3.850 \(\int \frac {1}{x^6 \sqrt {a-b x^4}} \, dx\)

Optimal. Leaf size=158 \[ \frac {3 b^{5/4} \sqrt {1-\frac {b x^4}{a}} F\left (\left .\sin ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )\right |-1\right )}{5 a^{5/4} \sqrt {a-b x^4}}-\frac {3 b^{5/4} \sqrt {1-\frac {b x^4}{a}} E\left (\left .\sin ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )\right |-1\right )}{5 a^{5/4} \sqrt {a-b x^4}}-\frac {3 b \sqrt {a-b x^4}}{5 a^2 x}-\frac {\sqrt {a-b x^4}}{5 a x^5} \]

[Out]

-1/5*(-b*x^4+a)^(1/2)/a/x^5-3/5*b*(-b*x^4+a)^(1/2)/a^2/x-3/5*b^(5/4)*EllipticE(b^(1/4)*x/a^(1/4),I)*(1-b*x^4/a
)^(1/2)/a^(5/4)/(-b*x^4+a)^(1/2)+3/5*b^(5/4)*EllipticF(b^(1/4)*x/a^(1/4),I)*(1-b*x^4/a)^(1/2)/a^(5/4)/(-b*x^4+
a)^(1/2)

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Rubi [A]  time = 0.09, antiderivative size = 158, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.438, Rules used = {325, 307, 224, 221, 1200, 1199, 424} \[ \frac {3 b^{5/4} \sqrt {1-\frac {b x^4}{a}} F\left (\left .\sin ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )\right |-1\right )}{5 a^{5/4} \sqrt {a-b x^4}}-\frac {3 b^{5/4} \sqrt {1-\frac {b x^4}{a}} E\left (\left .\sin ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )\right |-1\right )}{5 a^{5/4} \sqrt {a-b x^4}}-\frac {3 b \sqrt {a-b x^4}}{5 a^2 x}-\frac {\sqrt {a-b x^4}}{5 a x^5} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^6*Sqrt[a - b*x^4]),x]

[Out]

-Sqrt[a - b*x^4]/(5*a*x^5) - (3*b*Sqrt[a - b*x^4])/(5*a^2*x) - (3*b^(5/4)*Sqrt[1 - (b*x^4)/a]*EllipticE[ArcSin
[(b^(1/4)*x)/a^(1/4)], -1])/(5*a^(5/4)*Sqrt[a - b*x^4]) + (3*b^(5/4)*Sqrt[1 - (b*x^4)/a]*EllipticF[ArcSin[(b^(
1/4)*x)/a^(1/4)], -1])/(5*a^(5/4)*Sqrt[a - b*x^4])

Rule 221

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

Rule 224

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Dist[Sqrt[1 + (b*x^4)/a]/Sqrt[a + b*x^4], Int[1/Sqrt[1 + (b*x^4)
/a], x], x] /; FreeQ[{a, b}, x] && NegQ[b/a] &&  !GtQ[a, 0]

Rule 307

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[-(b/a), 2]}, -Dist[q^(-1), Int[1/Sqrt[a + b*x^
4], x], x] + Dist[1/q, Int[(1 + q*x^2)/Sqrt[a + b*x^4], x], x]] /; FreeQ[{a, b}, x] && NegQ[b/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]

Rule 424

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

Rule 1199

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> Dist[d/Sqrt[a], Int[Sqrt[1 + (e*x^2)/d]/Sqrt
[1 - (e*x^2)/d], x], x] /; FreeQ[{a, c, d, e}, x] && NegQ[c/a] && EqQ[c*d^2 + a*e^2, 0] && GtQ[a, 0]

Rule 1200

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> Dist[Sqrt[1 + (c*x^4)/a]/Sqrt[a + c*x^4], In
t[(d + e*x^2)/Sqrt[1 + (c*x^4)/a], x], x] /; FreeQ[{a, c, d, e}, x] && NegQ[c/a] && EqQ[c*d^2 + a*e^2, 0] &&
!GtQ[a, 0]

Rubi steps

\begin {align*} \int \frac {1}{x^6 \sqrt {a-b x^4}} \, dx &=-\frac {\sqrt {a-b x^4}}{5 a x^5}+\frac {(3 b) \int \frac {1}{x^2 \sqrt {a-b x^4}} \, dx}{5 a}\\ &=-\frac {\sqrt {a-b x^4}}{5 a x^5}-\frac {3 b \sqrt {a-b x^4}}{5 a^2 x}-\frac {\left (3 b^2\right ) \int \frac {x^2}{\sqrt {a-b x^4}} \, dx}{5 a^2}\\ &=-\frac {\sqrt {a-b x^4}}{5 a x^5}-\frac {3 b \sqrt {a-b x^4}}{5 a^2 x}+\frac {\left (3 b^{3/2}\right ) \int \frac {1}{\sqrt {a-b x^4}} \, dx}{5 a^{3/2}}-\frac {\left (3 b^{3/2}\right ) \int \frac {1+\frac {\sqrt {b} x^2}{\sqrt {a}}}{\sqrt {a-b x^4}} \, dx}{5 a^{3/2}}\\ &=-\frac {\sqrt {a-b x^4}}{5 a x^5}-\frac {3 b \sqrt {a-b x^4}}{5 a^2 x}+\frac {\left (3 b^{3/2} \sqrt {1-\frac {b x^4}{a}}\right ) \int \frac {1}{\sqrt {1-\frac {b x^4}{a}}} \, dx}{5 a^{3/2} \sqrt {a-b x^4}}-\frac {\left (3 b^{3/2} \sqrt {1-\frac {b x^4}{a}}\right ) \int \frac {1+\frac {\sqrt {b} x^2}{\sqrt {a}}}{\sqrt {1-\frac {b x^4}{a}}} \, dx}{5 a^{3/2} \sqrt {a-b x^4}}\\ &=-\frac {\sqrt {a-b x^4}}{5 a x^5}-\frac {3 b \sqrt {a-b x^4}}{5 a^2 x}+\frac {3 b^{5/4} \sqrt {1-\frac {b x^4}{a}} F\left (\left .\sin ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )\right |-1\right )}{5 a^{5/4} \sqrt {a-b x^4}}-\frac {\left (3 b^{3/2} \sqrt {1-\frac {b x^4}{a}}\right ) \int \frac {\sqrt {1+\frac {\sqrt {b} x^2}{\sqrt {a}}}}{\sqrt {1-\frac {\sqrt {b} x^2}{\sqrt {a}}}} \, dx}{5 a^{3/2} \sqrt {a-b x^4}}\\ &=-\frac {\sqrt {a-b x^4}}{5 a x^5}-\frac {3 b \sqrt {a-b x^4}}{5 a^2 x}-\frac {3 b^{5/4} \sqrt {1-\frac {b x^4}{a}} E\left (\left .\sin ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )\right |-1\right )}{5 a^{5/4} \sqrt {a-b x^4}}+\frac {3 b^{5/4} \sqrt {1-\frac {b x^4}{a}} F\left (\left .\sin ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )\right |-1\right )}{5 a^{5/4} \sqrt {a-b x^4}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 52, normalized size = 0.33 \[ -\frac {\sqrt {1-\frac {b x^4}{a}} \, _2F_1\left (-\frac {5}{4},\frac {1}{2};-\frac {1}{4};\frac {b x^4}{a}\right )}{5 x^5 \sqrt {a-b x^4}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^6*Sqrt[a - b*x^4]),x]

[Out]

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

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fricas [F]  time = 0.86, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {-b x^{4} + a}}{b x^{10} - a x^{6}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-sqrt(-b*x^4 + a)/(b*x^10 - a*x^6), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {-b x^{4} + a} x^{6}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 126, normalized size = 0.80 \[ \frac {3 \sqrt {-\frac {\sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {\sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, \left (-\EllipticE \left (\sqrt {\frac {\sqrt {b}}{\sqrt {a}}}\, x , i\right )+\EllipticF \left (\sqrt {\frac {\sqrt {b}}{\sqrt {a}}}\, x , i\right )\right ) b^{\frac {3}{2}}}{5 \sqrt {\frac {\sqrt {b}}{\sqrt {a}}}\, \sqrt {-b \,x^{4}+a}\, a^{\frac {3}{2}}}-\frac {3 \sqrt {-b \,x^{4}+a}\, b}{5 a^{2} x}-\frac {\sqrt {-b \,x^{4}+a}}{5 a \,x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/5*(-b*x^4+a)^(1/2)/a/x^5-3/5*b*(-b*x^4+a)^(1/2)/a^2/x+3/5/a^(3/2)*b^(3/2)/(1/a^(1/2)*b^(1/2))^(1/2)*(-1/a^(
1/2)*b^(1/2)*x^2+1)^(1/2)*(1/a^(1/2)*b^(1/2)*x^2+1)^(1/2)/(-b*x^4+a)^(1/2)*(EllipticF((1/a^(1/2)*b^(1/2))^(1/2
)*x,I)-EllipticE((1/a^(1/2)*b^(1/2))^(1/2)*x,I))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {-b x^{4} + a} x^{6}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{x^6\,\sqrt {a-b\,x^4}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^6*(a - b*x^4)^(1/2)),x)

[Out]

int(1/(x^6*(a - b*x^4)^(1/2)), x)

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sympy [A]  time = 1.48, size = 39, normalized size = 0.25 \[ - \frac {i \Gamma \left (- \frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {a}{b x^{4}}} \right )}}{4 \sqrt {b} x^{7} \Gamma \left (- \frac {3}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-I*gamma(-7/4)*hyper((1/2, 7/4), (11/4,), a/(b*x**4))/(4*sqrt(b)*x**7*gamma(-3/4))

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