3.47 \(\int \frac{2+3 x+5 x^2}{(3-x+2 x^2)^2} \, dx\)

Optimal. Leaf size=43 \[ -\frac{11 (3 x+5)}{46 \left (2 x^2-x+3\right )}-\frac{82 \tan ^{-1}\left (\frac{1-4 x}{\sqrt{23}}\right )}{23 \sqrt{23}} \]

[Out]

(-11*(5 + 3*x))/(46*(3 - x + 2*x^2)) - (82*ArcTan[(1 - 4*x)/Sqrt[23]])/(23*Sqrt[23])

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Rubi [A]  time = 0.0260438, antiderivative size = 43, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {1660, 12, 618, 204} \[ -\frac{11 (3 x+5)}{46 \left (2 x^2-x+3\right )}-\frac{82 \tan ^{-1}\left (\frac{1-4 x}{\sqrt{23}}\right )}{23 \sqrt{23}} \]

Antiderivative was successfully verified.

[In]

Int[(2 + 3*x + 5*x^2)/(3 - x + 2*x^2)^2,x]

[Out]

(-11*(5 + 3*x))/(46*(3 - x + 2*x^2)) - (82*ArcTan[(1 - 4*x)/Sqrt[23]])/(23*Sqrt[23])

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
]

Rule 12

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

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{2+3 x+5 x^2}{\left (3-x+2 x^2\right )^2} \, dx &=-\frac{11 (5+3 x)}{46 \left (3-x+2 x^2\right )}+\frac{1}{23} \int \frac{41}{3-x+2 x^2} \, dx\\ &=-\frac{11 (5+3 x)}{46 \left (3-x+2 x^2\right )}+\frac{41}{23} \int \frac{1}{3-x+2 x^2} \, dx\\ &=-\frac{11 (5+3 x)}{46 \left (3-x+2 x^2\right )}-\frac{82}{23} \operatorname{Subst}\left (\int \frac{1}{-23-x^2} \, dx,x,-1+4 x\right )\\ &=-\frac{11 (5+3 x)}{46 \left (3-x+2 x^2\right )}-\frac{82 \tan ^{-1}\left (\frac{1-4 x}{\sqrt{23}}\right )}{23 \sqrt{23}}\\ \end{align*}

Mathematica [A]  time = 0.0143532, size = 43, normalized size = 1. \[ \frac{82 \tan ^{-1}\left (\frac{4 x-1}{\sqrt{23}}\right )}{23 \sqrt{23}}-\frac{11 (3 x+5)}{46 \left (2 x^2-x+3\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(2 + 3*x + 5*x^2)/(3 - x + 2*x^2)^2,x]

[Out]

(-11*(5 + 3*x))/(46*(3 - x + 2*x^2)) + (82*ArcTan[(-1 + 4*x)/Sqrt[23]])/(23*Sqrt[23])

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Maple [A]  time = 0.047, size = 34, normalized size = 0.8 \begin{align*}{ \left ( -{\frac{33\,x}{92}}-{\frac{55}{92}} \right ) \left ({x}^{2}-{\frac{x}{2}}+{\frac{3}{2}} \right ) ^{-1}}+{\frac{82\,\sqrt{23}}{529}\arctan \left ({\frac{ \left ( -1+4\,x \right ) \sqrt{23}}{23}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5*x^2+3*x+2)/(2*x^2-x+3)^2,x)

[Out]

(-33/92*x-55/92)/(x^2-1/2*x+3/2)+82/529*23^(1/2)*arctan(1/23*(-1+4*x)*23^(1/2))

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Maxima [A]  time = 1.4319, size = 49, normalized size = 1.14 \begin{align*} \frac{82}{529} \, \sqrt{23} \arctan \left (\frac{1}{23} \, \sqrt{23}{\left (4 \, x - 1\right )}\right ) - \frac{11 \,{\left (3 \, x + 5\right )}}{46 \,{\left (2 \, x^{2} - x + 3\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x^2+3*x+2)/(2*x^2-x+3)^2,x, algorithm="maxima")

[Out]

82/529*sqrt(23)*arctan(1/23*sqrt(23)*(4*x - 1)) - 11/46*(3*x + 5)/(2*x^2 - x + 3)

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Fricas [A]  time = 0.988498, size = 138, normalized size = 3.21 \begin{align*} \frac{164 \, \sqrt{23}{\left (2 \, x^{2} - x + 3\right )} \arctan \left (\frac{1}{23} \, \sqrt{23}{\left (4 \, x - 1\right )}\right ) - 759 \, x - 1265}{1058 \,{\left (2 \, x^{2} - x + 3\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x^2+3*x+2)/(2*x^2-x+3)^2,x, algorithm="fricas")

[Out]

1/1058*(164*sqrt(23)*(2*x^2 - x + 3)*arctan(1/23*sqrt(23)*(4*x - 1)) - 759*x - 1265)/(2*x^2 - x + 3)

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Sympy [A]  time = 0.205915, size = 41, normalized size = 0.95 \begin{align*} - \frac{33 x + 55}{92 x^{2} - 46 x + 138} + \frac{82 \sqrt{23} \operatorname{atan}{\left (\frac{4 \sqrt{23} x}{23} - \frac{\sqrt{23}}{23} \right )}}{529} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x**2+3*x+2)/(2*x**2-x+3)**2,x)

[Out]

-(33*x + 55)/(92*x**2 - 46*x + 138) + 82*sqrt(23)*atan(4*sqrt(23)*x/23 - sqrt(23)/23)/529

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Giac [A]  time = 1.13908, size = 49, normalized size = 1.14 \begin{align*} \frac{82}{529} \, \sqrt{23} \arctan \left (\frac{1}{23} \, \sqrt{23}{\left (4 \, x - 1\right )}\right ) - \frac{11 \,{\left (3 \, x + 5\right )}}{46 \,{\left (2 \, x^{2} - x + 3\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x^2+3*x+2)/(2*x^2-x+3)^2,x, algorithm="giac")

[Out]

82/529*sqrt(23)*arctan(1/23*sqrt(23)*(4*x - 1)) - 11/46*(3*x + 5)/(2*x^2 - x + 3)