3.11.26 \(\int \frac {\sqrt {-1+x^2}}{(-i+x)^2} \, dx\)

Optimal. Leaf size=78 \[ \frac {\sqrt {x^2-1}}{-x+i}-\log \left (\sqrt {x^2-1}-x\right )-\sqrt {2} \tanh ^{-1}\left (-\frac {i \sqrt {x^2-1}}{\sqrt {2}}+\frac {i x}{\sqrt {2}}+\frac {1}{\sqrt {2}}\right ) \]

________________________________________________________________________________________

Rubi [A]  time = 0.03, antiderivative size = 64, normalized size of antiderivative = 0.82, number of steps used = 6, number of rules used = 6, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.353, Rules used = {733, 844, 217, 206, 725, 204} \begin {gather*} \frac {\sqrt {x^2-1}}{-x+i}-\frac {i \tan ^{-1}\left (\frac {1-i x}{\sqrt {2} \sqrt {x^2-1}}\right )}{\sqrt {2}}+\tanh ^{-1}\left (\frac {x}{\sqrt {x^2-1}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[-1 + x^2]/(-I + x)^2,x]

[Out]

Sqrt[-1 + x^2]/(I - x) - (I*ArcTan[(1 - I*x)/(Sqrt[2]*Sqrt[-1 + x^2])])/Sqrt[2] + ArcTanh[x/Sqrt[-1 + x^2]]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 217

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]

Rule 725

Int[1/(((d_) + (e_.)*(x_))*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> -Subst[Int[1/(c*d^2 + a*e^2 - x^2), x], x,
(a*e - c*d*x)/Sqrt[a + c*x^2]] /; FreeQ[{a, c, d, e}, x]

Rule 733

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(a + c*x^2)^p)/(
e*(m + 1)), x] - Dist[(2*c*p)/(e*(m + 1)), Int[x*(d + e*x)^(m + 1)*(a + c*x^2)^(p - 1), x], x] /; FreeQ[{a, c,
d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && GtQ[p, 0] && (IntegerQ[p] || LtQ[m, -1]) && NeQ[m, -1] &&  !ILtQ[m +
2*p + 1, 0] && IntQuadraticQ[a, 0, c, d, e, m, p, x]

Rule 844

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[g/e, Int[(d
+ e*x)^(m + 1)*(a + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + c*x^2)^p, x], x] /; FreeQ[{a,
c, d, e, f, g, m, p}, x] && NeQ[c*d^2 + a*e^2, 0] &&  !IGtQ[m, 0]

Rubi steps

\begin {align*} \int \frac {\sqrt {-1+x^2}}{(-i+x)^2} \, dx &=\frac {\sqrt {-1+x^2}}{i-x}+\int \frac {x}{(-i+x) \sqrt {-1+x^2}} \, dx\\ &=\frac {\sqrt {-1+x^2}}{i-x}+i \int \frac {1}{(-i+x) \sqrt {-1+x^2}} \, dx+\int \frac {1}{\sqrt {-1+x^2}} \, dx\\ &=\frac {\sqrt {-1+x^2}}{i-x}-i \operatorname {Subst}\left (\int \frac {1}{-2-x^2} \, dx,x,\frac {-1+i x}{\sqrt {-1+x^2}}\right )+\operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\frac {x}{\sqrt {-1+x^2}}\right )\\ &=\frac {\sqrt {-1+x^2}}{i-x}-\frac {i \tan ^{-1}\left (\frac {1-i x}{\sqrt {2} \sqrt {-1+x^2}}\right )}{\sqrt {2}}+\tanh ^{-1}\left (\frac {x}{\sqrt {-1+x^2}}\right )\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.06, size = 59, normalized size = 0.76 \begin {gather*} -\frac {\sqrt {x^2-1}}{x-i}+\tanh ^{-1}\left (\frac {x}{\sqrt {x^2-1}}\right )-\frac {\tanh ^{-1}\left (\frac {x+i}{\sqrt {2} \sqrt {x^2-1}}\right )}{\sqrt {2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[-1 + x^2]/(-I + x)^2,x]

[Out]

-(Sqrt[-1 + x^2]/(-I + x)) + ArcTanh[x/Sqrt[-1 + x^2]] - ArcTanh[(I + x)/(Sqrt[2]*Sqrt[-1 + x^2])]/Sqrt[2]

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.37, size = 78, normalized size = 1.00 \begin {gather*} \frac {\sqrt {-1+x^2}}{i-x}-\sqrt {2} \tanh ^{-1}\left (\frac {1}{\sqrt {2}}+\frac {i x}{\sqrt {2}}-\frac {i \sqrt {-1+x^2}}{\sqrt {2}}\right )-\log \left (-x+\sqrt {-1+x^2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[Sqrt[-1 + x^2]/(-I + x)^2,x]

[Out]

Sqrt[-1 + x^2]/(I - x) - Sqrt[2]*ArcTanh[1/Sqrt[2] + (I*x)/Sqrt[2] - (I*Sqrt[-1 + x^2])/Sqrt[2]] - Log[-x + Sq
rt[-1 + x^2]]

________________________________________________________________________________________

fricas [A]  time = 0.54, size = 89, normalized size = 1.14 \begin {gather*} -\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 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2-1)^(1/2)/(-I+x)^2,x, algorithm="fricas")

[Out]

-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)

________________________________________________________________________________________

giac [A]  time = 0.38, size = 84, normalized size = 1.08 \begin {gather*} 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 ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2-1)^(1/2)/(-I+x)^2,x, algorithm="giac")

[Out]

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)))

________________________________________________________________________________________

maple [A]  time = 0.15, size = 65, normalized size = 0.83

method	result	size
risch	\(-\frac {\sqrt {x^{2}-1}}{-i+x}+\ln \left (x +\sqrt {x^{2}-1}\right )+\frac {i \sqrt {2}\, \arctan \left (\frac {\left (-4+2 i \left (-i+x \right )\right ) \sqrt {2}}{4 \sqrt {\left (-i+x \right )^{2}+2 i \left (-i+x \right )-2}}\right )}{2}\)	\(65\)
default	\(\frac {\left (\left (-i+x \right )^{2}+2 i \left (-i+x \right )-2\right )^{\frac {3}{2}}}{-2 i+2 x}-\frac {i \left (\sqrt {\left (-i+x \right )^{2}+2 i \left (-i+x \right )-2}+i \ln \left (x +\sqrt {\left (-i+x \right )^{2}+2 i \left (-i+x \right )-2}\right )-\sqrt {2}\, \arctan \left (\frac {\left (-4+2 i \left (-i+x \right )\right ) \sqrt {2}}{4 \sqrt {\left (-i+x \right )^{2}+2 i \left (-i+x \right )-2}}\right )\right )}{2}-\frac {x \sqrt {\left (-i+x \right )^{2}+2 i \left (-i+x \right )-2}}{2}+\frac {\ln \left (x +\sqrt {\left (-i+x \right )^{2}+2 i \left (-i+x \right )-2}\right )}{2}\)	\(150\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2-1)^(1/2)/(-I+x)^2,x,method=_RETURNVERBOSE)

[Out]

-(x^2-1)^(1/2)/(-I+x)+ln(x+(x^2-1)^(1/2))+1/2*I*2^(1/2)*arctan(1/4*(-4+2*I*(-I+x))*2^(1/2)/((-I+x)^2+2*I*(-I+x
)-2)^(1/2))

________________________________________________________________________________________

maxima [A]  time = 0.44, size = 53, normalized size = 0.68 \begin {gather*} \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 ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2-1)^(1/2)/(-I+x)^2,x, algorithm="maxima")

[Out]

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))

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\sqrt {x^2-1}}{{\left (x-\mathrm {i}\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2 - 1)^(1/2)/(x - 1i)^2,x)

[Out]

int((x^2 - 1)^(1/2)/(x - 1i)^2, x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {\left (x - 1\right ) \left (x + 1\right )}}{\left (x - i\right )^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**2-1)**(1/2)/(-I+x)**2,x)

[Out]

Integral(sqrt((x - 1)*(x + 1))/(x - I)**2, x)

________________________________________________________________________________________