\(\int \frac {1}{(2^{2/3}+x) \sqrt {1+x^3}} \, dx\) [118]

Optimal result
Mathematica [C] (warning: unable to verify)
Rubi [A] (verified)
Maple [A] (verified)
Fricas [A] (verification not implemented)
Sympy [F]
Maxima [F]
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 19, antiderivative size = 145 \[ \int \frac {1}{\left (2^{2/3}+x\right ) \sqrt {1+x^3}} \, dx=\frac {2 \arctan \left (\frac {\sqrt {3} \left (1+\sqrt [3]{2} x\right )}{\sqrt {1+x^3}}\right )}{3 \sqrt {3}}+\frac {2 \sqrt [3]{2} \sqrt {2+\sqrt {3}} (1+x) \sqrt {\frac {1-x+x^2}{\left (1+\sqrt {3}+x\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {1-\sqrt {3}+x}{1+\sqrt {3}+x}\right ),-7-4 \sqrt {3}\right )}{3 \sqrt [4]{3} \sqrt {\frac {1+x}{\left (1+\sqrt {3}+x\right )^2}} \sqrt {1+x^3}} \] Output:

2/9*arctan(3^(1/2)*(1+2^(1/3)*x)/(x^3+1)^(1/2))*3^(1/2)+2/9*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^(3/4)/((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 = 20.28 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.02 \[ \int \frac {1}{\left (2^{2/3}+x\right ) \sqrt {1+x^3}} \, dx=\frac {4 i \sqrt {2} \sqrt {\frac {i (1+x)}{3 i+\sqrt {3}}} \sqrt {1-x+x^2} \operatorname {EllipticPi}\left (\frac {2 \sqrt {3}}{i+2 i 2^{2/3}+\sqrt {3}},\arcsin \left (\frac {\sqrt {i+\sqrt {3}-2 i x}}{\sqrt {2} \sqrt [4]{3}}\right ),\frac {2 \sqrt {3}}{3 i+\sqrt {3}}\right )}{\left (1+2\ 2^{2/3}-i \sqrt {3}\right ) \sqrt {1+x^3}} \] Input:

Integrate[1/((2^(2/3) + x)*Sqrt[1 + x^3]),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.65 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {2559, 759, 2562, 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.

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

\(\Big \downarrow \) 2559

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

\(\Big \downarrow \) 759

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

\(\Big \downarrow \) 2562

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

\(\Big \downarrow \) 216

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

Input:

Int[1/((2^(2/3) + x)*Sqrt[1 + x^3]),x]
 

Output:

(2*ArcTan[(Sqrt[3]*(1 + 2^(1/3)*x))/Sqrt[1 + x^3]])/(3*Sqrt[3]) + (2*2^(1/ 
3)*Sqrt[2 + Sqrt[3]]*(1 + x)*Sqrt[(1 - x + x^2)/(1 + Sqrt[3] + x)^2]*Ellip 
ticF[ArcSin[(1 - Sqrt[3] + x)/(1 + Sqrt[3] + x)], -7 - 4*Sqrt[3]])/(3*3^(1 
/4)*Sqrt[(1 + x)/(1 + Sqrt[3] + x)^2]*Sqrt[1 + x^3])
 

Defintions of rubi rules used

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 2559
Int[1/(((c_) + (d_.)*(x_))*Sqrt[(a_) + (b_.)*(x_)^3]), x_Symbol] :> Simp[2/ 
(3*c)   Int[1/Sqrt[a + b*x^3], x], x] + Simp[1/(3*c)   Int[(c - 2*d*x)/((c 
+ d*x)*Sqrt[a + b*x^3]), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c^3 - 4* 
a*d^3, 0]
 

rule 2562
Int[((e_) + (f_.)*(x_))/(((c_) + (d_.)*(x_))*Sqrt[(a_) + (b_.)*(x_)^3]), x_ 
Symbol] :> Simp[2*(e/d)   Subst[Int[1/(1 + 3*a*x^2), x], x, (1 + 2*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*c^3 - 4*a*d^3, 0] && EqQ[2*d*e + c*f, 0]
 
Maple [A] (verified)

Time = 1.67 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.96

method result size
default \(\frac {2 \left (\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {\frac {x +1}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}-\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \operatorname {EllipticPi}\left (\sqrt {\frac {x +1}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \frac {-\frac {3}{2}+\frac {i \sqrt {3}}{2}}{2^{\frac {2}{3}}-1}, \sqrt {\frac {-\frac {3}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right )}{\sqrt {x^{3}+1}\, \left (2^{\frac {2}{3}}-1\right )}\) \(139\)
elliptic \(\frac {2 \left (\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {\frac {x +1}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}-\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \operatorname {EllipticPi}\left (\sqrt {\frac {x +1}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \frac {-\frac {3}{2}+\frac {i \sqrt {3}}{2}}{2^{\frac {2}{3}}-1}, \sqrt {\frac {-\frac {3}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right )}{\sqrt {x^{3}+1}\, \left (2^{\frac {2}{3}}-1\right )}\) \(139\)

Input:

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

Output:

2*(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)/(2^(2/3)-1)*EllipticPi(((x+1)/(3/2-1/2*I*3^(1/2))) 
^(1/2),(-3/2+1/2*I*3^(1/2))/(2^(2/3)-1),((-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 = 74, normalized size of antiderivative = 0.51 \[ \int \frac {1}{\left (2^{2/3}+x\right ) \sqrt {1+x^3}} \, dx=-\frac {1}{9} \, \sqrt {3} \arctan \left (-\frac {\sqrt {3} {\left (5 \, x^{3} - 2^{\frac {2}{3}} {\left (x^{5} + x^{2}\right )} + 2^{\frac {1}{3}} {\left (7 \, x^{4} + 4 \, x\right )} + 2\right )} \sqrt {x^{3} + 1}}{6 \, {\left (2 \, x^{6} + 3 \, x^{3} + 1\right )}}\right ) + \frac {2}{3} \cdot 2^{\frac {1}{3}} {\rm weierstrassPInverse}\left (0, -4, x\right ) \] Input:

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

Output:

-1/9*sqrt(3)*arctan(-1/6*sqrt(3)*(5*x^3 - 2^(2/3)*(x^5 + x^2) + 2^(1/3)*(7 
*x^4 + 4*x) + 2)*sqrt(x^3 + 1)/(2*x^6 + 3*x^3 + 1)) + 2/3*2^(1/3)*weierstr 
assPInverse(0, -4, x)
 

Sympy [F]

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

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

Output:

Integral(1/(sqrt((x + 1)*(x**2 - x + 1))*(x + 2**(2/3))), x)
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\left (2^{2/3}+x\right ) \sqrt {1+x^3}} \, dx=\int \frac {1}{\sqrt {x^3+1}\,\left (x+2^{2/3}\right )} \,d x \] Input:

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

Output:

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

Reduce [F]

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

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

Output:

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