\(\int \sqrt {(2-x) x (4-2 x+x^2)} \, dx\) [56]

Optimal result

Integrand size = 19, antiderivative size = 68 \[ \int \sqrt {(2-x) x \left (4-2 x+x^2\right )} \, dx=-\frac {1}{3} \sqrt {3-2 (1-x)^2-(1-x)^4} (1-x)+\frac {2 E\left (\arcsin (1-x)\left |-\frac {1}{3}\right .\right )}{\sqrt {3}}-\frac {4 \operatorname {EllipticF}\left (\arcsin (1-x),-\frac {1}{3}\right )}{\sqrt {3}} \] Output:

-1/3*(3-2*(1-x)^2-(1-x)^4)^(1/2)*(1-x)-2/3*3^(1/2)*EllipticE(-1+x,1/3*I*3^ 
(1/2))+4/3*3^(1/2)*EllipticF(-1+x,1/3*I*3^(1/2))
 

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 20.86 (sec) , antiderivative size = 256, normalized size of antiderivative = 3.76 \[ \int \sqrt {(2-x) x \left (4-2 x+x^2\right )} \, dx=\frac {\sqrt {-x \left (-8+8 x-4 x^2+x^3\right )} \left (\sqrt {\frac {4-2 x+x^2}{x^2}} \left (-4+4 x-3 x^2+x^3\right )+2 \sqrt {2} \left (-i+\sqrt {3}\right ) \sqrt {-\frac {i (-2+x)}{\left (-i+\sqrt {3}\right ) x}} E\left (\arcsin \left (\frac {\sqrt {i+\sqrt {3}-\frac {4 i}{x}}}{\sqrt {2} \sqrt [4]{3}}\right )|\frac {2 \sqrt {3}}{-i+\sqrt {3}}\right )+8 i \sqrt {2} \sqrt {-\frac {i (-2+x)}{\left (-i+\sqrt {3}\right ) x}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {i+\sqrt {3}-\frac {4 i}{x}}}{\sqrt {2} \sqrt [4]{3}}\right ),\frac {2 \sqrt {3}}{-i+\sqrt {3}}\right )\right )}{3 (-2+x) x \sqrt {\frac {4-2 x+x^2}{x^2}}} \] Input:

Integrate[Sqrt[(2 - x)*x*(4 - 2*x + x^2)],x]
 

Output:

(Sqrt[-(x*(-8 + 8*x - 4*x^2 + x^3))]*(Sqrt[(4 - 2*x + x^2)/x^2]*(-4 + 4*x 
- 3*x^2 + x^3) + 2*Sqrt[2]*(-I + Sqrt[3])*Sqrt[((-I)*(-2 + x))/((-I + Sqrt 
[3])*x)]*EllipticE[ArcSin[Sqrt[I + Sqrt[3] - (4*I)/x]/(Sqrt[2]*3^(1/4))], 
(2*Sqrt[3])/(-I + Sqrt[3])] + (8*I)*Sqrt[2]*Sqrt[((-I)*(-2 + x))/((-I + Sq 
rt[3])*x)]*EllipticF[ArcSin[Sqrt[I + Sqrt[3] - (4*I)/x]/(Sqrt[2]*3^(1/4))] 
, (2*Sqrt[3])/(-I + Sqrt[3])]))/(3*(-2 + x)*x*Sqrt[(4 - 2*x + x^2)/x^2])
 

Rubi [A] (verified)

Time = 0.22 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.97, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.421, Rules used = {2458, 1404, 27, 1494, 27, 399, 321, 327}

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.

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

\(\Big \downarrow \) 2458

\(\displaystyle \int \sqrt {-(x-1)^4-2 (x-1)^2+3}d(x-1)\)

\(\Big \downarrow \) 1404

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1494

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 399

\(\displaystyle \frac {2}{3} \left (6 \int \frac {1}{\sqrt {1-(x-1)^2} \sqrt {(x-1)^2+3}}d(x-1)-\int \frac {\sqrt {(x-1)^2+3}}{\sqrt {1-(x-1)^2}}d(x-1)\right )+\frac {1}{3} \sqrt {-(x-1)^4-2 (x-1)^2+3} (x-1)\)

\(\Big \downarrow \) 321

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

\(\Big \downarrow \) 327

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

Input:

Int[Sqrt[(2 - x)*x*(4 - 2*x + x^2)],x]
 

Output:

(Sqrt[3 - 2*(-1 + x)^2 - (-1 + x)^4]*(-1 + x))/3 + (2*(Sqrt[3]*EllipticE[A 
rcSin[1 - x], -1/3] - 2*Sqrt[3]*EllipticF[ArcSin[1 - x], -1/3]))/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 321

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> S 
imp[(1/(Sqrt[a]*Sqrt[c]*Rt[-d/c, 2]))*EllipticF[ArcSin[Rt[-d/c, 2]*x], b*(c 
/(a*d))], x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && GtQ[a, 
0] &&  !(NegQ[b/a] && SimplerSqrtQ[-b/a, -d/c])
 

rule 327

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[ 
(Sqrt[a]/(Sqrt[c]*Rt[-d/c, 2]))*EllipticE[ArcSin[Rt[-d/c, 2]*x], b*(c/(a*d) 
)], x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && GtQ[a, 0]
 

rule 399

Int[((e_) + (f_.)*(x_)^2)/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_) 
^2]), x_Symbol] :> Simp[f/b   Int[Sqrt[a + b*x^2]/Sqrt[c + d*x^2], x], x] + 
 Simp[(b*e - a*f)/b   Int[1/(Sqrt[a + b*x^2]*Sqrt[c + d*x^2]), x], x] /; Fr 
eeQ[{a, b, c, d, e, f}, x] &&  !((PosQ[b/a] && PosQ[d/c]) || (NegQ[b/a] && 
(PosQ[d/c] || (GtQ[a, 0] && ( !GtQ[c, 0] || SimplerSqrtQ[-b/a, -d/c])))))
 

rule 1404

Int[((a_.) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[x*((a + b 
*x^2 + c*x^4)^p/(4*p + 1)), x] + Simp[2*(p/(4*p + 1))   Int[(2*a + b*x^2)*( 
a + b*x^2 + c*x^4)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a* 
c, 0] && GtQ[p, 0] && IntegerQ[2*p]
 

rule 1494

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbo 
l] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[2*Sqrt[-c]   Int[(d + e*x^2)/(Sqr 
t[b + q + 2*c*x^2]*Sqrt[-b + q - 2*c*x^2]), x], x]] /; FreeQ[{a, b, c, d, e 
}, x] && GtQ[b^2 - 4*a*c, 0] && LtQ[c, 0]
 

rule 2458

Int[(Pn_)^(p_.), x_Symbol] :> With[{S = Coeff[Pn, x, Expon[Pn, x] - 1]/(Exp 
on[Pn, x]*Coeff[Pn, x, Expon[Pn, x]])}, Subst[Int[ExpandToSum[Pn /. x -> x 
- S, x]^p, x], x, x + S] /; BinomialQ[Pn /. x -> x - S, x] || (IntegerQ[Exp 
on[Pn, x]/2] && TrinomialQ[Pn /. x -> x - S, x])] /; FreeQ[p, x] && PolyQ[P 
n, x] && GtQ[Expon[Pn, x], 2] && NeQ[Coeff[Pn, x, Expon[Pn, x] - 1], 0]
 

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 931 vs. \(2 (60 ) = 120\).

Time = 1.27 (sec) , antiderivative size = 932, normalized size of antiderivative = 13.71

method	result	size
risch	\(\text {Expression too large to display}\)	\(932\)
default	\(\text {Expression too large to display}\)	\(946\)
elliptic	\(\text {Expression too large to display}\)	\(946\)

Input:

int(((2-x)*x*(x^2-2*x+4))^(1/2),x,method=_RETURNVERBOSE)
 

Output:

-1/3*(x-1)*x*(x-2)*(x^2-2*x+4)/(-x*(x-2)*(x^2-2*x+4))^(1/2)-2/3*(x*(x-1+I* 
3^(1/2))*(x-1-I*3^(1/2))+2*(-1-I*3^(1/2))*((-1+I*3^(1/2))*x/(1+I*3^(1/2))/ 
(x-2))^(1/2)*(x-2)^2*((x-1+I*3^(1/2))/(1-I*3^(1/2))/(x-2))^(1/2)*((x-1-I*3 
^(1/2))/(1+I*3^(1/2))/(x-2))^(1/2)*(1/2*(6+2*I*3^(1/2))/(-1+I*3^(1/2))*Ell 
ipticF(((-1+I*3^(1/2))*x/(1+I*3^(1/2))/(x-2))^(1/2),((1+I*3^(1/2))*(-1-I*3 
^(1/2))/(-1+I*3^(1/2))/(1-I*3^(1/2)))^(1/2))+1/2*(-1+I*3^(1/2))*EllipticE( 
((-1+I*3^(1/2))*x/(1+I*3^(1/2))/(x-2))^(1/2),((1+I*3^(1/2))*(-1-I*3^(1/2)) 
/(-1+I*3^(1/2))/(1-I*3^(1/2)))^(1/2))-4/(-1+I*3^(1/2))*EllipticPi(((-1+I*3 
^(1/2))*x/(1+I*3^(1/2))/(x-2))^(1/2),(-1-I*3^(1/2))/(1-I*3^(1/2)),((1+I*3^ 
(1/2))*(-1-I*3^(1/2))/(-1+I*3^(1/2))/(1-I*3^(1/2)))^(1/2))))/(-x*(x-2)*(x- 
1+I*3^(1/2))*(x-1-I*3^(1/2)))^(1/2)+8/3*(-1-I*3^(1/2))*((-1+I*3^(1/2))*x/( 
1+I*3^(1/2))/(x-2))^(1/2)*(x-2)^2*((x-1+I*3^(1/2))/(1-I*3^(1/2))/(x-2))^(1 
/2)*((x-1-I*3^(1/2))/(1+I*3^(1/2))/(x-2))^(1/2)/(-1+I*3^(1/2))/(-x*(x-2)*( 
x-1+I*3^(1/2))*(x-1-I*3^(1/2)))^(1/2)*EllipticF(((-1+I*3^(1/2))*x/(1+I*3^( 
1/2))/(x-2))^(1/2),((1+I*3^(1/2))*(-1-I*3^(1/2))/(-1+I*3^(1/2))/(1-I*3^(1/ 
2)))^(1/2))+8/3*(-1-I*3^(1/2))*((-1+I*3^(1/2))*x/(1+I*3^(1/2))/(x-2))^(1/2 
)*(x-2)^2*((x-1+I*3^(1/2))/(1-I*3^(1/2))/(x-2))^(1/2)*((x-1-I*3^(1/2))/(1+ 
I*3^(1/2))/(x-2))^(1/2)/(-1+I*3^(1/2))/(-x*(x-2)*(x-1+I*3^(1/2))*(x-1-I*3^ 
(1/2)))^(1/2)*(2*EllipticF(((-1+I*3^(1/2))*x/(1+I*3^(1/2))/(x-2))^(1/2),(( 
1+I*3^(1/2))*(-1-I*3^(1/2))/(-1+I*3^(1/2))/(1-I*3^(1/2)))^(1/2))-2*Elli...
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.01 \[ \int \sqrt {(2-x) x \left (4-2 x+x^2\right )} \, dx=-\frac {2 \, {\left (-i \, x + i\right )} E(\arcsin \left (\frac {1}{x - 1}\right )\,|\,-3) + 4 \, {\left (-i \, x + i\right )} F(\arcsin \left (\frac {1}{x - 1}\right )\,|\,-3) - \sqrt {-x^{4} + 4 \, x^{3} - 8 \, x^{2} + 8 \, x} {\left (x^{2} - 2 \, x + 3\right )}}{3 \, {\left (x - 1\right )}} \] Input:

integrate(((2-x)*x*(x^2-2*x+4))^(1/2),x, algorithm="fricas")
 

Output:

-1/3*(2*(-I*x + I)*elliptic_e(arcsin(1/(x - 1)), -3) + 4*(-I*x + I)*ellipt 
ic_f(arcsin(1/(x - 1)), -3) - sqrt(-x^4 + 4*x^3 - 8*x^2 + 8*x)*(x^2 - 2*x 
+ 3))/(x - 1)
 

Sympy [F]

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

integrate(((2-x)*x*(x**2-2*x+4))**(1/2),x)
 

Output:

Integral(sqrt(x*(2 - x)*(x**2 - 2*x + 4)), x)
 

Maxima [F]

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

integrate(((2-x)*x*(x^2-2*x+4))^(1/2),x, algorithm="maxima")
 

Output:

integrate(sqrt(-(x^2 - 2*x + 4)*(x - 2)*x), x)
 

Giac [F]

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

integrate(((2-x)*x*(x^2-2*x+4))^(1/2),x, algorithm="giac")
 

Output:

integrate(sqrt(-(x^2 - 2*x + 4)*(x - 2)*x), x)
 

Mupad [F(-1)]

Timed out. \[ \int \sqrt {(2-x) x \left (4-2 x+x^2\right )} \, dx=\int \sqrt {-x\,\left (x-2\right )\,\left (x^2-2\,x+4\right )} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \sqrt {(2-x) x \left (4-2 x+x^2\right )} \, dx=\frac {\sqrt {x}\, \sqrt {-x^{3}+4 x^{2}-8 x +8}\, x}{3}-\frac {4 \sqrt {x}\, \sqrt {-x^{3}+4 x^{2}-8 x +8}}{9}+\frac {2 \left (\int \frac {\sqrt {x}\, \sqrt {-x^{3}+4 x^{2}-8 x +8}\, x^{2}}{x^{3}-4 x^{2}+8 x -8}d x \right )}{9}-\frac {16 \left (\int \frac {\sqrt {x}\, \sqrt {-x^{3}+4 x^{2}-8 x +8}}{x^{4}-4 x^{3}+8 x^{2}-8 x}d x \right )}{9}-\frac {4 \left (\int \frac {\sqrt {x}\, \sqrt {-x^{3}+4 x^{2}-8 x +8}}{x^{3}-4 x^{2}+8 x -8}d x \right )}{9} \] Input:

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

Output:

(3*sqrt(x)*sqrt( - x**3 + 4*x**2 - 8*x + 8)*x - 4*sqrt(x)*sqrt( - x**3 + 4 
*x**2 - 8*x + 8) + 2*int((sqrt(x)*sqrt( - x**3 + 4*x**2 - 8*x + 8)*x**2)/( 
x**3 - 4*x**2 + 8*x - 8),x) - 16*int((sqrt(x)*sqrt( - x**3 + 4*x**2 - 8*x 
+ 8))/(x**4 - 4*x**3 + 8*x**2 - 8*x),x) - 4*int((sqrt(x)*sqrt( - x**3 + 4* 
x**2 - 8*x + 8))/(x**3 - 4*x**2 + 8*x - 8),x))/9