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

Optimal. Leaf size=76 \[ -\frac {\left (a-b x^2\right )^{3/4}}{x}-\frac {3 \sqrt {a} \sqrt {b} \sqrt [4]{1-\frac {b x^2}{a}} E\left (\left .\frac {1}{2} \sin ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{\sqrt [4]{a-b x^2}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.02, antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {277, 229, 228} \[ -\frac {\left (a-b x^2\right )^{3/4}}{x}-\frac {3 \sqrt {a} \sqrt {b} \sqrt [4]{1-\frac {b x^2}{a}} E\left (\left .\frac {1}{2} \sin ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{\sqrt [4]{a-b x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(a - b*x^2)^(3/4)/x^2,x]

[Out]

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

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]

Rule 277

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^p)/(c*(m +
1)), x] - Dist[(b*n*p)/(c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] &&
IGtQ[n, 0] && GtQ[p, 0] && LtQ[m, -1] &&  !ILtQ[(m + n*p + n + 1)/n, 0] && IntBinomialQ[a, b, c, n, m, p, x]

Rubi steps

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

________________________________________________________________________________________

Mathematica [C]  time = 0.01, size = 50, normalized size = 0.66 \[ -\frac {\left (a-b x^2\right )^{3/4} \, _2F_1\left (-\frac {3}{4},-\frac {1}{2};\frac {1}{2};\frac {b x^2}{a}\right )}{x \left (1-\frac {b x^2}{a}\right )^{3/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a - b*x^2)^(3/4)/x^2,x]

[Out]

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

________________________________________________________________________________________

fricas [F]  time = 1.30, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (-b x^{2} + a\right )}^{\frac {3}{4}}}{x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((-b*x^2 + a)^(3/4)/x^2, x)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (-b x^{2} + a\right )}^{\frac {3}{4}}}{x^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((-b*x^2 + a)^(3/4)/x^2, x)

________________________________________________________________________________________

maple [F]  time = 0.29, size = 0, normalized size = 0.00 \[ \int \frac {\left (-b \,x^{2}+a \right )^{\frac {3}{4}}}{x^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (-b x^{2} + a\right )}^{\frac {3}{4}}}{x^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((-b*x^2 + a)^(3/4)/x^2, x)

________________________________________________________________________________________

mupad [B]  time = 5.17, size = 41, normalized size = 0.54 \[ \frac {2\,{\left (a-b\,x^2\right )}^{3/4}\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},-\frac {1}{4};\ \frac {3}{4};\ \frac {a}{b\,x^2}\right )}{x\,{\left (1-\frac {a}{b\,x^2}\right )}^{3/4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a - b*x^2)^(3/4)/x^2,x)

[Out]

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

________________________________________________________________________________________

sympy [C]  time = 1.05, size = 31, normalized size = 0.41 \[ - \frac {a^{\frac {3}{4}} {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, - \frac {1}{2} \\ \frac {1}{2} \end {matrix}\middle | {\frac {b x^{2} e^{2 i \pi }}{a}} \right )}}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________