3.21.74 \(\int \frac {1}{(-2+x) \sqrt {3-4 x+x^2}} \, dx\)

Optimal. Leaf size=13 \[ \tan ^{-1}\left (\sqrt {x^2-4 x+3}\right ) \]

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Rubi [A]  time = 0.01, antiderivative size = 13, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {688, 203} \begin {gather*} \tan ^{-1}\left (\sqrt {x^2-4 x+3}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((-2 + x)*Sqrt[3 - 4*x + x^2]),x]

[Out]

ArcTan[Sqrt[3 - 4*x + x^2]]

Rule 203

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

Rule 688

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

Rubi steps

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

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Mathematica [A]  time = 0.00, size = 12, normalized size = 0.92 \begin {gather*} \tan ^{-1}\left (\sqrt {(x-2)^2-1}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((-2 + x)*Sqrt[3 - 4*x + x^2]),x]

[Out]

ArcTan[Sqrt[-1 + (-2 + x)^2]]

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IntegrateAlgebraic [A]  time = 0.16, size = 21, normalized size = 1.62 \begin {gather*} -2 \tan ^{-1}\left (\frac {\sqrt {x^2-4 x+3}}{x-3}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/((-2 + x)*Sqrt[3 - 4*x + x^2]),x]

[Out]

-2*ArcTan[Sqrt[3 - 4*x + x^2]/(-3 + x)]

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fricas [A]  time = 0.40, size = 18, normalized size = 1.38 \begin {gather*} 2 \, \arctan \left (-x + \sqrt {x^{2} - 4 \, x + 3} + 2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-2+x)/(x^2-4*x+3)^(1/2),x, algorithm="fricas")

[Out]

2*arctan(-x + sqrt(x^2 - 4*x + 3) + 2)

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giac [A]  time = 0.17, size = 18, normalized size = 1.38 \begin {gather*} 2 \, \arctan \left (-x + \sqrt {x^{2} - 4 \, x + 3} + 2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-2+x)/(x^2-4*x+3)^(1/2),x, algorithm="giac")

[Out]

2*arctan(-x + sqrt(x^2 - 4*x + 3) + 2)

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maple [A]  time = 0.06, size = 13, normalized size = 1.00 \begin {gather*} -\arctan \left (\frac {1}{\sqrt {\left (x -2\right )^{2}-1}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x-2)/(x^2-4*x+3)^(1/2),x)

[Out]

-arctan(1/((x-2)^2-1)^(1/2))

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maxima [A]  time = 1.94, size = 9, normalized size = 0.69 \begin {gather*} -\arcsin \left (\frac {1}{{\left | x - 2 \right |}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-2+x)/(x^2-4*x+3)^(1/2),x, algorithm="maxima")

[Out]

-arcsin(1/abs(x - 2))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.08 \begin {gather*} \int \frac {1}{\left (x-2\right )\,\sqrt {x^2-4\,x+3}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((x - 2)*(x^2 - 4*x + 3)^(1/2)),x)

[Out]

int(1/((x - 2)*(x^2 - 4*x + 3)^(1/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt {\left (x - 3\right ) \left (x - 1\right )} \left (x - 2\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-2+x)/(x**2-4*x+3)**(1/2),x)

[Out]

Integral(1/(sqrt((x - 3)*(x - 1))*(x - 2)), x)

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