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

Optimal. Leaf size=43 \[ \frac {11 (13 x+7)}{155 \left (5 x^2+3 x+2\right )}+\frac {82 \tan ^{-1}\left (\frac {10 x+3}{\sqrt {31}}\right )}{31 \sqrt {31}} \]

[Out]

11/155*(7+13*x)/(5*x^2+3*x+2)+82/961*arctan(1/31*(3+10*x)*31^(1/2))*31^(1/2)

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Rubi [A]  time = 0.03, antiderivative size = 43, normalized size of antiderivative = 1.00, 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} \[ \frac {11 (13 x+7)}{155 \left (5 x^2+3 x+2\right )}+\frac {82 \tan ^{-1}\left (\frac {10 x+3}{\sqrt {31}}\right )}{31 \sqrt {31}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(11*(7 + 13*x))/(155*(2 + 3*x + 5*x^2)) + (82*ArcTan[(3 + 10*x)/Sqrt[31]])/(31*Sqrt[31])

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 {3-x+2 x^2}{\left (2+3 x+5 x^2\right )^2} \, dx &=\frac {11 (7+13 x)}{155 \left (2+3 x+5 x^2\right )}+\frac {1}{31} \int \frac {41}{2+3 x+5 x^2} \, dx\\ &=\frac {11 (7+13 x)}{155 \left (2+3 x+5 x^2\right )}+\frac {41}{31} \int \frac {1}{2+3 x+5 x^2} \, dx\\ &=\frac {11 (7+13 x)}{155 \left (2+3 x+5 x^2\right )}-\frac {82}{31} \operatorname {Subst}\left (\int \frac {1}{-31-x^2} \, dx,x,3+10 x\right )\\ &=\frac {11 (7+13 x)}{155 \left (2+3 x+5 x^2\right )}+\frac {82 \tan ^{-1}\left (\frac {3+10 x}{\sqrt {31}}\right )}{31 \sqrt {31}}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 43, normalized size = 1.00 \[ \frac {11 (13 x+7)}{155 \left (5 x^2+3 x+2\right )}+\frac {82 \tan ^{-1}\left (\frac {10 x+3}{\sqrt {31}}\right )}{31 \sqrt {31}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(11*(7 + 13*x))/(155*(2 + 3*x + 5*x^2)) + (82*ArcTan[(3 + 10*x)/Sqrt[31]])/(31*Sqrt[31])

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fricas [A]  time = 0.73, size = 45, normalized size = 1.05 \[ \frac {410 \, \sqrt {31} {\left (5 \, x^{2} + 3 \, x + 2\right )} \arctan \left (\frac {1}{31} \, \sqrt {31} {\left (10 \, x + 3\right )}\right ) + 4433 \, x + 2387}{4805 \, {\left (5 \, x^{2} + 3 \, x + 2\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4805*(410*sqrt(31)*(5*x^2 + 3*x + 2)*arctan(1/31*sqrt(31)*(10*x + 3)) + 4433*x + 2387)/(5*x^2 + 3*x + 2)

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giac [A]  time = 0.21, size = 36, normalized size = 0.84 \[ \frac {82}{961} \, \sqrt {31} \arctan \left (\frac {1}{31} \, \sqrt {31} {\left (10 \, x + 3\right )}\right ) + \frac {11 \, {\left (13 \, x + 7\right )}}{155 \, {\left (5 \, x^{2} + 3 \, x + 2\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

82/961*sqrt(31)*arctan(1/31*sqrt(31)*(10*x + 3)) + 11/155*(13*x + 7)/(5*x^2 + 3*x + 2)

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maple [A]  time = 0.00, size = 34, normalized size = 0.79 \[ \frac {82 \sqrt {31}\, \arctan \left (\frac {\left (10 x +3\right ) \sqrt {31}}{31}\right )}{961}+\frac {\frac {143 x}{775}+\frac {77}{775}}{x^{2}+\frac {3}{5} x +\frac {2}{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(143/775*x+77/775)/(x^2+3/5*x+2/5)+82/961*31^(1/2)*arctan(1/31*(10*x+3)*31^(1/2))

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maxima [A]  time = 0.97, size = 36, normalized size = 0.84 \[ \frac {82}{961} \, \sqrt {31} \arctan \left (\frac {1}{31} \, \sqrt {31} {\left (10 \, x + 3\right )}\right ) + \frac {11 \, {\left (13 \, x + 7\right )}}{155 \, {\left (5 \, x^{2} + 3 \, x + 2\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

82/961*sqrt(31)*arctan(1/31*sqrt(31)*(10*x + 3)) + 11/155*(13*x + 7)/(5*x^2 + 3*x + 2)

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mupad [B]  time = 0.04, size = 35, normalized size = 0.81 \[ \frac {\frac {143\,x}{775}+\frac {77}{775}}{x^2+\frac {3\,x}{5}+\frac {2}{5}}+\frac {82\,\sqrt {31}\,\mathrm {atan}\left (\frac {10\,\sqrt {31}\,x}{31}+\frac {3\,\sqrt {31}}{31}\right )}{961} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((143*x)/775 + 77/775)/((3*x)/5 + x^2 + 2/5) + (82*31^(1/2)*atan((10*31^(1/2)*x)/31 + (3*31^(1/2))/31))/961

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sympy [A]  time = 0.16, size = 42, normalized size = 0.98 \[ \frac {143 x + 77}{775 x^{2} + 465 x + 310} + \frac {82 \sqrt {31} \operatorname {atan}{\left (\frac {10 \sqrt {31} x}{31} + \frac {3 \sqrt {31}}{31} \right )}}{961} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(143*x + 77)/(775*x**2 + 465*x + 310) + 82*sqrt(31)*atan(10*sqrt(31)*x/31 + 3*sqrt(31)/31)/961

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