3.164 \(\int \frac {1}{(13-4 x+x^2)^3} \, dx\)

Optimal. Leaf size=51 \[ -\frac {2-x}{216 \left (x^2-4 x+13\right )}-\frac {2-x}{36 \left (x^2-4 x+13\right )^2}+\frac {1}{648} \tan ^{-1}\left (\frac {x-2}{3}\right ) \]

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Rubi [A]  time = 0.01, antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {614, 618, 204} \[ -\frac {2-x}{216 \left (x^2-4 x+13\right )}-\frac {2-x}{36 \left (x^2-4 x+13\right )^2}+\frac {1}{648} \tan ^{-1}\left (\frac {x-2}{3}\right ) \]

Antiderivative was successfully verified.

[In]

Int[(13 - 4*x + x^2)^(-3),x]

[Out]

-(2 - x)/(36*(13 - 4*x + x^2)^2) - (2 - x)/(216*(13 - 4*x + x^2)) + ArcTan[(-2 + x)/3]/648

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 614

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

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]

Rubi steps

\begin {align*} \int \frac {1}{\left (13-4 x+x^2\right )^3} \, dx &=-\frac {2-x}{36 \left (13-4 x+x^2\right )^2}+\frac {1}{12} \int \frac {1}{\left (13-4 x+x^2\right )^2} \, dx\\ &=-\frac {2-x}{36 \left (13-4 x+x^2\right )^2}-\frac {2-x}{216 \left (13-4 x+x^2\right )}+\frac {1}{216} \int \frac {1}{13-4 x+x^2} \, dx\\ &=-\frac {2-x}{36 \left (13-4 x+x^2\right )^2}-\frac {2-x}{216 \left (13-4 x+x^2\right )}-\frac {1}{108} \operatorname {Subst}\left (\int \frac {1}{-36-x^2} \, dx,x,-4+2 x\right )\\ &=-\frac {2-x}{36 \left (13-4 x+x^2\right )^2}-\frac {2-x}{216 \left (13-4 x+x^2\right )}+\frac {1}{648} \tan ^{-1}\left (\frac {1}{3} (-2+x)\right )\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 36, normalized size = 0.71 \[ \frac {1}{648} \left (\frac {3 (x-2) \left (x^2-4 x+19\right )}{\left (x^2-4 x+13\right )^2}+\tan ^{-1}\left (\frac {x-2}{3}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(13 - 4*x + x^2)^(-3),x]

[Out]

((3*(-2 + x)*(19 - 4*x + x^2))/(13 - 4*x + x^2)^2 + ArcTan[(-2 + x)/3])/648

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IntegrateAlgebraic [A]  time = 0.02, size = 40, normalized size = 0.78 \[ \frac {(x-2) \left (x^2-4 x+19\right )}{216 \left (x^2-4 x+13\right )^2}-\frac {1}{648} \tan ^{-1}\left (\frac {2}{3}-\frac {x}{3}\right ) \]

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(13 - 4*x + x^2)^(-3),x]

[Out]

((-2 + x)*(19 - 4*x + x^2))/(216*(13 - 4*x + x^2)^2) - ArcTan[2/3 - x/3]/648

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fricas [A]  time = 1.03, size = 62, normalized size = 1.22 \[ \frac {3 \, x^{3} - 18 \, x^{2} + {\left (x^{4} - 8 \, x^{3} + 42 \, x^{2} - 104 \, x + 169\right )} \arctan \left (\frac {1}{3} \, x - \frac {2}{3}\right ) + 81 \, x - 114}{648 \, {\left (x^{4} - 8 \, x^{3} + 42 \, x^{2} - 104 \, x + 169\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(x^2-4*x+13)^3,x, algorithm="fricas")

[Out]

1/648*(3*x^3 - 18*x^2 + (x^4 - 8*x^3 + 42*x^2 - 104*x + 169)*arctan(1/3*x - 2/3) + 81*x - 114)/(x^4 - 8*x^3 +
42*x^2 - 104*x + 169)

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giac [A]  time = 0.85, size = 34, normalized size = 0.67 \[ \frac {x^{3} - 6 \, x^{2} + 27 \, x - 38}{216 \, {\left (x^{2} - 4 \, x + 13\right )}^{2}} + \frac {1}{648} \, \arctan \left (\frac {1}{3} \, x - \frac {2}{3}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(x^2-4*x+13)^3,x, algorithm="giac")

[Out]

1/216*(x^3 - 6*x^2 + 27*x - 38)/(x^2 - 4*x + 13)^2 + 1/648*arctan(1/3*x - 2/3)

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maple [A]  time = 0.37, size = 36, normalized size = 0.71

method	result	size
risch	\(\frac {\frac {1}{216} x^{3}-\frac {1}{36} x^{2}+\frac {1}{8} x -\frac {19}{108}}{\left (x^{2}-4 x +13\right )^{2}}+\frac {\arctan \left (-\frac {2}{3}+\frac {x}{3}\right )}{648}\)	\(36\)
default	\(\frac {2 x -4}{72 \left (x^{2}-4 x +13\right )^{2}}+\frac {2 x -4}{432 x^{2}-1728 x +5616}+\frac {\arctan \left (-\frac {2}{3}+\frac {x}{3}\right )}{648}\)	\(44\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^2-4*x+13)^3,x,method=_RETURNVERBOSE)

[Out]

(1/216*x^3-1/36*x^2+1/8*x-19/108)/(x^2-4*x+13)^2+1/648*arctan(-2/3+1/3*x)

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maxima [A]  time = 0.96, size = 44, normalized size = 0.86 \[ \frac {x^{3} - 6 \, x^{2} + 27 \, x - 38}{216 \, {\left (x^{4} - 8 \, x^{3} + 42 \, x^{2} - 104 \, x + 169\right )}} + \frac {1}{648} \, \arctan \left (\frac {1}{3} \, x - \frac {2}{3}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(x^2-4*x+13)^3,x, algorithm="maxima")

[Out]

1/216*(x^3 - 6*x^2 + 27*x - 38)/(x^4 - 8*x^3 + 42*x^2 - 104*x + 169) + 1/648*arctan(1/3*x - 2/3)

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mupad [B]  time = 0.20, size = 39, normalized size = 0.76 \[ \frac {\mathrm {atan}\left (\frac {x}{3}-\frac {2}{3}\right )}{648}+6\,\left (x-2\right )\,\left (\frac {1}{1296\,\left (x^2-4\,x+13\right )}+\frac {1}{216\,{\left (x^2-4\,x+13\right )}^2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^2 - 4*x + 13)^3,x)

[Out]

atan(x/3 - 2/3)/648 + 6*(x - 2)*(1/(1296*(x^2 - 4*x + 13)) + 1/(216*(x^2 - 4*x + 13)^2))

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sympy [A]  time = 0.17, size = 42, normalized size = 0.82 \[ \frac {x^{3} - 6 x^{2} + 27 x - 38}{216 x^{4} - 1728 x^{3} + 9072 x^{2} - 22464 x + 36504} + \frac {\operatorname {atan}{\left (\frac {x}{3} - \frac {2}{3} \right )}}{648} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(x**2-4*x+13)**3,x)

[Out]

(x**3 - 6*x**2 + 27*x - 38)/(216*x**4 - 1728*x**3 + 9072*x**2 - 22464*x + 36504) + atan(x/3 - 2/3)/648

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