3.771 \(\int \frac{1}{\sqrt{(2-x) x (4-2 x+x^2)}} \, dx\)

Optimal. Leaf size=17 \[ -\frac{F\left (\sin ^{-1}(1-x)|-\frac{1}{3}\right )}{\sqrt{3}} \]

[Out]

-(EllipticF[ArcSin[1 - x], -1/3]/Sqrt[3])

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Rubi [A]  time = 0.0118359, antiderivative size = 17, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {1106, 1095, 419} \[ -\frac{F\left (\sin ^{-1}(1-x)|-\frac{1}{3}\right )}{\sqrt{3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(EllipticF[ArcSin[1 - x], -1/3]/Sqrt[3])

Rule 1106

Int[(P4_)^(p_), x_Symbol] :> With[{a = Coeff[P4, x, 0], b = Coeff[P4, x, 1], c = Coeff[P4, x, 2], d = Coeff[P4
, x, 3], e = Coeff[P4, x, 4]}, Subst[Int[SimplifyIntegrand[(a + d^4/(256*e^3) - (b*d)/(8*e) + (c - (3*d^2)/(8*
e))*x^2 + e*x^4)^p, x], x], x, d/(4*e) + x] /; EqQ[d^3 - 4*c*d*e + 8*b*e^2, 0] && NeQ[d, 0]] /; FreeQ[p, x] &&
PolyQ[P4, x, 4] && NeQ[p, 2] && NeQ[p, 3]

Rule 1095

Int[1/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Dist[2*Sqrt[-c], I
nt[1/(Sqrt[b + q + 2*c*x^2]*Sqrt[-b + q - 2*c*x^2]), x], x]] /; FreeQ[{a, b, c}, x] && GtQ[b^2 - 4*a*c, 0] &&
LtQ[c, 0]

Rule 419

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

Rubi steps

\begin{align*} \int \frac{1}{\sqrt{(2-x) x \left (4-2 x+x^2\right )}} \, dx &=\operatorname{Subst}\left (\int \frac{1}{\sqrt{3-2 x^2-x^4}} \, dx,x,-1+x\right )\\ &=2 \operatorname{Subst}\left (\int \frac{1}{\sqrt{2-2 x^2} \sqrt{6+2 x^2}} \, dx,x,-1+x\right )\\ &=-\frac{F\left (\sin ^{-1}(1-x)|-\frac{1}{3}\right )}{\sqrt{3}}\\ \end{align*}

Mathematica [C]  time = 0.262311, size = 100, normalized size = 5.88 \[ -\frac{\sqrt [3]{-1} (x-2)^2 \sqrt{\frac{x \left (x+i \sqrt{3}-1\right )}{(x-2)^2}} \sqrt{\frac{-\sqrt [3]{-1} x+x-2}{x-2}} F\left (\sin ^{-1}\left (\sqrt{-\frac{(-1)^{2/3} x}{x-2}}\right )|(-1)^{2/3}\right )}{\sqrt{-x \left (x^3-4 x^2+8 x-8\right )}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((-1)^(1/3)*(-2 + x)^2*Sqrt[(x*(-1 + I*Sqrt[3] + x))/(-2 + x)^2]*Sqrt[(-2 + x - (-1)^(1/3)*x)/(-2 + x)]*Elli
pticF[ArcSin[Sqrt[-(((-1)^(2/3)*x)/(-2 + x))]], (-1)^(2/3)])/Sqrt[-(x*(-8 + 8*x - 4*x^2 + x^3))])

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Maple [B]  time = 0.025, size = 200, normalized size = 11.8 \begin{align*} 2\,{\frac{ \left ( -i\sqrt{3}-1 \right ) \left ( -2+x \right ) ^{2}}{ \left ( i\sqrt{3}-1 \right ) \sqrt{-x \left ( -2+x \right ) \left ( x-1+i\sqrt{3} \right ) \left ( x-1-i\sqrt{3} \right ) }}\sqrt{{\frac{ \left ( i\sqrt{3}-1 \right ) x}{ \left ( 1+i\sqrt{3} \right ) \left ( -2+x \right ) }}}\sqrt{{\frac{x-1+i\sqrt{3}}{ \left ( 1-i\sqrt{3} \right ) \left ( -2+x \right ) }}}\sqrt{{\frac{x-1-i\sqrt{3}}{ \left ( 1+i\sqrt{3} \right ) \left ( -2+x \right ) }}}{\it EllipticF} \left ( \sqrt{{\frac{ \left ( i\sqrt{3}-1 \right ) x}{ \left ( 1+i\sqrt{3} \right ) \left ( -2+x \right ) }}},\sqrt{{\frac{ \left ( 1+i\sqrt{3} \right ) \left ( -i\sqrt{3}-1 \right ) }{ \left ( i\sqrt{3}-1 \right ) \left ( 1-i\sqrt{3} \right ) }}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{-{\left (x^{2} - 2 \, x + 4\right )}{\left (x - 2\right )} x}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{\sqrt{-x^{4} + 4 \, x^{3} - 8 \, x^{2} + 8 \, x}}{x^{4} - 4 \, x^{3} + 8 \, x^{2} - 8 \, x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-sqrt(-x^4 + 4*x^3 - 8*x^2 + 8*x)/(x^4 - 4*x^3 + 8*x^2 - 8*x), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{x \left (2 - x\right ) \left (x^{2} - 2 x + 4\right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{-{\left (x^{2} - 2 \, x + 4\right )}{\left (x - 2\right )} x}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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