3.850 \(\int \frac{x^6}{(a-b x^2)^{5/4}} \, dx\)

Optimal. Leaf size=124 \[ -\frac{16 a^{5/2} \sqrt [4]{1-\frac{b x^2}{a}} E\left (\left .\frac{1}{2} \sin ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )\right |2\right )}{3 b^{7/2} \sqrt [4]{a-b x^2}}+\frac{20 x^3 \left (a-b x^2\right )^{3/4}}{9 b^2}+\frac{8 a x \left (a-b x^2\right )^{3/4}}{3 b^3}+\frac{2 x^5}{b \sqrt [4]{a-b x^2}} \]

[Out]

(2*x^5)/(b*(a - b*x^2)^(1/4)) + (8*a*x*(a - b*x^2)^(3/4))/(3*b^3) + (20*x^3*(a - b*x^2)^(3/4))/(9*b^2) - (16*a
^(5/2)*(1 - (b*x^2)/a)^(1/4)*EllipticE[ArcSin[(Sqrt[b]*x)/Sqrt[a]]/2, 2])/(3*b^(7/2)*(a - b*x^2)^(1/4))

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Rubi [A]  time = 0.0474167, antiderivative size = 124, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {288, 321, 229, 228} \[ -\frac{16 a^{5/2} \sqrt [4]{1-\frac{b x^2}{a}} E\left (\left .\frac{1}{2} \sin ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )\right |2\right )}{3 b^{7/2} \sqrt [4]{a-b x^2}}+\frac{20 x^3 \left (a-b x^2\right )^{3/4}}{9 b^2}+\frac{8 a x \left (a-b x^2\right )^{3/4}}{3 b^3}+\frac{2 x^5}{b \sqrt [4]{a-b x^2}} \]

Antiderivative was successfully verified.

[In]

Int[x^6/(a - b*x^2)^(5/4),x]

[Out]

(2*x^5)/(b*(a - b*x^2)^(1/4)) + (8*a*x*(a - b*x^2)^(3/4))/(3*b^3) + (20*x^3*(a - b*x^2)^(3/4))/(9*b^2) - (16*a
^(5/2)*(1 - (b*x^2)/a)^(1/4)*EllipticE[ArcSin[(Sqrt[b]*x)/Sqrt[a]]/2, 2])/(3*b^(7/2)*(a - b*x^2)^(1/4))

Rule 288

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*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

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 228

Int[((a_) + (b_.)*(x_)^2)^(-1/4), x_Symbol] :> Simp[(2*EllipticE[(1*ArcSin[Rt[-(b/a), 2]*x])/2, 2])/(a^(1/4)*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 )^{5/4}} \, dx &=\frac{2 x^5}{b \sqrt [4]{a-b x^2}}-\frac{10 \int \frac{x^4}{\sqrt [4]{a-b x^2}} \, dx}{b}\\ &=\frac{2 x^5}{b \sqrt [4]{a-b x^2}}+\frac{20 x^3 \left (a-b x^2\right )^{3/4}}{9 b^2}-\frac{(20 a) \int \frac{x^2}{\sqrt [4]{a-b x^2}} \, dx}{3 b^2}\\ &=\frac{2 x^5}{b \sqrt [4]{a-b x^2}}+\frac{8 a x \left (a-b x^2\right )^{3/4}}{3 b^3}+\frac{20 x^3 \left (a-b x^2\right )^{3/4}}{9 b^2}-\frac{\left (8 a^2\right ) \int \frac{1}{\sqrt [4]{a-b x^2}} \, dx}{3 b^3}\\ &=\frac{2 x^5}{b \sqrt [4]{a-b x^2}}+\frac{8 a x \left (a-b x^2\right )^{3/4}}{3 b^3}+\frac{20 x^3 \left (a-b x^2\right )^{3/4}}{9 b^2}-\frac{\left (8 a^2 \sqrt [4]{1-\frac{b x^2}{a}}\right ) \int \frac{1}{\sqrt [4]{1-\frac{b x^2}{a}}} \, dx}{3 b^3 \sqrt [4]{a-b x^2}}\\ &=\frac{2 x^5}{b \sqrt [4]{a-b x^2}}+\frac{8 a x \left (a-b x^2\right )^{3/4}}{3 b^3}+\frac{20 x^3 \left (a-b x^2\right )^{3/4}}{9 b^2}-\frac{16 a^{5/2} \sqrt [4]{1-\frac{b x^2}{a}} E\left (\left .\frac{1}{2} \sin ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )\right |2\right )}{3 b^{7/2} \sqrt [4]{a-b x^2}}\\ \end{align*}

Mathematica [C]  time = 0.0287999, size = 78, normalized size = 0.63 \[ -\frac{2 x \left (12 a^2 \sqrt [4]{1-\frac{b x^2}{a}} \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{3}{2};\frac{b x^2}{a}\right )-12 a^2+2 a b x^2+b^2 x^4\right )}{9 b^3 \sqrt [4]{a-b x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^6/(a - b*x^2)^(5/4),x]

[Out]

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

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Maple [F]  time = 0.054, size = 0, normalized size = 0. \begin{align*} \int{{x}^{6} \left ( -b{x}^{2}+a \right ) ^{-{\frac{5}{4}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^6/(-b*x^2+a)^(5/4),x)

[Out]

int(x^6/(-b*x^2+a)^(5/4),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{5}{4}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6/(-b*x^2+a)^(5/4),x, algorithm="maxima")

[Out]

integrate(x^6/(-b*x^2 + a)^(5/4), 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{3}{4}} x^{6}}{b^{2} x^{4} - 2 \, a b x^{2} + a^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6/(-b*x^2+a)^(5/4),x, algorithm="fricas")

[Out]

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

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Sympy [C]  time = 0.93804, size = 29, normalized size = 0.23 \begin{align*} \frac{x^{7}{{}_{2}F_{1}\left (\begin{matrix} \frac{5}{4}, \frac{7}{2} \\ \frac{9}{2} \end{matrix}\middle |{\frac{b x^{2} e^{2 i \pi }}{a}} \right )}}{7 a^{\frac{5}{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**6/(-b*x**2+a)**(5/4),x)

[Out]

x**7*hyper((5/4, 7/2), (9/2,), b*x**2*exp_polar(2*I*pi)/a)/(7*a**(5/4))

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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{5}{4}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6/(-b*x^2+a)^(5/4),x, algorithm="giac")

[Out]

integrate(x^6/(-b*x^2 + a)^(5/4), x)