3.464 \(\int \frac {x^6}{\sqrt {1-x^3}} \, dx\)

Optimal. Leaf size=152 \[ -\frac {16}{55} \sqrt {1-x^3} x-\frac {2}{11} \sqrt {1-x^3} x^4-\frac {32 \sqrt {2+\sqrt {3}} (1-x) \sqrt {\frac {x^2+x+1}{\left (-x+\sqrt {3}+1\right )^2}} F\left (\sin ^{-1}\left (\frac {-x-\sqrt {3}+1}{-x+\sqrt {3}+1}\right )|-7-4 \sqrt {3}\right )}{55 \sqrt [4]{3} \sqrt {\frac {1-x}{\left (-x+\sqrt {3}+1\right )^2}} \sqrt {1-x^3}} \]

[Out]

-16/55*x*(-x^3+1)^(1/2)-2/11*x^4*(-x^3+1)^(1/2)-32/165*(1-x)*EllipticF((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)*3^(3/4)/(-x^3+1)^(1/2)/((1-x)/(1-x+3^(1/2))^2)
^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.04, antiderivative size = 152, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {321, 218} \[ -\frac {2}{11} \sqrt {1-x^3} x^4-\frac {16}{55} \sqrt {1-x^3} x-\frac {32 \sqrt {2+\sqrt {3}} (1-x) \sqrt {\frac {x^2+x+1}{\left (-x+\sqrt {3}+1\right )^2}} F\left (\sin ^{-1}\left (\frac {-x-\sqrt {3}+1}{-x+\sqrt {3}+1}\right )|-7-4 \sqrt {3}\right )}{55 \sqrt [4]{3} \sqrt {\frac {1-x}{\left (-x+\sqrt {3}+1\right )^2}} \sqrt {1-x^3}} \]

Antiderivative was successfully verified.

[In]

Int[x^6/Sqrt[1 - x^3],x]

[Out]

(-16*x*Sqrt[1 - x^3])/55 - (2*x^4*Sqrt[1 - x^3])/11 - (32*Sqrt[2 + Sqrt[3]]*(1 - x)*Sqrt[(1 + x + x^2)/(1 + Sq
rt[3] - x)^2]*EllipticF[ArcSin[(1 - Sqrt[3] - x)/(1 + Sqrt[3] - x)], -7 - 4*Sqrt[3]])/(55*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 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]

Rubi steps

\begin {align*} \int \frac {x^6}{\sqrt {1-x^3}} \, dx &=-\frac {2}{11} x^4 \sqrt {1-x^3}+\frac {8}{11} \int \frac {x^3}{\sqrt {1-x^3}} \, dx\\ &=-\frac {16}{55} x \sqrt {1-x^3}-\frac {2}{11} x^4 \sqrt {1-x^3}+\frac {16}{55} \int \frac {1}{\sqrt {1-x^3}} \, dx\\ &=-\frac {16}{55} x \sqrt {1-x^3}-\frac {2}{11} x^4 \sqrt {1-x^3}-\frac {32 \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 )}{55 \sqrt [4]{3} \sqrt {\frac {1-x}{\left (1+\sqrt {3}-x\right )^2}} \sqrt {1-x^3}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 0.03, size = 40, normalized size = 0.26 \[ -\frac {2}{55} x \left (\sqrt {1-x^3} \left (5 x^3+8\right )-8 \, _2F_1\left (\frac {1}{3},\frac {1}{2};\frac {4}{3};x^3\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^6/Sqrt[1 - x^3],x]

[Out]

(-2*x*(Sqrt[1 - x^3]*(8 + 5*x^3) - 8*Hypergeometric2F1[1/3, 1/2, 4/3, x^3]))/55

________________________________________________________________________________________

fricas [F]  time = 0.82, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {-x^{3} + 1} x^{6}}{x^{3} - 1}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6/(-x^3+1)^(1/2),x, algorithm="fricas")

[Out]

integral(-sqrt(-x^3 + 1)*x^6/(x^3 - 1), x)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{6}}{\sqrt {-x^{3} + 1}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6/(-x^3+1)^(1/2),x, algorithm="giac")

[Out]

integrate(x^6/sqrt(-x^3 + 1), x)

________________________________________________________________________________________

maple [A]  time = 0.02, size = 134, normalized size = 0.88 \[ -\frac {2 \sqrt {-x^{3}+1}\, x^{4}}{11}-\frac {16 \sqrt {-x^{3}+1}\, x}{55}-\frac {32 i \sqrt {3}\, \sqrt {i \left (x +\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}\, \sqrt {\frac {x -1}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \sqrt {-i \left (x +\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}\, \EllipticF \left (\frac {\sqrt {3}\, \sqrt {i \left (x +\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}}{3}, \sqrt {\frac {i \sqrt {3}}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\right )}{165 \sqrt {-x^{3}+1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^6/(-x^3+1)^(1/2),x)

[Out]

-2/11*x^4*(-x^3+1)^(1/2)-16/55*x*(-x^3+1)^(1/2)-32/165*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)*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))

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{6}}{\sqrt {-x^{3} + 1}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6/(-x^3+1)^(1/2),x, algorithm="maxima")

[Out]

integrate(x^6/sqrt(-x^3 + 1), x)

________________________________________________________________________________________

mupad [B]  time = 0.04, size = 199, normalized size = 1.31 \[ -\frac {2\,x^4\,\sqrt {1-x^3}}{11}-\frac {16\,x\,\sqrt {1-x^3}}{55}-\frac {32\,\left (\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\sqrt {x^3-1}\,\sqrt {-\frac {x+\frac {1}{2}-\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{-\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\,\sqrt {\frac {x+\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\,\sqrt {-\frac {x-1}{\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\,\mathrm {F}\left (\mathrm {asin}\left (\sqrt {-\frac {x-1}{\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\right )\middle |-\frac {\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{-\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}\right )}{55\,\sqrt {1-x^3}\,\sqrt {x^3+\left (-\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )-1\right )\,x+\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^6/(1 - x^3)^(1/2),x)

[Out]

- (2*x^4*(1 - x^3)^(1/2))/11 - (16*x*(1 - x^3)^(1/2))/55 - (32*((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)*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)))/(55*(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))

________________________________________________________________________________________

sympy [A]  time = 1.69, size = 31, normalized size = 0.20 \[ \frac {x^{7} \Gamma \left (\frac {7}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {7}{3} \\ \frac {10}{3} \end {matrix}\middle | {x^{3} e^{2 i \pi }} \right )}}{3 \Gamma \left (\frac {10}{3}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**6/(-x**3+1)**(1/2),x)

[Out]

x**7*gamma(7/3)*hyper((1/2, 7/3), (10/3,), x**3*exp_polar(2*I*pi))/(3*gamma(10/3))

________________________________________________________________________________________