3.768 \(\int \frac {1}{(8 x-8 x^2+4 x^3-x^4)^{5/2}} \, dx\)
Optimal. Leaf size=109 \[ \frac {\left (7 (x-1)^2+26\right ) (x-1)}{432 \sqrt {-(x-1)^4-2 (x-1)^2+3}}+\frac {\left ((x-1)^2+5\right ) (x-1)}{72 \left (-(x-1)^4-2 (x-1)^2+3\right )^{3/2}}-\frac {11 F\left (\sin ^{-1}(1-x)|-\frac {1}{3}\right )}{144 \sqrt {3}}+\frac {7 E\left (\sin ^{-1}(1-x)|-\frac {1}{3}\right )}{144 \sqrt {3}} \]
[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 = 0.07, antiderivative size = 109, normalized size of antiderivative = 1.00,
number of steps used = 7, number of rules used = 7, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used =
{1106, 1092, 1178, 1180, 524, 424, 419} \[ \frac {\left (7 (x-1)^2+26\right ) (x-1)}{432 \sqrt {-(x-1)^4-2 (x-1)^2+3}}+\frac {\left ((x-1)^2+5\right ) (x-1)}{72 \left (-(x-1)^4-2 (x-1)^2+3\right )^{3/2}}-\frac {11 F\left (\sin ^{-1}(1-x)|-\frac {1}{3}\right )}{144 \sqrt {3}}+\frac {7 E\left (\sin ^{-1}(1-x)|-\frac {1}{3}\right )}{144 \sqrt {3}} \]
Antiderivative was successfully verified.
[In]
Int[(8*x - 8*x^2 + 4*x^3 - x^4)^(-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 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 1178
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 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 (8 x-8 x^2+4 x^3-x^4\right )^{5/2}} \, dx &=\operatorname {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} \operatorname {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 {\operatorname {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 {\operatorname {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} \operatorname {Subst}\left (\int \frac {\sqrt {6+2 x^2}}{\sqrt {2-2 x^2}} \, dx,x,-1+x\right )+\frac {11}{72} \operatorname {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] time = 1.08, size = 298, normalized size = 2.73 \[ \frac {\frac {7 i \sqrt {2} (x-2) \sqrt {\frac {x^2-2 x+4}{x^2}} x^2 E\left (\sin ^{-1}\left (\frac {\sqrt {\sqrt {3}+i-\frac {4 i}{x}}}{\sqrt {2} \sqrt [4]{3}}\right )|\frac {2 \sqrt {3}}{-i+\sqrt {3}}\right )}{\sqrt {-\frac {i (x-2)}{\left (\sqrt {3}-i\right ) x}}}+\frac {7 x^6-37 x^5+115 x^4-226 x^3+274 x^2-19 i \sqrt {2} \sqrt {-\frac {i (x-2)}{\left (\sqrt {3}-i\right ) x}} \sqrt {\frac {x^2-2 x+4}{x^2}} \left (x^3-4 x^2+8 x-8\right ) x^3 F\left (\sin ^{-1}\left (\frac {\sqrt {\sqrt {3}+i-\frac {4 i}{x}}}{\sqrt {2} \sqrt [4]{3}}\right )|\frac {2 \sqrt {3}}{-i+\sqrt {3}}\right )-232 x+36}{x^3-4 x^2+8 x-8}}{432 x \sqrt {-x \left (x^3-4 x^2+8 x-8\right )}} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[(8*x - 8*x^2 + 4*x^3 - x^4)^(-5/2),x]
[Out]
(((7*I)*Sqrt[2]*(-2 + x)*x^2*Sqrt[(4 - 2*x + x^2)/x^2]*EllipticE[ArcSin[Sqrt[I + Sqrt[3] - (4*I)/x]/(Sqrt[2]*3
^(1/4))], (2*Sqrt[3])/(-I + Sqrt[3])])/Sqrt[((-I)*(-2 + x))/((-I + Sqrt[3])*x)] + (36 - 232*x + 274*x^2 - 226*
x^3 + 115*x^4 - 37*x^5 + 7*x^6 - (19*I)*Sqrt[2]*Sqrt[((-I)*(-2 + x))/((-I + Sqrt[3])*x)]*x^3*Sqrt[(4 - 2*x + x
^2)/x^2]*(-8 + 8*x - 4*x^2 + x^3)*EllipticF[ArcSin[Sqrt[I + Sqrt[3] - (4*I)/x]/(Sqrt[2]*3^(1/4))], (2*Sqrt[3])
/(-I + Sqrt[3])])/(-8 + 8*x - 4*x^2 + x^3))/(432*x*Sqrt[-(x*(-8 + 8*x - 4*x^2 + x^3))])
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fricas [F] time = 0.42, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {-x^{4} + 4 \, x^{3} - 8 \, x^{2} + 8 \, x}}{x^{12} - 12 \, x^{11} + 72 \, x^{10} - 280 \, x^{9} + 768 \, x^{8} - 1536 \, x^{7} + 2240 \, x^{6} - 2304 \, x^{5} + 1536 \, x^{4} - 512 \, x^{3}}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(-x^4+4*x^3-8*x^2+8*x)^(5/2),x, algorithm="fricas")
[Out]
integral(-sqrt(-x^4 + 4*x^3 - 8*x^2 + 8*x)/(x^12 - 12*x^11 + 72*x^10 - 280*x^9 + 768*x^8 - 1536*x^7 + 2240*x^6
- 2304*x^5 + 1536*x^4 - 512*x^3), x)
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (-x^{4} + 4 \, x^{3} - 8 \, x^{2} + 8 \, x\right )}^{\frac {5}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(-x^4+4*x^3-8*x^2+8*x)^(5/2),x, algorithm="giac")
[Out]
integrate((-x^4 + 4*x^3 - 8*x^2 + 8*x)^(-5/2), x)
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maple [B] time = 0.08, size = 1039, normalized size = 9.53 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(1/(-x^4+4*x^3-8*x^2+8*x)^(5/2),x)
[Out]
-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^3+4*x^2-8*x+8)*x)^(1/2)+(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^3-4*x^2+8*x-8)
*x)^(1/2)+5/216*(-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)*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))+7/108*(-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)))-7/432*((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)*EllipticE(((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 \[ \int \frac {1}{{\left (-x^{4} + 4 \, x^{3} - 8 \, x^{2} + 8 \, x\right )}^{\frac {5}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(-x^4+4*x^3-8*x^2+8*x)^(5/2),x, algorithm="maxima")
[Out]
integrate((-x^4 + 4*x^3 - 8*x^2 + 8*x)^(-5/2), x)
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\left (-x^4+4\,x^3-8\,x^2+8\,x\right )}^{5/2}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(1/(8*x - 8*x^2 + 4*x^3 - x^4)^(5/2),x)
[Out]
int(1/(8*x - 8*x^2 + 4*x^3 - x^4)^(5/2), x)
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (- x^{4} + 4 x^{3} - 8 x^{2} + 8 x\right )^{\frac {5}{2}}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(-x**4+4*x**3-8*x**2+8*x)**(5/2),x)
[Out]
Integral((-x**4 + 4*x**3 - 8*x**2 + 8*x)**(-5/2), x)
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