3.1.92 \(\int \frac {\sqrt {-1-x+x^2}}{1-x^2} \, dx\) [92]

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

[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 = 0.03, 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 = {1004, 635, 212, 1047, 738, 210} \begin {gather*} -\frac {1}{2} \text {ArcTan}\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 ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[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

Rule 210

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])

Rule 212

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] && (GtQ[a, 0] || LtQ[b, 0])

Rule 635

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 738

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 1004

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 1047

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 \text {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 )+\text {Subst}\left (\int \frac {1}{-4-x^2} \, dx,x,\frac {3-x}{\sqrt {-1-x+x^2}}\right )+\text {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 = 0.09, size = 57, normalized size = 0.76 \begin {gather*} \tan ^{-1}\left (1-x+\sqrt {-1-x+x^2}\right )+\tanh ^{-1}\left (1+x-\sqrt {-1-x+x^2}\right )+\log \left (1-2 x+2 \sqrt {-1-x+x^2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

ArcTan[1 - x + Sqrt[-1 - x + x^2]] + ArcTanh[1 + x - Sqrt[-1 - x + x^2]] + Log[1 - 2*x + 2*Sqrt[-1 - x + x^2]]

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Maple [A]
time = 0.18, size = 102, normalized size = 1.36

method	result	size
default	\(-\frac {\sqrt {\left (-1+x \right )^{2}-2+x}}{2}-\frac {\ln \left (-\frac {1}{2}+x +\sqrt {\left (-1+x \right )^{2}-2+x}\right )}{4}+\frac {\arctan \left (\frac {x -3}{2 \sqrt {\left (-1+x \right )^{2}-2+x}}\right )}{2}+\frac {\sqrt {\left (1+x \right )^{2}-2-3 x}}{2}-\frac {3 \ln \left (-\frac {1}{2}+x +\sqrt {\left (1+x \right )^{2}-2-3 x}\right )}{4}-\frac {\arctanh \left (\frac {-1-3 x}{2 \sqrt {\left (1+x \right )^{2}-2-3 x}}\right )}{2}\)	\(102\)
trager	\(\frac {\RootOf \left (\textit {\_Z}^{2}+1\right ) \ln \left (-\frac {-x \RootOf \left (\textit {\_Z}^{2}+1\right )+2 \sqrt {x^{2}-x -1}+3 \RootOf \left (\textit {\_Z}^{2}+1\right )}{-1+x}\right )}{2}-\frac {\ln \left (\frac {8 \sqrt {x^{2}-x -1}\, x^{2}+8 x^{3}+12 \sqrt {x^{2}-x -1}\, x +8 x^{2}+2 \sqrt {x^{2}-x -1}-9 x -11}{1+x}\right )}{2}\)	\(112\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.49, size = 83, normalized size = 1.11 \begin {gather*} \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 ) \end {gather*}

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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Fricas [A]
time = 0.35, size = 70, normalized size = 0.93 \begin {gather*} \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 ) \end {gather*}

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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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \int \frac {\sqrt {x^{2} - x - 1}}{x^{2} - 1}\, dx \end {gather*}

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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Giac [A]
time = 5.64, size = 73, normalized size = 0.97 \begin {gather*} \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 ) \end {gather*}

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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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} -\int \frac {\sqrt {x^2-x-1}}{x^2-1} \,d x \end {gather*}

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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