9/2,), b*x**2*exp_polar(I*pi)/a)/(7*a**(1/4))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{6}}{{\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^6/(b*x^2+a)^(1/4),x, algorithm="giac")

[Out]

integrate(x^6/(b*x^2 + a)^(1/4), x)