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

Optimal. Leaf size=79 \[ \frac {b^{3/4} \sqrt {1-\frac {b x^4}{a}} F\left (\left .\sin ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )\right |-1\right )}{3 a^{3/4} \sqrt {a-b x^4}}-\frac {\sqrt {a-b x^4}}{3 a x^3} \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 79, 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, 224, 221} \[ \frac {b^{3/4} \sqrt {1-\frac {b x^4}{a}} F\left (\left .\sin ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )\right |-1\right )}{3 a^{3/4} \sqrt {a-b x^4}}-\frac {\sqrt {a-b x^4}}{3 a x^3} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^4*Sqrt[a - b*x^4]),x]

[Out]

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

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Mathematica [C]  time = 0.03, size = 52, normalized size = 0.66 \[ -\frac {\sqrt {1-\frac {b x^4}{a}} \, _2F_1\left (-\frac {3}{4},\frac {1}{2};\frac {1}{4};\frac {b x^4}{a}\right )}{3 x^3 \sqrt {a-b x^4}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^4*Sqrt[a - b*x^4]),x]

[Out]

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

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fricas [F]  time = 0.66, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {-b x^{4} + a}}{b x^{8} - a x^{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-sqrt(-b*x^4 + a)/(b*x^8 - a*x^4), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 88, normalized size = 1.11 \[ \frac {\sqrt {-\frac {\sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {\sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, b \EllipticF \left (\sqrt {\frac {\sqrt {b}}{\sqrt {a}}}\, x , i\right )}{3 \sqrt {\frac {\sqrt {b}}{\sqrt {a}}}\, \sqrt {-b \,x^{4}+a}\, a}-\frac {\sqrt {-b \,x^{4}+a}}{3 a \,x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^4*(a - b*x^4)^(1/2)),x)

[Out]

int(1/(x^4*(a - b*x^4)^(1/2)), x)

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sympy [A]  time = 1.98, size = 42, normalized size = 0.53 \[ \frac {\Gamma \left (- \frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, \frac {1}{2} \\ \frac {1}{4} \end {matrix}\middle | {\frac {b x^{4} e^{2 i \pi }}{a}} \right )}}{4 \sqrt {a} x^{3} \Gamma \left (\frac {1}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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