3.349 \(\int \frac{1}{(4-4 x+x^2) (5-4 x+x^2)} \, dx\)

Optimal. Leaf size=14 \[ \frac{1}{2-x}+\tan ^{-1}(2-x) \]

[Out]

(2 - x)^(-1) + ArcTan[2 - x]

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Rubi [A]  time = 0.0106012, antiderivative size = 14, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {27, 693, 618, 204} \[ \frac{1}{2-x}+\tan ^{-1}(2-x) \]

Antiderivative was successfully verified.

[In]

Int[1/((4 - 4*x + x^2)*(5 - 4*x + x^2)),x]

[Out]

(2 - x)^(-1) + ArcTan[2 - x]

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 693

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(-2*b*d*(d + e*x)^(m
+ 1)*(a + b*x + c*x^2)^(p + 1))/(d^2*(m + 1)*(b^2 - 4*a*c)), x] + Dist[(b^2*(m + 2*p + 3))/(d^2*(m + 1)*(b^2
- 4*a*c)), Int[(d + e*x)^(m + 2)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b^2 - 4*a*
c, 0] && EqQ[2*c*d - b*e, 0] && NeQ[m + 2*p + 3, 0] && LtQ[m, -1] && (IntegerQ[2*p] || (IntegerQ[m] && Rationa
lQ[p]) || IntegerQ[(m + 2*p + 3)/2])

Rule 618

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

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{1}{\left (4-4 x+x^2\right ) \left (5-4 x+x^2\right )} \, dx &=\int \frac{1}{(-2+x)^2 \left (5-4 x+x^2\right )} \, dx\\ &=\frac{1}{2-x}-\int \frac{1}{5-4 x+x^2} \, dx\\ &=\frac{1}{2-x}+2 \operatorname{Subst}\left (\int \frac{1}{-4-x^2} \, dx,x,-4+2 x\right )\\ &=\frac{1}{2-x}+\tan ^{-1}(2-x)\\ \end{align*}

Mathematica [A]  time = 0.0085638, size = 14, normalized size = 1. \[ \tan ^{-1}(2-x)-\frac{1}{x-2} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((4 - 4*x + x^2)*(5 - 4*x + x^2)),x]

[Out]

-(-2 + x)^(-1) + ArcTan[2 - x]

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Maple [A]  time = 0.005, size = 15, normalized size = 1.1 \begin{align*} -\arctan \left ( -2+x \right ) - \left ( -2+x \right ) ^{-1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^2-4*x+4)/(x^2-4*x+5),x)

[Out]

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

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Maxima [A]  time = 2.3651, size = 19, normalized size = 1.36 \begin{align*} -\frac{1}{x - 2} - \arctan \left (x - 2\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(x^2-4*x+4)/(x^2-4*x+5),x, algorithm="maxima")

[Out]

-1/(x - 2) - arctan(x - 2)

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Fricas [A]  time = 1.47842, size = 51, normalized size = 3.64 \begin{align*} -\frac{{\left (x - 2\right )} \arctan \left (x - 2\right ) + 1}{x - 2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(x^2-4*x+4)/(x^2-4*x+5),x, algorithm="fricas")

[Out]

-((x - 2)*arctan(x - 2) + 1)/(x - 2)

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Sympy [A]  time = 0.122933, size = 10, normalized size = 0.71 \begin{align*} - \operatorname{atan}{\left (x - 2 \right )} - \frac{1}{x - 2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(x**2-4*x+4)/(x**2-4*x+5),x)

[Out]

-atan(x - 2) - 1/(x - 2)

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Giac [A]  time = 1.16776, size = 19, normalized size = 1.36 \begin{align*} -\frac{1}{x - 2} - \arctan \left (x - 2\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(x^2-4*x+4)/(x^2-4*x+5),x, algorithm="giac")

[Out]

-1/(x - 2) - arctan(x - 2)