3.853 \(\int \frac{1}{(a-b x^2)^{5/4}} \, dx\)

Optimal. Leaf size=77 \[ \frac{2 x}{a \sqrt [4]{a-b x^2}}-\frac{2 \sqrt [4]{1-\frac{b x^2}{a}} E\left (\left .\frac{1}{2} \sin ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )\right |2\right )}{\sqrt{a} \sqrt{b} \sqrt [4]{a-b x^2}} \]

[Out]

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

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Rubi [A]  time = 0.0193488, antiderivative size = 77, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {199, 229, 228} \[ \frac{2 x}{a \sqrt [4]{a-b x^2}}-\frac{2 \sqrt [4]{1-\frac{b x^2}{a}} E\left (\left .\frac{1}{2} \sin ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )\right |2\right )}{\sqrt{a} \sqrt{b} \sqrt [4]{a-b x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(a - b*x^2)^(-5/4),x]

[Out]

(2*x)/(a*(a - b*x^2)^(1/4)) - (2*(1 - (b*x^2)/a)^(1/4)*EllipticE[ArcSin[(Sqrt[b]*x)/Sqrt[a]]/2, 2])/(Sqrt[a]*S
qrt[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 )^{5/4}} \, dx &=\frac{2 x}{a \sqrt [4]{a-b x^2}}-\frac{\int \frac{1}{\sqrt [4]{a-b x^2}} \, dx}{a}\\ &=\frac{2 x}{a \sqrt [4]{a-b x^2}}-\frac{\sqrt [4]{1-\frac{b x^2}{a}} \int \frac{1}{\sqrt [4]{1-\frac{b x^2}{a}}} \, dx}{a \sqrt [4]{a-b x^2}}\\ &=\frac{2 x}{a \sqrt [4]{a-b x^2}}-\frac{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 )}{\sqrt{a} \sqrt{b} \sqrt [4]{a-b x^2}}\\ \end{align*}

Mathematica [C]  time = 0.0130012, size = 56, normalized size = 0.73 \[ \frac{2 x-x \sqrt [4]{1-\frac{b x^2}{a}} \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{3}{2};\frac{b x^2}{a}\right )}{a \sqrt [4]{a-b x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a - b*x^2)^(-5/4),x]

[Out]

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

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Maple [F]  time = 0.035, size = 0, normalized size = 0. \begin{align*} \int \left ( -b{x}^{2}+a \right ) ^{-{\frac{5}{4}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(-b*x^2+a)^(5/4),x)

[Out]

int(1/(-b*x^2+a)^(5/4),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{5}{4}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-b*x^2+a)^(5/4),x, algorithm="maxima")

[Out]

integrate((-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}}}{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(1/(-b*x^2+a)^(5/4),x, algorithm="fricas")

[Out]

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

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Sympy [C]  time = 0.801041, size = 26, normalized size = 0.34 \begin{align*} \frac{x{{}_{2}F_{1}\left (\begin{matrix} \frac{1}{2}, \frac{5}{4} \\ \frac{3}{2} \end{matrix}\middle |{\frac{b x^{2} e^{2 i \pi }}{a}} \right )}}{a^{\frac{5}{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-b*x**2+a)**(5/4),x)

[Out]

x*hyper((1/2, 5/4), (3/2,), b*x**2*exp_polar(2*I*pi)/a)/a**(5/4)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-b*x^2+a)^(5/4),x, algorithm="giac")

[Out]

integrate((-b*x^2 + a)^(-5/4), x)