Integrand size = 20, antiderivative size = 61 \[ \int \frac {1+x+x^2}{x^2 \left (1-x+x^2\right )^2} \, dx=-\frac {1}{x}+\frac {2 (1+x)}{3 \left (1-x+x^2\right )}-\frac {7 \arctan \left (\frac {1-2 x}{\sqrt {3}}\right )}{3 \sqrt {3}}+3 \log (x)-\frac {3}{2} \log \left (1-x+x^2\right ) \] Output:
-1/x+2*(1+x)/(3*x^2-3*x+3)-7/9*arctan(1/3*(1-2*x)*3^(1/2))*3^(1/2)+3*ln(x) -3/2*ln(x^2-x+1)
Time = 0.03 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.00 \[ \int \frac {1+x+x^2}{x^2 \left (1-x+x^2\right )^2} \, dx=-\frac {1}{x}+\frac {2 (1+x)}{3 \left (1-x+x^2\right )}+\frac {7 \arctan \left (\frac {-1+2 x}{\sqrt {3}}\right )}{3 \sqrt {3}}+3 \log (x)-\frac {3}{2} \log \left (1-x+x^2\right ) \] Input:
Integrate[(1 + x + x^2)/(x^2*(1 - x + x^2)^2),x]
Output:
-x^(-1) + (2*(1 + x))/(3*(1 - x + x^2)) + (7*ArcTan[(-1 + 2*x)/Sqrt[3]])/( 3*Sqrt[3]) + 3*Log[x] - (3*Log[1 - x + x^2])/2
Time = 0.29 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.05, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {2177, 2159, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {x^2+x+1}{x^2 \left (x^2-x+1\right )^2} \, dx\) |
\(\Big \downarrow \) 2177 |
\(\displaystyle \frac {1}{3} \int \frac {2 x^2+6 x+3}{x^2 \left (x^2-x+1\right )}dx+\frac {2 (x+1)}{3 \left (x^2-x+1\right )}\) |
\(\Big \downarrow \) 2159 |
\(\displaystyle \frac {1}{3} \int \left (\frac {8-9 x}{x^2-x+1}+\frac {9}{x}+\frac {3}{x^2}\right )dx+\frac {2 (x+1)}{3 \left (x^2-x+1\right )}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{3} \left (-\frac {7 \arctan \left (\frac {1-2 x}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {9}{2} \log \left (x^2-x+1\right )-\frac {3}{x}+9 \log (x)\right )+\frac {2 (x+1)}{3 \left (x^2-x+1\right )}\) |
Input:
Int[(1 + x + x^2)/(x^2*(1 - x + x^2)^2),x]
Output:
(2*(1 + x))/(3*(1 - x + x^2)) + (-3/x - (7*ArcTan[(1 - 2*x)/Sqrt[3]])/Sqrt [3] + 9*Log[x] - (9*Log[1 - x + x^2])/2)/3
Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p _.), x_Symbol] :> Int[ExpandIntegrand[(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]
Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p _), x_Symbol] :> With[{Qx = PolynomialQuotient[(d + e*x)^m*Pq, a + b*x + c* x^2, x], R = Coeff[PolynomialRemainder[(d + e*x)^m*Pq, a + b*x + c*x^2, x], x, 0], S = Coeff[PolynomialRemainder[(d + e*x)^m*Pq, a + b*x + c*x^2, x], x, 1]}, Simp[(b*R - 2*a*S + (2*c*R - b*S)*x)*((a + b*x + c*x^2)^(p + 1)/((p + 1)*(b^2 - 4*a*c))), x] + Simp[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)*Qx)/(d + e*x )^m - ((2*p + 3)*(2*c*R - b*S))/(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]
Time = 0.11 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.90
method | result | size |
default | \(-\frac {1}{x}+3 \ln \left (x \right )-\frac {-\frac {2 x}{3}-\frac {2}{3}}{x^{2}-x +1}-\frac {3 \ln \left (x^{2}-x +1\right )}{2}+\frac {7 \sqrt {3}\, \arctan \left (\frac {\left (2 x -1\right ) \sqrt {3}}{3}\right )}{9}\) | \(55\) |
risch | \(\frac {-\frac {1}{3} x^{2}+\frac {5}{3} x -1}{x \left (x^{2}-x +1\right )}+3 \ln \left (x \right )-\frac {3 \ln \left (49 x^{2}-49 x +49\right )}{2}+\frac {7 \sqrt {3}\, \arctan \left (\frac {2 \left (7 x -\frac {7}{2}\right ) \sqrt {3}}{21}\right )}{9}\) | \(59\) |
Input:
int((x^2+x+1)/x^2/(x^2-x+1)^2,x,method=_RETURNVERBOSE)
Output:
-1/x+3*ln(x)-(-2/3*x-2/3)/(x^2-x+1)-3/2*ln(x^2-x+1)+7/9*3^(1/2)*arctan(1/3 *(2*x-1)*3^(1/2))
Time = 0.07 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.39 \[ \int \frac {1+x+x^2}{x^2 \left (1-x+x^2\right )^2} \, dx=\frac {14 \, \sqrt {3} {\left (x^{3} - x^{2} + x\right )} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) - 6 \, x^{2} - 27 \, {\left (x^{3} - x^{2} + x\right )} \log \left (x^{2} - x + 1\right ) + 54 \, {\left (x^{3} - x^{2} + x\right )} \log \left (x\right ) + 30 \, x - 18}{18 \, {\left (x^{3} - x^{2} + x\right )}} \] Input:
integrate((x^2+x+1)/x^2/(x^2-x+1)^2,x, algorithm="fricas")
Output:
1/18*(14*sqrt(3)*(x^3 - x^2 + x)*arctan(1/3*sqrt(3)*(2*x - 1)) - 6*x^2 - 2 7*(x^3 - x^2 + x)*log(x^2 - x + 1) + 54*(x^3 - x^2 + x)*log(x) + 30*x - 18 )/(x^3 - x^2 + x)
Time = 0.08 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.07 \[ \int \frac {1+x+x^2}{x^2 \left (1-x+x^2\right )^2} \, dx=\frac {- x^{2} + 5 x - 3}{3 x^{3} - 3 x^{2} + 3 x} + 3 \log {\left (x \right )} - \frac {3 \log {\left (x^{2} - x + 1 \right )}}{2} + \frac {7 \sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} x}{3} - \frac {\sqrt {3}}{3} \right )}}{9} \] Input:
integrate((x**2+x+1)/x**2/(x**2-x+1)**2,x)
Output:
(-x**2 + 5*x - 3)/(3*x**3 - 3*x**2 + 3*x) + 3*log(x) - 3*log(x**2 - x + 1) /2 + 7*sqrt(3)*atan(2*sqrt(3)*x/3 - sqrt(3)/3)/9
Time = 0.11 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.89 \[ \int \frac {1+x+x^2}{x^2 \left (1-x+x^2\right )^2} \, dx=\frac {7}{9} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) - \frac {x^{2} - 5 \, x + 3}{3 \, {\left (x^{3} - x^{2} + x\right )}} - \frac {3}{2} \, \log \left (x^{2} - x + 1\right ) + 3 \, \log \left (x\right ) \] Input:
integrate((x^2+x+1)/x^2/(x^2-x+1)^2,x, algorithm="maxima")
Output:
7/9*sqrt(3)*arctan(1/3*sqrt(3)*(2*x - 1)) - 1/3*(x^2 - 5*x + 3)/(x^3 - x^2 + x) - 3/2*log(x^2 - x + 1) + 3*log(x)
Time = 0.19 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.90 \[ \int \frac {1+x+x^2}{x^2 \left (1-x+x^2\right )^2} \, dx=\frac {7}{9} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) - \frac {x^{2} - 5 \, x + 3}{3 \, {\left (x^{3} - x^{2} + x\right )}} - \frac {3}{2} \, \log \left (x^{2} - x + 1\right ) + 3 \, \log \left ({\left | x \right |}\right ) \] Input:
integrate((x^2+x+1)/x^2/(x^2-x+1)^2,x, algorithm="giac")
Output:
7/9*sqrt(3)*arctan(1/3*sqrt(3)*(2*x - 1)) - 1/3*(x^2 - 5*x + 3)/(x^3 - x^2 + x) - 3/2*log(x^2 - x + 1) + 3*log(abs(x))
Time = 10.24 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.11 \[ \int \frac {1+x+x^2}{x^2 \left (1-x+x^2\right )^2} \, dx=3\,\ln \left (x\right )-\frac {\frac {x^2}{3}-\frac {5\,x}{3}+1}{x^3-x^2+x}-\ln \left (x-\frac {1}{2}-\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {3}{2}+\frac {\sqrt {3}\,7{}\mathrm {i}}{18}\right )+\ln \left (x-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (-\frac {3}{2}+\frac {\sqrt {3}\,7{}\mathrm {i}}{18}\right ) \] Input:
int((x + x^2 + 1)/(x^2*(x^2 - x + 1)^2),x)
Output:
3*log(x) - (x^2/3 - (5*x)/3 + 1)/(x - x^2 + x^3) - log(x - (3^(1/2)*1i)/2 - 1/2)*((3^(1/2)*7i)/18 + 3/2) + log(x + (3^(1/2)*1i)/2 - 1/2)*((3^(1/2)*7 i)/18 - 3/2)
Time = 0.17 (sec) , antiderivative size = 136, normalized size of antiderivative = 2.23 \[ \int \frac {1+x+x^2}{x^2 \left (1-x+x^2\right )^2} \, dx=\frac {14 \sqrt {3}\, \mathit {atan} \left (\frac {2 x -1}{\sqrt {3}}\right ) x^{3}-14 \sqrt {3}\, \mathit {atan} \left (\frac {2 x -1}{\sqrt {3}}\right ) x^{2}+14 \sqrt {3}\, \mathit {atan} \left (\frac {2 x -1}{\sqrt {3}}\right ) x -27 \,\mathrm {log}\left (x^{2}-x +1\right ) x^{3}+27 \,\mathrm {log}\left (x^{2}-x +1\right ) x^{2}-27 \,\mathrm {log}\left (x^{2}-x +1\right ) x +54 \,\mathrm {log}\left (x \right ) x^{3}-54 \,\mathrm {log}\left (x \right ) x^{2}+54 \,\mathrm {log}\left (x \right ) x -6 x^{3}+24 x -18}{18 x \left (x^{2}-x +1\right )} \] Input:
int((x^2+x+1)/x^2/(x^2-x+1)^2,x)
Output:
(14*sqrt(3)*atan((2*x - 1)/sqrt(3))*x**3 - 14*sqrt(3)*atan((2*x - 1)/sqrt( 3))*x**2 + 14*sqrt(3)*atan((2*x - 1)/sqrt(3))*x - 27*log(x**2 - x + 1)*x** 3 + 27*log(x**2 - x + 1)*x**2 - 27*log(x**2 - x + 1)*x + 54*log(x)*x**3 - 54*log(x)*x**2 + 54*log(x)*x - 6*x**3 + 24*x - 18)/(18*x*(x**2 - x + 1))