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

Optimal. Leaf size=64 \[ \frac{\sqrt{x^2-1}}{-x+i}-\frac{i \tan ^{-1}\left (\frac{1-i x}{\sqrt{2} \sqrt{x^2-1}}\right )}{\sqrt{2}}+\tanh ^{-1}\left (\frac{x}{\sqrt{x^2-1}}\right ) \]

[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]]

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Rubi [A]  time = 0.0260134, antiderivative size = 64, normalized size of antiderivative = 1., 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} \[ \frac{\sqrt{x^2-1}}{-x+i}-\frac{i \tan ^{-1}\left (\frac{1-i x}{\sqrt{2} \sqrt{x^2-1}}\right )}{\sqrt{2}}+\tanh ^{-1}\left (\frac{x}{\sqrt{x^2-1}}\right ) \]

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 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]

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 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 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 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])

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 = 0.0594532, size = 59, normalized size = 0.92 \[ -\frac{\sqrt{x^2-1}}{x-i}+\tanh ^{-1}\left (\frac{x}{\sqrt{x^2-1}}\right )-\frac{\tanh ^{-1}\left (\frac{x+i}{\sqrt{2} \sqrt{x^2-1}}\right )}{\sqrt{2}} \]

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]

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Maple [B]  time = 0.025, size = 125, normalized size = 2. \begin{align*}{\frac{1}{2\,x-2\,i} \left ( \left ( x-i \right ) ^{2}-2+2\,i \left ( x-i \right ) \right ) ^{{\frac{3}{2}}}}+\ln \left ( x+\sqrt{ \left ( x-i \right ) ^{2}-2+2\,i \left ( x-i \right ) } \right ) +{\frac{i}{2}}\sqrt{2}\arctan \left ({\frac{ \left ( -4+2\,i \left ( x-i \right ) \right ) \sqrt{2}}{4}{\frac{1}{\sqrt{ \left ( x-i \right ) ^{2}-2+2\,i \left ( x-i \right ) }}}} \right ) -{\frac{i}{2}}\sqrt{ \left ( x-i \right ) ^{2}-2+2\,i \left ( x-i \right ) }-{\frac{x}{2}\sqrt{ \left ( x-i \right ) ^{2}-2+2\,i \left ( x-i \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2-1)^(1/2)/(x-I)^2,x)

[Out]

1/2/(x-I)*((x-I)^2-2+2*I*(x-I))^(3/2)+ln(x+((x-I)^2-2+2*I*(x-I))^(1/2))+1/2*I*2^(1/2)*arctan(1/4*(-4+2*I*(x-I)
)*2^(1/2)/((x-I)^2-2+2*I*(x-I))^(1/2))-1/2*I*((x-I)^2-2+2*I*(x-I))^(1/2)-1/2*x*((x-I)^2-2+2*I*(x-I))^(1/2)

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Maxima [A]  time = 1.429, size = 72, normalized size = 1.12 \begin{align*} \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{align*}

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))

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Fricas [A]  time = 2.19194, size = 267, normalized size = 4.17 \begin{align*} -\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 ) +{\left (2 \, x - 2 i\right )} \log \left (-x + \sqrt{x^{2} - 1}\right ) + 2 \, x + 2 \, \sqrt{x^{2} - 1} - 2 i}{2 \, x - 2 i} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(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 - 2*I)*log(-x + sqrt(x^2 - 1)) + 2*x + 2*sqrt(x^2 - 1) - 2*I)/(2*x - 2*I)

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

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)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{x^{2} - 1}}{{\left (x - i\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(x^2 - 1)/(x - I)^2, x)