3.9.62 \(\int \frac {1}{(a-b x^2)^{11/4}} \, dx\) [862]

Optimal. Leaf size=101 \[ \frac {2 x}{7 a \left (a-b x^2\right )^{7/4}}+\frac {10 x}{21 a^2 \left (a-b x^2\right )^{3/4}}+\frac {10 \left (1-\frac {b x^2}{a}\right )^{3/4} F\left (\left .\frac {1}{2} \sin ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{21 a^{3/2} \sqrt {b} \left (a-b x^2\right )^{3/4}} \]

[Out]

2/7*x/a/(-b*x^2+a)^(7/4)+10/21*x/a^2/(-b*x^2+a)^(3/4)+10/21*(1-b*x^2/a)^(3/4)*(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)))*EllipticF(sin(1/2*arcsin(x*b^(1/2)/a^(1/2))),2^(1/2))/a^(3/2)/
(-b*x^2+a)^(3/4)/b^(1/2)

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Rubi [A]
time = 0.02, 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 = {205, 239, 238} \begin {gather*} \frac {10 \left (1-\frac {b x^2}{a}\right )^{3/4} F\left (\left .\frac {1}{2} \text {ArcSin}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{21 a^{3/2} \sqrt {b} \left (a-b x^2\right )^{3/4}}+\frac {10 x}{21 a^2 \left (a-b x^2\right )^{3/4}}+\frac {2 x}{7 a \left (a-b x^2\right )^{7/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a - b*x^2)^(-11/4),x]

[Out]

(2*x)/(7*a*(a - b*x^2)^(7/4)) + (10*x)/(21*a^2*(a - b*x^2)^(3/4)) + (10*(1 - (b*x^2)/a)^(3/4)*EllipticF[ArcSin
[(Sqrt[b]*x)/Sqrt[a]]/2, 2])/(21*a^(3/2)*Sqrt[b]*(a - b*x^2)^(3/4))

Rule 205

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] && (
IntegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[
p])

Rule 238

Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Simp[(2/(a^(3/4)*Rt[-b/a, 2]))*EllipticF[(1/2)*ArcSin[Rt[-b/a,
2]*x], 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && NegQ[b/a]

Rule 239

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]

Rubi steps

\begin {align*} \int \frac {1}{\left (a-b x^2\right )^{11/4}} \, dx &=\frac {2 x}{7 a \left (a-b x^2\right )^{7/4}}+\frac {5 \int \frac {1}{\left (a-b x^2\right )^{7/4}} \, dx}{7 a}\\ &=\frac {2 x}{7 a \left (a-b x^2\right )^{7/4}}+\frac {10 x}{21 a^2 \left (a-b x^2\right )^{3/4}}+\frac {5 \int \frac {1}{\left (a-b x^2\right )^{3/4}} \, dx}{21 a^2}\\ &=\frac {2 x}{7 a \left (a-b x^2\right )^{7/4}}+\frac {10 x}{21 a^2 \left (a-b x^2\right )^{3/4}}+\frac {\left (5 \left (1-\frac {b x^2}{a}\right )^{3/4}\right ) \int \frac {1}{\left (1-\frac {b x^2}{a}\right )^{3/4}} \, dx}{21 a^2 \left (a-b x^2\right )^{3/4}}\\ &=\frac {2 x}{7 a \left (a-b x^2\right )^{7/4}}+\frac {10 x}{21 a^2 \left (a-b x^2\right )^{3/4}}+\frac {10 \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 )}{21 a^{3/2} \sqrt {b} \left (a-b x^2\right )^{3/4}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 8.88, size = 77, normalized size = 0.76 \begin {gather*} \frac {2 x \left (8 a-5 b x^2\right )+5 x \left (a-b x^2\right ) \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 )}{21 a^2 \left (a-b x^2\right )^{7/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a - b*x^2)^(-11/4),x]

[Out]

(2*x*(8*a - 5*b*x^2) + 5*x*(a - b*x^2)*(1 - (b*x^2)/a)^(3/4)*Hypergeometric2F1[1/2, 3/4, 3/2, (b*x^2)/a])/(21*
a^2*(a - b*x^2)^(7/4))

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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {1}{\left (-b \,x^{2}+a \right )^{\frac {11}{4}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(-b*x^2+a)^(11/4),x)

[Out]

int(1/(-b*x^2+a)^(11/4),x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-b*x^2+a)^(11/4),x, algorithm="maxima")

[Out]

integrate((-b*x^2 + a)^(-11/4), x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-b*x^2+a)^(11/4),x, algorithm="fricas")

[Out]

integral(-(-b*x^2 + a)^(1/4)/(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] Result contains complex when optimal does not.
time = 1.01, size = 26, normalized size = 0.26 \begin {gather*} \frac {x {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {11}{4} \\ \frac {3}{2} \end {matrix}\middle | {\frac {b x^{2} e^{2 i \pi }}{a}} \right )}}{a^{\frac {11}{4}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-b*x**2+a)**(11/4),x)

[Out]

x*hyper((1/2, 11/4), (3/2,), b*x**2*exp_polar(2*I*pi)/a)/a**(11/4)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-b*x^2+a)^(11/4),x, algorithm="giac")

[Out]

integrate((-b*x^2 + a)^(-11/4), x)

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Mupad [B]
time = 4.81, size = 38, normalized size = 0.38 \begin {gather*} \frac {x\,{\left (1-\frac {b\,x^2}{a}\right )}^{11/4}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {11}{4};\ \frac {3}{2};\ \frac {b\,x^2}{a}\right )}{{\left (a-b\,x^2\right )}^{11/4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a - b*x^2)^(11/4),x)

[Out]

(x*(1 - (b*x^2)/a)^(11/4)*hypergeom([1/2, 11/4], 3/2, (b*x^2)/a))/(a - b*x^2)^(11/4)

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