\(\int \frac {x}{\sqrt {-1-x^3}} \, dx\) [507]

Optimal result

Integrand size = 13, antiderivative size = 239 \[ \int \frac {x}{\sqrt {-1-x^3}} \, dx=-\frac {2 \sqrt {-1-x^3}}{1-\sqrt {3}+x}+\frac {\sqrt [4]{3} \sqrt {2+\sqrt {3}} (1+x) \sqrt {\frac {1-x+x^2}{\left (1-\sqrt {3}+x\right )^2}} E\left (\arcsin \left (\frac {1+\sqrt {3}+x}{1-\sqrt {3}+x}\right )|-7+4 \sqrt {3}\right )}{\sqrt {-\frac {1+x}{\left (1-\sqrt {3}+x\right )^2}} \sqrt {-1-x^3}}-\frac {2 \sqrt {2} (1+x) \sqrt {\frac {1-x+x^2}{\left (1-\sqrt {3}+x\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {1+\sqrt {3}+x}{1-\sqrt {3}+x}\right ),-7+4 \sqrt {3}\right )}{\sqrt [4]{3} \sqrt {-\frac {1+x}{\left (1-\sqrt {3}+x\right )^2}} \sqrt {-1-x^3}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 239, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {310, 225, 1893} \[ \int \frac {x}{\sqrt {-1-x^3}} \, dx=-\frac {2 \sqrt {2} (x+1) \sqrt {\frac {x^2-x+1}{\left (x-\sqrt {3}+1\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {x+\sqrt {3}+1}{x-\sqrt {3}+1}\right ),-7+4 \sqrt {3}\right )}{\sqrt [4]{3} \sqrt {-\frac {x+1}{\left (x-\sqrt {3}+1\right )^2}} \sqrt {-x^3-1}}+\frac {\sqrt [4]{3} \sqrt {2+\sqrt {3}} (x+1) \sqrt {\frac {x^2-x+1}{\left (x-\sqrt {3}+1\right )^2}} E\left (\arcsin \left (\frac {x+\sqrt {3}+1}{x-\sqrt {3}+1}\right )|-7+4 \sqrt {3}\right )}{\sqrt {-\frac {x+1}{\left (x-\sqrt {3}+1\right )^2}} \sqrt {-x^3-1}}-\frac {2 \sqrt {-x^3-1}}{x-\sqrt {3}+1} \]

[In]

Int[x/Sqrt[-1 - x^3],x]

[Out]

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

Rule 225

Int[1/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Simp[2*Sqrt
[2 - Sqrt[3]]*(s + r*x)*(Sqrt[(s^2 - r*s*x + r^2*x^2)/((1 - Sqrt[3])*s + r*x)^2]/(3^(1/4)*r*Sqrt[a + b*x^3]*Sq
rt[(-s)*((s + r*x)/((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]], x]] /; FreeQ[{a, b}, x] && NegQ[a]

Rule 310

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

Rule 1893

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]/(r^2
*Sqrt[a + b*x^3]*Sqrt[(-s)*((s + r*x)/((1 - Sqrt[3])*s + r*x)^2)]))*EllipticE[ArcSin[((1 + Sqrt[3])*s + r*x)/(
(1 - Sqrt[3])*s + r*x)], -7 + 4*Sqrt[3]], x]] /; FreeQ[{a, b, c, d}, x] && NegQ[a] && EqQ[b*c^3 - 2*(5 + 3*Sqr
t[3])*a*d^3, 0]

Rubi steps \begin{align*} \text {integral}& = \left (-1-\sqrt {3}\right ) \int \frac {1}{\sqrt {-1-x^3}} \, dx+\int \frac {1+\sqrt {3}+x}{\sqrt {-1-x^3}} \, dx \\ & = -\frac {2 \sqrt {-1-x^3}}{1-\sqrt {3}+x}+\frac {\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 )}{\sqrt {-\frac {1+x}{\left (1-\sqrt {3}+x\right )^2}} \sqrt {-1-x^3}}-\frac {2 \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 )}{\sqrt [4]{3} \sqrt {-\frac {1+x}{\left (1-\sqrt {3}+x\right )^2}} \sqrt {-1-x^3}} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 10.02 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.18 \[ \int \frac {x}{\sqrt {-1-x^3}} \, dx=\frac {x^2 \sqrt {1+x^3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {5}{3},-x^3\right )}{2 \sqrt {-1-x^3}} \]

[In]

Integrate[x/Sqrt[-1 - x^3],x]

[Out]

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

Maple [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4.

Time = 3.88 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.08

method	result	size
meijerg	\(-\frac {i x^{2} {}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (\frac {1}{2},\frac {2}{3};\frac {5}{3};-x^{3}\right )}{2}\)	\(18\)
default	\(-\frac {2 i \sqrt {3}\, \sqrt {i \left (x -\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}\, \sqrt {\frac {1+x}{\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \sqrt {-i \left (x -\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}\, \left (\left (\frac {3}{2}+\frac {i \sqrt {3}}{2}\right ) E\left (\frac {\sqrt {3}\, \sqrt {i \left (x -\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}}{3}, \sqrt {\frac {i \sqrt {3}}{\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\right )-F\left (\frac {\sqrt {3}\, \sqrt {i \left (x -\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}}{3}, \sqrt {\frac {i \sqrt {3}}{\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\right )\right )}{3 \sqrt {-x^{3}-1}}\)	\(160\)
elliptic	\(-\frac {2 i \sqrt {3}\, \sqrt {i \left (x -\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}\, \sqrt {\frac {1+x}{\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \sqrt {-i \left (x -\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}\, \left (\left (\frac {3}{2}+\frac {i \sqrt {3}}{2}\right ) E\left (\frac {\sqrt {3}\, \sqrt {i \left (x -\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}}{3}, \sqrt {\frac {i \sqrt {3}}{\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\right )-F\left (\frac {\sqrt {3}\, \sqrt {i \left (x -\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}}{3}, \sqrt {\frac {i \sqrt {3}}{\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\right )\right )}{3 \sqrt {-x^{3}-1}}\)	\(160\)

[In]

int(x/(-x^3-1)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-1/2*I*x^2*hypergeom([1/2,2/3],[5/3],-x^3)

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.09 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.04 \[ \int \frac {x}{\sqrt {-1-x^3}} \, dx=2 i \, {\rm weierstrassZeta}\left (0, -4, {\rm weierstrassPInverse}\left (0, -4, x\right )\right ) \]

[In]

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

[Out]

2*I*weierstrassZeta(0, -4, weierstrassPInverse(0, -4, x))

Sympy [A] (verification not implemented)

Time = 0.35 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.13 \[ \int \frac {x}{\sqrt {-1-x^3}} \, dx=- \frac {i x^{2} \Gamma \left (\frac {2}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {2}{3} \\ \frac {5}{3} \end {matrix}\middle | {x^{3} e^{i \pi }} \right )}}{3 \Gamma \left (\frac {5}{3}\right )} \]

[In]

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

[Out]

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

Maxima [F]

\[ \int \frac {x}{\sqrt {-1-x^3}} \, dx=\int { \frac {x}{\sqrt {-x^{3} - 1}} \,d x } \]

[In]

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

[Out]

integrate(x/sqrt(-x^3 - 1), x)

Giac [F]

\[ \int \frac {x}{\sqrt {-1-x^3}} \, dx=\int { \frac {x}{\sqrt {-x^{3} - 1}} \,d x } \]

[In]

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

[Out]

integrate(x/sqrt(-x^3 - 1), x)

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 230, normalized size of antiderivative = 0.96 \[ \int \frac {x}{\sqrt {-1-x^3}} \, dx=-\frac {2\,\left (\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\mathrm {F}\left (\mathrm {asin}\left (\sqrt {\frac {x+1}{\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\right )\middle |-\frac {\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{-\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}\right )-\left (-\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\mathrm {E}\left (\mathrm {asin}\left (\sqrt {\frac {x+1}{\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\right )\middle |-\frac {\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{-\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}\right )\right )\,\left (\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\sqrt {x^3+1}\,\sqrt {\frac {x-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{-\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\,\sqrt {\frac {x+1}{\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\,\sqrt {\frac {\frac {1}{2}-x+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}}{\sqrt {-x^3-1}\,\sqrt {x^3+\left (-\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )-1\right )\,x-\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}} \]

[In]

int(x/(- x^3 - 1)^(1/2),x)

[Out]

-(2*(((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 + 1)/((3^(1/2)*1i)/2 + 3/2))^(1/2)*(((3^(1/2)*1i)/2 - x + 1/2)/((3^(1/2)*1i)/
2 + 3/2))^(1/2))/((- x^3 - 1)^(1/2)*(x^3 - x*(((3^(1/2)*1i)/2 - 1/2)*((3^(1/2)*1i)/2 + 1/2) + 1) - ((3^(1/2)*1
i)/2 - 1/2)*((3^(1/2)*1i)/2 + 1/2))^(1/2))