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

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

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Rubi [A]  time = 0.12, antiderivative size = 63, normalized size of antiderivative = 1.00, 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 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 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]

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 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 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]

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*}

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Mathematica [A]  time = 0.04, size = 63, normalized size = 1.00 \[ -\frac {5 x+7}{3 \left (x^2+x+1\right )}+\frac {1}{2} \log \left (x^2+x+1\right )-\frac {2}{x+1}-\log (x+1)-\frac {25 \tan ^{-1}\left (\frac {2 x+1}{\sqrt {3}}\right )}{3 \sqrt {3}} \]

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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IntegrateAlgebraic [A]  time = 0.05, size = 69, normalized size = 1.10 \[ \frac {-11 x^2-18 x-13}{3 (x+1) \left (x^2+x+1\right )}+\frac {1}{2} \log \left (x^2+x+1\right )-\log (x+1)-\frac {25 \tan ^{-1}\left (\frac {2 x}{\sqrt {3}}+\frac {1}{\sqrt {3}}\right )}{3 \sqrt {3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-13 - 18*x - 11*x^2)/(3*(1 + x)*(1 + x + x^2)) - (25*ArcTan[1/Sqrt[3] + (2*x)/Sqrt[3]])/(3*Sqrt[3]) - Log[1 +
x] + Log[1 + x + x^2]/2

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fricas [A]  time = 1.04, size = 97, normalized size = 1.54 \[ -\frac {50 \, \sqrt {3} {\left (x^{3} + 2 \, x^{2} + 2 \, x + 1\right )} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) + 66 \, x^{2} - 9 \, {\left (x^{3} + 2 \, x^{2} + 2 \, x + 1\right )} \log \left (x^{2} + x + 1\right ) + 18 \, {\left (x^{3} + 2 \, x^{2} + 2 \, x + 1\right )} \log \left (x + 1\right ) + 108 \, x + 78}{18 \, {\left (x^{3} + 2 \, x^{2} + 2 \, x + 1\right )}} \]

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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giac [A]  time = 0.90, size = 72, normalized size = 1.14 \[ -\frac {25}{9} \, \sqrt {3} \arctan \left (-\frac {1}{3} \, \sqrt {3} {\left (\frac {2}{x + 1} - 1\right )}\right ) + \frac {\frac {7}{x + 1} - 2}{3 \, {\left (\frac {1}{x + 1} - \frac {1}{{\left (x + 1\right )}^{2}} - 1\right )}} - \frac {2}{x + 1} + \frac {1}{2} \, \log \left (-\frac {1}{x + 1} + \frac {1}{{\left (x + 1\right )}^{2}} + 1\right ) \]

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)

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maple [A]  time = 0.40, size = 54, normalized size = 0.86

method	result	size
default	\(\frac {-\frac {5 x}{3}-\frac {7}{3}}{x^{2}+x +1}+\frac {\ln \left (x^{2}+x +1\right )}{2}-\frac {25 \arctan \left (\frac {\left (1+2 x \right ) \sqrt {3}}{3}\right ) \sqrt {3}}{9}-\frac {2}{1+x}-\ln \left (1+x \right )\)	\(54\)
risch	\(\frac {-\frac {11}{3} x^{2}-6 x -\frac {13}{3}}{\left (x^{2}+x +1\right ) \left (1+x \right )}-\ln \left (1+x \right )+\frac {\ln \left (625 x^{2}+625 x +625\right )}{2}-\frac {25 \sqrt {3}\, \arctan \left (\frac {2 \left (25 x +\frac {25}{2}\right ) \sqrt {3}}{75}\right )}{9}\)	\(61\)

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,method=_RETURNVERBOSE)

[Out]

(-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)-2/(1+x)-ln(1+x)

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maxima [A]  time = 1.43, size = 59, normalized size = 0.94 \[ -\frac {25}{9} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) - \frac {11 \, x^{2} + 18 \, x + 13}{3 \, {\left (x^{3} + 2 \, x^{2} + 2 \, x + 1\right )}} + \frac {1}{2} \, \log \left (x^{2} + x + 1\right ) - \log \left (x + 1\right ) \]

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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mupad [B]  time = 0.25, size = 73, normalized size = 1.16 \[ -\ln \left (x+1\right )-\frac {\frac {11\,x^2}{3}+6\,x+\frac {13}{3}}{x^3+2\,x^2+2\,x+1}+\ln \left (x+\frac {1}{2}-\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,25{}\mathrm {i}}{18}\right )-\ln \left (x+\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,25{}\mathrm {i}}{18}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x - (3^(1/2)*1i)/2 + 1/2)*((3^(1/2)*25i)/18 + 1/2) - (6*x + (11*x^2)/3 + 13/3)/(2*x + 2*x^2 + x^3 + 1) - l
og(x + 1) - log(x + (3^(1/2)*1i)/2 + 1/2)*((3^(1/2)*25i)/18 - 1/2)

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sympy [A]  time = 0.20, size = 68, normalized size = 1.08 \[ \frac {- 11 x^{2} - 18 x - 13}{3 x^{3} + 6 x^{2} + 6 x + 3} - \log {\left (x + 1 \right )} + \frac {\log {\left (x^{2} + x + 1 \right )}}{2} - \frac {25 \sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} x}{3} + \frac {\sqrt {3}}{3} \right )}}{9} \]

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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