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

Optimal. Leaf size=120 \[ \frac{2}{5} x \sqrt{x^3+1}-\frac{4 \sqrt{2+\sqrt{3}} (x+1) \sqrt{\frac{x^2-x+1}{\left (x+\sqrt{3}+1\right )^2}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{x-\sqrt{3}+1}{x+\sqrt{3}+1}\right ),-7-4 \sqrt{3}\right )}{5 \sqrt [4]{3} \sqrt{\frac{x+1}{\left (x+\sqrt{3}+1\right )^2}} \sqrt{x^3+1}} \]

[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[ArcSin[
(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 = 0.0169843, antiderivative size = 120, 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 = {321, 218} \[ \frac{2}{5} x \sqrt{x^3+1}-\frac{4 \sqrt{2+\sqrt{3}} (x+1) \sqrt{\frac{x^2-x+1}{\left (x+\sqrt{3}+1\right )^2}} F\left (\sin ^{-1}\left (\frac{x-\sqrt{3}+1}{x+\sqrt{3}+1}\right )|-7-4 \sqrt{3}\right )}{5 \sqrt [4]{3} \sqrt{\frac{x+1}{\left (x+\sqrt{3}+1\right )^2}} \sqrt{x^3+1}} \]

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[ArcSin[
(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 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{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 = 0.0058072, size = 32, normalized size = 0.27 \[ \frac{2}{5} x \left (\sqrt{x^3+1}-\, _2F_1\left (\frac{1}{3},\frac{1}{2};\frac{4}{3};-x^3\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^3/Sqrt[1 + x^3],x]

[Out]

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

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Maple [A]  time = 0.017, size = 127, normalized size = 1.1 \begin{align*}{\frac{2\,x}{5}\sqrt{{x}^{3}+1}}-{\frac{6-2\,i\sqrt{3}}{5}\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(x^3/(x^3+1)^(1/2),x)

[Out]

2/5*x*(x^3+1)^(1/2)-4/5*(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))

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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{x^{3}}{\sqrt{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(x^3/sqrt(x^3 + 1), x)

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Sympy [A]  time = 0.894425, size = 29, normalized size = 0.24 \begin{align*} \frac{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]

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)