3.854 \(\int \frac {1}{x^2 (a-b x^2)^{5/4}} \, dx\)

Optimal. Leaf size=99 \[ -\frac {3 \sqrt {b} \sqrt [4]{1-\frac {b x^2}{a}} E\left (\left .\frac {1}{2} \sin ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{a^{3/2} \sqrt [4]{a-b x^2}}-\frac {3 \left (a-b x^2\right )^{3/4}}{a^2 x}+\frac {2}{a x \sqrt [4]{a-b x^2}} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 99, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {290, 325, 229, 228} \[ -\frac {3 \left (a-b x^2\right )^{3/4}}{a^2 x}-\frac {3 \sqrt {b} \sqrt [4]{1-\frac {b x^2}{a}} E\left (\left .\frac {1}{2} \sin ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{a^{3/2} \sqrt [4]{a-b x^2}}+\frac {2}{a x \sqrt [4]{a-b x^2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^2*(a - b*x^2)^(5/4)),x]

[Out]

2/(a*x*(a - b*x^2)^(1/4)) - (3*(a - b*x^2)^(3/4))/(a^2*x) - (3*Sqrt[b]*(1 - (b*x^2)/a)^(1/4)*EllipticE[ArcSin[
(Sqrt[b]*x)/Sqrt[a]]/2, 2])/(a^(3/2)*(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 290

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

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

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Mathematica [C]  time = 0.01, size = 53, normalized size = 0.54 \[ -\frac {\sqrt [4]{1-\frac {b x^2}{a}} \, _2F_1\left (-\frac {1}{2},\frac {5}{4};\frac {1}{2};\frac {b x^2}{a}\right )}{a x \sqrt [4]{a-b x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((1 - (b*x^2)/a)^(1/4)*Hypergeometric2F1[-1/2, 5/4, 1/2, (b*x^2)/a])/(a*x*(a - b*x^2)^(1/4)))

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fricas [F]  time = 0.60, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (-b x^{2} + a\right )}^{\frac {3}{4}}}{b^{2} x^{6} - 2 \, a b x^{4} + a^{2} x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(-b*x^2+a)^(5/4),x, algorithm="fricas")

[Out]

integral((-b*x^2 + a)^(3/4)/(b^2*x^6 - 2*a*b*x^4 + a^2*x^2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (-b x^{2} + a\right )}^{\frac {5}{4}} x^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(-b*x^2+a)^(5/4),x, algorithm="giac")

[Out]

integrate(1/((-b*x^2 + a)^(5/4)*x^2), x)

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maple [F]  time = 0.32, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (-b \,x^{2}+a \right )^{\frac {5}{4}} x^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^2/(-b*x^2+a)^(5/4),x)

[Out]

int(1/x^2/(-b*x^2+a)^(5/4),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (-b x^{2} + a\right )}^{\frac {5}{4}} x^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(-b*x^2+a)^(5/4),x, algorithm="maxima")

[Out]

integrate(1/((-b*x^2 + a)^(5/4)*x^2), x)

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mupad [B]  time = 5.10, size = 41, normalized size = 0.41 \[ -\frac {2\,{\left (1-\frac {a}{b\,x^2}\right )}^{5/4}\,{{}}_2{\mathrm {F}}_1\left (\frac {5}{4},\frac {7}{4};\ \frac {11}{4};\ \frac {a}{b\,x^2}\right )}{7\,x\,{\left (a-b\,x^2\right )}^{5/4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^2*(a - b*x^2)^(5/4)),x)

[Out]

-(2*(1 - a/(b*x^2))^(5/4)*hypergeom([5/4, 7/4], 11/4, a/(b*x^2)))/(7*x*(a - b*x^2)^(5/4))

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sympy [C]  time = 1.09, size = 29, normalized size = 0.29 \[ - \frac {{{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {5}{4} \\ \frac {1}{2} \end {matrix}\middle | {\frac {b x^{2} e^{2 i \pi }}{a}} \right )}}{a^{\frac {5}{4}} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**2/(-b*x**2+a)**(5/4),x)

[Out]

-hyper((-1/2, 5/4), (1/2,), b*x**2*exp_polar(2*I*pi)/a)/(a**(5/4)*x)

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