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

Optimal. Leaf size=75 \[ -\frac {1}{2} \tan ^{-1}\left (\frac {3-x}{2 \sqrt {x^2-x-1}}\right )+\tanh ^{-1}\left (\frac {1-2 x}{2 \sqrt {x^2-x-1}}\right )+\frac {1}{2} \tanh ^{-1}\left (\frac {3 x+1}{2 \sqrt {x^2-x-1}}\right ) \]

[Out]

-1/2*arctan(1/2*(3-x)/(x^2-x-1)^(1/2))+arctanh(1/2*(1-2*x)/(x^2-x-1)^(1/2))+1/2*arctanh(1/2*(1+3*x)/(x^2-x-1)^
(1/2))

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Rubi [A]  time = 0.06, antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {990, 621, 206, 1033, 724, 204} \[ -\frac {1}{2} \tan ^{-1}\left (\frac {3-x}{2 \sqrt {x^2-x-1}}\right )+\tanh ^{-1}\left (\frac {1-2 x}{2 \sqrt {x^2-x-1}}\right )+\frac {1}{2} \tanh ^{-1}\left (\frac {3 x+1}{2 \sqrt {x^2-x-1}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[-1 - x + x^2]/(1 - x^2),x]

[Out]

-ArcTan[(3 - x)/(2*Sqrt[-1 - x + x^2])]/2 + ArcTanh[(1 - 2*x)/(2*Sqrt[-1 - x + x^2])] + ArcTanh[(1 + 3*x)/(2*S
qrt[-1 - x + x^2])]/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 621

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(4*c - x^2), x], x, (b + 2*c*x)
/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 724

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[-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] && NeQ[b^2 - 4*a*c, 0] && NeQ[2*c*d - b*e, 0]

Rule 990

Int[Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2]/((d_) + (f_.)*(x_)^2), x_Symbol] :> Dist[c/f, Int[1/Sqrt[a + b*x +
c*x^2], x], x] - Dist[1/f, Int[(c*d - a*f - b*f*x)/(Sqrt[a + b*x + c*x^2]*(d + f*x^2)), x], x] /; FreeQ[{a, b,
c, d, f}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 1033

Int[((g_.) + (h_.)*(x_))/(((a_) + (c_.)*(x_)^2)*Sqrt[(d_.) + (e_.)*(x_) + (f_.)*(x_)^2]), x_Symbol] :> With[{q
= Rt[-(a*c), 2]}, Dist[h/2 + (c*g)/(2*q), Int[1/((-q + c*x)*Sqrt[d + e*x + f*x^2]), x], x] + Dist[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)]

Rubi steps

\begin {align*} \int \frac {\sqrt {-1-x+x^2}}{1-x^2} \, dx &=-\int \frac {1}{\sqrt {-1-x+x^2}} \, dx-\int \frac {x}{\left (1-x^2\right ) \sqrt {-1-x+x^2}} \, dx\\ &=-\left (\frac {1}{2} \int \frac {1}{(-1-x) \sqrt {-1-x+x^2}} \, dx\right )-\frac {1}{2} \int \frac {1}{(1-x) \sqrt {-1-x+x^2}} \, dx-2 \operatorname {Subst}\left (\int \frac {1}{4-x^2} \, dx,x,\frac {-1+2 x}{\sqrt {-1-x+x^2}}\right )\\ &=\tanh ^{-1}\left (\frac {1-2 x}{2 \sqrt {-1-x+x^2}}\right )+\operatorname {Subst}\left (\int \frac {1}{-4-x^2} \, dx,x,\frac {3-x}{\sqrt {-1-x+x^2}}\right )+\operatorname {Subst}\left (\int \frac {1}{4-x^2} \, dx,x,\frac {1+3 x}{\sqrt {-1-x+x^2}}\right )\\ &=-\frac {1}{2} \tan ^{-1}\left (\frac {3-x}{2 \sqrt {-1-x+x^2}}\right )+\tanh ^{-1}\left (\frac {1-2 x}{2 \sqrt {-1-x+x^2}}\right )+\frac {1}{2} \tanh ^{-1}\left (\frac {1+3 x}{2 \sqrt {-1-x+x^2}}\right )\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 75, normalized size = 1.00 \[ -\frac {1}{2} \tan ^{-1}\left (\frac {3-x}{2 \sqrt {x^2-x-1}}\right )+\tanh ^{-1}\left (\frac {1-2 x}{2 \sqrt {x^2-x-1}}\right )+\frac {1}{2} \tanh ^{-1}\left (\frac {3 x+1}{2 \sqrt {x^2-x-1}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*ArcTan[(3 - x)/(2*Sqrt[-1 - x + x^2])] + ArcTanh[(1 - 2*x)/(2*Sqrt[-1 - x + x^2])] + ArcTanh[(1 + 3*x)/(2
*Sqrt[-1 - x + x^2])]/2

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fricas [A]  time = 1.39, size = 70, normalized size = 0.93 \[ \arctan \left (-x + \sqrt {x^{2} - x - 1} + 1\right ) - \frac {1}{2} \, \log \left (-x + \sqrt {x^{2} - x - 1}\right ) + \frac {1}{2} \, \log \left (-x + \sqrt {x^{2} - x - 1} - 2\right ) + \log \left (-2 \, x + 2 \, \sqrt {x^{2} - x - 1} + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

arctan(-x + sqrt(x^2 - x - 1) + 1) - 1/2*log(-x + sqrt(x^2 - x - 1)) + 1/2*log(-x + sqrt(x^2 - x - 1) - 2) + l
og(-2*x + 2*sqrt(x^2 - x - 1) + 1)

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giac [A]  time = 0.20, size = 73, normalized size = 0.97 \[ \arctan \left (-x + \sqrt {x^{2} - x - 1} + 1\right ) - \frac {1}{2} \, \log \left ({\left | -x + \sqrt {x^{2} - x - 1} \right |}\right ) + \frac {1}{2} \, \log \left ({\left | -x + \sqrt {x^{2} - x - 1} - 2 \right |}\right ) + \log \left ({\left | -2 \, x + 2 \, \sqrt {x^{2} - x - 1} + 1 \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.02, size = 102, normalized size = 1.36 \[ -\frac {\arctanh \left (\frac {-3 x -1}{2 \sqrt {-3 x +\left (x +1\right )^{2}-2}}\right )}{2}+\frac {\arctan \left (\frac {x -3}{2 \sqrt {x +\left (x -1\right )^{2}-2}}\right )}{2}-\frac {3 \ln \left (x -\frac {1}{2}+\sqrt {-3 x +\left (x +1\right )^{2}-2}\right )}{4}-\frac {\ln \left (x -\frac {1}{2}+\sqrt {x +\left (x -1\right )^{2}-2}\right )}{4}-\frac {\sqrt {x +\left (x -1\right )^{2}-2}}{2}+\frac {\sqrt {-3 x +\left (x +1\right )^{2}-2}}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.97, size = 83, normalized size = 1.11 \[ \frac {1}{2} \, \arcsin \left (\frac {2 \, \sqrt {5} x}{5 \, {\left | 2 \, x - 2 \right |}} - \frac {6 \, \sqrt {5}}{5 \, {\left | 2 \, x - 2 \right |}}\right ) - \log \left (x + \sqrt {x^{2} - x - 1} - \frac {1}{2}\right ) - \frac {1}{2} \, \log \left (\frac {2 \, \sqrt {x^{2} - x - 1}}{{\left | 2 \, x + 2 \right |}} + \frac {2}{{\left | 2 \, x + 2 \right |}} - \frac {3}{2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*arcsin(2/5*sqrt(5)*x/abs(2*x - 2) - 6/5*sqrt(5)/abs(2*x - 2)) - log(x + sqrt(x^2 - x - 1) - 1/2) - 1/2*log
(2*sqrt(x^2 - x - 1)/abs(2*x + 2) + 2/abs(2*x + 2) - 3/2)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ -\int \frac {\sqrt {x^2-x-1}}{x^2-1} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(x^2 - x - 1)^(1/2)/(x^2 - 1),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \int \frac {\sqrt {x^{2} - x - 1}}{x^{2} - 1}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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