3.861 \(\int \frac{1}{(a-b x^2)^{9/4}} \, dx\)

Optimal. Leaf size=101 \[ \frac{6 x}{5 a^2 \sqrt [4]{a-b x^2}}-\frac{6 \sqrt [4]{1-\frac{b x^2}{a}} E\left (\left .\frac{1}{2} \sin ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )\right |2\right )}{5 a^{3/2} \sqrt{b} \sqrt [4]{a-b x^2}}+\frac{2 x}{5 a \left (a-b x^2\right )^{5/4}} \]

[Out]

(2*x)/(5*a*(a - b*x^2)^(5/4)) + (6*x)/(5*a^2*(a - b*x^2)^(1/4)) - (6*(1 - (b*x^2)/a)^(1/4)*EllipticE[ArcSin[(S
qrt[b]*x)/Sqrt[a]]/2, 2])/(5*a^(3/2)*Sqrt[b]*(a - b*x^2)^(1/4))

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Rubi [A]  time = 0.02682, antiderivative size = 101, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {199, 229, 228} \[ \frac{6 x}{5 a^2 \sqrt [4]{a-b x^2}}-\frac{6 \sqrt [4]{1-\frac{b x^2}{a}} E\left (\left .\frac{1}{2} \sin ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )\right |2\right )}{5 a^{3/2} \sqrt{b} \sqrt [4]{a-b x^2}}+\frac{2 x}{5 a \left (a-b x^2\right )^{5/4}} \]

Antiderivative was successfully verified.

[In]

Int[(a - b*x^2)^(-9/4),x]

[Out]

(2*x)/(5*a*(a - b*x^2)^(5/4)) + (6*x)/(5*a^2*(a - b*x^2)^(1/4)) - (6*(1 - (b*x^2)/a)^(1/4)*EllipticE[ArcSin[(S
qrt[b]*x)/Sqrt[a]]/2, 2])/(5*a^(3/2)*Sqrt[b]*(a - b*x^2)^(1/4))

Rule 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

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{1}{\left (a-b x^2\right )^{9/4}} \, dx &=\frac{2 x}{5 a \left (a-b x^2\right )^{5/4}}+\frac{3 \int \frac{1}{\left (a-b x^2\right )^{5/4}} \, dx}{5 a}\\ &=\frac{2 x}{5 a \left (a-b x^2\right )^{5/4}}+\frac{6 x}{5 a^2 \sqrt [4]{a-b x^2}}-\frac{3 \int \frac{1}{\sqrt [4]{a-b x^2}} \, dx}{5 a^2}\\ &=\frac{2 x}{5 a \left (a-b x^2\right )^{5/4}}+\frac{6 x}{5 a^2 \sqrt [4]{a-b x^2}}-\frac{\left (3 \sqrt [4]{1-\frac{b x^2}{a}}\right ) \int \frac{1}{\sqrt [4]{1-\frac{b x^2}{a}}} \, dx}{5 a^2 \sqrt [4]{a-b x^2}}\\ &=\frac{2 x}{5 a \left (a-b x^2\right )^{5/4}}+\frac{6 x}{5 a^2 \sqrt [4]{a-b x^2}}-\frac{6 \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 )}{5 a^{3/2} \sqrt{b} \sqrt [4]{a-b x^2}}\\ \end{align*}

Mathematica [C]  time = 0.0332388, size = 74, normalized size = 0.73 \[ \frac{-3 x \left (a-b x^2\right ) \sqrt [4]{1-\frac{b x^2}{a}} \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{3}{2};\frac{b x^2}{a}\right )+8 a x-6 b x^3}{5 a^2 \left (a-b x^2\right )^{5/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a - b*x^2)^(-9/4),x]

[Out]

(8*a*x - 6*b*x^3 - 3*x*(a - b*x^2)*(1 - (b*x^2)/a)^(1/4)*Hypergeometric2F1[1/4, 1/2, 3/2, (b*x^2)/a])/(5*a^2*(
a - b*x^2)^(5/4))

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Maple [F]  time = 0.036, size = 0, normalized size = 0. \begin{align*} \int \left ( -b{x}^{2}+a \right ) ^{-{\frac{9}{4}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(-b*x^2+a)^(9/4),x)

[Out]

int(1/(-b*x^2+a)^(9/4),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (-b x^{2} + a\right )}^{\frac{9}{4}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-b*x^2+a)^(9/4),x, algorithm="maxima")

[Out]

integrate((-b*x^2 + a)^(-9/4), 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}}}{b^{3} x^{6} - 3 \, a b^{2} x^{4} + 3 \, a^{2} b x^{2} - a^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-b*x^2+a)^(9/4),x, algorithm="fricas")

[Out]

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

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Sympy [C]  time = 1.78914, size = 26, normalized size = 0.26 \begin{align*} \frac{x{{}_{2}F_{1}\left (\begin{matrix} \frac{1}{2}, \frac{9}{4} \\ \frac{3}{2} \end{matrix}\middle |{\frac{b x^{2} e^{2 i \pi }}{a}} \right )}}{a^{\frac{9}{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-b*x**2+a)**(9/4),x)

[Out]

x*hyper((1/2, 9/4), (3/2,), b*x**2*exp_polar(2*I*pi)/a)/a**(9/4)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (-b x^{2} + a\right )}^{\frac{9}{4}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-b*x^2+a)^(9/4),x, algorithm="giac")

[Out]

integrate((-b*x^2 + a)^(-9/4), x)