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

Optimal. Leaf size=64 \[ \frac{121 (69 x+61)}{7750 \left (5 x^2+3 x+2\right )^2}+\frac{11 (45710 x+17557)}{240250 \left (5 x^2+3 x+2\right )}+\frac{4330 \tan ^{-1}\left (\frac{10 x+3}{\sqrt{31}}\right )}{961 \sqrt{31}} \]

[Out]

(121*(61 + 69*x))/(7750*(2 + 3*x + 5*x^2)^2) + (11*(17557 + 45710*x))/(240250*(2 + 3*x + 5*x^2)) + (4330*ArcTa
n[(3 + 10*x)/Sqrt[31]])/(961*Sqrt[31])

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Rubi [A]  time = 0.0516227, antiderivative size = 64, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {1660, 12, 618, 204} \[ \frac{121 (69 x+61)}{7750 \left (5 x^2+3 x+2\right )^2}+\frac{11 (45710 x+17557)}{240250 \left (5 x^2+3 x+2\right )}+\frac{4330 \tan ^{-1}\left (\frac{10 x+3}{\sqrt{31}}\right )}{961 \sqrt{31}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(121*(61 + 69*x))/(7750*(2 + 3*x + 5*x^2)^2) + (11*(17557 + 45710*x))/(240250*(2 + 3*x + 5*x^2)) + (4330*ArcTa
n[(3 + 10*x)/Sqrt[31]])/(961*Sqrt[31])

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{\left (3-x+2 x^2\right )^2}{\left (2+3 x+5 x^2\right )^3} \, dx &=\frac{121 (61+69 x)}{7750 \left (2+3 x+5 x^2\right )^2}+\frac{1}{62} \int \frac{\frac{48669}{125}-\frac{1984 x}{25}+\frac{248 x^2}{5}}{\left (2+3 x+5 x^2\right )^2} \, dx\\ &=\frac{121 (61+69 x)}{7750 \left (2+3 x+5 x^2\right )^2}+\frac{11 (17557+45710 x)}{240250 \left (2+3 x+5 x^2\right )}+\frac{\int \frac{4330}{2+3 x+5 x^2} \, dx}{1922}\\ &=\frac{121 (61+69 x)}{7750 \left (2+3 x+5 x^2\right )^2}+\frac{11 (17557+45710 x)}{240250 \left (2+3 x+5 x^2\right )}+\frac{2165}{961} \int \frac{1}{2+3 x+5 x^2} \, dx\\ &=\frac{121 (61+69 x)}{7750 \left (2+3 x+5 x^2\right )^2}+\frac{11 (17557+45710 x)}{240250 \left (2+3 x+5 x^2\right )}-\frac{4330}{961} \operatorname{Subst}\left (\int \frac{1}{-31-x^2} \, dx,x,3+10 x\right )\\ &=\frac{121 (61+69 x)}{7750 \left (2+3 x+5 x^2\right )^2}+\frac{11 (17557+45710 x)}{240250 \left (2+3 x+5 x^2\right )}+\frac{4330 \tan ^{-1}\left (\frac{3+10 x}{\sqrt{31}}\right )}{961 \sqrt{31}}\\ \end{align*}

Mathematica [A]  time = 0.0233833, size = 53, normalized size = 0.83 \[ \frac{11 \left (45710 x^3+44983 x^2+33524 x+11183\right )}{48050 \left (5 x^2+3 x+2\right )^2}+\frac{4330 \tan ^{-1}\left (\frac{10 x+3}{\sqrt{31}}\right )}{961 \sqrt{31}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(11*(11183 + 33524*x + 44983*x^2 + 45710*x^3))/(48050*(2 + 3*x + 5*x^2)^2) + (4330*ArcTan[(3 + 10*x)/Sqrt[31]]
)/(961*Sqrt[31])

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Maple [A]  time = 0.046, size = 47, normalized size = 0.7 \begin{align*} 25\,{\frac{1}{ \left ( 5\,{x}^{2}+3\,x+2 \right ) ^{2}} \left ({\frac{50281\,{x}^{3}}{120125}}+{\frac{494813\,{x}^{2}}{1201250}}+{\frac{184382\,x}{600625}}+{\frac{123013}{1201250}} \right ) }+{\frac{4330\,\sqrt{31}}{29791}\arctan \left ({\frac{ \left ( 3+10\,x \right ) \sqrt{31}}{31}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

25*(50281/120125*x^3+494813/1201250*x^2+184382/600625*x+123013/1201250)/(5*x^2+3*x+2)^2+4330/29791*arctan(1/31
*(3+10*x)*31^(1/2))*31^(1/2)

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Maxima [A]  time = 1.4875, size = 76, normalized size = 1.19 \begin{align*} \frac{4330}{29791} \, \sqrt{31} \arctan \left (\frac{1}{31} \, \sqrt{31}{\left (10 \, x + 3\right )}\right ) + \frac{11 \,{\left (45710 \, x^{3} + 44983 \, x^{2} + 33524 \, x + 11183\right )}}{48050 \,{\left (25 \, x^{4} + 30 \, x^{3} + 29 \, x^{2} + 12 \, x + 4\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4330/29791*sqrt(31)*arctan(1/31*sqrt(31)*(10*x + 3)) + 11/48050*(45710*x^3 + 44983*x^2 + 33524*x + 11183)/(25*
x^4 + 30*x^3 + 29*x^2 + 12*x + 4)

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Fricas [A]  time = 0.948494, size = 258, normalized size = 4.03 \begin{align*} \frac{15587110 \, x^{3} + 216500 \, \sqrt{31}{\left (25 \, x^{4} + 30 \, x^{3} + 29 \, x^{2} + 12 \, x + 4\right )} \arctan \left (\frac{1}{31} \, \sqrt{31}{\left (10 \, x + 3\right )}\right ) + 15339203 \, x^{2} + 11431684 \, x + 3813403}{1489550 \,{\left (25 \, x^{4} + 30 \, x^{3} + 29 \, x^{2} + 12 \, x + 4\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1489550*(15587110*x^3 + 216500*sqrt(31)*(25*x^4 + 30*x^3 + 29*x^2 + 12*x + 4)*arctan(1/31*sqrt(31)*(10*x + 3
)) + 15339203*x^2 + 11431684*x + 3813403)/(25*x^4 + 30*x^3 + 29*x^2 + 12*x + 4)

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Sympy [A]  time = 0.298147, size = 63, normalized size = 0.98 \begin{align*} \frac{502810 x^{3} + 494813 x^{2} + 368764 x + 123013}{1201250 x^{4} + 1441500 x^{3} + 1393450 x^{2} + 576600 x + 192200} + \frac{4330 \sqrt{31} \operatorname{atan}{\left (\frac{10 \sqrt{31} x}{31} + \frac{3 \sqrt{31}}{31} \right )}}{29791} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(502810*x**3 + 494813*x**2 + 368764*x + 123013)/(1201250*x**4 + 1441500*x**3 + 1393450*x**2 + 576600*x + 19220
0) + 4330*sqrt(31)*atan(10*sqrt(31)*x/31 + 3*sqrt(31)/31)/29791

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Giac [A]  time = 1.23913, size = 62, normalized size = 0.97 \begin{align*} \frac{4330}{29791} \, \sqrt{31} \arctan \left (\frac{1}{31} \, \sqrt{31}{\left (10 \, x + 3\right )}\right ) + \frac{11 \,{\left (45710 \, x^{3} + 44983 \, x^{2} + 33524 \, x + 11183\right )}}{48050 \,{\left (5 \, x^{2} + 3 \, x + 2\right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4330/29791*sqrt(31)*arctan(1/31*sqrt(31)*(10*x + 3)) + 11/48050*(45710*x^3 + 44983*x^2 + 33524*x + 11183)/(5*x
^2 + 3*x + 2)^2