\(\int \frac {1}{(1+\sqrt {3}+x) \sqrt {1+x^3}} \, dx\) [127]

Optimal result

Integrand size = 20, antiderivative size = 146 \[ \int \frac {1}{\left (1+\sqrt {3}+x\right ) \sqrt {1+x^3}} \, dx=\frac {\arctan \left (\frac {\sqrt {3+2 \sqrt {3}} (1+x)}{\sqrt {1+x^3}}\right )}{\sqrt {3 \left (3+2 \sqrt {3}\right )}}+\frac {\sqrt {2+\sqrt {3}} (1+x) \sqrt {\frac {1-x+x^2}{\left (1+\sqrt {3}+x\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {1-\sqrt {3}+x}{1+\sqrt {3}+x}\right ),-7-4 \sqrt {3}\right )}{3^{3/4} \sqrt {\frac {1+x}{\left (1+\sqrt {3}+x\right )^2}} \sqrt {1+x^3}} \] Output:

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

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 20.32 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.93 \[ \int \frac {1}{\left (1+\sqrt {3}+x\right ) \sqrt {1+x^3}} \, dx=-\frac {4 \sqrt {2} \sqrt {\frac {i (1+x)}{3 i+\sqrt {3}}} \sqrt {1-x+x^2} \operatorname {EllipticPi}\left (\frac {2 \sqrt {3}}{3 i+(1+2 i) \sqrt {3}},\arcsin \left (\frac {\sqrt {i+\sqrt {3}-2 i x}}{\sqrt {2} \sqrt [4]{3}}\right ),\frac {2 \sqrt {3}}{3 i+\sqrt {3}}\right )}{\left (3 i+(1+2 i) \sqrt {3}\right ) \sqrt {1+x^3}} \] 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]*EllipticPi 
[(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)*Sqrt[ 
3])*Sqrt[1 + x^3])
 

Rubi [A] (verified)

Time = 0.67 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2560, 27, 759, 2565, 216}

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

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

\(\Big \downarrow \) 759

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

\(\Big \downarrow \) 2565

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

\(\Big \downarrow \) 216

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

Input:

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

Output:

ArcTan[(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]*Ell 
ipticF[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 216

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 0])
 

rule 759

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] & 
& PosQ[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 = 1.08 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.90

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 = 0.16 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.39 \[ \int \frac {1}{\left (1+\sqrt {3}+x\right ) \sqrt {1+x^3}} \, dx=-\frac {1}{6} \, \sqrt {2 \, \sqrt {3} - 3} \arctan \left (\frac {{\left (\sqrt {3} {\left (x^{2} - 4 \, x - 2\right )} - 6 \, x - 6\right )} \sqrt {2 \, \sqrt {3} - 3}}{6 \, \sqrt {x^{3} + 1}}\right ) + \frac {1}{3} \, \sqrt {3} {\rm weierstrassPInverse}\left (0, -4, x\right ) \] Input:

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

Output:

-1/6*sqrt(2*sqrt(3) - 3)*arctan(1/6*(sqrt(3)*(x^2 - 4*x - 2) - 6*x - 6)*sq 
rt(2*sqrt(3) - 3)/sqrt(x^3 + 1)) + 1/3*sqrt(3)*weierstrassPInverse(0, -4, 
x)
 

Sympy [F]

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

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

Output:

Integral(1/(sqrt((x + 1)*(x**2 - x + 1))*(x + 1 + sqrt(3))), 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. \[ \int \frac {1}{\left (1+\sqrt {3}+x\right ) \sqrt {1+x^3}} \, dx=\text {Exception raised: RuntimeError} \] Input:

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

Output:

Exception raised: RuntimeError >> an error occurred running a Giac command 
:INPUT:sage2OUTPUT:Unable to divide, perhaps due to rounding error%%%{1,[2 
]%%%} / %%%{%%{[2,4]:[1,0,-3]%%},[2]%%%} Error: Bad Argument Value
 

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

Output:

\text{Hanged}
 

Reduce [F]

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