3.503 \(\int \frac{1}{x^3 \sqrt{-1-x^3}} \, dx\)

Optimal. Leaf size=133 \[ \frac{\sqrt{-x^3-1}}{2 x^2}-\frac{\sqrt{2-\sqrt{3}} (x+1) \sqrt{\frac{x^2-x+1}{\left (x-\sqrt{3}+1\right )^2}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{x+\sqrt{3}+1}{x-\sqrt{3}+1}\right ),4 \sqrt{3}-7\right )}{2 \sqrt [4]{3} \sqrt{-\frac{x+1}{\left (x-\sqrt{3}+1\right )^2}} \sqrt{-x^3-1}} \]

[Out]

Sqrt[-1 - x^3]/(2*x^2) - (Sqrt[2 - Sqrt[3]]*(1 + x)*Sqrt[(1 - x + x^2)/(1 - Sqrt[3] + x)^2]*EllipticF[ArcSin[(
1 + Sqrt[3] + x)/(1 - Sqrt[3] + x)], -7 + 4*Sqrt[3]])/(2*3^(1/4)*Sqrt[-((1 + x)/(1 - Sqrt[3] + x)^2)]*Sqrt[-1
- x^3])

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Rubi [A]  time = 0.0183783, antiderivative size = 133, 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, 219} \[ \frac{\sqrt{-x^3-1}}{2 x^2}-\frac{\sqrt{2-\sqrt{3}} (x+1) \sqrt{\frac{x^2-x+1}{\left (x-\sqrt{3}+1\right )^2}} F\left (\sin ^{-1}\left (\frac{x+\sqrt{3}+1}{x-\sqrt{3}+1}\right )|-7+4 \sqrt{3}\right )}{2 \sqrt [4]{3} \sqrt{-\frac{x+1}{\left (x-\sqrt{3}+1\right )^2}} \sqrt{-x^3-1}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^3*Sqrt[-1 - x^3]),x]

[Out]

Sqrt[-1 - x^3]/(2*x^2) - (Sqrt[2 - Sqrt[3]]*(1 + x)*Sqrt[(1 - x + x^2)/(1 - Sqrt[3] + x)^2]*EllipticF[ArcSin[(
1 + Sqrt[3] + x)/(1 - Sqrt[3] + x)], -7 + 4*Sqrt[3]])/(2*3^(1/4)*Sqrt[-((1 + x)/(1 - Sqrt[3] + x)^2)]*Sqrt[-1
- x^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 219

Int[1/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Simp[(2*Sqr
t[2 - Sqrt[3]]*(s + r*x)*Sqrt[(s^2 - r*s*x + r^2*x^2)/((1 - Sqrt[3])*s + r*x)^2]*EllipticF[ArcSin[((1 + Sqrt[3
])*s + r*x)/((1 - Sqrt[3])*s + r*x)], -7 + 4*Sqrt[3]])/(3^(1/4)*r*Sqrt[a + b*x^3]*Sqrt[-((s*(s + r*x))/((1 - S
qrt[3])*s + r*x)^2)]), x]] /; FreeQ[{a, b}, x] && NegQ[a]

Rubi steps

\begin{align*} \int \frac{1}{x^3 \sqrt{-1-x^3}} \, dx &=\frac{\sqrt{-1-x^3}}{2 x^2}-\frac{1}{4} \int \frac{1}{\sqrt{-1-x^3}} \, dx\\ &=\frac{\sqrt{-1-x^3}}{2 x^2}-\frac{\sqrt{2-\sqrt{3}} (1+x) \sqrt{\frac{1-x+x^2}{\left (1-\sqrt{3}+x\right )^2}} F\left (\sin ^{-1}\left (\frac{1+\sqrt{3}+x}{1-\sqrt{3}+x}\right )|-7+4 \sqrt{3}\right )}{2 \sqrt [4]{3} \sqrt{-\frac{1+x}{\left (1-\sqrt{3}+x\right )^2}} \sqrt{-1-x^3}}\\ \end{align*}

Mathematica [C]  time = 0.0052278, size = 42, normalized size = 0.32 \[ -\frac{\sqrt{x^3+1} \, _2F_1\left (-\frac{2}{3},\frac{1}{2};\frac{1}{3};-x^3\right )}{2 x^2 \sqrt{-x^3-1}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^3*Sqrt[-1 - x^3]),x]

[Out]

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

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Maple [A]  time = 0.023, size = 122, normalized size = 0.9 \begin{align*}{\frac{1}{2\,{x}^{2}}\sqrt{-{x}^{3}-1}}+{{\frac{i}{6}}\sqrt{3}\sqrt{i \left ( x-{\frac{1}{2}}-{\frac{i}{2}}\sqrt{3} \right ) \sqrt{3}}\sqrt{{\frac{1+x}{{\frac{3}{2}}+{\frac{i}{2}}\sqrt{3}}}}\sqrt{-i \left ( x-{\frac{1}{2}}+{\frac{i}{2}}\sqrt{3} \right ) \sqrt{3}}{\it EllipticF} \left ({\frac{\sqrt{3}}{3}\sqrt{i \left ( x-{\frac{1}{2}}-{\frac{i}{2}}\sqrt{3} \right ) \sqrt{3}}},\sqrt{{\frac{i\sqrt{3}}{{\frac{3}{2}}+{\frac{i}{2}}\sqrt{3}}}} \right ){\frac{1}{\sqrt{-{x}^{3}-1}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^3/(-x^3-1)^(1/2),x)

[Out]

1/2*(-x^3-1)^(1/2)/x^2+1/6*I*3^(1/2)*(I*(x-1/2-1/2*I*3^(1/2))*3^(1/2))^(1/2)*((1+x)/(3/2+1/2*I*3^(1/2)))^(1/2)
*(-I*(x-1/2+1/2*I*3^(1/2))*3^(1/2))^(1/2)/(-x^3-1)^(1/2)*EllipticF(1/3*3^(1/2)*(I*(x-1/2-1/2*I*3^(1/2))*3^(1/2
))^(1/2),(I*3^(1/2)/(3/2+1/2*I*3^(1/2)))^(1/2))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{-x^{3} - 1} x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(-x^3-1)^(1/2),x, algorithm="maxima")

[Out]

integrate(1/(sqrt(-x^3 - 1)*x^3), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{\sqrt{-x^{3} - 1}}{x^{6} + x^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(-x^3-1)^(1/2),x, algorithm="fricas")

[Out]

integral(-sqrt(-x^3 - 1)/(x^6 + x^3), x)

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Sympy [A]  time = 0.888085, size = 36, normalized size = 0.27 \begin{align*} - \frac{i \Gamma \left (- \frac{2}{3}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{2}{3}, \frac{1}{2} \\ \frac{1}{3} \end{matrix}\middle |{x^{3} e^{i \pi }} \right )}}{3 x^{2} \Gamma \left (\frac{1}{3}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**3/(-x**3-1)**(1/2),x)

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{-x^{3} - 1} x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(-x^3-1)^(1/2),x, algorithm="giac")

[Out]

integrate(1/(sqrt(-x^3 - 1)*x^3), x)