Integrand size = 17, antiderivative size = 64 \[ \int \frac {\sqrt {-1+x^2}}{(-i+x)^2} \, dx=\frac {\sqrt {-1+x^2}}{i-x}-\frac {i \arctan \left (\frac {1-i x}{\sqrt {2} \sqrt {-1+x^2}}\right )}{\sqrt {2}}+\text {arctanh}\left (\frac {x}{\sqrt {-1+x^2}}\right ) \]
arctanh(x/(x^2-1)^(1/2))-1/2*I*arctan(1/2*(1-I*x)*2^(1/2)/(x^2-1)^(1/2))*2 ^(1/2)+(x^2-1)^(1/2)/(I-x)
Time = 0.19 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.09 \[ \int \frac {\sqrt {-1+x^2}}{(-i+x)^2} \, dx=\frac {\sqrt {-1+x^2}}{i-x}-\sqrt {2} \text {arctanh}\left (\frac {1+i x-i \sqrt {-1+x^2}}{\sqrt {2}}\right )-\log \left (-x+\sqrt {-1+x^2}\right ) \]
Sqrt[-1 + x^2]/(I - x) - Sqrt[2]*ArcTanh[(1 + I*x - I*Sqrt[-1 + x^2])/Sqrt [2]] - Log[-x + Sqrt[-1 + x^2]]
Time = 0.18 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.412, Rules used = {492, 25, 605, 224, 219, 488, 217}
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.
\(\displaystyle \int \frac {\sqrt {x^2-1}}{(x-i)^2} \, dx\) |
\(\Big \downarrow \) 492 |
\(\displaystyle \int -\frac {x}{(i-x) \sqrt {x^2-1}}dx+\frac {\sqrt {x^2-1}}{-x+i}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\sqrt {x^2-1}}{-x+i}-\int \frac {x}{(i-x) \sqrt {x^2-1}}dx\) |
\(\Big \downarrow \) 605 |
\(\displaystyle \int \frac {1}{\sqrt {x^2-1}}dx-i \int \frac {1}{(i-x) \sqrt {x^2-1}}dx+\frac {\sqrt {x^2-1}}{-x+i}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle -i \int \frac {1}{(i-x) \sqrt {x^2-1}}dx+\int \frac {1}{1-\frac {x^2}{x^2-1}}d\frac {x}{\sqrt {x^2-1}}+\frac {\sqrt {x^2-1}}{-x+i}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle -i \int \frac {1}{(i-x) \sqrt {x^2-1}}dx+\text {arctanh}\left (\frac {x}{\sqrt {x^2-1}}\right )+\frac {\sqrt {x^2-1}}{-x+i}\) |
\(\Big \downarrow \) 488 |
\(\displaystyle i \int \frac {1}{-\frac {(1-i x)^2}{x^2-1}-2}d\frac {1-i x}{\sqrt {x^2-1}}+\text {arctanh}\left (\frac {x}{\sqrt {x^2-1}}\right )+\frac {\sqrt {x^2-1}}{-x+i}\) |
\(\Big \downarrow \) 217 |
\(\displaystyle -\frac {i \arctan \left (\frac {1-i x}{\sqrt {2} \sqrt {x^2-1}}\right )}{\sqrt {2}}+\text {arctanh}\left (\frac {x}{\sqrt {x^2-1}}\right )+\frac {\sqrt {x^2-1}}{-x+i}\) |
Sqrt[-1 + x^2]/(I - x) - (I*ArcTan[(1 - I*x)/(Sqrt[2]*Sqrt[-1 + x^2])])/Sq rt[2] + ArcTanh[x/Sqrt[-1 + x^2]]
3.1.6.3.1 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_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] && !GtQ[a, 0]
Int[1/(((c_) + (d_.)*(x_))*Sqrt[(a_) + (b_.)*(x_)^2]), x_Symbol] :> -Subst[ Int[1/(b*c^2 + a*d^2 - x^2), x], x, (a*d - b*c*x)/Sqrt[a + b*x^2]] /; FreeQ [{a, b, c, d}, x]
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ (c + d*x)^(n + 1)*((a + b*x^2)^p/(d*(n + 1))), x] - Simp[2*b*(p/(d*(n + 1)) ) Int[x*(c + d*x)^(n + 1)*(a + b*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, d, n}, x] && GtQ[p, 0] && (IntegerQ[p] || LtQ[n, -1]) && NeQ[n, -1] && !IL tQ[n + 2*p + 1, 0] && IntQuadraticQ[a, 0, b, c, d, n, p, x]
Int[((x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_))/((c_) + (d_.)*(x_)), x_Symbol] :> Simp[1/d Int[x^(m - 1)*(a + b*x^2)^p, x], x] - Simp[c/d Int[x^(m - 1 )*((a + b*x^2)^p/(c + d*x)), x], x] /; FreeQ[{a, b, c, d, p}, x] && IGtQ[m, 0] && LtQ[-1, p, 0]
Time = 0.42 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.02
method | result | size |
risch | \(-\frac {\sqrt {x^{2}-1}}{x -i}+\ln \left (x +\sqrt {x^{2}-1}\right )+\frac {i \sqrt {2}\, \arctan \left (\frac {\left (-4+2 i \left (x -i\right )\right ) \sqrt {2}}{4 \sqrt {\left (x -i\right )^{2}+2 i \left (x -i\right )-2}}\right )}{2}\) | \(65\) |
default | \(\frac {\left (\left (x -i\right )^{2}+2 i \left (x -i\right )-2\right )^{\frac {3}{2}}}{2 x -2 i}-\frac {i \left (\sqrt {\left (x -i\right )^{2}+2 i \left (x -i\right )-2}+i \ln \left (x +\sqrt {\left (x -i\right )^{2}+2 i \left (x -i\right )-2}\right )-\sqrt {2}\, \arctan \left (\frac {\left (-4+2 i \left (x -i\right )\right ) \sqrt {2}}{4 \sqrt {\left (x -i\right )^{2}+2 i \left (x -i\right )-2}}\right )\right )}{2}-\frac {x \sqrt {\left (x -i\right )^{2}+2 i \left (x -i\right )-2}}{2}+\frac {\ln \left (x +\sqrt {\left (x -i\right )^{2}+2 i \left (x -i\right )-2}\right )}{2}\) | \(150\) |
-(x^2-1)^(1/2)/(x-I)+ln(x+(x^2-1)^(1/2))+1/2*I*2^(1/2)*arctan(1/4*(-4+2*I* (x-I))*2^(1/2)/((x-I)^2+2*I*(x-I)-2)^(1/2))
Time = 0.24 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.39 \[ \int \frac {\sqrt {-1+x^2}}{(-i+x)^2} \, dx=-\frac {\sqrt {2} {\left (x - i\right )} \log \left (-x + i \, \sqrt {2} + \sqrt {x^{2} - 1} + i\right ) - \sqrt {2} {\left (x - i\right )} \log \left (-x - i \, \sqrt {2} + \sqrt {x^{2} - 1} + i\right ) + 2 \, {\left (x - i\right )} \log \left (-x + \sqrt {x^{2} - 1}\right ) + 2 \, x + 2 \, \sqrt {x^{2} - 1} - 2 i}{2 \, {\left (x - i\right )}} \]
-1/2*(sqrt(2)*(x - I)*log(-x + I*sqrt(2) + sqrt(x^2 - 1) + I) - sqrt(2)*(x - I)*log(-x - I*sqrt(2) + sqrt(x^2 - 1) + I) + 2*(x - I)*log(-x + sqrt(x^ 2 - 1)) + 2*x + 2*sqrt(x^2 - 1) - 2*I)/(x - I)
\[ \int \frac {\sqrt {-1+x^2}}{(-i+x)^2} \, dx=\int \frac {\sqrt {\left (x - 1\right ) \left (x + 1\right )}}{\left (x - i\right )^{2}}\, dx \]
Time = 0.27 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.83 \[ \int \frac {\sqrt {-1+x^2}}{(-i+x)^2} \, dx=\frac {1}{2} i \, \sqrt {2} \arcsin \left (\frac {i \, x}{{\left | x - i \right |}} - \frac {1}{{\left | x - i \right |}}\right ) - \frac {\sqrt {x^{2} - 1}}{x - i} + \log \left (2 \, x + 2 \, \sqrt {x^{2} - 1}\right ) \]
1/2*I*sqrt(2)*arcsin(I*x/abs(x - I) - 1/abs(x - I)) - sqrt(x^2 - 1)/(x - I ) + log(2*x + 2*sqrt(x^2 - 1))
Time = 0.29 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.31 \[ \int \frac {\sqrt {-1+x^2}}{(-i+x)^2} \, dx=i \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (x - \sqrt {x^{2} - 1} - i\right )}\right ) + \frac {2 \, {\left (i \, x - i \, \sqrt {x^{2} - 1} - 1\right )}}{{\left (x - \sqrt {x^{2} - 1}\right )}^{2} - 2 i \, x + 2 i \, \sqrt {x^{2} - 1} + 1} - \log \left ({\left | -x + \sqrt {x^{2} - 1} \right |}\right ) \]
I*sqrt(2)*arctan(-1/2*sqrt(2)*(x - sqrt(x^2 - 1) - I)) + 2*(I*x - I*sqrt(x ^2 - 1) - 1)/((x - sqrt(x^2 - 1))^2 - 2*I*x + 2*I*sqrt(x^2 - 1) + 1) - log (abs(-x + sqrt(x^2 - 1)))
Timed out. \[ \int \frac {\sqrt {-1+x^2}}{(-i+x)^2} \, dx=\int \frac {\sqrt {x^2-1}}{{\left (x-\mathrm {i}\right )}^2} \,d x \]
\[ \int \frac {\sqrt {-1+x^2}}{(-i+x)^2} \, dx=-\left (\int \frac {\sqrt {x^{2}-1}}{2 i x -x^{2}+1}d x \right ) \]