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

Optimal. Leaf size=63 \[ -\frac{5 x+7}{3 \left (x^2+x+1\right )}+\frac{1}{2} \log \left (x^2+x+1\right )-\frac{2}{x+1}-\log (x+1)-\frac{25 \tan ^{-1}\left (\frac{2 x+1}{\sqrt{3}}\right )}{3 \sqrt{3}} \]

[Out]

-2/(1 + x) - (7 + 5*x)/(3*(1 + x + x^2)) - (25*ArcTan[(1 + 2*x)/Sqrt[3]])/(3*Sqrt[3]) - Log[1 + x] + Log[1 + x
+ x^2]/2

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Rubi [A]  time = 0.121158, antiderivative size = 63, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {1646, 1628, 634, 618, 204, 628} \[ -\frac{5 x+7}{3 \left (x^2+x+1\right )}+\frac{1}{2} \log \left (x^2+x+1\right )-\frac{2}{x+1}-\log (x+1)-\frac{25 \tan ^{-1}\left (\frac{2 x+1}{\sqrt{3}}\right )}{3 \sqrt{3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-2/(1 + x) - (7 + 5*x)/(3*(1 + x + x^2)) - (25*ArcTan[(1 + 2*x)/Sqrt[3]])/(3*Sqrt[3]) - Log[1 + x] + Log[1 + x
+ x^2]/2

Rule 1646

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = Polynomi
alQuotient[(d + e*x)^m*Pq, a + b*x + c*x^2, x], f = Coeff[PolynomialRemainder[(d + e*x)^m*Pq, a + b*x + c*x^2,
x], x, 0], g = Coeff[PolynomialRemainder[(d + e*x)^m*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[(d
+ e*x)^m*(a + b*x + c*x^2)^(p + 1)*ExpandToSum[((p + 1)*(b^2 - 4*a*c)*Q)/(d + e*x)^m - ((2*p + 3)*(2*c*f - b*
g))/(d + e*x)^m, x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2
- b*d*e + a*e^2, 0] && LtQ[p, -1] && ILtQ[m, 0]

Rule 1628

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegra
nd[(d + e*x)^m*Pq*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, 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{-2-3 x+x^2}{(1+x)^2 \left (1+x+x^2\right )^2} \, dx &=-\frac{7+5 x}{3 \left (1+x+x^2\right )}+\frac{1}{3} \int \frac{-8-19 x-5 x^2}{(1+x)^2 \left (1+x+x^2\right )} \, dx\\ &=-\frac{7+5 x}{3 \left (1+x+x^2\right )}+\frac{1}{3} \int \left (\frac{6}{(1+x)^2}-\frac{3}{1+x}+\frac{-11+3 x}{1+x+x^2}\right ) \, dx\\ &=-\frac{2}{1+x}-\frac{7+5 x}{3 \left (1+x+x^2\right )}-\log (1+x)+\frac{1}{3} \int \frac{-11+3 x}{1+x+x^2} \, dx\\ &=-\frac{2}{1+x}-\frac{7+5 x}{3 \left (1+x+x^2\right )}-\log (1+x)+\frac{1}{2} \int \frac{1+2 x}{1+x+x^2} \, dx-\frac{25}{6} \int \frac{1}{1+x+x^2} \, dx\\ &=-\frac{2}{1+x}-\frac{7+5 x}{3 \left (1+x+x^2\right )}-\log (1+x)+\frac{1}{2} \log \left (1+x+x^2\right )+\frac{25}{3} \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,1+2 x\right )\\ &=-\frac{2}{1+x}-\frac{7+5 x}{3 \left (1+x+x^2\right )}-\frac{25 \tan ^{-1}\left (\frac{1+2 x}{\sqrt{3}}\right )}{3 \sqrt{3}}-\log (1+x)+\frac{1}{2} \log \left (1+x+x^2\right )\\ \end{align*}

Mathematica [A]  time = 0.0348852, size = 63, normalized size = 1. \[ -\frac{5 x+7}{3 \left (x^2+x+1\right )}+\frac{1}{2} \log \left (x^2+x+1\right )-\frac{2}{x+1}-\log (x+1)-\frac{25 \tan ^{-1}\left (\frac{2 x+1}{\sqrt{3}}\right )}{3 \sqrt{3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-2/(1 + x) - (7 + 5*x)/(3*(1 + x + x^2)) - (25*ArcTan[(1 + 2*x)/Sqrt[3]])/(3*Sqrt[3]) - Log[1 + x] + Log[1 + x
+ x^2]/2

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Maple [A]  time = 0.01, size = 54, normalized size = 0.9 \begin{align*} -2\, \left ( 1+x \right ) ^{-1}-\ln \left ( 1+x \right ) +{\frac{1}{{x}^{2}+x+1} \left ( -{\frac{5\,x}{3}}-{\frac{7}{3}} \right ) }+{\frac{\ln \left ({x}^{2}+x+1 \right ) }{2}}-{\frac{25\,\sqrt{3}}{9}\arctan \left ({\frac{ \left ( 1+2\,x \right ) \sqrt{3}}{3}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/(1+x)-ln(1+x)+(-5/3*x-7/3)/(x^2+x+1)+1/2*ln(x^2+x+1)-25/9*arctan(1/3*(1+2*x)*3^(1/2))*3^(1/2)

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Maxima [A]  time = 1.41475, size = 80, normalized size = 1.27 \begin{align*} -\frac{25}{9} \, \sqrt{3} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x + 1\right )}\right ) - \frac{11 \, x^{2} + 18 \, x + 13}{3 \,{\left (x^{3} + 2 \, x^{2} + 2 \, x + 1\right )}} + \frac{1}{2} \, \log \left (x^{2} + x + 1\right ) - \log \left (x + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-25/9*sqrt(3)*arctan(1/3*sqrt(3)*(2*x + 1)) - 1/3*(11*x^2 + 18*x + 13)/(x^3 + 2*x^2 + 2*x + 1) + 1/2*log(x^2 +
x + 1) - log(x + 1)

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Fricas [A]  time = 1.99066, size = 277, normalized size = 4.4 \begin{align*} -\frac{50 \, \sqrt{3}{\left (x^{3} + 2 \, x^{2} + 2 \, x + 1\right )} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x + 1\right )}\right ) + 66 \, x^{2} - 9 \,{\left (x^{3} + 2 \, x^{2} + 2 \, x + 1\right )} \log \left (x^{2} + x + 1\right ) + 18 \,{\left (x^{3} + 2 \, x^{2} + 2 \, x + 1\right )} \log \left (x + 1\right ) + 108 \, x + 78}{18 \,{\left (x^{3} + 2 \, x^{2} + 2 \, x + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/18*(50*sqrt(3)*(x^3 + 2*x^2 + 2*x + 1)*arctan(1/3*sqrt(3)*(2*x + 1)) + 66*x^2 - 9*(x^3 + 2*x^2 + 2*x + 1)*l
og(x^2 + x + 1) + 18*(x^3 + 2*x^2 + 2*x + 1)*log(x + 1) + 108*x + 78)/(x^3 + 2*x^2 + 2*x + 1)

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Sympy [A]  time = 0.171504, size = 66, normalized size = 1.05 \begin{align*} - \frac{11 x^{2} + 18 x + 13}{3 x^{3} + 6 x^{2} + 6 x + 3} - \log{\left (x + 1 \right )} + \frac{\log{\left (x^{2} + x + 1 \right )}}{2} - \frac{25 \sqrt{3} \operatorname{atan}{\left (\frac{2 \sqrt{3} x}{3} + \frac{\sqrt{3}}{3} \right )}}{9} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(11*x**2 + 18*x + 13)/(3*x**3 + 6*x**2 + 6*x + 3) - log(x + 1) + log(x**2 + x + 1)/2 - 25*sqrt(3)*atan(2*sqrt
(3)*x/3 + sqrt(3)/3)/9

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Giac [A]  time = 1.07256, size = 97, normalized size = 1.54 \begin{align*} -\frac{25}{9} \, \sqrt{3} \arctan \left (-\frac{1}{3} \, \sqrt{3}{\left (\frac{2}{x + 1} - 1\right )}\right ) + \frac{\frac{7}{x + 1} - 2}{3 \,{\left (\frac{1}{x + 1} - \frac{1}{{\left (x + 1\right )}^{2}} - 1\right )}} - \frac{2}{x + 1} + \frac{1}{2} \, \log \left (-\frac{1}{x + 1} + \frac{1}{{\left (x + 1\right )}^{2}} + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-25/9*sqrt(3)*arctan(-1/3*sqrt(3)*(2/(x + 1) - 1)) + 1/3*(7/(x + 1) - 2)/(1/(x + 1) - 1/(x + 1)^2 - 1) - 2/(x
+ 1) + 1/2*log(-1/(x + 1) + 1/(x + 1)^2 + 1)