3.1.52 \(\int \frac {2+3 x}{(2^{2/3}+x) \sqrt {1+x^3}} \, dx\) [52]

3.1.52.1 Optimal result
3.1.52.2 Mathematica [C] (warning: unable to verify)
3.1.52.3 Rubi [A] (verified)
3.1.52.4 Maple [B] (verified)
3.1.52.5 Fricas [C] (verification not implemented)
3.1.52.6 Sympy [F]
3.1.52.7 Maxima [F]
3.1.52.8 Giac [F(-2)]
3.1.52.9 Mupad [F(-1)]

3.1.52.1 Optimal result

Integrand size = 24, antiderivative size = 158 \[ \int \frac {2+3 x}{\left (2^{2/3}+x\right ) \sqrt {1+x^3}} \, dx=\frac {2 \left (2-3\ 2^{2/3}\right ) \arctan \left (\frac {\sqrt {3} \left (1+\sqrt [3]{2} x\right )}{\sqrt {1+x^3}}\right )}{3 \sqrt {3}}+\frac {2 \left (3+2 \sqrt [3]{2}\right ) \sqrt {2+\sqrt {3}} (1+x) \sqrt {\frac {1-x+x^2}{\left (1+\sqrt {3}+x\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {1-\sqrt {3}+x}{1+\sqrt {3}+x}\right ),-7-4 \sqrt {3}\right )}{3 \sqrt [4]{3} \sqrt {\frac {1+x}{\left (1+\sqrt {3}+x\right )^2}} \sqrt {1+x^3}} \]

output
2/9*(2-3*2^(2/3))*arctan((1+2^(1/3)*x)*3^(1/2)/(x^3+1)^(1/2))*3^(1/2)+2/9* 
(3+2*2^(1/3))*(1+x)*EllipticF((1+x-3^(1/2))/(1+x+3^(1/2)),I*3^(1/2)+2*I)*( 
1/2*6^(1/2)+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)
 
3.1.52.2 Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 20.43 (sec) , antiderivative size = 336, normalized size of antiderivative = 2.13 \[ \int \frac {2+3 x}{\left (2^{2/3}+x\right ) \sqrt {1+x^3}} \, dx=\frac {2 \sqrt [6]{2} \sqrt {\frac {i (1+x)}{3 i+\sqrt {3}}} \left (3 \sqrt {-i+\sqrt {3}+2 i x} \left (-6-3 \sqrt [3]{2}-2 i \sqrt {3}+i \sqrt [3]{2} \sqrt {3}+\left (3 \sqrt [3]{2}+4 i \sqrt {3}+i \sqrt [3]{2} \sqrt {3}\right ) x\right ) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {i+\sqrt {3}-2 i x}}{\sqrt {2} \sqrt [4]{3}}\right ),\frac {2 \sqrt {3}}{3 i+\sqrt {3}}\right )-4 \sqrt {3} \left (-3+\sqrt [3]{2}\right ) \sqrt {i+\sqrt {3}-2 i x} \sqrt {1-x+x^2} \operatorname {EllipticPi}\left (\frac {2 \sqrt {3}}{i+2 i 2^{2/3}+\sqrt {3}},\arcsin \left (\frac {\sqrt {i+\sqrt {3}-2 i x}}{\sqrt {2} \sqrt [4]{3}}\right ),\frac {2 \sqrt {3}}{3 i+\sqrt {3}}\right )\right )}{\sqrt {3} \left (i+2 i 2^{2/3}+\sqrt {3}\right ) \sqrt {i+\sqrt {3}-2 i x} \sqrt {1+x^3}} \]

input
Integrate[(2 + 3*x)/((2^(2/3) + x)*Sqrt[1 + x^3]),x]
 
output
(2*2^(1/6)*Sqrt[(I*(1 + x))/(3*I + Sqrt[3])]*(3*Sqrt[-I + Sqrt[3] + (2*I)* 
x]*(-6 - 3*2^(1/3) - (2*I)*Sqrt[3] + I*2^(1/3)*Sqrt[3] + (3*2^(1/3) + (4*I 
)*Sqrt[3] + I*2^(1/3)*Sqrt[3])*x)*EllipticF[ArcSin[Sqrt[I + Sqrt[3] - (2*I 
)*x]/(Sqrt[2]*3^(1/4))], (2*Sqrt[3])/(3*I + Sqrt[3])] - 4*Sqrt[3]*(-3 + 2^ 
(1/3))*Sqrt[I + Sqrt[3] - (2*I)*x]*Sqrt[1 - x + x^2]*EllipticPi[(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])]))/(Sqrt[3]*(I + (2*I)*2^(2/3) + 
 Sqrt[3])*Sqrt[I + Sqrt[3] - (2*I)*x]*Sqrt[1 + x^3])
 
3.1.52.3 Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.03, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2564, 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.

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

\(\Big \downarrow \) 2564

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

\(\Big \downarrow \) 759

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

\(\Big \downarrow \) 2562

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

\(\Big \downarrow \) 216

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

input
Int[(2 + 3*x)/((2^(2/3) + x)*Sqrt[1 + x^3]),x]
 
output
(-2*2^(2/3)*(3 - 2^(1/3))*ArcTan[(Sqrt[3]*(1 + 2^(1/3)*x))/Sqrt[1 + x^3]]) 
/(3*Sqrt[3]) + (2*(3 + 2*2^(1/3))*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^(1/4)*Sqrt[(1 + x)/(1 + Sqrt[3] + x)^2]*Sqrt[ 
1 + x^3])
 

3.1.52.3.1 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 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]
 

rule 2564
Int[((e_.) + (f_.)*(x_))/(((c_) + (d_.)*(x_))*Sqrt[(a_) + (b_.)*(x_)^3]), x 
_Symbol] :> Simp[(2*d*e + c*f)/(3*c*d)   Int[1/Sqrt[a + b*x^3], x], x] + Si 
mp[(d*e - c*f)/(3*c*d)   Int[(c - 2*d*x)/((c + d*x)*Sqrt[a + b*x^3]), x], 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[b*c^3 + 8*a*d^3, 0]) && NeQ[2*d*e + c*f, 0]
 
3.1.52.4 Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 261 vs. \(2 (125 ) = 250\).

Time = 4.34 (sec) , antiderivative size = 262, normalized size of antiderivative = 1.66

method result size
default \(\frac {6 \left (\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {\frac {x +1}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}-\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, F\left (\sqrt {\frac {x +1}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \sqrt {\frac {-\frac {3}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right )}{\sqrt {x^{3}+1}}+\frac {2 \left (2-3 \,2^{\frac {2}{3}}\right ) \left (\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {\frac {x +1}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}-\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \Pi \left (\sqrt {\frac {x +1}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \frac {-\frac {3}{2}+\frac {i \sqrt {3}}{2}}{2^{\frac {2}{3}}-1}, \sqrt {\frac {-\frac {3}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right )}{\sqrt {x^{3}+1}\, \left (2^{\frac {2}{3}}-1\right )}\) \(262\)
elliptic \(\frac {6 \left (\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {\frac {x +1}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}-\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, F\left (\sqrt {\frac {x +1}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \sqrt {\frac {-\frac {3}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right )}{\sqrt {x^{3}+1}}+\frac {2 \left (2-3 \,2^{\frac {2}{3}}\right ) \left (\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {\frac {x +1}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}-\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \Pi \left (\sqrt {\frac {x +1}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \frac {-\frac {3}{2}+\frac {i \sqrt {3}}{2}}{2^{\frac {2}{3}}-1}, \sqrt {\frac {-\frac {3}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right )}{\sqrt {x^{3}+1}\, \left (2^{\frac {2}{3}}-1\right )}\) \(262\)

input
int((3*x+2)/(2^(2/3)+x)/(x^3+1)^(1/2),x,method=_RETURNVERBOSE)
 
output
6*(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)*EllipticF(((x+1)/(3/2-1/2*I*3^(1/2)))^(1/2),((-3/2 
+1/2*I*3^(1/2))/(-3/2-1/2*I*3^(1/2)))^(1/2))+2*(2-3*2^(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)/(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) 
)
 
3.1.52.5 Fricas [C] (verification not implemented)

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

Time = 0.16 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.99 \[ \int \frac {2+3 x}{\left (2^{2/3}+x\right ) \sqrt {1+x^3}} \, dx=\frac {1}{9} \, \sqrt {3} \sqrt {-12 \cdot 2^{\frac {2}{3}} + 18 \cdot 2^{\frac {1}{3}} + 4} \arctan \left (\frac {\sqrt {3} {\left (18 \, x^{5} - 42 \, x^{4} - 10 \, x^{3} + 18 \, x^{2} + 2^{\frac {2}{3}} {\left (2 \, x^{5} - 63 \, x^{4} - 15 \, x^{3} + 2 \, x^{2} - 36 \, x - 6\right )} + 2^{\frac {1}{3}} {\left (6 \, x^{5} - 14 \, x^{4} - 45 \, x^{3} + 6 \, x^{2} - 8 \, x - 18\right )} - 24 \, x - 4\right )} \sqrt {x^{3} + 1} \sqrt {-12 \cdot 2^{\frac {2}{3}} + 18 \cdot 2^{\frac {1}{3}} + 4}}{300 \, {\left (2 \, x^{6} + 3 \, x^{3} + 1\right )}}\right ) + \frac {2}{3} \, {\left (2 \cdot 2^{\frac {1}{3}} + 3\right )} {\rm weierstrassPInverse}\left (0, -4, x\right ) \]

input
integrate((2+3*x)/(2^(2/3)+x)/(x^3+1)^(1/2),x, algorithm="fricas")
 
output
1/9*sqrt(3)*sqrt(-12*2^(2/3) + 18*2^(1/3) + 4)*arctan(1/300*sqrt(3)*(18*x^ 
5 - 42*x^4 - 10*x^3 + 18*x^2 + 2^(2/3)*(2*x^5 - 63*x^4 - 15*x^3 + 2*x^2 - 
36*x - 6) + 2^(1/3)*(6*x^5 - 14*x^4 - 45*x^3 + 6*x^2 - 8*x - 18) - 24*x - 
4)*sqrt(x^3 + 1)*sqrt(-12*2^(2/3) + 18*2^(1/3) + 4)/(2*x^6 + 3*x^3 + 1)) + 
 2/3*(2*2^(1/3) + 3)*weierstrassPInverse(0, -4, x)
 
3.1.52.6 Sympy [F]

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

input
integrate((2+3*x)/(2**(2/3)+x)/(x**3+1)**(1/2),x)
 
output
Integral((3*x + 2)/(sqrt((x + 1)*(x**2 - x + 1))*(x + 2**(2/3))), x)
 
3.1.52.7 Maxima [F]

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

input
integrate((2+3*x)/(2^(2/3)+x)/(x^3+1)^(1/2),x, algorithm="maxima")
 
output
integrate((3*x + 2)/(sqrt(x^3 + 1)*(x + 2^(2/3))), x)
 
3.1.52.8 Giac [F(-2)]

Exception generated. \[ \int \frac {2+3 x}{\left (2^{2/3}+x\right ) \sqrt {1+x^3}} \, dx=\text {Exception raised: TypeError} \]

input
integrate((2+3*x)/(2^(2/3)+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,[1]%%%} / %%%{%%{[1,0,0]:[1,0,0,-2]%%},[1]%%%} Error: Ba 
d Argumen
 
3.1.52.9 Mupad [F(-1)]

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

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