3.8.73 \(\int \frac {1}{((2-x) x (4-2 x+x^2))^{5/2}} \, dx\) [773]

Optimal. Leaf size=109 \[ \frac {\left (5+(-1+x)^2\right ) (-1+x)}{72 \left (3-2 (-1+x)^2-(-1+x)^4\right )^{3/2}}+\frac {\left (26+7 (-1+x)^2\right ) (-1+x)}{432 \sqrt {3-2 (-1+x)^2-(-1+x)^4}}+\frac {7 E\left (\sin ^{-1}(1-x)|-\frac {1}{3}\right )}{144 \sqrt {3}}-\frac {11 F\left (\sin ^{-1}(1-x)|-\frac {1}{3}\right )}{144 \sqrt {3}} \]

[Out]

1/72*(5+(-1+x)^2)*(-1+x)/(3-2*(-1+x)^2-(-1+x)^4)^(3/2)-7/432*EllipticE(-1+x,1/3*I*3^(1/2))*3^(1/2)+11/432*Elli
pticF(-1+x,1/3*I*3^(1/2))*3^(1/2)+1/432*(26+7*(-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 = 109, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {1120, 1106, 1192, 1194, 538, 435, 430} \begin {gather*} -\frac {11 F\left (\text {ArcSin}(1-x)\left |-\frac {1}{3}\right .\right )}{144 \sqrt {3}}+\frac {7 E\left (\text {ArcSin}(1-x)\left |-\frac {1}{3}\right .\right )}{144 \sqrt {3}}+\frac {\left (7 (x-1)^2+26\right ) (x-1)}{432 \sqrt {-(x-1)^4-2 (x-1)^2+3}}+\frac {\left ((x-1)^2+5\right ) (x-1)}{72 \left (-(x-1)^4-2 (x-1)^2+3\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((2 - x)*x*(4 - 2*x + x^2))^(-5/2),x]

[Out]

((5 + (-1 + x)^2)*(-1 + x))/(72*(3 - 2*(-1 + x)^2 - (-1 + x)^4)^(3/2)) + ((26 + 7*(-1 + x)^2)*(-1 + x))/(432*S
qrt[3 - 2*(-1 + x)^2 - (-1 + x)^4]) + (7*EllipticE[ArcSin[1 - x], -1/3])/(144*Sqrt[3]) - (11*EllipticF[ArcSin[
1 - x], -1/3])/(144*Sqrt[3])

Rule 430

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[(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] && Gt
Q[a, 0] &&  !(NegQ[b/a] && SimplerSqrtQ[-b/a, -d/c])

Rule 435

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[(Sqrt[a]/(Sqrt[c]*Rt[-d/c, 2]))*Ell
ipticE[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 538

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 1106

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 1120

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 1192

Int[((d_) + (e_.)*(x_)^2)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[x*(a*b*e - d*(b^2 - 2*a
*c) - c*(b*d - 2*a*e)*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[Simp[(2*p + 3)*d*b^2 - a*b*e - 2*a*c*d*(4*p + 5) + (4*p + 7)*(d*b - 2*a*e)*c*x^2, x]*(a +
b*x^2 + c*x^4)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e
^2, 0] && LtQ[p, -1] && IntegerQ[2*p]

Rule 1194

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 )^{5/2}} \, dx &=\text {Subst}\left (\int \frac {1}{\left (3-2 x^2-x^4\right )^{5/2}} \, dx,x,-1+x\right )\\ &=\frac {\left (5+(-1+x)^2\right ) (-1+x)}{72 \left (3-2 (-1+x)^2-(-1+x)^4\right )^{3/2}}-\frac {1}{144} \text {Subst}\left (\int \frac {-38-6 x^2}{\left (3-2 x^2-x^4\right )^{3/2}} \, dx,x,-1+x\right )\\ &=-\frac {\left (26+7 (1-x)^2\right ) (1-x)}{432 \sqrt {3-2 (1-x)^2-(1-x)^4}}+\frac {\left (5+(-1+x)^2\right ) (-1+x)}{72 \left (3-2 (-1+x)^2-(-1+x)^4\right )^{3/2}}+\frac {\text {Subst}\left (\int \frac {192-112 x^2}{\sqrt {3-2 x^2-x^4}} \, dx,x,-1+x\right )}{6912}\\ &=-\frac {\left (26+7 (1-x)^2\right ) (1-x)}{432 \sqrt {3-2 (1-x)^2-(1-x)^4}}+\frac {\left (5+(-1+x)^2\right ) (-1+x)}{72 \left (3-2 (-1+x)^2-(-1+x)^4\right )^{3/2}}+\frac {\text {Subst}\left (\int \frac {192-112 x^2}{\sqrt {2-2 x^2} \sqrt {6+2 x^2}} \, dx,x,-1+x\right )}{3456}\\ &=-\frac {\left (26+7 (1-x)^2\right ) (1-x)}{432 \sqrt {3-2 (1-x)^2-(1-x)^4}}+\frac {\left (5+(-1+x)^2\right ) (-1+x)}{72 \left (3-2 (-1+x)^2-(-1+x)^4\right )^{3/2}}-\frac {7}{432} \text {Subst}\left (\int \frac {\sqrt {6+2 x^2}}{\sqrt {2-2 x^2}} \, dx,x,-1+x\right )+\frac {11}{72} \text {Subst}\left (\int \frac {1}{\sqrt {2-2 x^2} \sqrt {6+2 x^2}} \, dx,x,-1+x\right )\\ &=-\frac {\left (26+7 (1-x)^2\right ) (1-x)}{432 \sqrt {3-2 (1-x)^2-(1-x)^4}}+\frac {\left (5+(-1+x)^2\right ) (-1+x)}{72 \left (3-2 (-1+x)^2-(-1+x)^4\right )^{3/2}}+\frac {7 E\left (\sin ^{-1}(1-x)|-\frac {1}{3}\right )}{144 \sqrt {3}}-\frac {11 F\left (\sin ^{-1}(1-x)|-\frac {1}{3}\right )}{144 \sqrt {3}}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 20.39, size = 327, normalized size = 3.00 \begin {gather*} \frac {(-2+x)^3 x^2 \left (4-2 x+x^2\right )^2 \left (-\frac {7 x \left (4-2 x+x^2\right )}{-2+x}+\frac {36+216 x-622 x^2+670 x^3-445 x^4+187 x^5-49 x^6+7 x^7}{(-2+x)^2 x \left (4-2 x+x^2\right )}+\frac {7 i \sqrt {2} x \sqrt {\frac {4-2 x+x^2}{(-2+x)^2}} E\left (\sin ^{-1}\left (\frac {\sqrt {-i+\sqrt {3}-\frac {4 i}{-2+x}}}{\sqrt {2} \sqrt [4]{3}}\right )|\frac {2 \sqrt {3}}{i+\sqrt {3}}\right )}{\sqrt {\frac {i x}{\left (i+\sqrt {3}\right ) (-2+x)}}}-19 i \sqrt {2} (-2+x) \sqrt {\frac {i x}{\left (i+\sqrt {3}\right ) (-2+x)}} \sqrt {\frac {4-2 x+x^2}{(-2+x)^2}} F\left (\sin ^{-1}\left (\frac {\sqrt {-i+\sqrt {3}-\frac {4 i}{-2+x}}}{\sqrt {2} \sqrt [4]{3}}\right )|\frac {2 \sqrt {3}}{i+\sqrt {3}}\right )\right )}{432 \left (-x \left (-8+8 x-4 x^2+x^3\right )\right )^{5/2}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[((2 - x)*x*(4 - 2*x + x^2))^(-5/2),x]

[Out]

((-2 + x)^3*x^2*(4 - 2*x + x^2)^2*((-7*x*(4 - 2*x + x^2))/(-2 + x) + (36 + 216*x - 622*x^2 + 670*x^3 - 445*x^4
+ 187*x^5 - 49*x^6 + 7*x^7)/((-2 + x)^2*x*(4 - 2*x + x^2)) + ((7*I)*Sqrt[2]*x*Sqrt[(4 - 2*x + x^2)/(-2 + x)^2
]*EllipticE[ArcSin[Sqrt[-I + Sqrt[3] - (4*I)/(-2 + x)]/(Sqrt[2]*3^(1/4))], (2*Sqrt[3])/(I + Sqrt[3])])/Sqrt[(I
*x)/((I + Sqrt[3])*(-2 + x))] - (19*I)*Sqrt[2]*(-2 + x)*Sqrt[(I*x)/((I + Sqrt[3])*(-2 + x))]*Sqrt[(4 - 2*x + x
^2)/(-2 + x)^2]*EllipticF[ArcSin[Sqrt[-I + Sqrt[3] - (4*I)/(-2 + x)]/(Sqrt[2]*3^(1/4))], (2*Sqrt[3])/(I + Sqrt
[3])]))/(432*(-(x*(-8 + 8*x - 4*x^2 + x^3)))^(5/2))

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Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1038 vs. \(2 (93 ) = 186\).
time = 0.51, size = 1039, normalized size = 9.53 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))^(5/2),x,method=_RETURNVERBOSE)

[Out]

(1/36+1/288*x^2-1/96*x)*(-x^4+4*x^3-8*x^2+8*x)^(1/2)/(x^3-4*x^2+8*x-8)^2+2*x*(53/3456+5/1728*x^2-19/4608*x)/(-
x*(x^3-4*x^2+8*x-8))^(1/2)-1/768*(-x^4+4*x^3-8*x^2+8*x)^(1/2)/x^2-1/96*(-x^3+4*x^2-8*x+8)/(x*(-x^3+4*x^2-8*x+8
))^(1/2)+5/216*(-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))+7/108*(-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*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)))-7/432*(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))*El
lipticF(((-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)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((2-x)*x*(x^2-2*x+4))^(5/2),x, algorithm="maxima")

[Out]

integrate((-(x^2 - 2*x + 4)*(x - 2)*x)^(-5/2), x)

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.08, size = 195, normalized size = 1.79 \begin {gather*} -\frac {43 \, \sqrt {2} {\left (x^{8} - 8 \, x^{7} + 32 \, x^{6} - 80 \, x^{5} + 128 \, x^{4} - 128 \, x^{3} + 64 \, x^{2}\right )} {\rm weierstrassPInverse}\left (-\frac {2}{3}, \frac {7}{54}, -\frac {x - 3}{3 \, x}\right ) - 84 \, \sqrt {2} {\left (x^{8} - 8 \, x^{7} + 32 \, x^{6} - 80 \, x^{5} + 128 \, x^{4} - 128 \, x^{3} + 64 \, x^{2}\right )} {\rm weierstrassZeta}\left (-\frac {2}{3}, \frac {7}{54}, {\rm weierstrassPInverse}\left (-\frac {2}{3}, \frac {7}{54}, -\frac {x - 3}{3 \, x}\right )\right ) + 6 \, {\left (7 \, x^{6} - 37 \, x^{5} + 115 \, x^{4} - 226 \, x^{3} + 274 \, x^{2} - 232 \, x + 36\right )} \sqrt {-x^{4} + 4 \, x^{3} - 8 \, x^{2} + 8 \, x}}{2592 \, {\left (x^{8} - 8 \, x^{7} + 32 \, x^{6} - 80 \, x^{5} + 128 \, x^{4} - 128 \, x^{3} + 64 \, x^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((2-x)*x*(x^2-2*x+4))^(5/2),x, algorithm="fricas")

[Out]

-1/2592*(43*sqrt(2)*(x^8 - 8*x^7 + 32*x^6 - 80*x^5 + 128*x^4 - 128*x^3 + 64*x^2)*weierstrassPInverse(-2/3, 7/5
4, -1/3*(x - 3)/x) - 84*sqrt(2)*(x^8 - 8*x^7 + 32*x^6 - 80*x^5 + 128*x^4 - 128*x^3 + 64*x^2)*weierstrassZeta(-
2/3, 7/54, weierstrassPInverse(-2/3, 7/54, -1/3*(x - 3)/x)) + 6*(7*x^6 - 37*x^5 + 115*x^4 - 226*x^3 + 274*x^2
- 232*x + 36)*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)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (x \left (2 - x\right ) \left (x^{2} - 2 x + 4\right )\right )^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((2-x)*x*(x**2-2*x+4))**(5/2),x)

[Out]

Integral((x*(2 - x)*(x**2 - 2*x + 4))**(-5/2), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((2-x)*x*(x^2-2*x+4))^(5/2),x, algorithm="giac")

[Out]

integrate((-(x^2 - 2*x + 4)*(x - 2)*x)^(-5/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{{\left (-x\,\left (x-2\right )\,\left (x^2-2\,x+4\right )\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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