\(\int \frac {1}{(1+\sqrt {3}-x) \sqrt {-1+x^3}} \, dx\) [129]

Optimal result

Integrand size = 22, antiderivative size = 167 \[ \int \frac {1}{\left (1+\sqrt {3}-x\right ) \sqrt {-1+x^3}} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {3+2 \sqrt {3}} (1-x)}{\sqrt {-1+x^3}}\right )}{\sqrt {3 \left (3+2 \sqrt {3}\right )}}-\frac {\sqrt {2-\sqrt {3}} (1-x) \sqrt {\frac {1+x+x^2}{\left (1-\sqrt {3}-x\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {1+\sqrt {3}-x}{1-\sqrt {3}-x}\right ),-7+4 \sqrt {3}\right )}{3^{3/4} \sqrt {-\frac {1-x}{\left (1-\sqrt {3}-x\right )^2}} \sqrt {-1+x^3}} \] Output:

-arctanh((3+2*3^(1/2))^(1/2)*(1-x)/(x^3-1)^(1/2))/(9+6*3^(1/2))^(1/2)-1/3* 
(1/2*6^(1/2)-1/2*2^(1/2))*(1-x)*((x^2+x+1)/(1-3^(1/2)-x)^2)^(1/2)*Elliptic 
F((1+3^(1/2)-x)/(1-3^(1/2)-x),2*I-I*3^(1/2))*3^(1/4)/(-(1-x)/(1-3^(1/2)-x) 
^2)^(1/2)/(x^3-1)^(1/2)
 

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 20.36 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.80 \[ \int \frac {1}{\left (1+\sqrt {3}-x\right ) \sqrt {-1+x^3}} \, dx=\frac {4 \sqrt {2} \sqrt {-\frac {i (-1+x)}{3 i+\sqrt {3}}} \sqrt {1+x+x^2} \operatorname {EllipticPi}\left (\frac {2 \sqrt {3}}{3 i+(1+2 i) \sqrt {3}},\arcsin \left (\frac {\sqrt {i+\sqrt {3}+2 i x}}{\sqrt {2} \sqrt [4]{3}}\right ),\frac {2 \sqrt {3}}{3 i+\sqrt {3}}\right )}{\left (3 i+(1+2 i) \sqrt {3}\right ) \sqrt {-1+x^3}} \] Input:

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

Output:

(4*Sqrt[2]*Sqrt[((-I)*(-1 + x))/(3*I + Sqrt[3])]*Sqrt[1 + x + x^2]*Ellipti 
cPi[(2*Sqrt[3])/(3*I + (1 + 2*I)*Sqrt[3]), ArcSin[Sqrt[I + Sqrt[3] + (2*I) 
*x]/(Sqrt[2]*3^(1/4))], (2*Sqrt[3])/(3*I + Sqrt[3])])/((3*I + (1 + 2*I)*Sq 
rt[3])*Sqrt[-1 + x^3])
 

Rubi [A] (verified)

Time = 0.68 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {2560, 27, 760, 2565, 219}

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.

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

\(\Big \downarrow \) 2560

\(\displaystyle \frac {\int \frac {1}{\sqrt {x^3-1}}dx}{2 \sqrt {3}}-\frac {\int \frac {6 \left (-x-\sqrt {3}+1\right )}{\left (-x+\sqrt {3}+1\right ) \sqrt {x^3-1}}dx}{12 \sqrt {3}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {1}{\sqrt {x^3-1}}dx}{2 \sqrt {3}}-\frac {\int \frac {-x-\sqrt {3}+1}{\left (-x+\sqrt {3}+1\right ) \sqrt {x^3-1}}dx}{2 \sqrt {3}}\)

\(\Big \downarrow \) 760

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

\(\Big \downarrow \) 2565

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

\(\Big \downarrow \) 219

\(\displaystyle -\frac {\sqrt {2-\sqrt {3}} (1-x) \sqrt {\frac {x^2+x+1}{\left (-x-\sqrt {3}+1\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {-x+\sqrt {3}+1}{-x-\sqrt {3}+1}\right ),-7+4 \sqrt {3}\right )}{3^{3/4} \sqrt {-\frac {1-x}{\left (-x-\sqrt {3}+1\right )^2}} \sqrt {x^3-1}}-\frac {\text {arctanh}\left (\frac {\sqrt {3+2 \sqrt {3}} (1-x)}{\sqrt {x^3-1}}\right )}{\sqrt {3 \left (3+2 \sqrt {3}\right )}}\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

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 2560

Int[1/(((c_) + (d_.)*(x_))*Sqrt[(a_) + (b_.)*(x_)^3]), x_Symbol] :> Simp[-6 
*a*(d^3/(c*(b*c^3 - 28*a*d^3)))   Int[1/Sqrt[a + b*x^3], x], x] + Simp[1/(c 
*(b*c^3 - 28*a*d^3))   Int[Simp[c*(b*c^3 - 22*a*d^3) + 6*a*d^4*x, x]/((c + 
d*x)*Sqrt[a + b*x^3]), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[b^2*c^6 - 20 
*a*b*c^3*d^3 - 8*a^2*d^6, 0]
 

rule 2565

Int[((e_) + (f_.)*(x_))/(((c_) + (d_.)*(x_))*Sqrt[(a_) + (b_.)*(x_)^3]), x_ 
Symbol] :> With[{k = Simplify[(d*e + 2*c*f)/(c*f)]}, Simp[(1 + k)*(e/d)   S 
ubst[Int[1/(1 + (3 + 2*k)*a*x^2), x], x, (1 + (1 + k)*d*(x/c))/Sqrt[a + b*x 
^3]], x]] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[d*e - c*f, 0] && EqQ[b^2*c 
^6 - 20*a*b*c^3*d^3 - 8*a^2*d^6, 0] && EqQ[6*a*d^4*e - c*f*(b*c^3 - 22*a*d^ 
3), 0]
 

Maple [A] (verified)

Time = 1.03 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.79

method	result	size
default	\(\frac {2 \left (-\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {\frac {x -1}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x +\frac {1}{2}-\frac {i \sqrt {3}}{2}}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x +\frac {1}{2}+\frac {i \sqrt {3}}{2}}{\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \sqrt {3}\, \operatorname {EllipticPi}\left (\sqrt {\frac {x -1}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, -\frac {\left (\frac {3}{2}+\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}{3}, \sqrt {\frac {\frac {3}{2}+\frac {i \sqrt {3}}{2}}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right )}{3 \sqrt {x^{3}-1}}\)	\(132\)
elliptic	\(\frac {2 \left (-\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {\frac {x -1}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x +\frac {1}{2}-\frac {i \sqrt {3}}{2}}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x +\frac {1}{2}+\frac {i \sqrt {3}}{2}}{\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \sqrt {3}\, \operatorname {EllipticPi}\left (\sqrt {\frac {x -1}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, -\frac {\left (\frac {3}{2}+\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}{3}, \sqrt {\frac {\frac {3}{2}+\frac {i \sqrt {3}}{2}}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right )}{3 \sqrt {x^{3}-1}}\)	\(132\)

Input:

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

Output:

2/3*(-3/2-1/2*I*3^(1/2))*((x-1)/(-3/2-1/2*I*3^(1/2)))^(1/2)*((x+1/2-1/2*I* 
3^(1/2))/(3/2-1/2*I*3^(1/2)))^(1/2)*((x+1/2+1/2*I*3^(1/2))/(3/2+1/2*I*3^(1 
/2)))^(1/2)/(x^3-1)^(1/2)*3^(1/2)*EllipticPi(((x-1)/(-3/2-1/2*I*3^(1/2)))^ 
(1/2),-1/3*(3/2+1/2*I*3^(1/2))*3^(1/2),((3/2+1/2*I*3^(1/2))/(3/2-1/2*I*3^( 
1/2)))^(1/2))
 

Fricas [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.26 \[ \int \frac {1}{\left (1+\sqrt {3}-x\right ) \sqrt {-1+x^3}} \, dx=\frac {1}{12} \, \sqrt {2 \, \sqrt {3} - 3} \log \left (\frac {x^{8} + 16 \, x^{7} + 112 \, x^{6} + 16 \, x^{5} + 112 \, x^{4} - 224 \, x^{3} + 64 \, x^{2} + 4 \, {\left (2 \, x^{6} + 18 \, x^{5} + 42 \, x^{4} + 8 \, x^{3} + \sqrt {3} {\left (x^{6} + 12 \, x^{5} + 18 \, x^{4} + 16 \, x^{3} - 12 \, x^{2} - 8\right )} - 24 \, x + 8\right )} \sqrt {x^{3} - 1} \sqrt {2 \, \sqrt {3} - 3} + 16 \, \sqrt {3} {\left (x^{7} + 2 \, x^{6} + 6 \, x^{5} - 5 \, x^{4} + 2 \, x^{3} - 6 \, x^{2} + 4 \, x - 4\right )} - 128 \, x + 112}{x^{8} - 8 \, x^{7} + 16 \, x^{6} + 16 \, x^{5} - 56 \, x^{4} - 32 \, x^{3} + 64 \, x^{2} + 64 \, x + 16}\right ) + \frac {1}{3} \, \sqrt {3} {\rm weierstrassPInverse}\left (0, 4, x\right ) \] Input:

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

Output:

1/12*sqrt(2*sqrt(3) - 3)*log((x^8 + 16*x^7 + 112*x^6 + 16*x^5 + 112*x^4 - 
224*x^3 + 64*x^2 + 4*(2*x^6 + 18*x^5 + 42*x^4 + 8*x^3 + sqrt(3)*(x^6 + 12* 
x^5 + 18*x^4 + 16*x^3 - 12*x^2 - 8) - 24*x + 8)*sqrt(x^3 - 1)*sqrt(2*sqrt( 
3) - 3) + 16*sqrt(3)*(x^7 + 2*x^6 + 6*x^5 - 5*x^4 + 2*x^3 - 6*x^2 + 4*x - 
4) - 128*x + 112)/(x^8 - 8*x^7 + 16*x^6 + 16*x^5 - 56*x^4 - 32*x^3 + 64*x^ 
2 + 64*x + 16)) + 1/3*sqrt(3)*weierstrassPInverse(0, 4, x)
 

Sympy [F]

\[ \int \frac {1}{\left (1+\sqrt {3}-x\right ) \sqrt {-1+x^3}} \, dx=- \int \frac {1}{x \sqrt {x^{3} - 1} - \sqrt {3} \sqrt {x^{3} - 1} - \sqrt {x^{3} - 1}}\, dx \] Input:

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F(-2)]

Exception generated. \[ \int \frac {1}{\left (1+\sqrt {3}-x\right ) \sqrt {-1+x^3}} \, dx=\text {Exception raised: TypeError} \] Input:

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Unable to divide, perhaps due to ro 
unding error%%%{1,[2]%%%} / %%%{%%{[2,4]:[1,0,-3]%%},[2]%%%} Error: Bad Ar 
gument Va
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\left (1+\sqrt {3}-x\right ) \sqrt {-1+x^3}} \, dx=\text {Hanged} \] Input:

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

Output:

\text{Hanged}
 

Reduce [F]

\[ \int \frac {1}{\left (1+\sqrt {3}-x\right ) \sqrt {-1+x^3}} \, dx=-\sqrt {3}\, \left (\int \frac {\sqrt {x^{3}-1}}{x^{5}-2 x^{4}-2 x^{3}-x^{2}+2 x +2}d x \right )+\int \frac {\sqrt {x^{3}-1}}{x^{5}-2 x^{4}-2 x^{3}-x^{2}+2 x +2}d x -\left (\int \frac {\sqrt {x^{3}-1}\, x}{x^{5}-2 x^{4}-2 x^{3}-x^{2}+2 x +2}d x \right ) \] Input:

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

Output:

 - sqrt(3)*int(sqrt(x**3 - 1)/(x**5 - 2*x**4 - 2*x**3 - x**2 + 2*x + 2),x) 
 + int(sqrt(x**3 - 1)/(x**5 - 2*x**4 - 2*x**3 - x**2 + 2*x + 2),x) - int(( 
sqrt(x**3 - 1)*x)/(x**5 - 2*x**4 - 2*x**3 - x**2 + 2*x + 2),x)