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

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

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Rubi [A]  time = 0.05, antiderivative size = 64, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {1660, 12, 618, 204} \begin {gather*} \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}} \end {gather*}

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 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 {\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*}

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Mathematica [A]  time = 0.02, size = 53, normalized size = 0.83 \begin {gather*} \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}} \end {gather*}

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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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (3-x+2 x^2\right )^2}{\left (2+3 x+5 x^2\right )^3} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.38, size = 75, normalized size = 1.17 \begin {gather*} \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 {gather*}

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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giac [A]  time = 0.18, size = 46, normalized size = 0.72 \begin {gather*} \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 {gather*}

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

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maple [A]  time = 0.01, size = 47, normalized size = 0.73 \begin {gather*} \frac {4330 \sqrt {31}\, \arctan \left (\frac {\left (10 x +3\right ) \sqrt {31}}{31}\right )}{29791}+\frac {\frac {50281}{4805} x^{3}+\frac {494813}{48050} x^{2}+\frac {184382}{24025} x +\frac {123013}{48050}}{\left (5 x^{2}+3 x +2\right )^{2}} \end {gather*}

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*31^(1/2)*ar
ctan(1/31*(10*x+3)*31^(1/2))

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maxima [A]  time = 0.97, size = 56, normalized size = 0.88 \begin {gather*} \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 {gather*}

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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mupad [B]  time = 3.44, size = 55, normalized size = 0.86 \begin {gather*} \frac {4330\,\sqrt {31}\,\mathrm {atan}\left (\frac {10\,\sqrt {31}\,x}{31}+\frac {3\,\sqrt {31}}{31}\right )}{29791}+\frac {\frac {50281\,x^3}{120125}+\frac {494813\,x^2}{1201250}+\frac {184382\,x}{600625}+\frac {123013}{1201250}}{x^4+\frac {6\,x^3}{5}+\frac {29\,x^2}{25}+\frac {12\,x}{25}+\frac {4}{25}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(4330*31^(1/2)*atan((10*31^(1/2)*x)/31 + (3*31^(1/2))/31))/29791 + ((184382*x)/600625 + (494813*x^2)/1201250 +
(50281*x^3)/120125 + 123013/1201250)/((12*x)/25 + (29*x^2)/25 + (6*x^3)/5 + x^4 + 4/25)

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sympy [A]  time = 0.20, size = 63, normalized size = 0.98 \begin {gather*} \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 {gather*}

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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