3.470 \(\int \frac{x^4}{\sqrt{1-x^3}} \, dx\)

Optimal. Leaf size=276 \[ \frac{8 \sqrt{2} (1-x) \sqrt{\frac{x^2+x+1}{\left (-x+\sqrt{3}+1\right )^2}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{-x-\sqrt{3}+1}{-x+\sqrt{3}+1}\right ),-7-4 \sqrt{3}\right )}{7 \sqrt [4]{3} \sqrt{\frac{1-x}{\left (-x+\sqrt{3}+1\right )^2}} \sqrt{1-x^3}}-\frac{2}{7} \sqrt{1-x^3} x^2+\frac{8 \sqrt{1-x^3}}{7 \left (-x+\sqrt{3}+1\right )}-\frac{4 \sqrt [4]{3} \sqrt{2-\sqrt{3}} (1-x) \sqrt{\frac{x^2+x+1}{\left (-x+\sqrt{3}+1\right )^2}} E\left (\sin ^{-1}\left (\frac{-x-\sqrt{3}+1}{-x+\sqrt{3}+1}\right )|-7-4 \sqrt{3}\right )}{7 \sqrt{\frac{1-x}{\left (-x+\sqrt{3}+1\right )^2}} \sqrt{1-x^3}} \]

[Out]

(8*Sqrt[1 - x^3])/(7*(1 + Sqrt[3] - x)) - (2*x^2*Sqrt[1 - x^3])/7 - (4*3^(1/4)*Sqrt[2 - Sqrt[3]]*(1 - x)*Sqrt[
(1 + x + x^2)/(1 + Sqrt[3] - x)^2]*EllipticE[ArcSin[(1 - Sqrt[3] - x)/(1 + Sqrt[3] - x)], -7 - 4*Sqrt[3]])/(7*
Sqrt[(1 - x)/(1 + Sqrt[3] - x)^2]*Sqrt[1 - x^3]) + (8*Sqrt[2]*(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]])/(7*3^(1/4)*Sqrt[(1 - x)/(1 + Sqrt[3] -
x)^2]*Sqrt[1 - x^3])

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Rubi [A]  time = 0.0608835, antiderivative size = 276, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.267, Rules used = {321, 303, 218, 1877} \[ -\frac{2}{7} \sqrt{1-x^3} x^2+\frac{8 \sqrt{1-x^3}}{7 \left (-x+\sqrt{3}+1\right )}+\frac{8 \sqrt{2} (1-x) \sqrt{\frac{x^2+x+1}{\left (-x+\sqrt{3}+1\right )^2}} F\left (\sin ^{-1}\left (\frac{-x-\sqrt{3}+1}{-x+\sqrt{3}+1}\right )|-7-4 \sqrt{3}\right )}{7 \sqrt [4]{3} \sqrt{\frac{1-x}{\left (-x+\sqrt{3}+1\right )^2}} \sqrt{1-x^3}}-\frac{4 \sqrt [4]{3} \sqrt{2-\sqrt{3}} (1-x) \sqrt{\frac{x^2+x+1}{\left (-x+\sqrt{3}+1\right )^2}} E\left (\sin ^{-1}\left (\frac{-x-\sqrt{3}+1}{-x+\sqrt{3}+1}\right )|-7-4 \sqrt{3}\right )}{7 \sqrt{\frac{1-x}{\left (-x+\sqrt{3}+1\right )^2}} \sqrt{1-x^3}} \]

Antiderivative was successfully verified.

[In]

Int[x^4/Sqrt[1 - x^3],x]

[Out]

(8*Sqrt[1 - x^3])/(7*(1 + Sqrt[3] - x)) - (2*x^2*Sqrt[1 - x^3])/7 - (4*3^(1/4)*Sqrt[2 - Sqrt[3]]*(1 - x)*Sqrt[
(1 + x + x^2)/(1 + Sqrt[3] - x)^2]*EllipticE[ArcSin[(1 - Sqrt[3] - x)/(1 + Sqrt[3] - x)], -7 - 4*Sqrt[3]])/(7*
Sqrt[(1 - x)/(1 + Sqrt[3] - x)^2]*Sqrt[1 - x^3]) + (8*Sqrt[2]*(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]])/(7*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 303

Int[(x_)/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Dist[(Sq
rt[2]*s)/(Sqrt[2 + Sqrt[3]]*r), Int[1/Sqrt[a + b*x^3], x], x] + Dist[1/r, Int[((1 - Sqrt[3])*s + r*x)/Sqrt[a +
b*x^3], x], x]] /; FreeQ[{a, b}, x] && PosQ[a]

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]

Rule 1877

Int[((c_) + (d_.)*(x_))/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Simplify[((1 - Sqrt[3])*d)/c]]
, s = Denom[Simplify[((1 - Sqrt[3])*d)/c]]}, Simp[(2*d*s^3*Sqrt[a + b*x^3])/(a*r^2*((1 + Sqrt[3])*s + r*x)), x
] - Simp[(3^(1/4)*Sqrt[2 - Sqrt[3]]*d*s*(s + r*x)*Sqrt[(s^2 - r*s*x + r^2*x^2)/((1 + Sqrt[3])*s + r*x)^2]*Elli
pticE[ArcSin[((1 - Sqrt[3])*s + r*x)/((1 + Sqrt[3])*s + r*x)], -7 - 4*Sqrt[3]])/(r^2*Sqrt[a + b*x^3]*Sqrt[(s*(
s + r*x))/((1 + Sqrt[3])*s + r*x)^2]), x]] /; FreeQ[{a, b, c, d}, x] && PosQ[a] && EqQ[b*c^3 - 2*(5 - 3*Sqrt[3
])*a*d^3, 0]

Rubi steps

\begin{align*} \int \frac{x^4}{\sqrt{1-x^3}} \, dx &=-\frac{2}{7} x^2 \sqrt{1-x^3}+\frac{4}{7} \int \frac{x}{\sqrt{1-x^3}} \, dx\\ &=-\frac{2}{7} x^2 \sqrt{1-x^3}-\frac{4}{7} \int \frac{1-\sqrt{3}-x}{\sqrt{1-x^3}} \, dx-\frac{1}{7} \left (4 \sqrt{2 \left (2-\sqrt{3}\right )}\right ) \int \frac{1}{\sqrt{1-x^3}} \, dx\\ &=\frac{8 \sqrt{1-x^3}}{7 \left (1+\sqrt{3}-x\right )}-\frac{2}{7} x^2 \sqrt{1-x^3}-\frac{4 \sqrt [4]{3} \sqrt{2-\sqrt{3}} (1-x) \sqrt{\frac{1+x+x^2}{\left (1+\sqrt{3}-x\right )^2}} E\left (\sin ^{-1}\left (\frac{1-\sqrt{3}-x}{1+\sqrt{3}-x}\right )|-7-4 \sqrt{3}\right )}{7 \sqrt{\frac{1-x}{\left (1+\sqrt{3}-x\right )^2}} \sqrt{1-x^3}}+\frac{8 \sqrt{2} (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 )}{7 \sqrt [4]{3} \sqrt{\frac{1-x}{\left (1+\sqrt{3}-x\right )^2}} \sqrt{1-x^3}}\\ \end{align*}

Mathematica [C]  time = 0.0058242, size = 34, normalized size = 0.12 \[ \frac{2}{7} x^2 \left (\, _2F_1\left (\frac{1}{2},\frac{2}{3};\frac{5}{3};x^3\right )-\sqrt{1-x^3}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^4/Sqrt[1 - x^3],x]

[Out]

(2*x^2*(-Sqrt[1 - x^3] + Hypergeometric2F1[1/2, 2/3, 5/3, x^3]))/7

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Maple [A]  time = 0.02, size = 173, normalized size = 0.6 \begin{align*} -{\frac{2\,{x}^{2}}{7}\sqrt{-{x}^{3}+1}}-{{\frac{8\,i}{21}}\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}} \left ( \left ( -{\frac{3}{2}}+{\frac{i}{2}}\sqrt{3} \right ){\it EllipticE} \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 ) +{\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 ) \right ){\frac{1}{\sqrt{-{x}^{3}+1}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/(-x^3+1)^(1/2),x)

[Out]

-2/7*x^2*(-x^3+1)^(1/2)-8/21*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)*((-3/2+1/2*I*3^(1/2))*EllipticE(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))+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^{4}}{\sqrt{-x^{3} + 1}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(-x^3+1)^(1/2),x, algorithm="maxima")

[Out]

integrate(x^4/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^{4}}{x^{3} - 1}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(-x^3+1)^(1/2),x, algorithm="fricas")

[Out]

integral(-sqrt(-x^3 + 1)*x^4/(x^3 - 1), x)

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Sympy [A]  time = 0.814475, size = 31, normalized size = 0.11 \begin{align*} \frac{x^{5} \Gamma \left (\frac{5}{3}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{1}{2}, \frac{5}{3} \\ \frac{8}{3} \end{matrix}\middle |{x^{3} e^{2 i \pi }} \right )}}{3 \Gamma \left (\frac{8}{3}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4/(-x**3+1)**(1/2),x)

[Out]

x**5*gamma(5/3)*hyper((1/2, 5/3), (8/3,), x**3*exp_polar(2*I*pi))/(3*gamma(8/3))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{4}}{\sqrt{-x^{3} + 1}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(-x^3+1)^(1/2),x, algorithm="giac")

[Out]

integrate(x^4/sqrt(-x^3 + 1), x)