\(\int \frac {\sqrt {1+x} \sqrt {1-x+x^2}}{x^3} \, dx\) [478]

Optimal result

Integrand size = 23, antiderivative size = 146 \[ \int \frac {\sqrt {1+x} \sqrt {1-x+x^2}}{x^3} \, dx=-\frac {\sqrt {1+x} \sqrt {1-x+x^2}}{2 x^2}+\frac {3^{3/4} \sqrt {2+\sqrt {3}} (1+x)^{3/2} \sqrt {1-x+x^2} \sqrt {\frac {1-x+x^2}{\left (1+\sqrt {3}+x\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {1-\sqrt {3}+x}{1+\sqrt {3}+x}\right ),-7-4 \sqrt {3}\right )}{2 \sqrt {\frac {1+x}{\left (1+\sqrt {3}+x\right )^2}} \left (1+x^3\right )} \] Output:

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 20.50 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.27 \[ \int \frac {\sqrt {1+x} \sqrt {1-x+x^2}}{x^3} \, dx=\frac {\sqrt {1+x} \left (-\frac {2 \left (1-x+x^2\right )}{x^2}-\frac {3 i \sqrt {2} \sqrt {\frac {i+\sqrt {3}-2 i x}{3 i+\sqrt {3}}} \sqrt {\frac {-i+\sqrt {3}+2 i x}{-3 i+\sqrt {3}}} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {2} \sqrt {-\frac {i (1+x)}{3 i+\sqrt {3}}}\right ),\frac {3 i+\sqrt {3}}{3 i-\sqrt {3}}\right )}{\sqrt {-\frac {i (1+x)}{3 i+\sqrt {3}}}}\right )}{4 \sqrt {1-x+x^2}} \] Input:

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

Output:

(Sqrt[1 + x]*((-2*(1 - x + x^2))/x^2 - ((3*I)*Sqrt[2]*Sqrt[(I + Sqrt[3] - 
(2*I)*x)/(3*I + Sqrt[3])]*Sqrt[(-I + Sqrt[3] + (2*I)*x)/(-3*I + Sqrt[3])]* 
EllipticF[I*ArcSinh[Sqrt[2]*Sqrt[((-I)*(1 + x))/(3*I + Sqrt[3])]], (3*I + 
Sqrt[3])/(3*I - Sqrt[3])])/Sqrt[((-I)*(1 + x))/(3*I + Sqrt[3])]))/(4*Sqrt[ 
1 - x + x^2])
 

Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.03, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {1210, 809, 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.

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

\(\Big \downarrow \) 1210

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

\(\Big \downarrow \) 809

\(\displaystyle \frac {\sqrt {x+1} \sqrt {x^2-x+1} \left (\frac {3}{4} \int \frac {1}{\sqrt {x^3+1}}dx-\frac {\sqrt {x^3+1}}{2 x^2}\right )}{\sqrt {x^3+1}}\)

\(\Big \downarrow \) 759

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

Input:

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

Output:

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

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 809

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c* 
x)^(m + 1)*((a + b*x^n)^p/(c*(m + 1))), x] - Simp[b*n*(p/(c^n*(m + 1)))   I 
nt[(c*x)^(m + n)*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && IGtQ 
[n, 0] && GtQ[p, 0] && LtQ[m, -1] &&  !ILtQ[(m + n*p + n + 1)/n, 0] && IntB 
inomialQ[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 = 2.87 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.09

method	result	size
elliptic	\(\frac {\sqrt {\left (x +1\right ) \left (x^{2}-x +1\right )}\, \left (-\frac {\sqrt {x^{3}+1}}{2 x^{2}}+\frac {3 \left (\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {\frac {x +1}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}-\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {x +1}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \sqrt {\frac {-\frac {3}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right )}{2 \sqrt {x^{3}+1}}\right )}{\sqrt {x +1}\, \sqrt {x^{2}-x +1}}\)	\(159\)
risch	\(-\frac {\sqrt {x +1}\, \sqrt {x^{2}-x +1}}{2 x^{2}}+\frac {3 \left (\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {\frac {x +1}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}-\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {x +1}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \sqrt {\frac {-\frac {3}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right ) \sqrt {\left (x +1\right ) \left (x^{2}-x +1\right )}}{2 \sqrt {x^{3}+1}\, \sqrt {x +1}\, \sqrt {x^{2}-x +1}}\)	\(166\)
default	\(-\frac {\sqrt {x +1}\, \sqrt {x^{2}-x +1}\, \left (3 i \sqrt {3}\, \sqrt {-\frac {2 \left (x +1\right )}{i \sqrt {3}-3}}\, \sqrt {\frac {i \sqrt {3}-2 x +1}{i \sqrt {3}+3}}\, \sqrt {\frac {i \sqrt {3}+2 x -1}{i \sqrt {3}-3}}\, \operatorname {EllipticF}\left (\sqrt {-\frac {2 \left (x +1\right )}{i \sqrt {3}-3}}, \sqrt {-\frac {i \sqrt {3}-3}{i \sqrt {3}+3}}\right ) x^{2}-9 \sqrt {-\frac {2 \left (x +1\right )}{i \sqrt {3}-3}}\, \sqrt {\frac {i \sqrt {3}-2 x +1}{i \sqrt {3}+3}}\, \sqrt {\frac {i \sqrt {3}+2 x -1}{i \sqrt {3}-3}}\, \operatorname {EllipticF}\left (\sqrt {-\frac {2 \left (x +1\right )}{i \sqrt {3}-3}}, \sqrt {-\frac {i \sqrt {3}-3}{i \sqrt {3}+3}}\right ) x^{2}+2 x^{3}+2\right )}{4 \left (x^{3}+1\right ) x^{2}}\)	\(259\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.22 \[ \int \frac {\sqrt {1+x} \sqrt {1-x+x^2}}{x^3} \, dx=\frac {3 \, x^{2} {\rm weierstrassPInverse}\left (0, -4, x\right ) - \sqrt {x^{2} - x + 1} \sqrt {x + 1}}{2 \, x^{2}} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\sqrt {1+x} \sqrt {1-x+x^2}}{x^3} \, dx=\frac {-2 \sqrt {x +1}\, \sqrt {x^{2}-x +1}-3 \left (\int \frac {\sqrt {x +1}\, \sqrt {x^{2}-x +1}}{x^{6}+x^{3}}d x \right ) x^{2}}{x^{2}} \] Input:

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

Output:

( - 2*sqrt(x + 1)*sqrt(x**2 - x + 1) - 3*int((sqrt(x + 1)*sqrt(x**2 - x + 
1))/(x**6 + x**3),x)*x**2)/x**2