Integrand size = 22, antiderivative size = 75 \[ \int \frac {\sqrt {-1-x+x^2}}{1-x^2} \, dx=-\frac {1}{2} \arctan \left (\frac {3-x}{2 \sqrt {-1-x+x^2}}\right )+\text {arctanh}\left (\frac {1-2 x}{2 \sqrt {-1-x+x^2}}\right )+\frac {1}{2} \text {arctanh}\left (\frac {1+3 x}{2 \sqrt {-1-x+x^2}}\right ) \]
-1/2*arctan(1/2*(3-x)/(x^2-x-1)^(1/2))+arctanh(1/2*(1-2*x)/(x^2-x-1)^(1/2) )+1/2*arctanh(1/2*(1+3*x)/(x^2-x-1)^(1/2))
Time = 0.09 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.76 \[ \int \frac {\sqrt {-1-x+x^2}}{1-x^2} \, dx=\arctan \left (1-x+\sqrt {-1-x+x^2}\right )+\text {arctanh}\left (1+x-\sqrt {-1-x+x^2}\right )+\log \left (1-2 x+2 \sqrt {-1-x+x^2}\right ) \]
ArcTan[1 - x + Sqrt[-1 - x + x^2]] + ArcTanh[1 + x - Sqrt[-1 - x + x^2]] + Log[1 - 2*x + 2*Sqrt[-1 - x + x^2]]
Time = 0.27 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.409, Rules used = {1321, 25, 1092, 219, 1366, 25, 1154, 217, 219}
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.
\(\displaystyle \int \frac {\sqrt {x^2-x-1}}{1-x^2} \, dx\) |
\(\Big \downarrow \) 1321 |
\(\displaystyle \int -\frac {x}{\left (1-x^2\right ) \sqrt {x^2-x-1}}dx-\int \frac {1}{\sqrt {x^2-x-1}}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \frac {1}{\sqrt {x^2-x-1}}dx-\int \frac {x}{\left (1-x^2\right ) \sqrt {x^2-x-1}}dx\) |
\(\Big \downarrow \) 1092 |
\(\displaystyle -\int \frac {x}{\left (1-x^2\right ) \sqrt {x^2-x-1}}dx-2 \int \frac {1}{4-\frac {(1-2 x)^2}{x^2-x-1}}d\left (-\frac {1-2 x}{\sqrt {x^2-x-1}}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \text {arctanh}\left (\frac {1-2 x}{2 \sqrt {x^2-x-1}}\right )-\int \frac {x}{\left (1-x^2\right ) \sqrt {x^2-x-1}}dx\) |
\(\Big \downarrow \) 1366 |
\(\displaystyle -\frac {1}{2} \int \frac {1}{(1-x) \sqrt {x^2-x-1}}dx-\frac {1}{2} \int -\frac {1}{(x+1) \sqrt {x^2-x-1}}dx+\text {arctanh}\left (\frac {1-2 x}{2 \sqrt {x^2-x-1}}\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {1}{2} \int \frac {1}{(1-x) \sqrt {x^2-x-1}}dx+\frac {1}{2} \int \frac {1}{(x+1) \sqrt {x^2-x-1}}dx+\text {arctanh}\left (\frac {1-2 x}{2 \sqrt {x^2-x-1}}\right )\) |
\(\Big \downarrow \) 1154 |
\(\displaystyle \int \frac {1}{-\frac {(3-x)^2}{x^2-x-1}-4}d\frac {3-x}{\sqrt {x^2-x-1}}-\int \frac {1}{4-\frac {(3 x+1)^2}{x^2-x-1}}d\left (-\frac {3 x+1}{\sqrt {x^2-x-1}}\right )+\text {arctanh}\left (\frac {1-2 x}{2 \sqrt {x^2-x-1}}\right )\) |
\(\Big \downarrow \) 217 |
\(\displaystyle -\int \frac {1}{4-\frac {(3 x+1)^2}{x^2-x-1}}d\left (-\frac {3 x+1}{\sqrt {x^2-x-1}}\right )-\frac {1}{2} \arctan \left (\frac {3-x}{2 \sqrt {x^2-x-1}}\right )+\text {arctanh}\left (\frac {1-2 x}{2 \sqrt {x^2-x-1}}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle -\frac {1}{2} \arctan \left (\frac {3-x}{2 \sqrt {x^2-x-1}}\right )+\text {arctanh}\left (\frac {1-2 x}{2 \sqrt {x^2-x-1}}\right )+\frac {1}{2} \text {arctanh}\left (\frac {3 x+1}{2 \sqrt {x^2-x-1}}\right )\) |
-1/2*ArcTan[(3 - x)/(2*Sqrt[-1 - x + x^2])] + ArcTanh[(1 - 2*x)/(2*Sqrt[-1 - x + x^2])] + ArcTanh[(1 + 3*x)/(2*Sqrt[-1 - x + x^2])]/2
3.1.92.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( -1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & & (LtQ[a, 0] || LtQ[b, 0])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Simp[2 Subst[I nt[1/(4*c - x^2), x], x, (b + 2*c*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a , b, c}, x]
Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Sym bol] :> Simp[-2 Subst[Int[1/(4*c*d^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, ( 2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c , d, e}, x]
Int[Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2]/((d_) + (f_.)*(x_)^2), x_Symbol] :> Simp[c/f Int[1/Sqrt[a + b*x + c*x^2], x], x] - Simp[1/f Int[(c*d - a*f - b*f*x)/(Sqrt[a + b*x + c*x^2]*(d + f*x^2)), x], x] /; FreeQ[{a, b, c, d, f}, x] && NeQ[b^2 - 4*a*c, 0]
Int[((g_.) + (h_.)*(x_))/(((a_) + (c_.)*(x_)^2)*Sqrt[(d_.) + (e_.)*(x_) + ( f_.)*(x_)^2]), x_Symbol] :> With[{q = Rt[(-a)*c, 2]}, Simp[(h/2 + c*(g/(2*q ))) Int[1/((-q + c*x)*Sqrt[d + e*x + f*x^2]), x], x] + Simp[(h/2 - c*(g/( 2*q))) Int[1/((q + c*x)*Sqrt[d + e*x + f*x^2]), x], x]] /; FreeQ[{a, c, d , e, f, g, h}, x] && NeQ[e^2 - 4*d*f, 0] && PosQ[(-a)*c]
Time = 0.69 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.36
method | result | size |
default | \(-\frac {\sqrt {\left (-1+x \right )^{2}-2+x}}{2}-\frac {\ln \left (-\frac {1}{2}+x +\sqrt {\left (-1+x \right )^{2}-2+x}\right )}{4}+\frac {\arctan \left (\frac {-3+x}{2 \sqrt {\left (-1+x \right )^{2}-2+x}}\right )}{2}+\frac {\sqrt {\left (1+x \right )^{2}-2-3 x}}{2}-\frac {3 \ln \left (-\frac {1}{2}+x +\sqrt {\left (1+x \right )^{2}-2-3 x}\right )}{4}-\frac {\operatorname {arctanh}\left (\frac {-1-3 x}{2 \sqrt {\left (1+x \right )^{2}-2-3 x}}\right )}{2}\) | \(102\) |
trager | \(-\frac {\ln \left (\frac {8 \sqrt {x^{2}-x -1}\, x^{2}+8 x^{3}+12 \sqrt {x^{2}-x -1}\, x +8 x^{2}+2 \sqrt {x^{2}-x -1}-9 x -11}{1+x}\right )}{2}+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (-\frac {-\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x +2 \sqrt {x^{2}-x -1}+3 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right )}{-1+x}\right )}{2}\) | \(112\) |
-1/2*((-1+x)^2-2+x)^(1/2)-1/4*ln(-1/2+x+((-1+x)^2-2+x)^(1/2))+1/2*arctan(1 /2*(-3+x)/((-1+x)^2-2+x)^(1/2))+1/2*((1+x)^2-2-3*x)^(1/2)-3/4*ln(-1/2+x+(( 1+x)^2-2-3*x)^(1/2))-1/2*arctanh(1/2*(-1-3*x)/((1+x)^2-2-3*x)^(1/2))
Time = 0.31 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.93 \[ \int \frac {\sqrt {-1-x+x^2}}{1-x^2} \, dx=\arctan \left (-x + \sqrt {x^{2} - x - 1} + 1\right ) - \frac {1}{2} \, \log \left (-x + \sqrt {x^{2} - x - 1}\right ) + \frac {1}{2} \, \log \left (-x + \sqrt {x^{2} - x - 1} - 2\right ) + \log \left (-2 \, x + 2 \, \sqrt {x^{2} - x - 1} + 1\right ) \]
arctan(-x + sqrt(x^2 - x - 1) + 1) - 1/2*log(-x + sqrt(x^2 - x - 1)) + 1/2 *log(-x + sqrt(x^2 - x - 1) - 2) + log(-2*x + 2*sqrt(x^2 - x - 1) + 1)
\[ \int \frac {\sqrt {-1-x+x^2}}{1-x^2} \, dx=- \int \frac {\sqrt {x^{2} - x - 1}}{x^{2} - 1}\, dx \]
Time = 0.27 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.11 \[ \int \frac {\sqrt {-1-x+x^2}}{1-x^2} \, dx=\frac {1}{2} \, \arcsin \left (\frac {2 \, \sqrt {5} x}{5 \, {\left | 2 \, x - 2 \right |}} - \frac {6 \, \sqrt {5}}{5 \, {\left | 2 \, x - 2 \right |}}\right ) - \log \left (x + \sqrt {x^{2} - x - 1} - \frac {1}{2}\right ) - \frac {1}{2} \, \log \left (\frac {2 \, \sqrt {x^{2} - x - 1}}{{\left | 2 \, x + 2 \right |}} + \frac {2}{{\left | 2 \, x + 2 \right |}} - \frac {3}{2}\right ) \]
1/2*arcsin(2/5*sqrt(5)*x/abs(2*x - 2) - 6/5*sqrt(5)/abs(2*x - 2)) - log(x + sqrt(x^2 - x - 1) - 1/2) - 1/2*log(2*sqrt(x^2 - x - 1)/abs(2*x + 2) + 2/ abs(2*x + 2) - 3/2)
Time = 0.28 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.97 \[ \int \frac {\sqrt {-1-x+x^2}}{1-x^2} \, dx=\arctan \left (-x + \sqrt {x^{2} - x - 1} + 1\right ) - \frac {1}{2} \, \log \left ({\left | -x + \sqrt {x^{2} - x - 1} \right |}\right ) + \frac {1}{2} \, \log \left ({\left | -x + \sqrt {x^{2} - x - 1} - 2 \right |}\right ) + \log \left ({\left | -2 \, x + 2 \, \sqrt {x^{2} - x - 1} + 1 \right |}\right ) \]
arctan(-x + sqrt(x^2 - x - 1) + 1) - 1/2*log(abs(-x + sqrt(x^2 - x - 1))) + 1/2*log(abs(-x + sqrt(x^2 - x - 1) - 2)) + log(abs(-2*x + 2*sqrt(x^2 - x - 1) + 1))
Timed out. \[ \int \frac {\sqrt {-1-x+x^2}}{1-x^2} \, dx=-\int \frac {\sqrt {x^2-x-1}}{x^2-1} \,d x \]