3.41 \(\int \frac{1}{(3-x+2 x^2) (2+3 x+5 x^2)} \, dx\)

Optimal. Leaf size=73 \[ -\frac{1}{44} \log \left (2 x^2-x+3\right )+\frac{1}{44} \log \left (5 x^2+3 x+2\right )+\frac{3 \tan ^{-1}\left (\frac{1-4 x}{\sqrt{23}}\right )}{22 \sqrt{23}}+\frac{13 \tan ^{-1}\left (\frac{10 x+3}{\sqrt{31}}\right )}{22 \sqrt{31}} \]

[Out]

(3*ArcTan[(1 - 4*x)/Sqrt[23]])/(22*Sqrt[23]) + (13*ArcTan[(3 + 10*x)/Sqrt[31]])/(22*Sqrt[31]) - Log[3 - x + 2*
x^2]/44 + Log[2 + 3*x + 5*x^2]/44

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Rubi [A]  time = 0.0533335, antiderivative size = 73, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {980, 634, 618, 204, 628} \[ -\frac{1}{44} \log \left (2 x^2-x+3\right )+\frac{1}{44} \log \left (5 x^2+3 x+2\right )+\frac{3 \tan ^{-1}\left (\frac{1-4 x}{\sqrt{23}}\right )}{22 \sqrt{23}}+\frac{13 \tan ^{-1}\left (\frac{10 x+3}{\sqrt{31}}\right )}{22 \sqrt{31}} \]

Antiderivative was successfully verified.

[In]

Int[1/((3 - x + 2*x^2)*(2 + 3*x + 5*x^2)),x]

[Out]

(3*ArcTan[(1 - 4*x)/Sqrt[23]])/(22*Sqrt[23]) + (13*ArcTan[(3 + 10*x)/Sqrt[31]])/(22*Sqrt[31]) - Log[3 - x + 2*
x^2]/44 + Log[2 + 3*x + 5*x^2]/44

Rule 980

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

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{1}{\left (3-x+2 x^2\right ) \left (2+3 x+5 x^2\right )} \, dx &=\frac{1}{242} \int \frac{-11-22 x}{3-x+2 x^2} \, dx+\frac{1}{242} \int \frac{88+55 x}{2+3 x+5 x^2} \, dx\\ &=-\left (\frac{1}{44} \int \frac{-1+4 x}{3-x+2 x^2} \, dx\right )+\frac{1}{44} \int \frac{3+10 x}{2+3 x+5 x^2} \, dx-\frac{3}{44} \int \frac{1}{3-x+2 x^2} \, dx+\frac{13}{44} \int \frac{1}{2+3 x+5 x^2} \, dx\\ &=-\frac{1}{44} \log \left (3-x+2 x^2\right )+\frac{1}{44} \log \left (2+3 x+5 x^2\right )+\frac{3}{22} \operatorname{Subst}\left (\int \frac{1}{-23-x^2} \, dx,x,-1+4 x\right )-\frac{13}{22} \operatorname{Subst}\left (\int \frac{1}{-31-x^2} \, dx,x,3+10 x\right )\\ &=\frac{3 \tan ^{-1}\left (\frac{1-4 x}{\sqrt{23}}\right )}{22 \sqrt{23}}+\frac{13 \tan ^{-1}\left (\frac{3+10 x}{\sqrt{31}}\right )}{22 \sqrt{31}}-\frac{1}{44} \log \left (3-x+2 x^2\right )+\frac{1}{44} \log \left (2+3 x+5 x^2\right )\\ \end{align*}

Mathematica [A]  time = 0.0307099, size = 73, normalized size = 1. \[ -\frac{1}{44} \log \left (2 x^2-x+3\right )+\frac{1}{44} \log \left (5 x^2+3 x+2\right )-\frac{3 \tan ^{-1}\left (\frac{4 x-1}{\sqrt{23}}\right )}{22 \sqrt{23}}+\frac{13 \tan ^{-1}\left (\frac{10 x+3}{\sqrt{31}}\right )}{22 \sqrt{31}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((3 - x + 2*x^2)*(2 + 3*x + 5*x^2)),x]

[Out]

(-3*ArcTan[(-1 + 4*x)/Sqrt[23]])/(22*Sqrt[23]) + (13*ArcTan[(3 + 10*x)/Sqrt[31]])/(22*Sqrt[31]) - Log[3 - x +
2*x^2]/44 + Log[2 + 3*x + 5*x^2]/44

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Maple [A]  time = 0.048, size = 60, normalized size = 0.8 \begin{align*}{\frac{\ln \left ( 5\,{x}^{2}+3\,x+2 \right ) }{44}}+{\frac{13\,\sqrt{31}}{682}\arctan \left ({\frac{ \left ( 3+10\,x \right ) \sqrt{31}}{31}} \right ) }-{\frac{\ln \left ( 2\,{x}^{2}-x+3 \right ) }{44}}-{\frac{3\,\sqrt{23}}{506}\arctan \left ({\frac{ \left ( -1+4\,x \right ) \sqrt{23}}{23}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(2*x^2-x+3)/(5*x^2+3*x+2),x)

[Out]

1/44*ln(5*x^2+3*x+2)+13/682*arctan(1/31*(3+10*x)*31^(1/2))*31^(1/2)-1/44*ln(2*x^2-x+3)-3/506*23^(1/2)*arctan(1
/23*(-1+4*x)*23^(1/2))

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Maxima [A]  time = 1.42659, size = 80, normalized size = 1.1 \begin{align*} \frac{13}{682} \, \sqrt{31} \arctan \left (\frac{1}{31} \, \sqrt{31}{\left (10 \, x + 3\right )}\right ) - \frac{3}{506} \, \sqrt{23} \arctan \left (\frac{1}{23} \, \sqrt{23}{\left (4 \, x - 1\right )}\right ) + \frac{1}{44} \, \log \left (5 \, x^{2} + 3 \, x + 2\right ) - \frac{1}{44} \, \log \left (2 \, x^{2} - x + 3\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2*x^2-x+3)/(5*x^2+3*x+2),x, algorithm="maxima")

[Out]

13/682*sqrt(31)*arctan(1/31*sqrt(31)*(10*x + 3)) - 3/506*sqrt(23)*arctan(1/23*sqrt(23)*(4*x - 1)) + 1/44*log(5
*x^2 + 3*x + 2) - 1/44*log(2*x^2 - x + 3)

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Fricas [A]  time = 1.00017, size = 207, normalized size = 2.84 \begin{align*} \frac{13}{682} \, \sqrt{31} \arctan \left (\frac{1}{31} \, \sqrt{31}{\left (10 \, x + 3\right )}\right ) - \frac{3}{506} \, \sqrt{23} \arctan \left (\frac{1}{23} \, \sqrt{23}{\left (4 \, x - 1\right )}\right ) + \frac{1}{44} \, \log \left (5 \, x^{2} + 3 \, x + 2\right ) - \frac{1}{44} \, \log \left (2 \, x^{2} - x + 3\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2*x^2-x+3)/(5*x^2+3*x+2),x, algorithm="fricas")

[Out]

13/682*sqrt(31)*arctan(1/31*sqrt(31)*(10*x + 3)) - 3/506*sqrt(23)*arctan(1/23*sqrt(23)*(4*x - 1)) + 1/44*log(5
*x^2 + 3*x + 2) - 1/44*log(2*x^2 - x + 3)

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Sympy [A]  time = 0.267898, size = 83, normalized size = 1.14 \begin{align*} - \frac{\log{\left (x^{2} - \frac{x}{2} + \frac{3}{2} \right )}}{44} + \frac{\log{\left (x^{2} + \frac{3 x}{5} + \frac{2}{5} \right )}}{44} - \frac{3 \sqrt{23} \operatorname{atan}{\left (\frac{4 \sqrt{23} x}{23} - \frac{\sqrt{23}}{23} \right )}}{506} + \frac{13 \sqrt{31} \operatorname{atan}{\left (\frac{10 \sqrt{31} x}{31} + \frac{3 \sqrt{31}}{31} \right )}}{682} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2*x**2-x+3)/(5*x**2+3*x+2),x)

[Out]

-log(x**2 - x/2 + 3/2)/44 + log(x**2 + 3*x/5 + 2/5)/44 - 3*sqrt(23)*atan(4*sqrt(23)*x/23 - sqrt(23)/23)/506 +
13*sqrt(31)*atan(10*sqrt(31)*x/31 + 3*sqrt(31)/31)/682

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Giac [A]  time = 1.26467, size = 80, normalized size = 1.1 \begin{align*} \frac{13}{682} \, \sqrt{31} \arctan \left (\frac{1}{31} \, \sqrt{31}{\left (10 \, x + 3\right )}\right ) - \frac{3}{506} \, \sqrt{23} \arctan \left (\frac{1}{23} \, \sqrt{23}{\left (4 \, x - 1\right )}\right ) + \frac{1}{44} \, \log \left (5 \, x^{2} + 3 \, x + 2\right ) - \frac{1}{44} \, \log \left (2 \, x^{2} - x + 3\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2*x^2-x+3)/(5*x^2+3*x+2),x, algorithm="giac")

[Out]

13/682*sqrt(31)*arctan(1/31*sqrt(31)*(10*x + 3)) - 3/506*sqrt(23)*arctan(1/23*sqrt(23)*(4*x - 1)) + 1/44*log(5
*x^2 + 3*x + 2) - 1/44*log(2*x^2 - x + 3)