3.839 \(\int \frac{1}{(a-b x^2)^{3/4}} \, dx\)

Optimal. Leaf size=58 \[ \frac{2 \sqrt{a} \left (1-\frac{b x^2}{a}\right )^{3/4} \text{EllipticF}\left (\frac{1}{2} \sin ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right ),2\right )}{\sqrt{b} \left (a-b x^2\right )^{3/4}} \]

[Out]

(2*Sqrt[a]*(1 - (b*x^2)/a)^(3/4)*EllipticF[ArcSin[(Sqrt[b]*x)/Sqrt[a]]/2, 2])/(Sqrt[b]*(a - b*x^2)^(3/4))

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Rubi [A]  time = 0.010857, antiderivative size = 58, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {233, 232} \[ \frac{2 \sqrt{a} \left (1-\frac{b x^2}{a}\right )^{3/4} F\left (\left .\frac{1}{2} \sin ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )\right |2\right )}{\sqrt{b} \left (a-b x^2\right )^{3/4}} \]

Antiderivative was successfully verified.

[In]

Int[(a - b*x^2)^(-3/4),x]

[Out]

(2*Sqrt[a]*(1 - (b*x^2)/a)^(3/4)*EllipticF[ArcSin[(Sqrt[b]*x)/Sqrt[a]]/2, 2])/(Sqrt[b]*(a - b*x^2)^(3/4))

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

Mathematica [C]  time = 0.0067723, size = 47, normalized size = 0.81 \[ \frac{x \left (1-\frac{b x^2}{a}\right )^{3/4} \, _2F_1\left (\frac{1}{2},\frac{3}{4};\frac{3}{2};\frac{b x^2}{a}\right )}{\left (a-b x^2\right )^{3/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a - b*x^2)^(-3/4),x]

[Out]

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

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Maple [F]  time = 0.032, size = 0, normalized size = 0. \begin{align*} \int \left ( -b{x}^{2}+a \right ) ^{-{\frac{3}{4}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(-b*x^2+a)^(3/4),x)

[Out]

int(1/(-b*x^2+a)^(3/4),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{3}{4}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-b*x^2+a)^(3/4),x, algorithm="maxima")

[Out]

integrate((-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}}}{b x^{2} - a}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-b*x^2+a)^(3/4),x, algorithm="fricas")

[Out]

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

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Sympy [C]  time = 0.692853, size = 26, normalized size = 0.45 \begin{align*} \frac{x{{}_{2}F_{1}\left (\begin{matrix} \frac{1}{2}, \frac{3}{4} \\ \frac{3}{2} \end{matrix}\middle |{\frac{b x^{2} e^{2 i \pi }}{a}} \right )}}{a^{\frac{3}{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-b*x**2+a)**(3/4),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-b*x^2+a)^(3/4),x, algorithm="giac")

[Out]

integrate((-b*x^2 + a)^(-3/4), x)