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

Optimal. Leaf size=63 \[ \frac{121 (69 x+61)}{3875 \left (5 x^2+3 x+2\right )}-\frac{22}{125} \log \left (5 x^2+3 x+2\right )+\frac{4 x}{25}+\frac{41932 \tan ^{-1}\left (\frac{10 x+3}{\sqrt{31}}\right )}{3875 \sqrt{31}} \]

[Out]

(4*x)/25 + (121*(61 + 69*x))/(3875*(2 + 3*x + 5*x^2)) + (41932*ArcTan[(3 + 10*x)/Sqrt[31]])/(3875*Sqrt[31]) -
(22*Log[2 + 3*x + 5*x^2])/125

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Rubi [A]  time = 0.0608414, antiderivative size = 63, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {1660, 1657, 634, 618, 204, 628} \[ \frac{121 (69 x+61)}{3875 \left (5 x^2+3 x+2\right )}-\frac{22}{125} \log \left (5 x^2+3 x+2\right )+\frac{4 x}{25}+\frac{41932 \tan ^{-1}\left (\frac{10 x+3}{\sqrt{31}}\right )}{3875 \sqrt{31}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*x)/25 + (121*(61 + 69*x))/(3875*(2 + 3*x + 5*x^2)) + (41932*ArcTan[(3 + 10*x)/Sqrt[31]])/(3875*Sqrt[31]) -
(22*Log[2 + 3*x + 5*x^2])/125

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 1657

Int[(Pq_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x + c*x^2)^p, x
], x] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rule 634

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

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])

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rubi steps

\begin{align*} \int \frac{\left (3-x+2 x^2\right )^2}{\left (2+3 x+5 x^2\right )^2} \, dx &=\frac{121 (61+69 x)}{3875 \left (2+3 x+5 x^2\right )}+\frac{1}{31} \int \frac{\frac{4032}{25}-\frac{992 x}{25}+\frac{124 x^2}{5}}{2+3 x+5 x^2} \, dx\\ &=\frac{121 (61+69 x)}{3875 \left (2+3 x+5 x^2\right )}+\frac{1}{31} \int \left (\frac{124}{25}+\frac{44 (86-31 x)}{25 \left (2+3 x+5 x^2\right )}\right ) \, dx\\ &=\frac{4 x}{25}+\frac{121 (61+69 x)}{3875 \left (2+3 x+5 x^2\right )}+\frac{44}{775} \int \frac{86-31 x}{2+3 x+5 x^2} \, dx\\ &=\frac{4 x}{25}+\frac{121 (61+69 x)}{3875 \left (2+3 x+5 x^2\right )}-\frac{22}{125} \int \frac{3+10 x}{2+3 x+5 x^2} \, dx+\frac{20966 \int \frac{1}{2+3 x+5 x^2} \, dx}{3875}\\ &=\frac{4 x}{25}+\frac{121 (61+69 x)}{3875 \left (2+3 x+5 x^2\right )}-\frac{22}{125} \log \left (2+3 x+5 x^2\right )-\frac{41932 \operatorname{Subst}\left (\int \frac{1}{-31-x^2} \, dx,x,3+10 x\right )}{3875}\\ &=\frac{4 x}{25}+\frac{121 (61+69 x)}{3875 \left (2+3 x+5 x^2\right )}+\frac{41932 \tan ^{-1}\left (\frac{3+10 x}{\sqrt{31}}\right )}{3875 \sqrt{31}}-\frac{22}{125} \log \left (2+3 x+5 x^2\right )\\ \end{align*}

Mathematica [A]  time = 0.0308515, size = 59, normalized size = 0.94 \[ \frac{\frac{3751 (69 x+61)}{5 x^2+3 x+2}-21142 \log \left (5 x^2+3 x+2\right )+19220 x+41932 \sqrt{31} \tan ^{-1}\left (\frac{10 x+3}{\sqrt{31}}\right )}{120125} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(19220*x + (3751*(61 + 69*x))/(2 + 3*x + 5*x^2) + 41932*Sqrt[31]*ArcTan[(3 + 10*x)/Sqrt[31]] - 21142*Log[2 + 3
*x + 5*x^2])/120125

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Maple [A]  time = 0.046, size = 51, normalized size = 0.8 \begin{align*}{\frac{4\,x}{25}}-{\frac{11}{25} \left ( -{\frac{759\,x}{775}}-{\frac{671}{775}} \right ) \left ({x}^{2}+{\frac{3\,x}{5}}+{\frac{2}{5}} \right ) ^{-1}}-{\frac{22\,\ln \left ( 5\,{x}^{2}+3\,x+2 \right ) }{125}}+{\frac{41932\,\sqrt{31}}{120125}\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)^2,x)

[Out]

4/25*x-11/25*(-759/775*x-671/775)/(x^2+3/5*x+2/5)-22/125*ln(5*x^2+3*x+2)+41932/120125*arctan(1/31*(3+10*x)*31^
(1/2))*31^(1/2)

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Maxima [A]  time = 1.45998, size = 70, normalized size = 1.11 \begin{align*} \frac{41932}{120125} \, \sqrt{31} \arctan \left (\frac{1}{31} \, \sqrt{31}{\left (10 \, x + 3\right )}\right ) + \frac{4}{25} \, x + \frac{121 \,{\left (69 \, x + 61\right )}}{3875 \,{\left (5 \, x^{2} + 3 \, x + 2\right )}} - \frac{22}{125} \, \log \left (5 \, x^{2} + 3 \, x + 2\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)^2,x, algorithm="maxima")

[Out]

41932/120125*sqrt(31)*arctan(1/31*sqrt(31)*(10*x + 3)) + 4/25*x + 121/3875*(69*x + 61)/(5*x^2 + 3*x + 2) - 22/
125*log(5*x^2 + 3*x + 2)

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Fricas [A]  time = 0.997507, size = 252, normalized size = 4. \begin{align*} \frac{96100 \, x^{3} + 41932 \, \sqrt{31}{\left (5 \, x^{2} + 3 \, x + 2\right )} \arctan \left (\frac{1}{31} \, \sqrt{31}{\left (10 \, x + 3\right )}\right ) + 57660 \, x^{2} - 21142 \,{\left (5 \, x^{2} + 3 \, x + 2\right )} \log \left (5 \, x^{2} + 3 \, x + 2\right ) + 297259 \, x + 228811}{120125 \,{\left (5 \, x^{2} + 3 \, x + 2\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)^2,x, algorithm="fricas")

[Out]

1/120125*(96100*x^3 + 41932*sqrt(31)*(5*x^2 + 3*x + 2)*arctan(1/31*sqrt(31)*(10*x + 3)) + 57660*x^2 - 21142*(5
*x^2 + 3*x + 2)*log(5*x^2 + 3*x + 2) + 297259*x + 228811)/(5*x^2 + 3*x + 2)

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Sympy [A]  time = 0.213602, size = 65, normalized size = 1.03 \begin{align*} \frac{4 x}{25} + \frac{8349 x + 7381}{19375 x^{2} + 11625 x + 7750} - \frac{22 \log{\left (x^{2} + \frac{3 x}{5} + \frac{2}{5} \right )}}{125} + \frac{41932 \sqrt{31} \operatorname{atan}{\left (\frac{10 \sqrt{31} x}{31} + \frac{3 \sqrt{31}}{31} \right )}}{120125} \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)**2,x)

[Out]

4*x/25 + (8349*x + 7381)/(19375*x**2 + 11625*x + 7750) - 22*log(x**2 + 3*x/5 + 2/5)/125 + 41932*sqrt(31)*atan(
10*sqrt(31)*x/31 + 3*sqrt(31)/31)/120125

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Giac [A]  time = 1.18491, size = 70, normalized size = 1.11 \begin{align*} \frac{41932}{120125} \, \sqrt{31} \arctan \left (\frac{1}{31} \, \sqrt{31}{\left (10 \, x + 3\right )}\right ) + \frac{4}{25} \, x + \frac{121 \,{\left (69 \, x + 61\right )}}{3875 \,{\left (5 \, x^{2} + 3 \, x + 2\right )}} - \frac{22}{125} \, \log \left (5 \, x^{2} + 3 \, x + 2\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)^2,x, algorithm="giac")

[Out]

41932/120125*sqrt(31)*arctan(1/31*sqrt(31)*(10*x + 3)) + 4/25*x + 121/3875*(69*x + 61)/(5*x^2 + 3*x + 2) - 22/
125*log(5*x^2 + 3*x + 2)