Integrand size = 19, antiderivative size = 82 \[ \int \frac {1}{\sqrt {8-8 x+4 x^2-x^3}} \, dx=-\frac {\sqrt {2-x} (4-x) \sqrt {\frac {4-2 x+x^2}{(4-x)^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt {2-x}}{\sqrt {2}}\right ),\frac {3}{4}\right )}{\sqrt {2} \sqrt {8-8 x+4 x^2-x^3}} \] Output:
-1/2*(2-x)^(1/2)*(4-x)*((x^2-2*x+4)/(4-x)^2)^(1/2)*InverseJacobiAM(2*arcta n(1/2*2^(1/2)*(2-x)^(1/2)),1/2*3^(1/2))*2^(1/2)/(-x^3+4*x^2-8*x+8)^(1/2)
Result contains complex when optimal does not.
Time = 10.06 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.71 \[ \int \frac {1}{\sqrt {8-8 x+4 x^2-x^3}} \, dx=\frac {2 \sqrt {-i+\sqrt {3}+i x} \sqrt {\frac {i (-2+x)}{-i+\sqrt {3}}} \left (-1+i \sqrt {3}+x\right ) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {i+\sqrt {3}-i x}}{\sqrt {2} \sqrt [4]{3}}\right ),\frac {2 \sqrt {3}}{-i+\sqrt {3}}\right )}{\sqrt {i+\sqrt {3}-i x} \sqrt {8-8 x+4 x^2-x^3}} \] Input:
Integrate[1/Sqrt[8 - 8*x + 4*x^2 - x^3],x]
Output:
(2*Sqrt[-I + Sqrt[3] + I*x]*Sqrt[(I*(-2 + x))/(-I + Sqrt[3])]*(-1 + I*Sqrt [3] + x)*EllipticF[ArcSin[Sqrt[I + Sqrt[3] - I*x]/(Sqrt[2]*3^(1/4))], (2*S qrt[3])/(-I + Sqrt[3])])/(Sqrt[I + Sqrt[3] - I*x]*Sqrt[8 - 8*x + 4*x^2 - x ^3])
Result contains complex when optimal does not.
Time = 0.29 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.21, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {2481, 2475, 27, 1172, 321}
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.
\(\displaystyle \int \frac {1}{\sqrt {-x^3+4 x^2-8 x+8}} \, dx\) |
\(\Big \downarrow \) 2481 |
\(\displaystyle \int \frac {1}{\sqrt {-\left (x-\frac {4}{3}\right )^3-\frac {8}{3} \left (x-\frac {4}{3}\right )+\frac {56}{27}}}d\left (x-\frac {4}{3}\right )\) |
\(\Big \downarrow \) 2475 |
\(\displaystyle \frac {\sqrt {2-3 \left (x-\frac {4}{3}\right )} \sqrt {9 \left (x-\frac {4}{3}\right )^2+6 \left (x-\frac {4}{3}\right )+28} \int \frac {3 \sqrt {3}}{\sqrt {2-3 \left (x-\frac {4}{3}\right )} \sqrt {9 \left (x-\frac {4}{3}\right )^2+6 \left (x-\frac {4}{3}\right )+28}}d\left (x-\frac {4}{3}\right )}{\sqrt {-27 \left (x-\frac {4}{3}\right )^3-72 \left (x-\frac {4}{3}\right )+56}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {3 \sqrt {3} \sqrt {2-3 \left (x-\frac {4}{3}\right )} \sqrt {9 \left (x-\frac {4}{3}\right )^2+6 \left (x-\frac {4}{3}\right )+28} \int \frac {1}{\sqrt {2-3 \left (x-\frac {4}{3}\right )} \sqrt {9 \left (x-\frac {4}{3}\right )^2+6 \left (x-\frac {4}{3}\right )+28}}d\left (x-\frac {4}{3}\right )}{\sqrt {-27 \left (x-\frac {4}{3}\right )^3-72 \left (x-\frac {4}{3}\right )+56}}\) |
\(\Big \downarrow \) 1172 |
\(\displaystyle \frac {2 i \sqrt {\frac {2-3 \left (x-\frac {4}{3}\right )}{1+i \sqrt {3}}} \sqrt {9 \left (x-\frac {4}{3}\right )^2+6 \left (x-\frac {4}{3}\right )+28} \int \frac {1}{\sqrt {\frac {i \left (3 \left (x-\frac {4}{3}\right )+3 i \sqrt {3}+1\right )}{6 \sqrt {3}}+1} \sqrt {\frac {i \left (3 \left (x-\frac {4}{3}\right )+3 i \sqrt {3}+1\right )}{\sqrt {3} \left (3-i \sqrt {3}\right )}+1}}d\frac {\sqrt {-i \left (3 \left (x-\frac {4}{3}\right )+3 i \sqrt {3}+1\right )}}{\sqrt {2} 3^{3/4}}}{\sqrt {-27 \left (x-\frac {4}{3}\right )^3-72 \left (x-\frac {4}{3}\right )+56}}\) |
\(\Big \downarrow \) 321 |
\(\displaystyle -\frac {2 i \sqrt {\frac {2-3 \left (x-\frac {4}{3}\right )}{1+i \sqrt {3}}} \sqrt {9 \left (x-\frac {4}{3}\right )^2+6 \left (x-\frac {4}{3}\right )+28} \operatorname {EllipticF}\left (\arcsin \left (\frac {4}{3}-x\right ),\frac {6}{3-i \sqrt {3}}\right )}{\sqrt {-27 \left (x-\frac {4}{3}\right )^3-72 \left (x-\frac {4}{3}\right )+56}}\) |
Input:
Int[1/Sqrt[8 - 8*x + 4*x^2 - x^3],x]
Output:
((-2*I)*Sqrt[(2 - 3*(-4/3 + x))/(1 + I*Sqrt[3])]*Sqrt[28 + 6*(-4/3 + x) + 9*(-4/3 + x)^2]*EllipticF[ArcSin[4/3 - x], 6/(3 - I*Sqrt[3])])/Sqrt[56 - 7 2*(-4/3 + x) - 27*(-4/3 + x)^3]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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])
Int[((d_.) + (e_.)*(x_))^(m_)/Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2], x_Sy mbol] :> Simp[2*Rt[b^2 - 4*a*c, 2]*(d + e*x)^m*(Sqrt[(-c)*((a + b*x + c*x^2 )/(b^2 - 4*a*c))]/(c*Sqrt[a + b*x + c*x^2]*(2*c*((d + e*x)/(2*c*d - b*e - e *Rt[b^2 - 4*a*c, 2])))^m)) Subst[Int[(1 + 2*e*Rt[b^2 - 4*a*c, 2]*(x^2/(2* c*d - b*e - e*Rt[b^2 - 4*a*c, 2])))^m/Sqrt[1 - x^2], x], x, Sqrt[(b + Rt[b^ 2 - 4*a*c, 2] + 2*c*x)/(2*Rt[b^2 - 4*a*c, 2])]], x] /; FreeQ[{a, b, c, d, e }, x] && EqQ[m^2, 1/4]
Int[((a_.) + (b_.)*(x_) + (d_.)*(x_)^3)^(p_), x_Symbol] :> With[{r = Rt[-9* a*d^2 + Sqrt[3]*d*Sqrt[4*b^3*d + 27*a^2*d^2], 3]}, Simp[(a + b*x + d*x^3)^p /(Simp[18^(1/3)*b*(d/(3*r)) - r/18^(1/3) + d*x, x]^p*Simp[b*(d/3) + 12^(1/3 )*b^2*(d^2/(3*r^2)) + r^2/(3*12^(1/3)) - d*(2^(1/3)*b*(d/(3^(1/3)*r)) - r/1 8^(1/3))*x + d^2*x^2, x]^p) Int[Simp[18^(1/3)*b*(d/(3*r)) - r/18^(1/3) + d*x, x]^p*Simp[b*(d/3) + 12^(1/3)*b^2*(d^2/(3*r^2)) + r^2/(3*12^(1/3)) - d* (2^(1/3)*b*(d/(3^(1/3)*r)) - r/18^(1/3))*x + d^2*x^2, x]^p, x], x]] /; Free Q[{a, b, d, p}, x] && NeQ[4*b^3 + 27*a^2*d, 0] && !IntegerQ[p]
Int[(Px_)^(p_), x_Symbol] :> With[{a = Coeff[Px, x, 0], b = Coeff[Px, x, 1] , c = Coeff[Px, x, 2], d = Coeff[Px, x, 3]}, Subst[Int[Simp[(2*c^3 - 9*b*c* d + 27*a*d^2)/(27*d^2) - (c^2 - 3*b*d)*(x/(3*d)) + d*x^3, x]^p, x], x, c/(3 *d) + x]] /; FreeQ[p, x] && PolyQ[Px, x, 3]
Result contains complex when optimal does not.
Time = 0.33 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.45
method | result | size |
default | \(-\frac {2 i \sqrt {3}\, \sqrt {i \left (x -1-i \sqrt {3}\right ) \sqrt {3}}\, \sqrt {\frac {x -2}{-1+i \sqrt {3}}}\, \sqrt {-i \left (x -1+i \sqrt {3}\right ) \sqrt {3}}\, \operatorname {EllipticF}\left (\frac {\sqrt {6}\, \sqrt {i \left (x -1-i \sqrt {3}\right ) \sqrt {3}}}{6}, \sqrt {2}\, \sqrt {\frac {i \sqrt {3}}{-1+i \sqrt {3}}}\right )}{3 \sqrt {-x^{3}+4 x^{2}-8 x +8}}\) | \(119\) |
elliptic | \(-\frac {2 i \sqrt {3}\, \sqrt {i \left (x -1-i \sqrt {3}\right ) \sqrt {3}}\, \sqrt {\frac {x -2}{-1+i \sqrt {3}}}\, \sqrt {-i \left (x -1+i \sqrt {3}\right ) \sqrt {3}}\, \operatorname {EllipticF}\left (\frac {\sqrt {6}\, \sqrt {i \left (x -1-i \sqrt {3}\right ) \sqrt {3}}}{6}, \sqrt {2}\, \sqrt {\frac {i \sqrt {3}}{-1+i \sqrt {3}}}\right )}{3 \sqrt {-x^{3}+4 x^{2}-8 x +8}}\) | \(119\) |
Input:
int(1/(-x^3+4*x^2-8*x+8)^(1/2),x,method=_RETURNVERBOSE)
Output:
-2/3*I*3^(1/2)*(I*(x-1-I*3^(1/2))*3^(1/2))^(1/2)*((x-2)/(-1+I*3^(1/2)))^(1 /2)*(-I*(x-1+I*3^(1/2))*3^(1/2))^(1/2)/(-x^3+4*x^2-8*x+8)^(1/2)*EllipticF( 1/6*6^(1/2)*(I*(x-1-I*3^(1/2))*3^(1/2))^(1/2),2^(1/2)*(I*3^(1/2)/(-1+I*3^( 1/2)))^(1/2))
Result contains complex when optimal does not.
Time = 0.10 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.10 \[ \int \frac {1}{\sqrt {8-8 x+4 x^2-x^3}} \, dx=-2 i \, {\rm weierstrassPInverse}\left (-\frac {32}{3}, \frac {224}{27}, x - \frac {4}{3}\right ) \] Input:
integrate(1/(-x^3+4*x^2-8*x+8)^(1/2),x, algorithm="fricas")
Output:
-2*I*weierstrassPInverse(-32/3, 224/27, x - 4/3)
\[ \int \frac {1}{\sqrt {8-8 x+4 x^2-x^3}} \, dx=\int \frac {1}{\sqrt {- x^{3} + 4 x^{2} - 8 x + 8}}\, dx \] Input:
integrate(1/(-x**3+4*x**2-8*x+8)**(1/2),x)
Output:
Integral(1/sqrt(-x**3 + 4*x**2 - 8*x + 8), x)
\[ \int \frac {1}{\sqrt {8-8 x+4 x^2-x^3}} \, dx=\int { \frac {1}{\sqrt {-x^{3} + 4 \, x^{2} - 8 \, x + 8}} \,d x } \] Input:
integrate(1/(-x^3+4*x^2-8*x+8)^(1/2),x, algorithm="maxima")
Output:
integrate(1/sqrt(-x^3 + 4*x^2 - 8*x + 8), x)
\[ \int \frac {1}{\sqrt {8-8 x+4 x^2-x^3}} \, dx=\int { \frac {1}{\sqrt {-x^{3} + 4 \, x^{2} - 8 \, x + 8}} \,d x } \] Input:
integrate(1/(-x^3+4*x^2-8*x+8)^(1/2),x, algorithm="giac")
Output:
integrate(1/sqrt(-x^3 + 4*x^2 - 8*x + 8), x)
Time = 0.14 (sec) , antiderivative size = 193, normalized size of antiderivative = 2.35 \[ \int \frac {1}{\sqrt {8-8 x+4 x^2-x^3}} \, dx=\frac {2\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )\,\sqrt {\frac {x-1+\sqrt {3}\,1{}\mathrm {i}}{1+\sqrt {3}\,1{}\mathrm {i}}}\,\sqrt {\frac {x-2}{-1+\sqrt {3}\,1{}\mathrm {i}}}\,\sqrt {\frac {1-x+\sqrt {3}\,1{}\mathrm {i}}{-1+\sqrt {3}\,1{}\mathrm {i}}}\,\mathrm {F}\left (\mathrm {asin}\left (\sqrt {\frac {x-2}{-1+\sqrt {3}\,1{}\mathrm {i}}}\right )\middle |-\frac {-1+\sqrt {3}\,1{}\mathrm {i}}{1+\sqrt {3}\,1{}\mathrm {i}}\right )\,\sqrt {x^3-4\,x^2+8\,x-8}}{\sqrt {-x^3+4\,x^2-8\,x+8}\,\sqrt {x^3-4\,x^2+\left (4-\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )\right )\,x+2\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}} \] Input:
int(1/(4*x^2 - 8*x - x^3 + 8)^(1/2),x)
Output:
(2*(3^(1/2)*1i - 1)*((x + 3^(1/2)*1i - 1)/(3^(1/2)*1i + 1))^(1/2)*((x - 2) /(3^(1/2)*1i - 1))^(1/2)*((3^(1/2)*1i - x + 1)/(3^(1/2)*1i - 1))^(1/2)*ell ipticF(asin(((x - 2)/(3^(1/2)*1i - 1))^(1/2)), -(3^(1/2)*1i - 1)/(3^(1/2)* 1i + 1))*(8*x - 4*x^2 + x^3 - 8)^(1/2))/((4*x^2 - 8*x - x^3 + 8)^(1/2)*(2* (3^(1/2)*1i - 1)*(3^(1/2)*1i + 1) - x*((3^(1/2)*1i - 1)*(3^(1/2)*1i + 1) - 4) - 4*x^2 + x^3)^(1/2))
\[ \int \frac {1}{\sqrt {8-8 x+4 x^2-x^3}} \, dx=-\left (\int \frac {\sqrt {-x^{3}+4 x^{2}-8 x +8}}{x^{3}-4 x^{2}+8 x -8}d x \right ) \] Input:
int(1/(-x^3+4*x^2-8*x+8)^(1/2),x)
Output:
- int(sqrt( - x**3 + 4*x**2 - 8*x + 8)/(x**3 - 4*x**2 + 8*x - 8),x)