3.836 \(\int \frac{x^6}{(a-b x^2)^{3/4}} \, dx\)

Optimal. Leaf size=129 \[ \frac{80 a^{7/2} \left (1-\frac{b x^2}{a}\right )^{3/4} \text{EllipticF}\left (\frac{1}{2} \sin ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right ),2\right )}{77 b^{7/2} \left (a-b x^2\right )^{3/4}}-\frac{40 a^2 x \sqrt [4]{a-b x^2}}{77 b^3}-\frac{20 a x^3 \sqrt [4]{a-b x^2}}{77 b^2}-\frac{2 x^5 \sqrt [4]{a-b x^2}}{11 b} \]

[Out]

(-40*a^2*x*(a - b*x^2)^(1/4))/(77*b^3) - (20*a*x^3*(a - b*x^2)^(1/4))/(77*b^2) - (2*x^5*(a - b*x^2)^(1/4))/(11
*b) + (80*a^(7/2)*(1 - (b*x^2)/a)^(3/4)*EllipticF[ArcSin[(Sqrt[b]*x)/Sqrt[a]]/2, 2])/(77*b^(7/2)*(a - b*x^2)^(
3/4))

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Rubi [A]  time = 0.0464821, antiderivative size = 129, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.188, Rules used = {321, 233, 232} \[ -\frac{40 a^2 x \sqrt [4]{a-b x^2}}{77 b^3}+\frac{80 a^{7/2} \left (1-\frac{b x^2}{a}\right )^{3/4} F\left (\left .\frac{1}{2} \sin ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )\right |2\right )}{77 b^{7/2} \left (a-b x^2\right )^{3/4}}-\frac{20 a x^3 \sqrt [4]{a-b x^2}}{77 b^2}-\frac{2 x^5 \sqrt [4]{a-b x^2}}{11 b} \]

Antiderivative was successfully verified.

[In]

Int[x^6/(a - b*x^2)^(3/4),x]

[Out]

(-40*a^2*x*(a - b*x^2)^(1/4))/(77*b^3) - (20*a*x^3*(a - b*x^2)^(1/4))/(77*b^2) - (2*x^5*(a - b*x^2)^(1/4))/(11
*b) + (80*a^(7/2)*(1 - (b*x^2)/a)^(3/4)*EllipticF[ArcSin[(Sqrt[b]*x)/Sqrt[a]]/2, 2])/(77*b^(7/2)*(a - b*x^2)^(
3/4))

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 233

Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Dist[(1 + (b*x^2)/a)^(3/4)/(a + b*x^2)^(3/4), Int[1/(1 + (b*x^2
)/a)^(3/4), x], x] /; FreeQ[{a, b}, x] && PosQ[a]

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]

Rubi steps

\begin{align*} \int \frac{x^6}{\left (a-b x^2\right )^{3/4}} \, dx &=-\frac{2 x^5 \sqrt [4]{a-b x^2}}{11 b}+\frac{(10 a) \int \frac{x^4}{\left (a-b x^2\right )^{3/4}} \, dx}{11 b}\\ &=-\frac{20 a x^3 \sqrt [4]{a-b x^2}}{77 b^2}-\frac{2 x^5 \sqrt [4]{a-b x^2}}{11 b}+\frac{\left (60 a^2\right ) \int \frac{x^2}{\left (a-b x^2\right )^{3/4}} \, dx}{77 b^2}\\ &=-\frac{40 a^2 x \sqrt [4]{a-b x^2}}{77 b^3}-\frac{20 a x^3 \sqrt [4]{a-b x^2}}{77 b^2}-\frac{2 x^5 \sqrt [4]{a-b x^2}}{11 b}+\frac{\left (40 a^3\right ) \int \frac{1}{\left (a-b x^2\right )^{3/4}} \, dx}{77 b^3}\\ &=-\frac{40 a^2 x \sqrt [4]{a-b x^2}}{77 b^3}-\frac{20 a x^3 \sqrt [4]{a-b x^2}}{77 b^2}-\frac{2 x^5 \sqrt [4]{a-b x^2}}{11 b}+\frac{\left (40 a^3 \left (1-\frac{b x^2}{a}\right )^{3/4}\right ) \int \frac{1}{\left (1-\frac{b x^2}{a}\right )^{3/4}} \, dx}{77 b^3 \left (a-b x^2\right )^{3/4}}\\ &=-\frac{40 a^2 x \sqrt [4]{a-b x^2}}{77 b^3}-\frac{20 a x^3 \sqrt [4]{a-b x^2}}{77 b^2}-\frac{2 x^5 \sqrt [4]{a-b x^2}}{11 b}+\frac{80 a^{7/2} \left (1-\frac{b x^2}{a}\right )^{3/4} F\left (\left .\frac{1}{2} \sin ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )\right |2\right )}{77 b^{7/2} \left (a-b x^2\right )^{3/4}}\\ \end{align*}

Mathematica [C]  time = 0.0351946, size = 91, normalized size = 0.71 \[ \frac{2 \left (20 a^3 x \left (1-\frac{b x^2}{a}\right )^{3/4} \, _2F_1\left (\frac{1}{2},\frac{3}{4};\frac{3}{2};\frac{b x^2}{a}\right )+10 a^2 b x^3-20 a^3 x+3 a b^2 x^5+7 b^3 x^7\right )}{77 b^3 \left (a-b x^2\right )^{3/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^6/(a - b*x^2)^(3/4),x]

[Out]

(2*(-20*a^3*x + 10*a^2*b*x^3 + 3*a*b^2*x^5 + 7*b^3*x^7 + 20*a^3*x*(1 - (b*x^2)/a)^(3/4)*Hypergeometric2F1[1/2,
 3/4, 3/2, (b*x^2)/a]))/(77*b^3*(a - b*x^2)^(3/4))

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Maple [F]  time = 0.014, size = 0, normalized size = 0. \begin{align*} \int{{x}^{6} \left ( -b{x}^{2}+a \right ) ^{-{\frac{3}{4}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^6/(-b*x^2+a)^(3/4),x)

[Out]

int(x^6/(-b*x^2+a)^(3/4),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{3}{4}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6/(-b*x^2+a)^(3/4),x, algorithm="maxima")

[Out]

integrate(x^6/(-b*x^2 + a)^(3/4), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6/(-b*x^2+a)^(3/4),x, algorithm="fricas")

[Out]

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

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Sympy [C]  time = 0.910052, size = 29, normalized size = 0.22 \begin{align*} \frac{x^{7}{{}_{2}F_{1}\left (\begin{matrix} \frac{3}{4}, \frac{7}{2} \\ \frac{9}{2} \end{matrix}\middle |{\frac{b x^{2} e^{2 i \pi }}{a}} \right )}}{7 a^{\frac{3}{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**6/(-b*x**2+a)**(3/4),x)

[Out]

x**7*hyper((3/4, 7/2), (9/2,), b*x**2*exp_polar(2*I*pi)/a)/(7*a**(3/4))

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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{3}{4}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6/(-b*x^2+a)^(3/4),x, algorithm="giac")

[Out]

integrate(x^6/(-b*x^2 + a)^(3/4), x)