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

Optimal. Leaf size=131 \[ -\frac{4 \sqrt{2-\sqrt{3}} (x+1) \sqrt{\frac{x^2-x+1}{\left (x-\sqrt{3}+1\right )^2}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{x+\sqrt{3}+1}{x-\sqrt{3}+1}\right ),4 \sqrt{3}-7\right )}{5 \sqrt [4]{3} \sqrt{-\frac{x+1}{\left (x-\sqrt{3}+1\right )^2}} \sqrt{-x^3-1}}-\frac{2}{5} \sqrt{-x^3-1} x \]

[Out]

(-2*x*Sqrt[-1 - x^3])/5 - (4*Sqrt[2 - Sqrt[3]]*(1 + x)*Sqrt[(1 - x + x^2)/(1 - Sqrt[3] + x)^2]*EllipticF[ArcSi
n[(1 + Sqrt[3] + x)/(1 - Sqrt[3] + x)], -7 + 4*Sqrt[3]])/(5*3^(1/4)*Sqrt[-((1 + x)/(1 - Sqrt[3] + x)^2)]*Sqrt[
-1 - x^3])

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Rubi [A]  time = 0.0184333, antiderivative size = 131, 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 = {321, 219} \[ -\frac{2}{5} \sqrt{-x^3-1} x-\frac{4 \sqrt{2-\sqrt{3}} (x+1) \sqrt{\frac{x^2-x+1}{\left (x-\sqrt{3}+1\right )^2}} F\left (\sin ^{-1}\left (\frac{x+\sqrt{3}+1}{x-\sqrt{3}+1}\right )|-7+4 \sqrt{3}\right )}{5 \sqrt [4]{3} \sqrt{-\frac{x+1}{\left (x-\sqrt{3}+1\right )^2}} \sqrt{-x^3-1}} \]

Antiderivative was successfully verified.

[In]

Int[x^3/Sqrt[-1 - x^3],x]

[Out]

(-2*x*Sqrt[-1 - x^3])/5 - (4*Sqrt[2 - Sqrt[3]]*(1 + x)*Sqrt[(1 - x + x^2)/(1 - Sqrt[3] + x)^2]*EllipticF[ArcSi
n[(1 + Sqrt[3] + x)/(1 - Sqrt[3] + x)], -7 + 4*Sqrt[3]])/(5*3^(1/4)*Sqrt[-((1 + x)/(1 - Sqrt[3] + x)^2)]*Sqrt[
-1 - x^3])

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && 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{x^3}{\sqrt{-1-x^3}} \, dx &=-\frac{2}{5} x \sqrt{-1-x^3}-\frac{2}{5} \int \frac{1}{\sqrt{-1-x^3}} \, dx\\ &=-\frac{2}{5} x \sqrt{-1-x^3}-\frac{4 \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 )}{5 \sqrt [4]{3} \sqrt{-\frac{1+x}{\left (1-\sqrt{3}+x\right )^2}} \sqrt{-1-x^3}}\\ \end{align*}

Mathematica [C]  time = 0.0077083, size = 47, normalized size = 0.36 \[ \frac{2 \left (-\sqrt{x^3+1} x \, _2F_1\left (\frac{1}{3},\frac{1}{2};\frac{4}{3};-x^3\right )+x^4+x\right )}{5 \sqrt{-x^3-1}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3/Sqrt[-1 - x^3],x]

[Out]

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

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Maple [A]  time = 0.019, size = 120, normalized size = 0.9 \begin{align*} -{\frac{2\,x}{5}\sqrt{-{x}^{3}-1}}+{{\frac{4\,i}{15}}\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(x^3/(-x^3-1)^(1/2),x)

[Out]

-2/5*x*(-x^3-1)^(1/2)+4/15*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{x^{3}}{\sqrt{-x^{3} - 1}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^3/sqrt(-x^3 - 1), 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^{3}}{x^{3} + 1}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.725496, size = 32, normalized size = 0.24 \begin{align*} - \frac{i x^{4} \Gamma \left (\frac{4}{3}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{1}{2}, \frac{4}{3} \\ \frac{7}{3} \end{matrix}\middle |{x^{3} e^{i \pi }} \right )}}{3 \Gamma \left (\frac{7}{3}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-I*x**4*gamma(4/3)*hyper((1/2, 4/3), (7/3,), x**3*exp_polar(I*pi))/(3*gamma(7/3))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^3/sqrt(-x^3 - 1), x)