Integrand size = 18, antiderivative size = 70 \[ \int \frac {1}{\left (-1+x^2\right )^2 \sqrt {-1+x+x^2}} \, dx=\frac {\sqrt {-1+x+x^2}}{2 \left (1-x^2\right )}-\frac {1}{8} \arctan \left (\frac {3+x}{2 \sqrt {-1+x+x^2}}\right )-\frac {5}{8} \text {arctanh}\left (\frac {1-3 x}{2 \sqrt {-1+x+x^2}}\right ) \]
-1/8*arctan(1/2*(3+x)/(x^2+x-1)^(1/2))-5/8*arctanh(1/2*(1-3*x)/(x^2+x-1)^( 1/2))+1/2*(x^2+x-1)^(1/2)/(-x^2+1)
Time = 0.17 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.89 \[ \int \frac {1}{\left (-1+x^2\right )^2 \sqrt {-1+x+x^2}} \, dx=-\frac {\sqrt {-1+x+x^2}}{2 \left (-1+x^2\right )}-\frac {1}{4} \arctan \left (1+x-\sqrt {-1+x+x^2}\right )+\frac {5}{4} \text {arctanh}\left (1-x+\sqrt {-1+x+x^2}\right ) \]
-1/2*Sqrt[-1 + x + x^2]/(-1 + x^2) - ArcTan[1 + x - Sqrt[-1 + x + x^2]]/4 + (5*ArcTanh[1 - x + Sqrt[-1 + x + x^2]])/4
Time = 0.23 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.07, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.389, Rules used = {1307, 25, 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.
\(\displaystyle \int \frac {1}{\left (x^2-1\right )^2 \sqrt {x^2+x-1}} \, dx\) |
\(\Big \downarrow \) 1307 |
\(\displaystyle \frac {\sqrt {x^2+x-1}}{2 \left (1-x^2\right )}-\frac {1}{4} \int -\frac {2 x+3}{\left (1-x^2\right ) \sqrt {x^2+x-1}}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{4} \int \frac {2 x+3}{\left (1-x^2\right ) \sqrt {x^2+x-1}}dx+\frac {\sqrt {x^2+x-1}}{2 \left (1-x^2\right )}\) |
\(\Big \downarrow \) 1366 |
\(\displaystyle \frac {1}{4} \left (\frac {5}{2} \int \frac {1}{(1-x) \sqrt {x^2+x-1}}dx-\frac {1}{2} \int -\frac {1}{(x+1) \sqrt {x^2+x-1}}dx\right )+\frac {\sqrt {x^2+x-1}}{2 \left (1-x^2\right )}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{4} \left (\frac {5}{2} \int \frac {1}{(1-x) \sqrt {x^2+x-1}}dx+\frac {1}{2} \int \frac {1}{(x+1) \sqrt {x^2+x-1}}dx\right )+\frac {\sqrt {x^2+x-1}}{2 \left (1-x^2\right )}\) |
\(\Big \downarrow \) 1154 |
\(\displaystyle \frac {1}{4} \left (-5 \int \frac {1}{4-\frac {(1-3 x)^2}{x^2+x-1}}d\frac {1-3 x}{\sqrt {x^2+x-1}}-\int \frac {1}{-\frac {(x+3)^2}{x^2+x-1}-4}d\left (-\frac {x+3}{\sqrt {x^2+x-1}}\right )\right )+\frac {\sqrt {x^2+x-1}}{2 \left (1-x^2\right )}\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {1}{4} \left (-5 \int \frac {1}{4-\frac {(1-3 x)^2}{x^2+x-1}}d\frac {1-3 x}{\sqrt {x^2+x-1}}-\frac {1}{2} \arctan \left (\frac {x+3}{2 \sqrt {x^2+x-1}}\right )\right )+\frac {\sqrt {x^2+x-1}}{2 \left (1-x^2\right )}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{4} \left (-\frac {1}{2} \arctan \left (\frac {x+3}{2 \sqrt {x^2+x-1}}\right )-\frac {5}{2} \text {arctanh}\left (\frac {1-3 x}{2 \sqrt {x^2+x-1}}\right )\right )+\frac {\sqrt {x^2+x-1}}{2 \left (1-x^2\right )}\) |
Sqrt[-1 + x + x^2]/(2*(1 - x^2)) + (-1/2*ArcTan[(3 + x)/(2*Sqrt[-1 + x + x ^2])] - (5*ArcTanh[(1 - 3*x)/(2*Sqrt[-1 + x + x^2])])/2)/4
3.1.12.3.1 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/(((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[((a_.) + (c_.)*(x_)^2)^(p_)*((d_.) + (e_.)*(x_) + (f_.)*(x_)^2)^(q_), x _Symbol] :> Simp[(2*a*c^2*e + c*(2*c^2*d - c*(2*a*f))*x)*(a + c*x^2)^(p + 1 )*((d + e*x + f*x^2)^(q + 1)/((-4*a*c)*(a*c*e^2 + (c*d - a*f)^2)*(p + 1))), x] - Simp[1/((-4*a*c)*(a*c*e^2 + (c*d - a*f)^2)*(p + 1)) Int[(a + c*x^2) ^(p + 1)*(d + e*x + f*x^2)^q*Simp[2*c*((c*d - a*f)^2 - ((-a)*e)*(c*e))*(p + 1) - (2*c^2*d - c*(2*a*f))*(a*f*(p + 1) - c*d*(p + 2)) - e*(-2*a*c^2*e)*(p + q + 2) + (2*f*(2*a*c^2*e)*(p + q + 2) - (2*c^2*d - c*(2*a*f))*((-c)*e*(2 *p + q + 4)))*x + c*f*(2*c^2*d - c*(2*a*f))*(2*p + 2*q + 5)*x^2, x], x], x] /; FreeQ[{a, c, d, e, f, q}, x] && NeQ[e^2 - 4*d*f, 0] && LtQ[p, -1] && Ne Q[a*c*e^2 + (c*d - a*f)^2, 0] && !( !IntegerQ[p] && ILtQ[q, -1]) && !IGtQ [q, 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 = 0.62 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.90
method | result | size |
risch | \(-\frac {\sqrt {x^{2}+x -1}}{2 \left (x^{2}-1\right )}+\frac {\arctan \left (\frac {-3-x}{2 \sqrt {\left (1+x \right )^{2}-2-x}}\right )}{8}+\frac {5 \,\operatorname {arctanh}\left (\frac {-1+3 x}{2 \sqrt {\left (-1+x \right )^{2}-2+3 x}}\right )}{8}\) | \(63\) |
default | \(-\frac {\sqrt {\left (-1+x \right )^{2}-2+3 x}}{4 \left (-1+x \right )}+\frac {5 \,\operatorname {arctanh}\left (\frac {-1+3 x}{2 \sqrt {\left (-1+x \right )^{2}-2+3 x}}\right )}{8}+\frac {\sqrt {\left (1+x \right )^{2}-2-x}}{4+4 x}+\frac {\arctan \left (\frac {-3-x}{2 \sqrt {\left (1+x \right )^{2}-2-x}}\right )}{8}\) | \(84\) |
trager | \(-\frac {\sqrt {x^{2}+x -1}}{2 \left (x^{2}-1\right )}+\frac {5 \ln \left (-\frac {2 \sqrt {x^{2}+x -1}-1+3 x}{-1+x}\right )}{8}+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x +2 \sqrt {x^{2}+x -1}+3 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right )}{1+x}\right )}{8}\) | \(86\) |
-1/2/(x^2-1)*(x^2+x-1)^(1/2)+1/8*arctan(1/2*(-3-x)/((1+x)^2-2-x)^(1/2))+5/ 8*arctanh(1/2*(-1+3*x)/((-1+x)^2-2+3*x)^(1/2))
Time = 0.33 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.17 \[ \int \frac {1}{\left (-1+x^2\right )^2 \sqrt {-1+x+x^2}} \, dx=\frac {2 \, {\left (x^{2} - 1\right )} \arctan \left (-x + \sqrt {x^{2} + x - 1} - 1\right ) + 5 \, {\left (x^{2} - 1\right )} \log \left (-x + \sqrt {x^{2} + x - 1} + 2\right ) - 5 \, {\left (x^{2} - 1\right )} \log \left (-x + \sqrt {x^{2} + x - 1}\right ) - 4 \, \sqrt {x^{2} + x - 1}}{8 \, {\left (x^{2} - 1\right )}} \]
1/8*(2*(x^2 - 1)*arctan(-x + sqrt(x^2 + x - 1) - 1) + 5*(x^2 - 1)*log(-x + sqrt(x^2 + x - 1) + 2) - 5*(x^2 - 1)*log(-x + sqrt(x^2 + x - 1)) - 4*sqrt (x^2 + x - 1))/(x^2 - 1)
\[ \int \frac {1}{\left (-1+x^2\right )^2 \sqrt {-1+x+x^2}} \, dx=\int \frac {1}{\left (x - 1\right )^{2} \left (x + 1\right )^{2} \sqrt {x^{2} + x - 1}}\, dx \]
\[ \int \frac {1}{\left (-1+x^2\right )^2 \sqrt {-1+x+x^2}} \, dx=\int { \frac {1}{\sqrt {x^{2} + x - 1} {\left (x^{2} - 1\right )}^{2}} \,d x } \]
Leaf count of result is larger than twice the leaf count of optimal. 143 vs. \(2 (52) = 104\).
Time = 0.30 (sec) , antiderivative size = 143, normalized size of antiderivative = 2.04 \[ \int \frac {1}{\left (-1+x^2\right )^2 \sqrt {-1+x+x^2}} \, dx=\frac {2 \, {\left (x - \sqrt {x^{2} + x - 1}\right )}^{3} + 3 \, {\left (x - \sqrt {x^{2} + x - 1}\right )}^{2} - x + \sqrt {x^{2} + x - 1} - 1}{2 \, {\left ({\left (x - \sqrt {x^{2} + x - 1}\right )}^{4} - 2 \, {\left (x - \sqrt {x^{2} + x - 1}\right )}^{2} - 4 \, x + 4 \, \sqrt {x^{2} + x - 1}\right )}} + \frac {1}{4} \, \arctan \left (-x + \sqrt {x^{2} + x - 1} - 1\right ) + \frac {5}{8} \, \log \left ({\left | -x + \sqrt {x^{2} + x - 1} + 2 \right |}\right ) - \frac {5}{8} \, \log \left ({\left | -x + \sqrt {x^{2} + x - 1} \right |}\right ) \]
1/2*(2*(x - sqrt(x^2 + x - 1))^3 + 3*(x - sqrt(x^2 + x - 1))^2 - x + sqrt( x^2 + x - 1) - 1)/((x - sqrt(x^2 + x - 1))^4 - 2*(x - sqrt(x^2 + x - 1))^2 - 4*x + 4*sqrt(x^2 + x - 1)) + 1/4*arctan(-x + sqrt(x^2 + x - 1) - 1) + 5 /8*log(abs(-x + sqrt(x^2 + x - 1) + 2)) - 5/8*log(abs(-x + sqrt(x^2 + x - 1)))
Timed out. \[ \int \frac {1}{\left (-1+x^2\right )^2 \sqrt {-1+x+x^2}} \, dx=\int \frac {1}{{\left (x^2-1\right )}^2\,\sqrt {x^2+x-1}} \,d x \]