3.9.56 \(\int \frac {1}{x^6 (a-b x^2)^{5/4}} \, dx\) [856]

Optimal. Leaf size=151 \[ \frac {2}{a x^5 \sqrt [4]{a-b x^2}}-\frac {11 \left (a-b x^2\right )^{3/4}}{5 a^2 x^5}-\frac {77 b \left (a-b x^2\right )^{3/4}}{30 a^3 x^3}-\frac {77 b^2 \left (a-b x^2\right )^{3/4}}{20 a^4 x}-\frac {77 b^{5/2} \sqrt [4]{1-\frac {b x^2}{a}} E\left (\left .\frac {1}{2} \sin ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{20 a^{7/2} \sqrt [4]{a-b x^2}} \]

[Out]

2/a/x^5/(-b*x^2+a)^(1/4)-11/5*(-b*x^2+a)^(3/4)/a^2/x^5-77/30*b*(-b*x^2+a)^(3/4)/a^3/x^3-77/20*b^2*(-b*x^2+a)^(
3/4)/a^4/x-77/20*b^(5/2)*(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^(7/2)/(-b*x^2+a)^(1/4)

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Rubi [A]
time = 0.04, antiderivative size = 151, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {296, 331, 235, 234} \begin {gather*} -\frac {77 b^{5/2} \sqrt [4]{1-\frac {b x^2}{a}} E\left (\left .\frac {1}{2} \text {ArcSin}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{20 a^{7/2} \sqrt [4]{a-b x^2}}-\frac {77 b^2 \left (a-b x^2\right )^{3/4}}{20 a^4 x}-\frac {77 b \left (a-b x^2\right )^{3/4}}{30 a^3 x^3}-\frac {11 \left (a-b x^2\right )^{3/4}}{5 a^2 x^5}+\frac {2}{a x^5 \sqrt [4]{a-b x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^6*(a - b*x^2)^(5/4)),x]

[Out]

2/(a*x^5*(a - b*x^2)^(1/4)) - (11*(a - b*x^2)^(3/4))/(5*a^2*x^5) - (77*b*(a - b*x^2)^(3/4))/(30*a^3*x^3) - (77
*b^2*(a - b*x^2)^(3/4))/(20*a^4*x) - (77*b^(5/2)*(1 - (b*x^2)/a)^(1/4)*EllipticE[ArcSin[(Sqrt[b]*x)/Sqrt[a]]/2
, 2])/(20*a^(7/2)*(a - b*x^2)^(1/4))

Rule 234

Int[((a_) + (b_.)*(x_)^2)^(-1/4), x_Symbol] :> Simp[(2/(a^(1/4)*Rt[-b/a, 2]))*EllipticE[(1/2)*ArcSin[Rt[-b/a,
2]*x], 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && NegQ[b/a]

Rule 235

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 296

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] /; Free
Q[{a, b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 331

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^6 \left (a-b x^2\right )^{5/4}} \, dx &=\frac {2}{a x^5 \sqrt [4]{a-b x^2}}+\frac {11 \int \frac {1}{x^6 \sqrt [4]{a-b x^2}} \, dx}{a}\\ &=\frac {2}{a x^5 \sqrt [4]{a-b x^2}}-\frac {11 \left (a-b x^2\right )^{3/4}}{5 a^2 x^5}+\frac {(77 b) \int \frac {1}{x^4 \sqrt [4]{a-b x^2}} \, dx}{10 a^2}\\ &=\frac {2}{a x^5 \sqrt [4]{a-b x^2}}-\frac {11 \left (a-b x^2\right )^{3/4}}{5 a^2 x^5}-\frac {77 b \left (a-b x^2\right )^{3/4}}{30 a^3 x^3}+\frac {\left (77 b^2\right ) \int \frac {1}{x^2 \sqrt [4]{a-b x^2}} \, dx}{20 a^3}\\ &=\frac {2}{a x^5 \sqrt [4]{a-b x^2}}-\frac {11 \left (a-b x^2\right )^{3/4}}{5 a^2 x^5}-\frac {77 b \left (a-b x^2\right )^{3/4}}{30 a^3 x^3}-\frac {77 b^2 \left (a-b x^2\right )^{3/4}}{20 a^4 x}-\frac {\left (77 b^3\right ) \int \frac {1}{\sqrt [4]{a-b x^2}} \, dx}{40 a^4}\\ &=\frac {2}{a x^5 \sqrt [4]{a-b x^2}}-\frac {11 \left (a-b x^2\right )^{3/4}}{5 a^2 x^5}-\frac {77 b \left (a-b x^2\right )^{3/4}}{30 a^3 x^3}-\frac {77 b^2 \left (a-b x^2\right )^{3/4}}{20 a^4 x}-\frac {\left (77 b^3 \sqrt [4]{1-\frac {b x^2}{a}}\right ) \int \frac {1}{\sqrt [4]{1-\frac {b x^2}{a}}} \, dx}{40 a^4 \sqrt [4]{a-b x^2}}\\ &=\frac {2}{a x^5 \sqrt [4]{a-b x^2}}-\frac {11 \left (a-b x^2\right )^{3/4}}{5 a^2 x^5}-\frac {77 b \left (a-b x^2\right )^{3/4}}{30 a^3 x^3}-\frac {77 b^2 \left (a-b x^2\right )^{3/4}}{20 a^4 x}-\frac {77 b^{5/2} \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 )}{20 a^{7/2} \sqrt [4]{a-b x^2}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 10.02, size = 55, normalized size = 0.36 \begin {gather*} -\frac {\sqrt [4]{1-\frac {b x^2}{a}} \, _2F_1\left (-\frac {5}{2},\frac {5}{4};-\frac {3}{2};\frac {b x^2}{a}\right )}{5 a x^5 \sqrt [4]{a-b x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^6*(a - b*x^2)^(5/4)),x]

[Out]

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

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Maple [F]
time = 0.00, size = 0, normalized size = 0.00 \[\int \frac {1}{x^{6} \left (-b \,x^{2}+a \right )^{\frac {5}{4}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^6/(-b*x^2+a)^(5/4),x)

[Out]

int(1/x^6/(-b*x^2+a)^(5/4),x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^6/(-b*x^2+a)^(5/4),x, algorithm="maxima")

[Out]

integrate(1/((-b*x^2 + a)^(5/4)*x^6), x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^6/(-b*x^2+a)^(5/4),x, algorithm="fricas")

[Out]

integral((-b*x^2 + a)^(3/4)/(b^2*x^10 - 2*a*b*x^8 + a^2*x^6), x)

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Sympy [C] Result contains complex when optimal does not.
time = 0.84, size = 34, normalized size = 0.23 \begin {gather*} - \frac {{{}_{2}F_{1}\left (\begin {matrix} - \frac {5}{2}, \frac {5}{4} \\ - \frac {3}{2} \end {matrix}\middle | {\frac {b x^{2} e^{2 i \pi }}{a}} \right )}}{5 a^{\frac {5}{4}} x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**6/(-b*x**2+a)**(5/4),x)

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^6/(-b*x^2+a)^(5/4),x, algorithm="giac")

[Out]

integrate(1/((-b*x^2 + a)^(5/4)*x^6), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{x^6\,{\left (a-b\,x^2\right )}^{5/4}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^6*(a - b*x^2)^(5/4)),x)

[Out]

int(1/(x^6*(a - b*x^2)^(5/4)), x)

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