3.860 \(\int \frac {1}{(a-b x^2)^{7/4}} \, dx\)

Optimal. Leaf size=81 \[ \frac {2 \left (1-\frac {b x^2}{a}\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \sin ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),2\right )}{3 \sqrt {a} \sqrt {b} \left (a-b x^2\right )^{3/4}}+\frac {2 x}{3 a \left (a-b x^2\right )^{3/4}} \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {199, 233, 232} \[ \frac {2 x}{3 a \left (a-b x^2\right )^{3/4}}+\frac {2 \left (1-\frac {b x^2}{a}\right )^{3/4} F\left (\left .\frac {1}{2} \sin ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{3 \sqrt {a} \sqrt {b} \left (a-b x^2\right )^{3/4}} \]

Antiderivative was successfully verified.

[In]

Int[(a - b*x^2)^(-7/4),x]

[Out]

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

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 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]

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]

Rubi steps

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

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Mathematica [C]  time = 0.02, size = 56, normalized size = 0.69 \[ \frac {x \left (\left (1-\frac {b x^2}{a}\right )^{3/4} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {3}{2};\frac {b x^2}{a}\right )+2\right )}{3 a \left (a-b x^2\right )^{3/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a - b*x^2)^(-7/4),x]

[Out]

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

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fricas [F]  time = 0.83, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (-b x^{2} + a\right )}^{\frac {1}{4}}}{b^{2} x^{4} - 2 \, a b x^{2} + a^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-b*x^2+a)^(7/4),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (-b x^{2} + a\right )}^{\frac {7}{4}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-b*x^2+a)^(7/4),x, algorithm="giac")

[Out]

integrate((-b*x^2 + a)^(-7/4), x)

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maple [F]  time = 0.32, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (-b \,x^{2}+a \right )^{\frac {7}{4}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(-b*x^2+a)^(7/4),x)

[Out]

int(1/(-b*x^2+a)^(7/4),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (-b x^{2} + a\right )}^{\frac {7}{4}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-b*x^2+a)^(7/4),x, algorithm="maxima")

[Out]

integrate((-b*x^2 + a)^(-7/4), x)

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mupad [B]  time = 4.87, size = 38, normalized size = 0.47 \[ \frac {x\,{\left (1-\frac {b\,x^2}{a}\right )}^{7/4}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {7}{4};\ \frac {3}{2};\ \frac {b\,x^2}{a}\right )}{{\left (a-b\,x^2\right )}^{7/4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a - b*x^2)^(7/4),x)

[Out]

(x*(1 - (b*x^2)/a)^(7/4)*hypergeom([1/2, 7/4], 3/2, (b*x^2)/a))/(a - b*x^2)^(7/4)

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sympy [C]  time = 1.06, size = 26, normalized size = 0.32 \[ \frac {x {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {7}{4} \\ \frac {3}{2} \end {matrix}\middle | {\frac {b x^{2} e^{2 i \pi }}{a}} \right )}}{a^{\frac {7}{4}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-b*x**2+a)**(7/4),x)

[Out]

x*hyper((1/2, 7/4), (3/2,), b*x**2*exp_polar(2*I*pi)/a)/a**(7/4)

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