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

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

[Out]

-2/7*x^2*(-x^3+1)^(1/2)+8/7*(-x^3+1)^(1/2)/(1-x+3^(1/2))+8/21*(1-x)*EllipticF((1-x-3^(1/2))/(1-x+3^(1/2)),I*3^
(1/2)+2*I)*2^(1/2)*((x^2+x+1)/(1-x+3^(1/2))^2)^(1/2)*3^(3/4)/(-x^3+1)^(1/2)/((1-x)/(1-x+3^(1/2))^2)^(1/2)-4/7*
3^(1/4)*(1-x)*EllipticE((1-x-3^(1/2))/(1-x+3^(1/2)),I*3^(1/2)+2*I)*(1/2*6^(1/2)-1/2*2^(1/2))*((x^2+x+1)/(1-x+3
^(1/2))^2)^(1/2)/(-x^3+1)^(1/2)/((1-x)/(1-x+3^(1/2))^2)^(1/2)

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Rubi [A]  time = 0.06, antiderivative size = 276, normalized size of antiderivative = 1.00, 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 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 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 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 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*}

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Mathematica [C]  time = 0.01, 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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fricas [F]  time = 0.90, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {-x^{3} + 1} x^{4}}{x^{3} - 1}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{4}}{\sqrt {-x^{3} + 1}}\,{d x} \]

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)

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maple [A]  time = 0.02, size = 173, normalized size = 0.63 \[ -\frac {2 \sqrt {-x^{3}+1}\, x^{2}}{7}-\frac {8 i \sqrt {3}\, \sqrt {i \left (x +\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}\, \sqrt {\frac {x -1}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \sqrt {-i \left (x +\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}\, \left (\left (-\frac {3}{2}+\frac {i \sqrt {3}}{2}\right ) \EllipticE \left (\frac {\sqrt {3}\, \sqrt {i \left (x +\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}}{3}, \sqrt {\frac {i \sqrt {3}}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\right )+\EllipticF \left (\frac {\sqrt {3}\, \sqrt {i \left (x +\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}}{3}, \sqrt {\frac {i \sqrt {3}}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\right )\right )}{21 \sqrt {-x^{3}+1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/7*(-x^3+1)^(1/2)*x^2-8/21*I*3^(1/2)*(I*(x+1/2-1/2*I*3^(1/2))*3^(1/2))^(1/2)*((x-1)/(-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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{4}}{\sqrt {-x^{3} + 1}}\,{d x} \]

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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mupad [B]  time = 0.06, size = 246, normalized size = 0.89 \[ -\frac {2\,x^2\,\sqrt {1-x^3}}{7}-\frac {8\,\left (\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\mathrm {F}\left (\mathrm {asin}\left (\sqrt {-\frac {x-1}{\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\right )\middle |-\frac {\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{-\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}\right )-\left (-\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\mathrm {E}\left (\mathrm {asin}\left (\sqrt {-\frac {x-1}{\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\right )\middle |-\frac {\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{-\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}\right )\right )\,\left (\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\sqrt {x^3-1}\,\sqrt {-\frac {x+\frac {1}{2}-\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{-\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\,\sqrt {\frac {x+\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\,\sqrt {-\frac {x-1}{\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}}{7\,\sqrt {1-x^3}\,\sqrt {x^3+\left (-\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )-1\right )\,x+\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/(1 - x^3)^(1/2),x)

[Out]

- (2*x^2*(1 - x^3)^(1/2))/7 - (8*(((3^(1/2)*1i)/2 - 1/2)*ellipticF(asin((-(x - 1)/((3^(1/2)*1i)/2 + 3/2))^(1/2
)), -((3^(1/2)*1i)/2 + 3/2)/((3^(1/2)*1i)/2 - 3/2)) - ((3^(1/2)*1i)/2 - 3/2)*ellipticE(asin((-(x - 1)/((3^(1/2
)*1i)/2 + 3/2))^(1/2)), -((3^(1/2)*1i)/2 + 3/2)/((3^(1/2)*1i)/2 - 3/2)))*((3^(1/2)*1i)/2 + 3/2)*(x^3 - 1)^(1/2
)*(-(x - (3^(1/2)*1i)/2 + 1/2)/((3^(1/2)*1i)/2 - 3/2))^(1/2)*((x + (3^(1/2)*1i)/2 + 1/2)/((3^(1/2)*1i)/2 + 3/2
))^(1/2)*(-(x - 1)/((3^(1/2)*1i)/2 + 3/2))^(1/2))/(7*(1 - x^3)^(1/2)*(((3^(1/2)*1i)/2 - 1/2)*((3^(1/2)*1i)/2 +
1/2) - x*(((3^(1/2)*1i)/2 - 1/2)*((3^(1/2)*1i)/2 + 1/2) + 1) + x^3)^(1/2))

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sympy [A]  time = 1.02, size = 31, normalized size = 0.11 \[ \frac {x^{5} \Gamma \left (\frac {5}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {5}{3} \\ \frac {8}{3} \end {matrix}\middle | {x^{3} e^{2 i \pi }} \right )}}{3 \Gamma \left (\frac {8}{3}\right )} \]

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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