3.772 \(\int \frac {1}{((2-x) x (4-2 x+x^2))^{3/2}} \, dx\)

Optimal. Leaf size=73 \[ \frac {\left ((x-1)^2+5\right ) (x-1)}{24 \sqrt {-(x-1)^4-2 (x-1)^2+3}}-\frac {F\left (\sin ^{-1}(1-x)|-\frac {1}{3}\right )}{4 \sqrt {3}}+\frac {E\left (\sin ^{-1}(1-x)|-\frac {1}{3}\right )}{8 \sqrt {3}} \]

[Out]

-1/24*EllipticE(-1+x,1/3*I*3^(1/2))*3^(1/2)+1/12*EllipticF(-1+x,1/3*I*3^(1/2))*3^(1/2)+1/24*(5+(-1+x)^2)*(-1+x
)/(3-2*(-1+x)^2-(-1+x)^4)^(1/2)

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Rubi [A]  time = 0.05, antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {1106, 1092, 1180, 524, 424, 419} \[ \frac {\left ((x-1)^2+5\right ) (x-1)}{24 \sqrt {-(x-1)^4-2 (x-1)^2+3}}-\frac {F\left (\sin ^{-1}(1-x)|-\frac {1}{3}\right )}{4 \sqrt {3}}+\frac {E\left (\sin ^{-1}(1-x)|-\frac {1}{3}\right )}{8 \sqrt {3}} \]

Antiderivative was successfully verified.

[In]

Int[((2 - x)*x*(4 - 2*x + x^2))^(-3/2),x]

[Out]

((5 + (-1 + x)^2)*(-1 + x))/(24*Sqrt[3 - 2*(-1 + x)^2 - (-1 + x)^4]) + EllipticE[ArcSin[1 - x], -1/3]/(8*Sqrt[
3]) - EllipticF[ArcSin[1 - x], -1/3]/(4*Sqrt[3])

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)])

Rule 424

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

Rule 524

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

Rule 1092

Int[((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> -Simp[(x*(b^2 - 2*a*c + b*c*x^2)*(a + b*x^2 + c*x^
4)^(p + 1))/(2*a*(p + 1)*(b^2 - 4*a*c)), x] + Dist[1/(2*a*(p + 1)*(b^2 - 4*a*c)), Int[(b^2 - 2*a*c + 2*(p + 1)
*(b^2 - 4*a*c) + b*c*(4*p + 7)*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] && LtQ[p, -1] && IntegerQ[2*p]

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 1180

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}
, Dist[2*Sqrt[-c], Int[(d + e*x^2)/(Sqrt[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]

Rubi steps

\begin {align*} \int \frac {1}{\left ((2-x) x \left (4-2 x+x^2\right )\right )^{3/2}} \, dx &=\operatorname {Subst}\left (\int \frac {1}{\left (3-2 x^2-x^4\right )^{3/2}} \, dx,x,-1+x\right )\\ &=\frac {\left (5+(-1+x)^2\right ) (-1+x)}{24 \sqrt {3-2 (-1+x)^2-(-1+x)^4}}-\frac {1}{48} \operatorname {Subst}\left (\int \frac {-6+2 x^2}{\sqrt {3-2 x^2-x^4}} \, dx,x,-1+x\right )\\ &=\frac {\left (5+(-1+x)^2\right ) (-1+x)}{24 \sqrt {3-2 (-1+x)^2-(-1+x)^4}}-\frac {1}{24} \operatorname {Subst}\left (\int \frac {-6+2 x^2}{\sqrt {2-2 x^2} \sqrt {6+2 x^2}} \, dx,x,-1+x\right )\\ &=\frac {\left (5+(-1+x)^2\right ) (-1+x)}{24 \sqrt {3-2 (-1+x)^2-(-1+x)^4}}-\frac {1}{24} \operatorname {Subst}\left (\int \frac {\sqrt {6+2 x^2}}{\sqrt {2-2 x^2}} \, dx,x,-1+x\right )+\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{\sqrt {2-2 x^2} \sqrt {6+2 x^2}} \, dx,x,-1+x\right )\\ &=\frac {\left (5+(-1+x)^2\right ) (-1+x)}{24 \sqrt {3-2 (-1+x)^2-(-1+x)^4}}+\frac {E\left (\sin ^{-1}(1-x)|-\frac {1}{3}\right )}{8 \sqrt {3}}-\frac {F\left (\sin ^{-1}(1-x)|-\frac {1}{3}\right )}{4 \sqrt {3}}\\ \end {align*}

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Mathematica [C]  time = 0.99, size = 298, normalized size = 4.08 \[ \frac {(x-2)^2 x \left (x^2-2 x+4\right ) \left (-\frac {3 \left (x^2-2 x+4\right ) x}{x-2}-3 \left (x^2-2 x+4\right )-4 (2-x) \sqrt {\frac {x^2-2 x+4}{(x-2)^2}} \left (\sqrt {\frac {x^2-2 x+4}{(x-2)^2}} x+4 i \sqrt {2} \sqrt {\frac {i x}{\left (\sqrt {3}+i\right ) (x-2)}} F\left (\sin ^{-1}\left (\frac {\sqrt {\sqrt {3}-i-\frac {4 i}{x-2}}}{\sqrt {2} \sqrt [4]{3}}\right )|\frac {2 \sqrt {3}}{i+\sqrt {3}}\right )-\sqrt {2} \left (\sqrt {3}+i\right ) \sqrt {\frac {i x}{\left (\sqrt {3}+i\right ) (x-2)}} E\left (\sin ^{-1}\left (\frac {\sqrt {\sqrt {3}-i-\frac {4 i}{x-2}}}{\sqrt {2} \sqrt [4]{3}}\right )|\frac {2 \sqrt {3}}{i+\sqrt {3}}\right )\right )+2 (x-1) x\right )}{96 \left (-x \left (x^3-4 x^2+8 x-8\right )\right )^{3/2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[((2 - x)*x*(4 - 2*x + x^2))^(-3/2),x]

[Out]

((-2 + x)^2*x*(4 - 2*x + x^2)*(2*(-1 + x)*x - 3*(4 - 2*x + x^2) - (3*x*(4 - 2*x + x^2))/(-2 + x) - 4*(2 - x)*S
qrt[(4 - 2*x + x^2)/(-2 + x)^2]*(x*Sqrt[(4 - 2*x + x^2)/(-2 + x)^2] - Sqrt[2]*(I + Sqrt[3])*Sqrt[(I*x)/((I + S
qrt[3])*(-2 + x))]*EllipticE[ArcSin[Sqrt[-I + Sqrt[3] - (4*I)/(-2 + x)]/(Sqrt[2]*3^(1/4))], (2*Sqrt[3])/(I + S
qrt[3])] + (4*I)*Sqrt[2]*Sqrt[(I*x)/((I + Sqrt[3])*(-2 + x))]*EllipticF[ArcSin[Sqrt[-I + Sqrt[3] - (4*I)/(-2 +
x)]/(Sqrt[2]*3^(1/4))], (2*Sqrt[3])/(I + Sqrt[3])])))/(96*(-(x*(-8 + 8*x - 4*x^2 + x^3)))^(3/2))

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fricas [F]  time = 0.41, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {-x^{4} + 4 \, x^{3} - 8 \, x^{2} + 8 \, x}}{x^{8} - 8 \, x^{7} + 32 \, x^{6} - 80 \, x^{5} + 128 \, x^{4} - 128 \, x^{3} + 64 \, x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((2-x)*x*(x^2-2*x+4))^(3/2),x, algorithm="fricas")

[Out]

integral(sqrt(-x^4 + 4*x^3 - 8*x^2 + 8*x)/(x^8 - 8*x^7 + 32*x^6 - 80*x^5 + 128*x^4 - 128*x^3 + 64*x^2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (-{\left (x^{2} - 2 \, x + 4\right )} {\left (x - 2\right )} x\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((2-x)*x*(x^2-2*x+4))^(3/2),x, algorithm="giac")

[Out]

integrate((-(x^2 - 2*x + 4)*(x - 2)*x)^(-3/2), x)

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maple [B]  time = 0.08, size = 963, normalized size = 13.19 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(1/192*x^2+1/24)/(-(x^3-4*x^2+8*x-8)*x)^(1/2)*x-1/32*(-x^3+4*x^2-8*x+8)/((-x^3+4*x^2-8*x+8)*x)^(1/2)+1/6*(-1
-I*3^(1/2))*((I*3^(1/2)-1)/(1+I*3^(1/2))/(x-2)*x)^(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)/(I*3^(1/2)-1)/(-(x-2)*(x-1+I*3^(1/2))*(x-1-I*3^(1/2))*x)^(1/2)*Ellip
ticF(((I*3^(1/2)-1)/(1+I*3^(1/2))/(x-2)*x)^(1/2),((1+I*3^(1/2))*(-1-I*3^(1/2))/(I*3^(1/2)-1)/(1-I*3^(1/2)))^(1
/2))+1/6*(-1-I*3^(1/2))*((I*3^(1/2)-1)/(1+I*3^(1/2))/(x-2)*x)^(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)/(I*3^(1/2)-1)/(-(x-2)*(x-1+I*3^(1/2))*(x-1-I*3^(1/2))*x)
^(1/2)*(2*EllipticF(((I*3^(1/2)-1)/(1+I*3^(1/2))/(x-2)*x)^(1/2),((1+I*3^(1/2))*(-1-I*3^(1/2))/(I*3^(1/2)-1)/(1
-I*3^(1/2)))^(1/2))-2*EllipticPi(((I*3^(1/2)-1)/(1+I*3^(1/2))/(x-2)*x)^(1/2),(1+I*3^(1/2))/(I*3^(1/2)-1),((1+I
*3^(1/2))*(-1-I*3^(1/2))/(I*3^(1/2)-1)/(1-I*3^(1/2)))^(1/2)))-1/24*((x-1+I*3^(1/2))*(x-1-I*3^(1/2))*x+2*(-1-I*
3^(1/2))*((I*3^(1/2)-1)/(1+I*3^(1/2))/(x-2)*x)^(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))/(I*3^(1/2)-1)*EllipticF(((I*3^(1/2)-1)/(1+I*3^(1/2
))/(x-2)*x)^(1/2),((1+I*3^(1/2))*(-1-I*3^(1/2))/(I*3^(1/2)-1)/(1-I*3^(1/2)))^(1/2))+1/2*(I*3^(1/2)-1)*Elliptic
E(((I*3^(1/2)-1)/(1+I*3^(1/2))/(x-2)*x)^(1/2),((1+I*3^(1/2))*(-1-I*3^(1/2))/(I*3^(1/2)-1)/(1-I*3^(1/2)))^(1/2)
)-4/(I*3^(1/2)-1)*EllipticPi(((I*3^(1/2)-1)/(1+I*3^(1/2))/(x-2)*x)^(1/2),(-1-I*3^(1/2))/(1-I*3^(1/2)),((1+I*3^
(1/2))*(-1-I*3^(1/2))/(I*3^(1/2)-1)/(1-I*3^(1/2)))^(1/2))))/(-(x-2)*(x-1+I*3^(1/2))*(x-1-I*3^(1/2))*x)^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (-{\left (x^{2} - 2 \, x + 4\right )} {\left (x - 2\right )} x\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((2-x)*x*(x^2-2*x+4))^(3/2),x, algorithm="maxima")

[Out]

integrate((-(x^2 - 2*x + 4)*(x - 2)*x)^(-3/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\left (-x\,\left (x-2\right )\,\left (x^2-2\,x+4\right )\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (x \left (2 - x\right ) \left (x^{2} - 2 x + 4\right )\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((x*(2 - x)*(x**2 - 2*x + 4))**(-3/2), x)

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