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

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

[Out]

2/5*x/a/(-b*x^2+a)^(5/4)+6/5*x/a^2/(-b*x^2+a)^(1/4)-6/5*(1-b*x^2/a)^(1/4)*(cos(1/2*arcsin(x*b^(1/2)/a^(1/2)))^
2)^(1/2)/cos(1/2*arcsin(x*b^(1/2)/a^(1/2)))*EllipticE(sin(1/2*arcsin(x*b^(1/2)/a^(1/2))),2^(1/2))/a^(3/2)/(-b*
x^2+a)^(1/4)/b^(1/2)

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Rubi [A]  time = 0.03, antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, 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 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]

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]

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*}

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Mathematica [C]  time = 0.04, 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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fricas [F]  time = 0.57, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {{\left (-b x^{2} + a\right )}^{\frac {3}{4}}}{b^{3} x^{6} - 3 \, a b^{2} x^{4} + 3 \, a^{2} b x^{2} - a^{3}}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (-b x^{2} + a\right )}^{\frac {9}{4}}}\,{d x} \]

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)

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maple [F]  time = 0.32, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (-b \,x^{2}+a \right )^{\frac {9}{4}}}\, dx \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (-b x^{2} + a\right )}^{\frac {9}{4}}}\,{d x} \]

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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mupad [B]  time = 4.86, size = 38, normalized size = 0.38 \[ \frac {x\,{\left (1-\frac {b\,x^2}{a}\right )}^{9/4}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {9}{4};\ \frac {3}{2};\ \frac {b\,x^2}{a}\right )}{{\left (a-b\,x^2\right )}^{9/4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a - b*x^2)^(9/4),x)

[Out]

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

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sympy [C]  time = 1.36, size = 26, normalized size = 0.26 \[ \frac {x {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {9}{4} \\ \frac {3}{2} \end {matrix}\middle | {\frac {b x^{2} e^{2 i \pi }}{a}} \right )}}{a^{\frac {9}{4}}} \]

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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