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

Optimal. Leaf size=42 \[ -\frac{11}{50} \log \left (5 x^2+3 x+2\right )+\frac{2 x}{5}+\frac{143 \tan ^{-1}\left (\frac{10 x+3}{\sqrt{31}}\right )}{25 \sqrt{31}} \]

[Out]

(2*x)/5 + (143*ArcTan[(3 + 10*x)/Sqrt[31]])/(25*Sqrt[31]) - (11*Log[2 + 3*x + 5*x^2])/50

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Rubi [A]  time = 0.0402874, antiderivative size = 42, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {1657, 634, 618, 204, 628} \[ -\frac{11}{50} \log \left (5 x^2+3 x+2\right )+\frac{2 x}{5}+\frac{143 \tan ^{-1}\left (\frac{10 x+3}{\sqrt{31}}\right )}{25 \sqrt{31}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x)/5 + (143*ArcTan[(3 + 10*x)/Sqrt[31]])/(25*Sqrt[31]) - (11*Log[2 + 3*x + 5*x^2])/50

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{3-x+2 x^2}{2+3 x+5 x^2} \, dx &=\int \left (\frac{2}{5}+\frac{11 (1-x)}{5 \left (2+3 x+5 x^2\right )}\right ) \, dx\\ &=\frac{2 x}{5}+\frac{11}{5} \int \frac{1-x}{2+3 x+5 x^2} \, dx\\ &=\frac{2 x}{5}-\frac{11}{50} \int \frac{3+10 x}{2+3 x+5 x^2} \, dx+\frac{143}{50} \int \frac{1}{2+3 x+5 x^2} \, dx\\ &=\frac{2 x}{5}-\frac{11}{50} \log \left (2+3 x+5 x^2\right )-\frac{143}{25} \operatorname{Subst}\left (\int \frac{1}{-31-x^2} \, dx,x,3+10 x\right )\\ &=\frac{2 x}{5}+\frac{143 \tan ^{-1}\left (\frac{3+10 x}{\sqrt{31}}\right )}{25 \sqrt{31}}-\frac{11}{50} \log \left (2+3 x+5 x^2\right )\\ \end{align*}

Mathematica [A]  time = 0.0154453, size = 42, normalized size = 1. \[ -\frac{11}{50} \log \left (5 x^2+3 x+2\right )+\frac{2 x}{5}+\frac{143 \tan ^{-1}\left (\frac{10 x+3}{\sqrt{31}}\right )}{25 \sqrt{31}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x)/5 + (143*ArcTan[(3 + 10*x)/Sqrt[31]])/(25*Sqrt[31]) - (11*Log[2 + 3*x + 5*x^2])/50

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Maple [A]  time = 0.049, size = 34, normalized size = 0.8 \begin{align*}{\frac{2\,x}{5}}-{\frac{11\,\ln \left ( 5\,{x}^{2}+3\,x+2 \right ) }{50}}+{\frac{143\,\sqrt{31}}{775}\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),x)

[Out]

2/5*x-11/50*ln(5*x^2+3*x+2)+143/775*arctan(1/31*(3+10*x)*31^(1/2))*31^(1/2)

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Maxima [A]  time = 1.4844, size = 45, normalized size = 1.07 \begin{align*} \frac{143}{775} \, \sqrt{31} \arctan \left (\frac{1}{31} \, \sqrt{31}{\left (10 \, x + 3\right )}\right ) + \frac{2}{5} \, x - \frac{11}{50} \, \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)/(5*x^2+3*x+2),x, algorithm="maxima")

[Out]

143/775*sqrt(31)*arctan(1/31*sqrt(31)*(10*x + 3)) + 2/5*x - 11/50*log(5*x^2 + 3*x + 2)

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Fricas [A]  time = 0.83428, size = 119, normalized size = 2.83 \begin{align*} \frac{143}{775} \, \sqrt{31} \arctan \left (\frac{1}{31} \, \sqrt{31}{\left (10 \, x + 3\right )}\right ) + \frac{2}{5} \, x - \frac{11}{50} \, \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)/(5*x^2+3*x+2),x, algorithm="fricas")

[Out]

143/775*sqrt(31)*arctan(1/31*sqrt(31)*(10*x + 3)) + 2/5*x - 11/50*log(5*x^2 + 3*x + 2)

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Sympy [A]  time = 0.15425, size = 49, normalized size = 1.17 \begin{align*} \frac{2 x}{5} - \frac{11 \log{\left (x^{2} + \frac{3 x}{5} + \frac{2}{5} \right )}}{50} + \frac{143 \sqrt{31} \operatorname{atan}{\left (\frac{10 \sqrt{31} x}{31} + \frac{3 \sqrt{31}}{31} \right )}}{775} \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),x)

[Out]

2*x/5 - 11*log(x**2 + 3*x/5 + 2/5)/50 + 143*sqrt(31)*atan(10*sqrt(31)*x/31 + 3*sqrt(31)/31)/775

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Giac [A]  time = 1.16842, size = 45, normalized size = 1.07 \begin{align*} \frac{143}{775} \, \sqrt{31} \arctan \left (\frac{1}{31} \, \sqrt{31}{\left (10 \, x + 3\right )}\right ) + \frac{2}{5} \, x - \frac{11}{50} \, \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)/(5*x^2+3*x+2),x, algorithm="giac")

[Out]

143/775*sqrt(31)*arctan(1/31*sqrt(31)*(10*x + 3)) + 2/5*x - 11/50*log(5*x^2 + 3*x + 2)