\(\int \frac {x^3}{\sqrt {1+x} \sqrt {1-x+x^2}} \, dx\) [486]

Optimal result

Integrand size = 23, antiderivative size = 142 \[ \int \frac {x^3}{\sqrt {1+x} \sqrt {1-x+x^2}} \, dx=\frac {2 x \left (1+x^3\right )}{5 \sqrt {1+x} \sqrt {1-x+x^2}}-\frac {4 \sqrt {2+\sqrt {3}} \sqrt {1+x} \sqrt {\frac {1-x+x^2}{\left (1+\sqrt {3}+x\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {1-\sqrt {3}+x}{1+\sqrt {3}+x}\right ),-7-4 \sqrt {3}\right )}{5 \sqrt [4]{3} \sqrt {\frac {1+x}{\left (1+\sqrt {3}+x\right )^2}} \sqrt {1-x+x^2}} \] Output:

2/5*x*(x^3+1)/(1+x)^(1/2)/(x^2-x+1)^(1/2)-4/15*(1/2*6^(1/2)+1/2*2^(1/2))*( 
1+x)^(1/2)*((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^2-x+1)^(1 
/2)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 21.75 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.19 \[ \int \frac {x^3}{\sqrt {1+x} \sqrt {1-x+x^2}} \, dx=\frac {6 x \sqrt {1+x} \left (1-x+x^2\right )-\frac {2 i (1+x) \sqrt {1+\frac {6 i}{\left (-3 i+\sqrt {3}\right ) (1+x)}} \sqrt {6-\frac {36 i}{\left (3 i+\sqrt {3}\right ) (1+x)}} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\frac {\sqrt {-\frac {6 i}{3 i+\sqrt {3}}}}{\sqrt {1+x}}\right ),\frac {3 i+\sqrt {3}}{3 i-\sqrt {3}}\right )}{\sqrt {-\frac {i}{3 i+\sqrt {3}}}}}{15 \sqrt {1-x+x^2}} \] Input:

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

Output:

(6*x*Sqrt[1 + x]*(1 - x + x^2) - ((2*I)*(1 + x)*Sqrt[1 + (6*I)/((-3*I + Sq 
rt[3])*(1 + x))]*Sqrt[6 - (36*I)/((3*I + Sqrt[3])*(1 + x))]*EllipticF[I*Ar 
cSinh[Sqrt[(-6*I)/(3*I + Sqrt[3])]/Sqrt[1 + x]], (3*I + Sqrt[3])/(3*I - Sq 
rt[3])])/Sqrt[(-I)/(3*I + Sqrt[3])])/(15*Sqrt[1 - x + x^2])
 

Rubi [A] (verified)

Time = 0.41 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.05, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {1210, 843, 759}

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.

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

\(\Big \downarrow \) 1210

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

\(\Big \downarrow \) 843

\(\displaystyle \frac {\sqrt {x^3+1} \left (\frac {2}{5} x \sqrt {x^3+1}-\frac {2}{5} \int \frac {1}{\sqrt {x^3+1}}dx\right )}{\sqrt {x+1} \sqrt {x^2-x+1}}\)

\(\Big \downarrow \) 759

\(\displaystyle \frac {\sqrt {x^3+1} \left (\frac {2}{5} x \sqrt {x^3+1}-\frac {4 \sqrt {2+\sqrt {3}} (x+1) \sqrt {\frac {x^2-x+1}{\left (x+\sqrt {3}+1\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {x-\sqrt {3}+1}{x+\sqrt {3}+1}\right ),-7-4 \sqrt {3}\right )}{5 \sqrt [4]{3} \sqrt {\frac {x+1}{\left (x+\sqrt {3}+1\right )^2}} \sqrt {x^3+1}}\right )}{\sqrt {x+1} \sqrt {x^2-x+1}}\)

Input:

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

Output:

(Sqrt[1 + x^3]*((2*x*Sqrt[1 + x^3])/5 - (4*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]])/(5*3^(1/4)*Sqrt[(1 + x)/(1 + Sqrt[3] + x) 
^2]*Sqrt[1 + x^3])))/(Sqrt[1 + x]*Sqrt[1 - x + x^2])
 

Defintions of rubi rules used

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 843

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n 
 - 1)*(c*x)^(m - n + 1)*((a + b*x^n)^(p + 1)/(b*(m + n*p + 1))), x] - Simp[ 
a*c^n*((m - n + 1)/(b*(m + n*p + 1)))   Int[(c*x)^(m - n)*(a + b*x^n)^p, x] 
, x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n* 
p + 1, 0] && IntBinomialQ[a, b, c, n, m, p, x]
 

rule 1210

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_.)*((a_) + (b_.)*(x_) 
 + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d + e*x)^FracPart[p]*((a + b*x + 
c*x^2)^FracPart[p]/(a*d + c*e*x^3)^FracPart[p])   Int[(f + g*x)^n*(a*d + c* 
e*x^3)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[b*d + a*e, 
 0] && EqQ[c*d + b*e, 0] && EqQ[m, p]
 

Maple [A] (verified)

Time = 3.02 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.11

method	result	size
elliptic	\(\frac {\sqrt {\left (x +1\right ) \left (x^{2}-x +1\right )}\, \left (\frac {2 x \sqrt {x^{3}+1}}{5}-\frac {4 \left (\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {\frac {x +1}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}-\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {x +1}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \sqrt {\frac {-\frac {3}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right )}{5 \sqrt {x^{3}+1}}\right )}{\sqrt {x +1}\, \sqrt {x^{2}-x +1}}\)	\(157\)
risch	\(\frac {2 x \sqrt {x +1}\, \sqrt {x^{2}-x +1}}{5}-\frac {4 \left (\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {\frac {x +1}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}-\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {x +1}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \sqrt {\frac {-\frac {3}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right ) \sqrt {\left (x +1\right ) \left (x^{2}-x +1\right )}}{5 \sqrt {x^{3}+1}\, \sqrt {x +1}\, \sqrt {x^{2}-x +1}}\)	\(164\)
default	\(\frac {2 \sqrt {x +1}\, \sqrt {x^{2}-x +1}\, \left (i \sqrt {3}\, \sqrt {-\frac {2 \left (x +1\right )}{i \sqrt {3}-3}}\, \sqrt {\frac {i \sqrt {3}-2 x +1}{i \sqrt {3}+3}}\, \sqrt {\frac {i \sqrt {3}+2 x -1}{i \sqrt {3}-3}}\, \operatorname {EllipticF}\left (\sqrt {-\frac {2 \left (x +1\right )}{i \sqrt {3}-3}}, \sqrt {-\frac {i \sqrt {3}-3}{i \sqrt {3}+3}}\right )-3 \sqrt {-\frac {2 \left (x +1\right )}{i \sqrt {3}-3}}\, \sqrt {\frac {i \sqrt {3}-2 x +1}{i \sqrt {3}+3}}\, \sqrt {\frac {i \sqrt {3}+2 x -1}{i \sqrt {3}-3}}\, \operatorname {EllipticF}\left (\sqrt {-\frac {2 \left (x +1\right )}{i \sqrt {3}-3}}, \sqrt {-\frac {i \sqrt {3}-3}{i \sqrt {3}+3}}\right )+x^{4}+x \right )}{5 \left (x^{3}+1\right )}\)	\(248\)

Input:

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

Output:

((x+1)*(x^2-x+1))^(1/2)/(x+1)^(1/2)/(x^2-x+1)^(1/2)*(2/5*x*(x^3+1)^(1/2)-4 
/5*(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)))
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.18 \[ \int \frac {x^3}{\sqrt {1+x} \sqrt {1-x+x^2}} \, dx=\frac {2}{5} \, \sqrt {x^{2} - x + 1} \sqrt {x + 1} x - \frac {4}{5} \, {\rm weierstrassPInverse}\left (0, -4, x\right ) \] Input:

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

Output:

2/5*sqrt(x^2 - x + 1)*sqrt(x + 1)*x - 4/5*weierstrassPInverse(0, -4, x)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {x^3}{\sqrt {1+x} \sqrt {1-x+x^2}} \, dx=\int \frac {x^3}{\sqrt {x+1}\,\sqrt {x^2-x+1}} \,d x \] Input:

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

Output:

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

Reduce [F]

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

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

Output:

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