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

Optimal. Leaf size=126 \[ -\frac {7 b^{3/2} \sqrt [4]{1-\frac {b x^2}{a}} E\left (\left .\frac {1}{2} \sin ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{2 a^{5/2} \sqrt [4]{a-b x^2}}-\frac {7 b \left (a-b x^2\right )^{3/4}}{2 a^3 x}-\frac {7 \left (a-b x^2\right )^{3/4}}{3 a^2 x^3}+\frac {2}{a x^3 \sqrt [4]{a-b x^2}} \]

[Out]

2/a/x^3/(-b*x^2+a)^(1/4)-7/3*(-b*x^2+a)^(3/4)/a^2/x^3-7/2*b*(-b*x^2+a)^(3/4)/a^3/x-7/2*b^(3/2)*(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^(5/2)/(-b*x^2+a)^(1/4)

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Rubi [A]  time = 0.05, antiderivative size = 126, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {290, 325, 229, 228} \[ -\frac {7 b^{3/2} \sqrt [4]{1-\frac {b x^2}{a}} E\left (\left .\frac {1}{2} \sin ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{2 a^{5/2} \sqrt [4]{a-b x^2}}-\frac {7 b \left (a-b x^2\right )^{3/4}}{2 a^3 x}-\frac {7 \left (a-b x^2\right )^{3/4}}{3 a^2 x^3}+\frac {2}{a x^3 \sqrt [4]{a-b x^2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^4*(a - b*x^2)^(5/4)),x]

[Out]

2/(a*x^3*(a - b*x^2)^(1/4)) - (7*(a - b*x^2)^(3/4))/(3*a^2*x^3) - (7*b*(a - b*x^2)^(3/4))/(2*a^3*x) - (7*b^(3/
2)*(1 - (b*x^2)/a)^(1/4)*EllipticE[ArcSin[(Sqrt[b]*x)/Sqrt[a]]/2, 2])/(2*a^(5/2)*(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^4 \left (a-b x^2\right )^{5/4}} \, dx &=\frac {2}{a x^3 \sqrt [4]{a-b x^2}}+\frac {7 \int \frac {1}{x^4 \sqrt [4]{a-b x^2}} \, dx}{a}\\ &=\frac {2}{a x^3 \sqrt [4]{a-b x^2}}-\frac {7 \left (a-b x^2\right )^{3/4}}{3 a^2 x^3}+\frac {(7 b) \int \frac {1}{x^2 \sqrt [4]{a-b x^2}} \, dx}{2 a^2}\\ &=\frac {2}{a x^3 \sqrt [4]{a-b x^2}}-\frac {7 \left (a-b x^2\right )^{3/4}}{3 a^2 x^3}-\frac {7 b \left (a-b x^2\right )^{3/4}}{2 a^3 x}-\frac {\left (7 b^2\right ) \int \frac {1}{\sqrt [4]{a-b x^2}} \, dx}{4 a^3}\\ &=\frac {2}{a x^3 \sqrt [4]{a-b x^2}}-\frac {7 \left (a-b x^2\right )^{3/4}}{3 a^2 x^3}-\frac {7 b \left (a-b x^2\right )^{3/4}}{2 a^3 x}-\frac {\left (7 b^2 \sqrt [4]{1-\frac {b x^2}{a}}\right ) \int \frac {1}{\sqrt [4]{1-\frac {b x^2}{a}}} \, dx}{4 a^3 \sqrt [4]{a-b x^2}}\\ &=\frac {2}{a x^3 \sqrt [4]{a-b x^2}}-\frac {7 \left (a-b x^2\right )^{3/4}}{3 a^2 x^3}-\frac {7 b \left (a-b x^2\right )^{3/4}}{2 a^3 x}-\frac {7 b^{3/2} \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 )}{2 a^{5/2} \sqrt [4]{a-b x^2}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 55, normalized size = 0.44 \[ -\frac {\sqrt [4]{1-\frac {b x^2}{a}} \, _2F_1\left (-\frac {3}{2},\frac {5}{4};-\frac {1}{2};\frac {b x^2}{a}\right )}{3 a x^3 \sqrt [4]{a-b x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.94, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (-b x^{2} + a\right )}^{\frac {3}{4}}}{b^{2} x^{8} - 2 \, a b x^{6} + a^{2} x^{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((-b*x^2 + a)^(3/4)/(b^2*x^8 - 2*a*b*x^6 + a^2*x^4), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.31, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (-b \,x^{2}+a \right )^{\frac {5}{4}} x^{4}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{x^4\,{\left (a-b\,x^2\right )}^{5/4}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^4*(a - b*x^2)^(5/4)),x)

[Out]

int(1/(x^4*(a - b*x^2)^(5/4)), x)

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sympy [C]  time = 1.25, size = 34, normalized size = 0.27 \[ - \frac {{{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{2}, \frac {5}{4} \\ - \frac {1}{2} \end {matrix}\middle | {\frac {b x^{2} e^{2 i \pi }}{a}} \right )}}{3 a^{\frac {5}{4}} x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-hyper((-3/2, 5/4), (-1/2,), b*x**2*exp_polar(2*I*pi)/a)/(3*a**(5/4)*x**3)

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