\(\int \frac {x}{\sqrt {-1+x^3}} \, dx\) [283]

Optimal result
Mathematica [C] (verified)
Rubi [A] (warning: unable to verify)
Maple [C] (warning: unable to verify)
Fricas [A] (verification not implemented)
Sympy [A] (verification not implemented)
Maxima [F]
Giac [F]
Mupad [B] (verification not implemented)
Reduce [F]

Optimal result

Integrand size = 11, antiderivative size = 255 \[ \int \frac {x}{\sqrt {-1+x^3}} \, dx=-\frac {2 \sqrt {-1+x^3}}{1-\sqrt {3}-x}+\frac {\sqrt [4]{3} \sqrt {2+\sqrt {3}} (1-x) \sqrt {\frac {1+x+x^2}{\left (1-\sqrt {3}-x\right )^2}} E\left (\arcsin \left (\frac {1+\sqrt {3}-x}{1-\sqrt {3}-x}\right )|-7+4 \sqrt {3}\right )}{\sqrt {-\frac {1-x}{\left (1-\sqrt {3}-x\right )^2}} \sqrt {-1+x^3}}-\frac {2 \sqrt {2} (1-x) \sqrt {\frac {1+x+x^2}{\left (1-\sqrt {3}-x\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {1+\sqrt {3}-x}{1-\sqrt {3}-x}\right ),-7+4 \sqrt {3}\right )}{\sqrt [4]{3} \sqrt {-\frac {1-x}{\left (1-\sqrt {3}-x\right )^2}} \sqrt {-1+x^3}} \] Output:

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

Mathematica [C] (verified)

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

Time = 10.01 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.16 \[ \int \frac {x}{\sqrt {-1+x^3}} \, dx=\frac {x^2 \sqrt {1-x^3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {5}{3},x^3\right )}{2 \sqrt {-1+x^3}} \] Input:

Integrate[x/Sqrt[-1 + x^3],x]
 

Output:

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

Rubi [A] (warning: unable to verify)

Time = 0.57 (sec) , antiderivative size = 270, normalized size of antiderivative = 1.06, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {833, 760, 2418}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 833

\(\displaystyle \left (1+\sqrt {3}\right ) \int \frac {1}{\sqrt {x^3-1}}dx-\int \frac {-x+\sqrt {3}+1}{\sqrt {x^3-1}}dx\)

\(\Big \downarrow \) 760

\(\displaystyle -\int \frac {-x+\sqrt {3}+1}{\sqrt {x^3-1}}dx-\frac {2 \sqrt {2-\sqrt {3}} \left (1+\sqrt {3}\right ) (1-x) \sqrt {\frac {x^2+x+1}{\left (-x-\sqrt {3}+1\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {-x+\sqrt {3}+1}{-x-\sqrt {3}+1}\right ),-7+4 \sqrt {3}\right )}{\sqrt [4]{3} \sqrt {-\frac {1-x}{\left (-x-\sqrt {3}+1\right )^2}} \sqrt {x^3-1}}\)

\(\Big \downarrow \) 2418

\(\displaystyle -\frac {2 \sqrt {2-\sqrt {3}} \left (1+\sqrt {3}\right ) (1-x) \sqrt {\frac {x^2+x+1}{\left (-x-\sqrt {3}+1\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {-x+\sqrt {3}+1}{-x-\sqrt {3}+1}\right ),-7+4 \sqrt {3}\right )}{\sqrt [4]{3} \sqrt {-\frac {1-x}{\left (-x-\sqrt {3}+1\right )^2}} \sqrt {x^3-1}}+\frac {\sqrt [4]{3} \sqrt {2+\sqrt {3}} (1-x) \sqrt {\frac {x^2+x+1}{\left (-x-\sqrt {3}+1\right )^2}} E\left (\arcsin \left (\frac {-x+\sqrt {3}+1}{-x-\sqrt {3}+1}\right )|-7+4 \sqrt {3}\right )}{\sqrt {-\frac {1-x}{\left (-x-\sqrt {3}+1\right )^2}} \sqrt {x^3-1}}-\frac {2 \sqrt {x^3-1}}{-x-\sqrt {3}+1}\)

Input:

Int[x/Sqrt[-1 + x^3],x]
 

Output:

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

Defintions of rubi rules used

rule 760
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]*Sqrt[(- 
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 833
Int[(x_)/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Rt[b/a, 3] 
], s = Denom[Rt[b/a, 3]]}, Simp[(-(1 + Sqrt[3]))*(s/r)   Int[1/Sqrt[a + b*x 
^3], x], x] + Simp[1/r   Int[((1 + Sqrt[3])*s + r*x)/Sqrt[a + b*x^3], x], x 
]] /; FreeQ[{a, b}, x] && NegQ[a]
 

rule 2418
Int[((c_) + (d_.)*(x_))/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = N 
umer[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] + S 
imp[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 - S 
qrt[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*Sqrt[3])*a*d^3, 0]
 
Maple [C] (warning: unable to verify)

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

Time = 0.77 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.13

method result size
meijerg \(\frac {\sqrt {-\operatorname {signum}\left (x^{3}-1\right )}\, x^{2} \operatorname {hypergeom}\left (\left [\frac {1}{2}, \frac {2}{3}\right ], \left [\frac {5}{3}\right ], x^{3}\right )}{2 \sqrt {\operatorname {signum}\left (x^{3}-1\right )}}\) \(33\)
default \(\frac {2 \left (-\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {\frac {-1+x}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x +\frac {1}{2}-\frac {i \sqrt {3}}{2}}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x +\frac {1}{2}+\frac {i \sqrt {3}}{2}}{\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \left (\left (\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) \operatorname {EllipticE}\left (\sqrt {\frac {-1+x}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \sqrt {\frac {\frac {3}{2}+\frac {i \sqrt {3}}{2}}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right )+\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \operatorname {EllipticF}\left (\sqrt {\frac {-1+x}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \sqrt {\frac {\frac {3}{2}+\frac {i \sqrt {3}}{2}}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right )\right )}{\sqrt {x^{3}-1}}\) \(173\)
elliptic \(\frac {2 \left (-\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {\frac {-1+x}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x +\frac {1}{2}-\frac {i \sqrt {3}}{2}}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x +\frac {1}{2}+\frac {i \sqrt {3}}{2}}{\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \left (\left (\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) \operatorname {EllipticE}\left (\sqrt {\frac {-1+x}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \sqrt {\frac {\frac {3}{2}+\frac {i \sqrt {3}}{2}}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right )+\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \operatorname {EllipticF}\left (\sqrt {\frac {-1+x}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \sqrt {\frac {\frac {3}{2}+\frac {i \sqrt {3}}{2}}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right )\right )}{\sqrt {x^{3}-1}}\) \(173\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.04 \[ \int \frac {x}{\sqrt {-1+x^3}} \, dx=-2 \, {\rm weierstrassZeta}\left (0, 4, {\rm weierstrassPInverse}\left (0, 4, x\right )\right ) \] Input:

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

Output:

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

Sympy [A] (verification not implemented)

Time = 0.40 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.11 \[ \int \frac {x}{\sqrt {-1+x^3}} \, dx=- \frac {i x^{2} \Gamma \left (\frac {2}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {2}{3} \\ \frac {5}{3} \end {matrix}\middle | {x^{3}} \right )}}{3 \Gamma \left (\frac {5}{3}\right )} \] Input:

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

Output:

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

Maxima [F]

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

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

Output:

integrate(x/sqrt(x^3 - 1), x)
 

Giac [F]

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

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

Output:

integrate(x/sqrt(x^3 - 1), x)
 

Mupad [B] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 215, normalized size of antiderivative = 0.84 \[ \int \frac {x}{\sqrt {-1+x^3}} \, dx=-\frac {2\,\left (\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\mathrm {F}\left (\mathrm {asin}\left (\sqrt {-\frac {x-1}{\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\right )\middle |-\frac {\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{-\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}\right )-\left (-\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\mathrm {E}\left (\mathrm {asin}\left (\sqrt {-\frac {x-1}{\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\right )\middle |-\frac {\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{-\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}\right )\right )\,\left (\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\sqrt {-\frac {x+\frac {1}{2}-\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{-\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\,\sqrt {\frac {x+\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\,\sqrt {-\frac {x-1}{\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}}{\sqrt {x^3+\left (-\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )-1\right )\,x+\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}} \] Input:

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

Output:

-(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 
^(1/2)*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)) 
/(((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)
 

Reduce [F]

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

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

Output:

int((sqrt(x**3 - 1)*x)/(x**3 - 1),x)