3.449 \(\int \frac{1}{x^3 \sqrt{1+x^3}} \, dx\)

Optimal. Leaf size=122 \[ -\frac{\sqrt{2+\sqrt{3}} (x+1) \sqrt{\frac{x^2-x+1}{\left (x+\sqrt{3}+1\right )^2}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{x-\sqrt{3}+1}{x+\sqrt{3}+1}\right ),-7-4 \sqrt{3}\right )}{2 \sqrt [4]{3} \sqrt{\frac{x+1}{\left (x+\sqrt{3}+1\right )^2}} \sqrt{x^3+1}}-\frac{\sqrt{x^3+1}}{2 x^2} \]

[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])

________________________________________________________________________________________

Rubi [A]  time = 0.0172496, antiderivative size = 122, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {325, 218} \[ -\frac{\sqrt{x^3+1}}{2 x^2}-\frac{\sqrt{2+\sqrt{3}} (x+1) \sqrt{\frac{x^2-x+1}{\left (x+\sqrt{3}+1\right )^2}} F\left (\sin ^{-1}\left (\frac{x-\sqrt{3}+1}{x+\sqrt{3}+1}\right )|-7-4 \sqrt{3}\right )}{2 \sqrt [4]{3} \sqrt{\frac{x+1}{\left (x+\sqrt{3}+1\right )^2}} \sqrt{x^3+1}} \]

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 218

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 + Sqr
t[3])*s + r*x)^2]), x]] /; FreeQ[{a, b}, x] && PosQ[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 = 0.0023798, size = 22, normalized size = 0.18 \[ -\frac{\, _2F_1\left (-\frac{2}{3},\frac{1}{2};\frac{1}{3};-x^3\right )}{2 x^2} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^3*Sqrt[1 + x^3]),x]

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.017, size = 129, normalized size = 1.1 \begin{align*} -{\frac{1}{2\,{x}^{2}}\sqrt{{x}^{3}+1}}-{\frac{{\frac{3}{2}}-{\frac{i}{2}}\sqrt{3}}{2}\sqrt{{\frac{1+x}{{\frac{3}{2}}-{\frac{i}{2}}\sqrt{3}}}}\sqrt{{\frac{1}{-{\frac{3}{2}}-{\frac{i}{2}}\sqrt{3}} \left ( x-{\frac{1}{2}}-{\frac{i}{2}}\sqrt{3} \right ) }}\sqrt{{\frac{1}{-{\frac{3}{2}}+{\frac{i}{2}}\sqrt{3}} \left ( x-{\frac{1}{2}}+{\frac{i}{2}}\sqrt{3} \right ) }}{\it EllipticF} \left ( \sqrt{{\frac{1+x}{{\frac{3}{2}}-{\frac{i}{2}}\sqrt{3}}}},\sqrt{{\frac{-{\frac{3}{2}}+{\frac{i}{2}}\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^2-1/2*(3/2-1/2*I*3^(1/2))*((1+x)/(3/2-1/2*I*3^(1/2)))^(1/2)*((x-1/2-1/2*I*3^(1/2))/(-3/2-
1/2*I*3^(1/2)))^(1/2)*((x-1/2+1/2*I*3^(1/2))/(-3/2+1/2*I*3^(1/2)))^(1/2)/(x^3+1)^(1/2)*EllipticF(((1+x)/(3/2-1
/2*I*3^(1/2)))^(1/2),((-3/2+1/2*I*3^(1/2))/(-3/2-1/2*I*3^(1/2)))^(1/2))

________________________________________________________________________________________

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)

________________________________________________________________________________________

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)

________________________________________________________________________________________

Sympy [A]  time = 0.890315, size = 32, normalized size = 0.26 \begin{align*} \frac{\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]

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

________________________________________________________________________________________

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)