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

Optimal. Leaf size=64 \[ -\frac {131 (1-4 x)}{2116 \left (2 x^2-x+3\right )}-\frac {11 (3 x+5)}{92 \left (2 x^2-x+3\right )^2}-\frac {262 \tan ^{-1}\left (\frac {1-4 x}{\sqrt {23}}\right )}{529 \sqrt {23}} \]

[Out]

-11/92*(5+3*x)/(2*x^2-x+3)^2-131/2116*(1-4*x)/(2*x^2-x+3)-262/12167*arctan(1/23*(1-4*x)*23^(1/2))*23^(1/2)

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Rubi [A]  time = 0.03, antiderivative size = 64, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {1660, 12, 614, 618, 204} \[ -\frac {131 (1-4 x)}{2116 \left (2 x^2-x+3\right )}-\frac {11 (3 x+5)}{92 \left (2 x^2-x+3\right )^2}-\frac {262 \tan ^{-1}\left (\frac {1-4 x}{\sqrt {23}}\right )}{529 \sqrt {23}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-11*(5 + 3*x))/(92*(3 - x + 2*x^2)^2) - (131*(1 - 4*x))/(2116*(3 - x + 2*x^2)) - (262*ArcTan[(1 - 4*x)/Sqrt[2
3]])/(529*Sqrt[23])

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 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]

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 {2+3 x+5 x^2}{\left (3-x+2 x^2\right )^3} \, dx &=-\frac {11 (5+3 x)}{92 \left (3-x+2 x^2\right )^2}+\frac {1}{46} \int \frac {131}{2 \left (3-x+2 x^2\right )^2} \, dx\\ &=-\frac {11 (5+3 x)}{92 \left (3-x+2 x^2\right )^2}+\frac {131}{92} \int \frac {1}{\left (3-x+2 x^2\right )^2} \, dx\\ &=-\frac {11 (5+3 x)}{92 \left (3-x+2 x^2\right )^2}-\frac {131 (1-4 x)}{2116 \left (3-x+2 x^2\right )}+\frac {131}{529} \int \frac {1}{3-x+2 x^2} \, dx\\ &=-\frac {11 (5+3 x)}{92 \left (3-x+2 x^2\right )^2}-\frac {131 (1-4 x)}{2116 \left (3-x+2 x^2\right )}-\frac {262}{529} \operatorname {Subst}\left (\int \frac {1}{-23-x^2} \, dx,x,-1+4 x\right )\\ &=-\frac {11 (5+3 x)}{92 \left (3-x+2 x^2\right )^2}-\frac {131 (1-4 x)}{2116 \left (3-x+2 x^2\right )}-\frac {262 \tan ^{-1}\left (\frac {1-4 x}{\sqrt {23}}\right )}{529 \sqrt {23}}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 51, normalized size = 0.80 \[ \frac {\frac {46 \left (524 x^3-393 x^2+472 x-829\right )}{\left (-2 x^2+x-3\right )^2}+1048 \sqrt {23} \tan ^{-1}\left (\frac {4 x-1}{\sqrt {23}}\right )}{48668} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((46*(-829 + 472*x - 393*x^2 + 524*x^3))/(-3 + x - 2*x^2)^2 + 1048*Sqrt[23]*ArcTan[(-1 + 4*x)/Sqrt[23]])/48668

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fricas [A]  time = 0.92, size = 75, normalized size = 1.17 \[ \frac {12052 \, x^{3} + 524 \, \sqrt {23} {\left (4 \, x^{4} - 4 \, x^{3} + 13 \, x^{2} - 6 \, x + 9\right )} \arctan \left (\frac {1}{23} \, \sqrt {23} {\left (4 \, x - 1\right )}\right ) - 9039 \, x^{2} + 10856 \, x - 19067}{24334 \, {\left (4 \, x^{4} - 4 \, x^{3} + 13 \, x^{2} - 6 \, x + 9\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24334*(12052*x^3 + 524*sqrt(23)*(4*x^4 - 4*x^3 + 13*x^2 - 6*x + 9)*arctan(1/23*sqrt(23)*(4*x - 1)) - 9039*x^
2 + 10856*x - 19067)/(4*x^4 - 4*x^3 + 13*x^2 - 6*x + 9)

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giac [A]  time = 0.21, size = 46, normalized size = 0.72 \[ \frac {262}{12167} \, \sqrt {23} \arctan \left (\frac {1}{23} \, \sqrt {23} {\left (4 \, x - 1\right )}\right ) + \frac {524 \, x^{3} - 393 \, x^{2} + 472 \, x - 829}{1058 \, {\left (2 \, x^{2} - x + 3\right )}^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

262/12167*sqrt(23)*arctan(1/23*sqrt(23)*(4*x - 1)) + 1/1058*(524*x^3 - 393*x^2 + 472*x - 829)/(2*x^2 - x + 3)^
2

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maple [A]  time = 0.01, size = 47, normalized size = 0.73 \[ \frac {262 \sqrt {23}\, \arctan \left (\frac {\left (4 x -1\right ) \sqrt {23}}{23}\right )}{12167}+\frac {\frac {262}{529} x^{3}-\frac {393}{1058} x^{2}+\frac {236}{529} x -\frac {829}{1058}}{\left (2 x^{2}-x +3\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4*(131/1058*x^3-393/4232*x^2+59/529*x-829/4232)/(2*x^2-x+3)^2+262/12167*23^(1/2)*arctan(1/23*(4*x-1)*23^(1/2))

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maxima [A]  time = 0.96, size = 56, normalized size = 0.88 \[ \frac {262}{12167} \, \sqrt {23} \arctan \left (\frac {1}{23} \, \sqrt {23} {\left (4 \, x - 1\right )}\right ) + \frac {524 \, x^{3} - 393 \, x^{2} + 472 \, x - 829}{1058 \, {\left (4 \, x^{4} - 4 \, x^{3} + 13 \, x^{2} - 6 \, x + 9\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

262/12167*sqrt(23)*arctan(1/23*sqrt(23)*(4*x - 1)) + 1/1058*(524*x^3 - 393*x^2 + 472*x - 829)/(4*x^4 - 4*x^3 +
13*x^2 - 6*x + 9)

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mupad [B]  time = 0.04, size = 55, normalized size = 0.86 \[ \frac {262\,\sqrt {23}\,\mathrm {atan}\left (\frac {4\,\sqrt {23}\,x}{23}-\frac {\sqrt {23}}{23}\right )}{12167}+\frac {\frac {131\,x^3}{1058}-\frac {393\,x^2}{4232}+\frac {59\,x}{529}-\frac {829}{4232}}{x^4-x^3+\frac {13\,x^2}{4}-\frac {3\,x}{2}+\frac {9}{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(262*23^(1/2)*atan((4*23^(1/2)*x)/23 - 23^(1/2)/23))/12167 + ((59*x)/529 - (393*x^2)/4232 + (131*x^3)/1058 - 8
29/4232)/((13*x^2)/4 - (3*x)/2 - x^3 + x^4 + 9/4)

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sympy [A]  time = 0.18, size = 61, normalized size = 0.95 \[ \frac {524 x^{3} - 393 x^{2} + 472 x - 829}{4232 x^{4} - 4232 x^{3} + 13754 x^{2} - 6348 x + 9522} + \frac {262 \sqrt {23} \operatorname {atan}{\left (\frac {4 \sqrt {23} x}{23} - \frac {\sqrt {23}}{23} \right )}}{12167} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(524*x**3 - 393*x**2 + 472*x - 829)/(4232*x**4 - 4232*x**3 + 13754*x**2 - 6348*x + 9522) + 262*sqrt(23)*atan(4
*sqrt(23)*x/23 - sqrt(23)/23)/12167

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