3.847 \(\int \frac{x^6}{\sqrt{a-b x^4}} \, dx\)

Optimal. Leaf size=135 \[ -\frac{3 a^{7/4} \sqrt{1-\frac{b x^4}{a}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),-1\right )}{5 b^{7/4} \sqrt{a-b x^4}}+\frac{3 a^{7/4} \sqrt{1-\frac{b x^4}{a}} E\left (\left .\sin ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )\right |-1\right )}{5 b^{7/4} \sqrt{a-b x^4}}-\frac{x^3 \sqrt{a-b x^4}}{5 b} \]

[Out]

-(x^3*Sqrt[a - b*x^4])/(5*b) + (3*a^(7/4)*Sqrt[1 - (b*x^4)/a]*EllipticE[ArcSin[(b^(1/4)*x)/a^(1/4)], -1])/(5*b
^(7/4)*Sqrt[a - b*x^4]) - (3*a^(7/4)*Sqrt[1 - (b*x^4)/a]*EllipticF[ArcSin[(b^(1/4)*x)/a^(1/4)], -1])/(5*b^(7/4
)*Sqrt[a - b*x^4])

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Rubi [A]  time = 0.0770988, antiderivative size = 135, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.438, Rules used = {321, 307, 224, 221, 1200, 1199, 424} \[ -\frac{3 a^{7/4} \sqrt{1-\frac{b x^4}{a}} F\left (\left .\sin ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )\right |-1\right )}{5 b^{7/4} \sqrt{a-b x^4}}+\frac{3 a^{7/4} \sqrt{1-\frac{b x^4}{a}} E\left (\left .\sin ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )\right |-1\right )}{5 b^{7/4} \sqrt{a-b x^4}}-\frac{x^3 \sqrt{a-b x^4}}{5 b} \]

Antiderivative was successfully verified.

[In]

Int[x^6/Sqrt[a - b*x^4],x]

[Out]

-(x^3*Sqrt[a - b*x^4])/(5*b) + (3*a^(7/4)*Sqrt[1 - (b*x^4)/a]*EllipticE[ArcSin[(b^(1/4)*x)/a^(1/4)], -1])/(5*b
^(7/4)*Sqrt[a - b*x^4]) - (3*a^(7/4)*Sqrt[1 - (b*x^4)/a]*EllipticF[ArcSin[(b^(1/4)*x)/a^(1/4)], -1])/(5*b^(7/4
)*Sqrt[a - b*x^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 307

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[-(b/a), 2]}, -Dist[q^(-1), Int[1/Sqrt[a + b*x^
4], x], x] + Dist[1/q, Int[(1 + q*x^2)/Sqrt[a + b*x^4], x], x]] /; FreeQ[{a, b}, x] && NegQ[b/a]

Rule 224

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Dist[Sqrt[1 + (b*x^4)/a]/Sqrt[a + b*x^4], Int[1/Sqrt[1 + (b*x^4)
/a], x], x] /; FreeQ[{a, b}, x] && NegQ[b/a] &&  !GtQ[a, 0]

Rule 221

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Simp[EllipticF[ArcSin[(Rt[-b, 4]*x)/Rt[a, 4]], -1]/(Rt[a, 4]*Rt[
-b, 4]), x] /; FreeQ[{a, b}, x] && NegQ[b/a] && GtQ[a, 0]

Rule 1200

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> Dist[Sqrt[1 + (c*x^4)/a]/Sqrt[a + c*x^4], In
t[(d + e*x^2)/Sqrt[1 + (c*x^4)/a], x], x] /; FreeQ[{a, c, d, e}, x] && NegQ[c/a] && EqQ[c*d^2 + a*e^2, 0] &&
!GtQ[a, 0]

Rule 1199

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> Dist[d/Sqrt[a], Int[Sqrt[1 + (e*x^2)/d]/Sqrt
[1 - (e*x^2)/d], x], x] /; FreeQ[{a, c, d, e}, x] && NegQ[c/a] && EqQ[c*d^2 + a*e^2, 0] && GtQ[a, 0]

Rule 424

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[(Sqrt[a]*EllipticE[ArcSin[Rt[-(d/c)
, 2]*x], (b*c)/(a*d)])/(Sqrt[c]*Rt[-(d/c), 2]), x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && GtQ[
a, 0]

Rubi steps

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

Mathematica [C]  time = 0.0179395, size = 66, normalized size = 0.49 \[ \frac{x^3 \left (a \sqrt{1-\frac{b x^4}{a}} \, _2F_1\left (\frac{1}{2},\frac{3}{4};\frac{7}{4};\frac{b x^4}{a}\right )-a+b x^4\right )}{5 b \sqrt{a-b x^4}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^6/Sqrt[a - b*x^4],x]

[Out]

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

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Maple [A]  time = 0.008, size = 107, normalized size = 0.8 \begin{align*} -{\frac{{x}^{3}}{5\,b}\sqrt{-b{x}^{4}+a}}-{\frac{3}{5}{a}^{{\frac{3}{2}}}\sqrt{1-{{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}\sqrt{1+{{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}} \left ({\it EllipticF} \left ( x\sqrt{{\sqrt{b}{\frac{1}{\sqrt{a}}}}},i \right ) -{\it EllipticE} \left ( x\sqrt{{\sqrt{b}{\frac{1}{\sqrt{a}}}}},i \right ) \right ){b}^{-{\frac{3}{2}}}{\frac{1}{\sqrt{{\sqrt{b}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{-b{x}^{4}+a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/5*x^3*(-b*x^4+a)^(1/2)/b-3/5/b^(3/2)*a^(3/2)/(1/a^(1/2)*b^(1/2))^(1/2)*(1-x^2*b^(1/2)/a^(1/2))^(1/2)*(1+x^2
*b^(1/2)/a^(1/2))^(1/2)/(-b*x^4+a)^(1/2)*(EllipticF(x*(1/a^(1/2)*b^(1/2))^(1/2),I)-EllipticE(x*(1/a^(1/2)*b^(1
/2))^(1/2),I))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{6}}{\sqrt{-b x^{4} + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6/(-b*x^4+a)^(1/2),x, algorithm="maxima")

[Out]

integrate(x^6/sqrt(-b*x^4 + a), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6/(-b*x^4+a)^(1/2),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 1.15809, size = 39, normalized size = 0.29 \begin{align*} \frac{x^{7} \Gamma \left (\frac{7}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{1}{2}, \frac{7}{4} \\ \frac{11}{4} \end{matrix}\middle |{\frac{b x^{4} e^{2 i \pi }}{a}} \right )}}{4 \sqrt{a} \Gamma \left (\frac{11}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x**7*gamma(7/4)*hyper((1/2, 7/4), (11/4,), b*x**4*exp_polar(2*I*pi)/a)/(4*sqrt(a)*gamma(11/4))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{6}}{\sqrt{-b x^{4} + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6/(-b*x^4+a)^(1/2),x, algorithm="giac")

[Out]

integrate(x^6/sqrt(-b*x^4 + a), x)