Integrand size = 20, antiderivative size = 60 \[ \int \frac {\sqrt {-1+\frac {1}{x^2}} \left (-1+x^2\right )^2}{x} \, dx=-\frac {15}{8} \sqrt {-1+\frac {1}{x^2}}+\frac {5}{8} \left (-1+\frac {1}{x^2}\right )^{3/2} x^2+\frac {1}{4} \left (-1+\frac {1}{x^2}\right )^{5/2} x^4+\frac {15}{8} \arctan \left (\sqrt {-1+\frac {1}{x^2}}\right ) \]
5/8*(-1+1/x^2)^(3/2)*x^2+1/4*(-1+1/x^2)^(5/2)*x^4+15/8*arctan((-1+1/x^2)^( 1/2))-15/8*(-1+1/x^2)^(1/2)
Time = 0.08 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.08 \[ \int \frac {\sqrt {-1+\frac {1}{x^2}} \left (-1+x^2\right )^2}{x} \, dx=\frac {1}{8} \sqrt {-1+\frac {1}{x^2}} \left (-8-9 x^2+2 x^4\right )+\frac {15 \sqrt {-1+\frac {1}{x^2}} x \text {arctanh}\left (\frac {\sqrt {-1+x^2}}{-1+x}\right )}{4 \sqrt {-1+x^2}} \]
(Sqrt[-1 + x^(-2)]*(-8 - 9*x^2 + 2*x^4))/8 + (15*Sqrt[-1 + x^(-2)]*x*ArcTa nh[Sqrt[-1 + x^2]/(-1 + x)])/(4*Sqrt[-1 + x^2])
Time = 0.20 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.13, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {1016, 281, 798, 51, 51, 60, 73, 216}
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.
\(\displaystyle \int \frac {\sqrt {\frac {1}{x^2}-1} \left (x^2-1\right )^2}{x} \, dx\) |
\(\Big \downarrow \) 1016 |
\(\displaystyle \int \left (1-\frac {1}{x^2}\right )^2 \sqrt {\frac {1}{x^2}-1} x^3dx\) |
\(\Big \downarrow \) 281 |
\(\displaystyle \int \left (\frac {1}{x^2}-1\right )^{5/2} x^3dx\) |
\(\Big \downarrow \) 798 |
\(\displaystyle -\frac {1}{2} \int \left (\frac {1}{x^2}-1\right )^{5/2} x^6d\frac {1}{x^2}\) |
\(\Big \downarrow \) 51 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \left (\frac {1}{x^2}-1\right )^{5/2} x^4-\frac {5}{4} \int \left (\frac {1}{x^2}-1\right )^{3/2} x^4d\frac {1}{x^2}\right )\) |
\(\Big \downarrow \) 51 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \left (\frac {1}{x^2}-1\right )^{5/2} x^4-\frac {5}{4} \left (\frac {3}{2} \int \sqrt {\frac {1}{x^2}-1} x^2d\frac {1}{x^2}-\left (\frac {1}{x^2}-1\right )^{3/2} x^2\right )\right )\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \left (\frac {1}{x^2}-1\right )^{5/2} x^4-\frac {5}{4} \left (\frac {3}{2} \left (2 \sqrt {\frac {1}{x^2}-1}-\int \frac {x^2}{\sqrt {\frac {1}{x^2}-1}}d\frac {1}{x^2}\right )-\left (\frac {1}{x^2}-1\right )^{3/2} x^2\right )\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \left (\frac {1}{x^2}-1\right )^{5/2} x^4-\frac {5}{4} \left (\frac {3}{2} \left (2 \sqrt {\frac {1}{x^2}-1}-2 \int \frac {1}{1+\frac {1}{x^4}}d\sqrt {\frac {1}{x^2}-1}\right )-\left (\frac {1}{x^2}-1\right )^{3/2} x^2\right )\right )\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \left (\frac {1}{x^2}-1\right )^{5/2} x^4-\frac {5}{4} \left (\frac {3}{2} \left (2 \sqrt {\frac {1}{x^2}-1}-2 \arctan \left (\sqrt {\frac {1}{x^2}-1}\right )\right )-\left (\frac {1}{x^2}-1\right )^{3/2} x^2\right )\right )\) |
(((-1 + x^(-2))^(5/2)*x^4)/2 - (5*(-((-1 + x^(-2))^(3/2)*x^2) + (3*(2*Sqrt [-1 + x^(-2)] - 2*ArcTan[Sqrt[-1 + x^(-2)]]))/2))/4)/2
3.7.76.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + 1))), x] - Simp[d*(n/(b*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d, n}, x ] && ILtQ[m, -1] && FractionQ[n] && GtQ[n, 0]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( b*(m + n + 1))) Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !Integer Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinear Q[a, b, c, d, m, n, x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[(u_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_ Symbol] :> Simp[(b/d)^p Int[u*(c + d*x^n)^(p + q), x], x] /; FreeQ[{a, b, c, d, n, p, q}, x] && EqQ[b*c - a*d, 0] && IntegerQ[p] && !(IntegerQ[q] & & SimplerQ[a + b*x^n, c + d*x^n])
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[1/n Subst [Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]
Int[(x_)^(m_.)*((c_) + (d_.)*(x_)^(mn_.))^(q_.)*((a_) + (b_.)*(x_)^(n_.))^( p_.), x_Symbol] :> Int[x^(m - n*q)*(a + b*x^n)^p*(d + c*x^n)^q, x] /; FreeQ [{a, b, c, d, m, n, p}, x] && EqQ[mn, -n] && IntegerQ[q] && (PosQ[n] || !I ntegerQ[p])
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.99 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.97
method | result | size |
trager | \(2 \left (\frac {1}{8} x^{4}-\frac {9}{16} x^{2}-\frac {1}{2}\right ) \sqrt {-\frac {x^{2}-1}{x^{2}}}+\frac {15 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (\left (\sqrt {-\frac {x^{2}-1}{x^{2}}}+\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right )\right ) x \right )}{8}\) | \(58\) |
default | \(-\frac {\sqrt {-\frac {x^{2}-1}{x^{2}}}\, \left (2 x^{2} \left (-x^{2}+1\right )^{\frac {3}{2}}+8 \left (-x^{2}+1\right )^{\frac {3}{2}}+15 x^{2} \sqrt {-x^{2}+1}+15 \arcsin \left (x \right ) x \right )}{8 \sqrt {-x^{2}+1}}\) | \(69\) |
risch | \(\frac {\left (2 x^{6}-11 x^{4}+x^{2}+8\right ) \sqrt {-\frac {x^{2}-1}{x^{2}}}}{8 x^{2}-8}+\frac {15 \arcsin \left (x \right ) \sqrt {-\frac {x^{2}-1}{x^{2}}}\, x \sqrt {-x^{2}+1}}{8 \left (x^{2}-1\right )}\) | \(71\) |
2*(1/8*x^4-9/16*x^2-1/2)*(-(x^2-1)/x^2)^(1/2)+15/8*RootOf(_Z^2+1)*ln(((-(x ^2-1)/x^2)^(1/2)+RootOf(_Z^2+1))*x)
Time = 0.27 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.83 \[ \int \frac {\sqrt {-1+\frac {1}{x^2}} \left (-1+x^2\right )^2}{x} \, dx=\frac {1}{8} \, {\left (2 \, x^{4} - 9 \, x^{2} - 8\right )} \sqrt {-\frac {x^{2} - 1}{x^{2}}} + \frac {15}{4} \, \arctan \left (\frac {x \sqrt {-\frac {x^{2} - 1}{x^{2}}} - 1}{x}\right ) \]
Time = 23.56 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.00 \[ \int \frac {\sqrt {-1+\frac {1}{x^2}} \left (-1+x^2\right )^2}{x} \, dx=\frac {x^{4} \sqrt {-1 + \frac {1}{x^{2}}} \cdot \left (2 - \frac {1}{x^{2}}\right )}{8} - x^{2} \sqrt {-1 + \frac {1}{x^{2}}} - \sqrt {-1 + \frac {1}{x^{2}}} + \frac {15 \operatorname {atan}{\left (\sqrt {-1 + \frac {1}{x^{2}}} \right )}}{8} \]
x**4*sqrt(-1 + x**(-2))*(2 - 1/x**2)/8 - x**2*sqrt(-1 + x**(-2)) - sqrt(-1 + x**(-2)) + 15*atan(sqrt(-1 + x**(-2)))/8
Time = 0.27 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.12 \[ \int \frac {\sqrt {-1+\frac {1}{x^2}} \left (-1+x^2\right )^2}{x} \, dx=-x^{2} \sqrt {\frac {1}{x^{2}} - 1} - \sqrt {\frac {1}{x^{2}} - 1} - \frac {{\left (\frac {1}{x^{2}} - 1\right )}^{\frac {3}{2}} - \sqrt {\frac {1}{x^{2}} - 1}}{8 \, {\left ({\left (\frac {1}{x^{2}} - 1\right )}^{2} + \frac {2}{x^{2}} - 1\right )}} + \frac {15}{8} \, \arctan \left (\sqrt {\frac {1}{x^{2}} - 1}\right ) \]
-x^2*sqrt(1/x^2 - 1) - sqrt(1/x^2 - 1) - 1/8*((1/x^2 - 1)^(3/2) - sqrt(1/x ^2 - 1))/((1/x^2 - 1)^2 + 2/x^2 - 1) + 15/8*arctan(sqrt(1/x^2 - 1))
Time = 0.34 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.12 \[ \int \frac {\sqrt {-1+\frac {1}{x^2}} \left (-1+x^2\right )^2}{x} \, dx=\frac {1}{8} \, {\left (2 \, x^{2} \mathrm {sgn}\left (x\right ) - 9 \, \mathrm {sgn}\left (x\right )\right )} \sqrt {-x^{2} + 1} x - \frac {15}{8} \, \arcsin \left (x\right ) \mathrm {sgn}\left (x\right ) + \frac {x \mathrm {sgn}\left (x\right )}{2 \, {\left (\sqrt {-x^{2} + 1} - 1\right )}} - \frac {{\left (\sqrt {-x^{2} + 1} - 1\right )} \mathrm {sgn}\left (x\right )}{2 \, x} \]
1/8*(2*x^2*sgn(x) - 9*sgn(x))*sqrt(-x^2 + 1)*x - 15/8*arcsin(x)*sgn(x) + 1 /2*x*sgn(x)/(sqrt(-x^2 + 1) - 1) - 1/2*(sqrt(-x^2 + 1) - 1)*sgn(x)/x
Time = 17.65 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.73 \[ \int \frac {\sqrt {-1+\frac {1}{x^2}} \left (-1+x^2\right )^2}{x} \, dx=\frac {15\,\mathrm {atan}\left (\sqrt {\frac {1}{x^2}-1}\right )}{8}-\sqrt {\frac {1}{x^2}-1}-\frac {7\,x^4\,\sqrt {\frac {1}{x^2}-1}}{8}-\frac {9\,x^4\,{\left (\frac {1}{x^2}-1\right )}^{3/2}}{8} \]