3.176 \(\int \frac {1+x+x^2}{(3+2 x+x^2)^2} \, dx\)

Optimal. Leaf size=39 \[ \frac {1-x}{4 \left (x^2+2 x+3\right )}+\frac {3 \tan ^{-1}\left (\frac {x+1}{\sqrt {2}}\right )}{4 \sqrt {2}} \]

[Out]

1/4*(1-x)/(x^2+2*x+3)+3/8*arctan(1/2*(1+x)*2^(1/2))*2^(1/2)

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Rubi [A]  time = 0.02, antiderivative size = 39, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {1660, 12, 618, 204} \[ \frac {1-x}{4 \left (x^2+2 x+3\right )}+\frac {3 \tan ^{-1}\left (\frac {x+1}{\sqrt {2}}\right )}{4 \sqrt {2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 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 1660

Int[(Pq_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, a + b*x + c*
x^2, x], f = Coeff[PolynomialRemainder[Pq, a + b*x + c*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b
*x + c*x^2, x], x, 1]}, Simp[((b*f - 2*a*g + (2*c*f - b*g)*x)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c
)), x] + Dist[1/((p + 1)*(b^2 - 4*a*c)), Int[(a + b*x + c*x^2)^(p + 1)*ExpandToSum[(p + 1)*(b^2 - 4*a*c)*Q - (
2*p + 3)*(2*c*f - b*g), x], x], x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1
]

Rubi steps

\begin {align*} \int \frac {1+x+x^2}{\left (3+2 x+x^2\right )^2} \, dx &=\frac {1-x}{4 \left (3+2 x+x^2\right )}+\frac {1}{8} \int \frac {6}{3+2 x+x^2} \, dx\\ &=\frac {1-x}{4 \left (3+2 x+x^2\right )}+\frac {3}{4} \int \frac {1}{3+2 x+x^2} \, dx\\ &=\frac {1-x}{4 \left (3+2 x+x^2\right )}-\frac {3}{2} \operatorname {Subst}\left (\int \frac {1}{-8-x^2} \, dx,x,2+2 x\right )\\ &=\frac {1-x}{4 \left (3+2 x+x^2\right )}+\frac {3 \tan ^{-1}\left (\frac {1+x}{\sqrt {2}}\right )}{4 \sqrt {2}}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 39, normalized size = 1.00 \[ \frac {1-x}{4 \left (x^2+2 x+3\right )}+\frac {3 \tan ^{-1}\left (\frac {x+1}{\sqrt {2}}\right )}{4 \sqrt {2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.64, size = 39, normalized size = 1.00 \[ \frac {3 \, \sqrt {2} {\left (x^{2} + 2 \, x + 3\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (x + 1\right )}\right ) - 2 \, x + 2}{8 \, {\left (x^{2} + 2 \, x + 3\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(3*sqrt(2)*(x^2 + 2*x + 3)*arctan(1/2*sqrt(2)*(x + 1)) - 2*x + 2)/(x^2 + 2*x + 3)

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giac [A]  time = 0.16, size = 30, normalized size = 0.77 \[ \frac {3}{8} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (x + 1\right )}\right ) - \frac {x - 1}{4 \, {\left (x^{2} + 2 \, x + 3\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/8*sqrt(2)*arctan(1/2*sqrt(2)*(x + 1)) - 1/4*(x - 1)/(x^2 + 2*x + 3)

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maple [A]  time = 0.01, size = 34, normalized size = 0.87 \[ \frac {3 \sqrt {2}\, \arctan \left (\frac {\left (2 x +2\right ) \sqrt {2}}{4}\right )}{8}+\frac {-\frac {x}{4}+\frac {1}{4}}{x^{2}+2 x +3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-1/4*x+1/4)/(x^2+2*x+3)+3/8*2^(1/2)*arctan(1/4*(2*x+2)*2^(1/2))

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maxima [A]  time = 0.96, size = 30, normalized size = 0.77 \[ \frac {3}{8} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (x + 1\right )}\right ) - \frac {x - 1}{4 \, {\left (x^{2} + 2 \, x + 3\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/8*sqrt(2)*arctan(1/2*sqrt(2)*(x + 1)) - 1/4*(x - 1)/(x^2 + 2*x + 3)

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mupad [B]  time = 3.84, size = 36, normalized size = 0.92 \[ \frac {3\,\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,x}{2}+\frac {\sqrt {2}}{2}\right )}{8}-\frac {\frac {x}{4}-\frac {1}{4}}{x^2+2\,x+3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(3*2^(1/2)*atan((2^(1/2)*x)/2 + 2^(1/2)/2))/8 - (x/4 - 1/4)/(2*x + x^2 + 3)

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sympy [A]  time = 0.14, size = 37, normalized size = 0.95 \[ \frac {1 - x}{4 x^{2} + 8 x + 12} + \frac {3 \sqrt {2} \operatorname {atan}{\left (\frac {\sqrt {2} x}{2} + \frac {\sqrt {2}}{2} \right )}}{8} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(1 - x)/(4*x**2 + 8*x + 12) + 3*sqrt(2)*atan(sqrt(2)*x/2 + sqrt(2)/2)/8

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