\(\int \frac {x}{(2-x) \sqrt {1+x^3}} \, dx\) [138]

Optimal result

Integrand size = 18, antiderivative size = 129 \[ \int \frac {x}{(2-x) \sqrt {1+x^3}} \, dx=\frac {4}{9} \text {arctanh}\left (\frac {(1+x)^2}{3 \sqrt {1+x^3}}\right )-\frac {2 \sqrt {2+\sqrt {3}} (1+x) \sqrt {\frac {1-x+x^2}{\left (1+\sqrt {3}+x\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {1-\sqrt {3}+x}{1+\sqrt {3}+x}\right ),-7-4 \sqrt {3}\right )}{3 \sqrt [4]{3} \sqrt {\frac {1+x}{\left (1+\sqrt {3}+x\right )^2}} \sqrt {1+x^3}} \] Output:

4/9*arctanh(1/3*(1+x)^2/(x^3+1)^(1/2))-2/9*(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 = 10.26 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.50 \[ \int \frac {x}{(2-x) \sqrt {1+x^3}} \, dx=\frac {2 \sqrt {\frac {1+x}{1+\sqrt [3]{-1}}} \left (\frac {\left (\sqrt [3]{-1}-x\right ) \sqrt {\frac {\sqrt [3]{-1}-(-1)^{2/3} x}{1+\sqrt [3]{-1}}} \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\frac {1+(-1)^{2/3} x}{1+\sqrt [3]{-1}}}\right ),\sqrt [3]{-1}\right )}{\sqrt {\frac {1+(-1)^{2/3} x}{1+\sqrt [3]{-1}}}}+\frac {2 i \sqrt {1-x+x^2} \operatorname {EllipticPi}\left (\frac {2 \sqrt {3}}{3 i+\sqrt {3}},\arcsin \left (\sqrt {\frac {1+(-1)^{2/3} x}{1+\sqrt [3]{-1}}}\right ),\sqrt [3]{-1}\right )}{-2+\sqrt [3]{-1}}\right )}{\sqrt {1+x^3}} \] Input:

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

Output:

(2*Sqrt[(1 + x)/(1 + (-1)^(1/3))]*((((-1)^(1/3) - x)*Sqrt[((-1)^(1/3) - (- 
1)^(2/3)*x)/(1 + (-1)^(1/3))]*EllipticF[ArcSin[Sqrt[(1 + (-1)^(2/3)*x)/(1 
+ (-1)^(1/3))]], (-1)^(1/3)])/Sqrt[(1 + (-1)^(2/3)*x)/(1 + (-1)^(1/3))] + 
((2*I)*Sqrt[1 - x + x^2]*EllipticPi[(2*Sqrt[3])/(3*I + Sqrt[3]), ArcSin[Sq 
rt[(1 + (-1)^(2/3)*x)/(1 + (-1)^(1/3))]], (-1)^(1/3)])/(-2 + (-1)^(1/3)))) 
/Sqrt[1 + x^3]
 

Rubi [A] (verified)

Time = 0.60 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {2564, 27, 759, 2563, 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.

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

\(\Big \downarrow \) 2564

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 759

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

\(\Big \downarrow \) 2563

\(\displaystyle \frac {4}{3} \int \frac {1}{9-\frac {(x+1)^4}{x^3+1}}d\frac {(x+1)^2}{\sqrt {x^3+1}}-\frac {2 \sqrt {2+\sqrt {3}} (x+1) \sqrt {\frac {x^2-x+1}{\left (x+\sqrt {3}+1\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {x-\sqrt {3}+1}{x+\sqrt {3}+1}\right ),-7-4 \sqrt {3}\right )}{3 \sqrt [4]{3} \sqrt {\frac {x+1}{\left (x+\sqrt {3}+1\right )^2}} \sqrt {x^3+1}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {4}{9} \text {arctanh}\left (\frac {(x+1)^2}{3 \sqrt {x^3+1}}\right )-\frac {2 \sqrt {2+\sqrt {3}} (x+1) \sqrt {\frac {x^2-x+1}{\left (x+\sqrt {3}+1\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {x-\sqrt {3}+1}{x+\sqrt {3}+1}\right ),-7-4 \sqrt {3}\right )}{3 \sqrt [4]{3} \sqrt {\frac {x+1}{\left (x+\sqrt {3}+1\right )^2}} \sqrt {x^3+1}}\)

Input:

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

Output:

(4*ArcTanh[(1 + x)^2/(3*Sqrt[1 + x^3])])/9 - (2*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])
 

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 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 2563

Int[((e_) + (f_.)*(x_))/(((c_) + (d_.)*(x_))*Sqrt[(a_) + (b_.)*(x_)^3]), x_ 
Symbol] :> Simp[-2*(e/d)   Subst[Int[1/(9 - a*x^2), x], x, (1 + f*(x/e))^2/ 
Sqrt[a + b*x^3]], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[d*e - c*f, 0] & 
& EqQ[b*c^3 + 8*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]
 

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 239 vs. \(2 (104 ) = 208\).

Time = 0.44 (sec) , antiderivative size = 240, normalized size of antiderivative = 1.86

method	result	size
default	\(-\frac {2 \left (\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {\frac {x +1}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}-\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {x +1}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \sqrt {\frac {-\frac {3}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right )}{\sqrt {x^{3}+1}}+\frac {4 \left (\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {\frac {x +1}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}-\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \operatorname {EllipticPi}\left (\sqrt {\frac {x +1}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \frac {1}{2}-\frac {i \sqrt {3}}{6}, \sqrt {\frac {-\frac {3}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right )}{3 \sqrt {x^{3}+1}}\)	\(240\)
elliptic	\(-\frac {2 \left (\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {\frac {x +1}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}-\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {x +1}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \sqrt {\frac {-\frac {3}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right )}{\sqrt {x^{3}+1}}+\frac {4 \left (\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {\frac {x +1}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}-\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \operatorname {EllipticPi}\left (\sqrt {\frac {x +1}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \frac {1}{2}-\frac {i \sqrt {3}}{6}, \sqrt {\frac {-\frac {3}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right )}{3 \sqrt {x^{3}+1}}\)	\(240\)

Input:

int(x/(2-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)*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))+4/3*(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)*El 
lipticPi(((x+1)/(3/2-1/2*I*3^(1/2)))^(1/2),1/2-1/6*I*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 = 0.10 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.40 \[ \int \frac {x}{(2-x) \sqrt {1+x^3}} \, dx=\frac {2}{9} \, \log \left (\frac {x^{3} + 12 \, x^{2} + 6 \, \sqrt {x^{3} + 1} {\left (x + 1\right )} - 6 \, x + 10}{x^{3} - 6 \, x^{2} + 12 \, x - 8}\right ) - \frac {2}{3} \, {\rm weierstrassPInverse}\left (0, -4, x\right ) \] Input:

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

Output:

2/9*log((x^3 + 12*x^2 + 6*sqrt(x^3 + 1)*(x + 1) - 6*x + 10)/(x^3 - 6*x^2 + 
 12*x - 8)) - 2/3*weierstrassPInverse(0, -4, x)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.60 \[ \int \frac {x}{(2-x) \sqrt {1+x^3}} \, dx=-\frac {\left (3+\sqrt {3}\,1{}\mathrm {i}\right )\,\sqrt {\frac {x-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{-\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\,\sqrt {\frac {x+1}{\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\,\sqrt {\frac {\frac {1}{2}-x+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\,\left (3\,\mathrm {F}\left (\mathrm {asin}\left (\sqrt {\frac {x+1}{\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\right )\middle |-\frac {\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{-\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}\right )-2\,\Pi \left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6};\mathrm {asin}\left (\sqrt {\frac {x+1}{\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\right )\middle |-\frac {\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{-\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}\right )\right )}{3\,\sqrt {x^3+\left (-\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )-1\right )\,x-\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}} \] Input:

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

Output:

-((3^(1/2)*1i + 3)*((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 - x + 1/2)/((3^ 
(1/2)*1i)/2 + 3/2))^(1/2)*(3*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)) - 2*ellipticPi( 
(3^(1/2)*1i)/6 + 1/2, 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*(x^3 - x*(((3^(1/2)*1i)/2 - 
 1/2)*((3^(1/2)*1i)/2 + 1/2) + 1) - ((3^(1/2)*1i)/2 - 1/2)*((3^(1/2)*1i)/2 
 + 1/2))^(1/2))
 

Reduce [F]

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

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

Output:

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