3.887 \(\int \frac{1}{x^4 \sqrt{1-x^4}} \, dx\)

Optimal. Leaf size=27 \[ \frac{1}{3} \text{EllipticF}\left (\sin ^{-1}(x),-1\right )-\frac{\sqrt{1-x^4}}{3 x^3} \]

[Out]

-Sqrt[1 - x^4]/(3*x^3) + EllipticF[ArcSin[x], -1]/3

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Rubi [A]  time = 0.0048177, antiderivative size = 27, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {325, 221} \[ \frac{1}{3} F\left (\left .\sin ^{-1}(x)\right |-1\right )-\frac{\sqrt{1-x^4}}{3 x^3} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^4*Sqrt[1 - x^4]),x]

[Out]

-Sqrt[1 - x^4]/(3*x^3) + EllipticF[ArcSin[x], -1]/3

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 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]

Rubi steps

\begin{align*} \int \frac{1}{x^4 \sqrt{1-x^4}} \, dx &=-\frac{\sqrt{1-x^4}}{3 x^3}+\frac{1}{3} \int \frac{1}{\sqrt{1-x^4}} \, dx\\ &=-\frac{\sqrt{1-x^4}}{3 x^3}+\frac{1}{3} F\left (\left .\sin ^{-1}(x)\right |-1\right )\\ \end{align*}

Mathematica [C]  time = 0.0026633, size = 20, normalized size = 0.74 \[ -\frac{\, _2F_1\left (-\frac{3}{4},\frac{1}{2};\frac{1}{4};x^4\right )}{3 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^4*Sqrt[1 - x^4]),x]

[Out]

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

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Maple [B]  time = 0.011, size = 47, normalized size = 1.7 \begin{align*} -{\frac{1}{3\,{x}^{3}}\sqrt{-{x}^{4}+1}}+{\frac{{\it EllipticF} \left ( x,i \right ) }{3}\sqrt{-{x}^{2}+1}\sqrt{{x}^{2}+1}{\frac{1}{\sqrt{-{x}^{4}+1}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{-x^{4} + 1} x^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/(sqrt(-x^4 + 1)*x^4), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{\sqrt{-x^{4} + 1}}{x^{8} - x^{4}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-sqrt(-x^4 + 1)/(x^8 - x^4), x)

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Sympy [A]  time = 0.961365, size = 34, normalized size = 1.26 \begin{align*} \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 |{x^{4} e^{2 i \pi }} \right )}}{4 x^{3} \Gamma \left (\frac{1}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{-x^{4} + 1} x^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/(sqrt(-x^4 + 1)*x^4), x)