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

Optimal. Leaf size=64 \[ \frac{553 (10 x+3)}{9610 \left (5 x^2+3 x+2\right )}+\frac{11 (13 x+7)}{310 \left (5 x^2+3 x+2\right )^2}+\frac{1106 \tan ^{-1}\left (\frac{10 x+3}{\sqrt{31}}\right )}{961 \sqrt{31}} \]

[Out]

(11*(7 + 13*x))/(310*(2 + 3*x + 5*x^2)^2) + (553*(3 + 10*x))/(9610*(2 + 3*x + 5*x^2)) + (1106*ArcTan[(3 + 10*x
)/Sqrt[31]])/(961*Sqrt[31])

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Rubi [A]  time = 0.0370305, antiderivative size = 64, normalized size of antiderivative = 1., 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} \[ \frac{553 (10 x+3)}{9610 \left (5 x^2+3 x+2\right )}+\frac{11 (13 x+7)}{310 \left (5 x^2+3 x+2\right )^2}+\frac{1106 \tan ^{-1}\left (\frac{10 x+3}{\sqrt{31}}\right )}{961 \sqrt{31}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(11*(7 + 13*x))/(310*(2 + 3*x + 5*x^2)^2) + (553*(3 + 10*x))/(9610*(2 + 3*x + 5*x^2)) + (1106*ArcTan[(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 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 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{3-x+2 x^2}{\left (2+3 x+5 x^2\right )^3} \, dx &=\frac{11 (7+13 x)}{310 \left (2+3 x+5 x^2\right )^2}+\frac{1}{62} \int \frac{553}{5 \left (2+3 x+5 x^2\right )^2} \, dx\\ &=\frac{11 (7+13 x)}{310 \left (2+3 x+5 x^2\right )^2}+\frac{553}{310} \int \frac{1}{\left (2+3 x+5 x^2\right )^2} \, dx\\ &=\frac{11 (7+13 x)}{310 \left (2+3 x+5 x^2\right )^2}+\frac{553 (3+10 x)}{9610 \left (2+3 x+5 x^2\right )}+\frac{553}{961} \int \frac{1}{2+3 x+5 x^2} \, dx\\ &=\frac{11 (7+13 x)}{310 \left (2+3 x+5 x^2\right )^2}+\frac{553 (3+10 x)}{9610 \left (2+3 x+5 x^2\right )}-\frac{1106}{961} \operatorname{Subst}\left (\int \frac{1}{-31-x^2} \, dx,x,3+10 x\right )\\ &=\frac{11 (7+13 x)}{310 \left (2+3 x+5 x^2\right )^2}+\frac{553 (3+10 x)}{9610 \left (2+3 x+5 x^2\right )}+\frac{1106 \tan ^{-1}\left (\frac{3+10 x}{\sqrt{31}}\right )}{961 \sqrt{31}}\\ \end{align*}

Mathematica [A]  time = 0.0249325, size = 53, normalized size = 0.83 \[ \frac{\frac{31 \left (5530 x^3+4977 x^2+4094 x+1141\right )}{\left (5 x^2+3 x+2\right )^2}+2212 \sqrt{31} \tan ^{-1}\left (\frac{10 x+3}{\sqrt{31}}\right )}{59582} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((31*(1141 + 4094*x + 4977*x^2 + 5530*x^3))/(2 + 3*x + 5*x^2)^2 + 2212*Sqrt[31]*ArcTan[(3 + 10*x)/Sqrt[31]])/5
9582

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Maple [A]  time = 0.044, size = 47, normalized size = 0.7 \begin{align*} 25\,{\frac{1}{ \left ( 5\,{x}^{2}+3\,x+2 \right ) ^{2}} \left ({\frac{553\,{x}^{3}}{4805}}+{\frac{4977\,{x}^{2}}{48050}}+{\frac{2047\,x}{24025}}+{\frac{1141}{48050}} \right ) }+{\frac{1106\,\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)/(5*x^2+3*x+2)^3,x)

[Out]

25*(553/4805*x^3+4977/48050*x^2+2047/24025*x+1141/48050)/(5*x^2+3*x+2)^2+1106/29791*arctan(1/31*(3+10*x)*31^(1
/2))*31^(1/2)

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Maxima [A]  time = 1.47233, size = 76, normalized size = 1.19 \begin{align*} \frac{1106}{29791} \, \sqrt{31} \arctan \left (\frac{1}{31} \, \sqrt{31}{\left (10 \, x + 3\right )}\right ) + \frac{5530 \, x^{3} + 4977 \, x^{2} + 4094 \, x + 1141}{1922 \,{\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)/(5*x^2+3*x+2)^3,x, algorithm="maxima")

[Out]

1106/29791*sqrt(31)*arctan(1/31*sqrt(31)*(10*x + 3)) + 1/1922*(5530*x^3 + 4977*x^2 + 4094*x + 1141)/(25*x^4 +
30*x^3 + 29*x^2 + 12*x + 4)

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Fricas [A]  time = 0.801471, size = 242, normalized size = 3.78 \begin{align*} \frac{171430 \, x^{3} + 2212 \, \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 ) + 154287 \, x^{2} + 126914 \, x + 35371}{59582 \,{\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)/(5*x^2+3*x+2)^3,x, algorithm="fricas")

[Out]

1/59582*(171430*x^3 + 2212*sqrt(31)*(25*x^4 + 30*x^3 + 29*x^2 + 12*x + 4)*arctan(1/31*sqrt(31)*(10*x + 3)) + 1
54287*x^2 + 126914*x + 35371)/(25*x^4 + 30*x^3 + 29*x^2 + 12*x + 4)

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Sympy [A]  time = 0.280638, size = 63, normalized size = 0.98 \begin{align*} \frac{5530 x^{3} + 4977 x^{2} + 4094 x + 1141}{48050 x^{4} + 57660 x^{3} + 55738 x^{2} + 23064 x + 7688} + \frac{1106 \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)/(5*x**2+3*x+2)**3,x)

[Out]

(5530*x**3 + 4977*x**2 + 4094*x + 1141)/(48050*x**4 + 57660*x**3 + 55738*x**2 + 23064*x + 7688) + 1106*sqrt(31
)*atan(10*sqrt(31)*x/31 + 3*sqrt(31)/31)/29791

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Giac [A]  time = 1.25107, size = 62, normalized size = 0.97 \begin{align*} \frac{1106}{29791} \, \sqrt{31} \arctan \left (\frac{1}{31} \, \sqrt{31}{\left (10 \, x + 3\right )}\right ) + \frac{5530 \, x^{3} + 4977 \, x^{2} + 4094 \, x + 1141}{1922 \,{\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)/(5*x^2+3*x+2)^3,x, algorithm="giac")

[Out]

1106/29791*sqrt(31)*arctan(1/31*sqrt(31)*(10*x + 3)) + 1/1922*(5530*x^3 + 4977*x^2 + 4094*x + 1141)/(5*x^2 + 3
*x + 2)^2