3.840 \(\int \frac {1}{x^2 (a-b x^2)^{3/4}} \, dx\)

Optimal. Leaf size=78 \[ \frac {\sqrt {b} \left (1-\frac {b x^2}{a}\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \sin ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),2\right )}{\sqrt {a} \left (a-b x^2\right )^{3/4}}-\frac {\sqrt [4]{a-b x^2}}{a x} \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 78, 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 = {325, 233, 232} \[ \frac {\sqrt {b} \left (1-\frac {b x^2}{a}\right )^{3/4} F\left (\left .\frac {1}{2} \sin ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{\sqrt {a} \left (a-b x^2\right )^{3/4}}-\frac {\sqrt [4]{a-b x^2}}{a x} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^2*(a - b*x^2)^(3/4)),x]

[Out]

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

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]

Rule 325

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

Rubi steps

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

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Mathematica [C]  time = 0.01, size = 50, normalized size = 0.64 \[ -\frac {\left (1-\frac {b x^2}{a}\right )^{3/4} \, _2F_1\left (-\frac {1}{2},\frac {3}{4};\frac {1}{2};\frac {b x^2}{a}\right )}{x \left (a-b x^2\right )^{3/4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 1.16, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {{\left (-b x^{2} + a\right )}^{\frac {1}{4}}}{b x^{4} - a x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(-b*x^2+a)^(3/4),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (-b x^{2} + a\right )}^{\frac {3}{4}} x^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(-b*x^2+a)^(3/4),x, algorithm="giac")

[Out]

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

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maple [F]  time = 0.04, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (-b \,x^{2}+a \right )^{\frac {3}{4}} x^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (-b x^{2} + a\right )}^{\frac {3}{4}} x^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(-b*x^2+a)^(3/4),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 5.10, size = 41, normalized size = 0.53 \[ -\frac {2\,{\left (1-\frac {a}{b\,x^2}\right )}^{3/4}\,{{}}_2{\mathrm {F}}_1\left (\frac {3}{4},\frac {5}{4};\ \frac {9}{4};\ \frac {a}{b\,x^2}\right )}{5\,x\,{\left (a-b\,x^2\right )}^{3/4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^2*(a - b*x^2)^(3/4)),x)

[Out]

-(2*(1 - a/(b*x^2))^(3/4)*hypergeom([3/4, 5/4], 9/4, a/(b*x^2)))/(5*x*(a - b*x^2)^(3/4))

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sympy [C]  time = 0.98, size = 29, normalized size = 0.37 \[ - \frac {{{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {3}{4} \\ \frac {1}{2} \end {matrix}\middle | {\frac {b x^{2} e^{2 i \pi }}{a}} \right )}}{a^{\frac {3}{4}} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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