Integrand size = 32, antiderivative size = 103 \[ \int \frac {1+x^2+x^3}{(-1+x) x \left (1+x^2\right )^3 \left (1+x+x^2\right )} \, dx=\frac {1+x}{8 \left (1+x^2\right )^2}-\frac {3 (1-x)}{8 \left (1+x^2\right )}+\frac {3 x}{16 \left (1+x^2\right )}+\frac {7 \arctan (x)}{16}-\frac {\arctan \left (\frac {1+2 x}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {1}{8} \log (1-x)-\log (x)+\frac {15}{16} \log \left (1+x^2\right )-\frac {1}{2} \log \left (1+x+x^2\right ) \]
1/8*(1+x)/(x^2+1)^2-3/8*(1-x)/(x^2+1)+3/16*x/(x^2+1)+7/16*arctan(x)+1/8*ln (1-x)-ln(x)+15/16*ln(x^2+1)-1/2*ln(x^2+x+1)-1/3*arctan(1/3*(1+2*x)*3^(1/2) )*3^(1/2)
Time = 0.05 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.90 \[ \int \frac {1+x^2+x^3}{(-1+x) x \left (1+x^2\right )^3 \left (1+x+x^2\right )} \, dx=\frac {1}{48} \left (\frac {6 (1+x)}{\left (1+x^2\right )^2}+\frac {9 (-2+3 x)}{1+x^2}+21 \arctan (x)-16 \sqrt {3} \arctan \left (\frac {1+2 x}{\sqrt {3}}\right )+20 \log (1-x)-48 \log (x)+45 \log \left (1+x^2\right )-10 \log \left (1+x+x^2\right )-14 \log \left (1-x^3\right )\right ) \]
((6*(1 + x))/(1 + x^2)^2 + (9*(-2 + 3*x))/(1 + x^2) + 21*ArcTan[x] - 16*Sq rt[3]*ArcTan[(1 + 2*x)/Sqrt[3]] + 20*Log[1 - x] - 48*Log[x] + 45*Log[1 + x ^2] - 10*Log[1 + x + x^2] - 14*Log[1 - x^3])/48
Time = 0.56 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {7279, 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.
\(\displaystyle \int \frac {x^3+x^2+1}{(x-1) x \left (x^2+1\right )^3 \left (x^2+x+1\right )} \, dx\) |
\(\Big \downarrow \) 7279 |
\(\displaystyle \int \left (\frac {-x-1}{x^2+x+1}+\frac {15 x-1}{8 \left (x^2+1\right )}+\frac {3 (x+1)}{4 \left (x^2+1\right )^2}+\frac {1-x}{2 \left (x^2+1\right )^3}+\frac {1}{8 (x-1)}-\frac {1}{x}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {7 \arctan (x)}{16}-\frac {\arctan \left (\frac {2 x+1}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {3 (1-x)}{8 \left (x^2+1\right )}+\frac {3 x}{16 \left (x^2+1\right )}+\frac {x+1}{8 \left (x^2+1\right )^2}+\frac {15}{16} \log \left (x^2+1\right )-\frac {1}{2} \log \left (x^2+x+1\right )+\frac {1}{8} \log (1-x)-\log (x)\) |
(1 + x)/(8*(1 + x^2)^2) - (3*(1 - x))/(8*(1 + x^2)) + (3*x)/(16*(1 + x^2)) + (7*ArcTan[x])/16 - ArcTan[(1 + 2*x)/Sqrt[3]]/Sqrt[3] + Log[1 - x]/8 - L og[x] + (15*Log[1 + x^2])/16 - Log[1 + x + x^2]/2
3.2.59.3.1 Defintions of rubi rules used
Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[ {v = RationalFunctionExpand[u/(a + b*x^n + c*x^(2*n)), x]}, Int[v, x] /; Su mQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]
Time = 0.45 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.70
method | result | size |
risch | \(\frac {\frac {9}{16} x^{3}-\frac {3}{8} x^{2}+\frac {11}{16} x -\frac {1}{4}}{\left (x^{2}+1\right )^{2}}+\frac {\ln \left (-1+x \right )}{8}+\frac {15 \ln \left (49 x^{2}+49\right )}{16}+\frac {7 \arctan \left (x \right )}{16}-\frac {\sqrt {3}\, \arctan \left (\frac {2 \sqrt {3}\, \left (x +\frac {1}{2}\right )}{3}\right )}{3}-\frac {\ln \left (x^{2}+x +1\right )}{2}-\ln \left (x \right )\) | \(72\) |
default | \(\frac {\ln \left (-1+x \right )}{8}+\frac {\frac {9}{2} x^{3}-3 x^{2}+\frac {11}{2} x -2}{8 \left (x^{2}+1\right )^{2}}+\frac {15 \ln \left (x^{2}+1\right )}{16}+\frac {7 \arctan \left (x \right )}{16}-\frac {\ln \left (x^{2}+x +1\right )}{2}-\frac {\arctan \left (\frac {\left (1+2 x \right ) \sqrt {3}}{3}\right ) \sqrt {3}}{3}-\ln \left (x \right )\) | \(73\) |
(9/16*x^3-3/8*x^2+11/16*x-1/4)/(x^2+1)^2+1/8*ln(-1+x)+15/16*ln(49*x^2+49)+ 7/16*arctan(x)-1/3*3^(1/2)*arctan(2/3*3^(1/2)*(x+1/2))-1/2*ln(x^2+x+1)-ln( x)
Time = 0.25 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.32 \[ \int \frac {1+x^2+x^3}{(-1+x) x \left (1+x^2\right )^3 \left (1+x+x^2\right )} \, dx=\frac {27 \, x^{3} - 16 \, \sqrt {3} {\left (x^{4} + 2 \, x^{2} + 1\right )} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) - 18 \, x^{2} + 21 \, {\left (x^{4} + 2 \, x^{2} + 1\right )} \arctan \left (x\right ) - 24 \, {\left (x^{4} + 2 \, x^{2} + 1\right )} \log \left (x^{2} + x + 1\right ) + 45 \, {\left (x^{4} + 2 \, x^{2} + 1\right )} \log \left (x^{2} + 1\right ) + 6 \, {\left (x^{4} + 2 \, x^{2} + 1\right )} \log \left (x - 1\right ) - 48 \, {\left (x^{4} + 2 \, x^{2} + 1\right )} \log \left (x\right ) + 33 \, x - 12}{48 \, {\left (x^{4} + 2 \, x^{2} + 1\right )}} \]
1/48*(27*x^3 - 16*sqrt(3)*(x^4 + 2*x^2 + 1)*arctan(1/3*sqrt(3)*(2*x + 1)) - 18*x^2 + 21*(x^4 + 2*x^2 + 1)*arctan(x) - 24*(x^4 + 2*x^2 + 1)*log(x^2 + x + 1) + 45*(x^4 + 2*x^2 + 1)*log(x^2 + 1) + 6*(x^4 + 2*x^2 + 1)*log(x - 1) - 48*(x^4 + 2*x^2 + 1)*log(x) + 33*x - 12)/(x^4 + 2*x^2 + 1)
Time = 0.30 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.85 \[ \int \frac {1+x^2+x^3}{(-1+x) x \left (1+x^2\right )^3 \left (1+x+x^2\right )} \, dx=- \log {\left (x \right )} + \frac {\log {\left (x - 1 \right )}}{8} + \frac {15 \log {\left (x^{2} + 1 \right )}}{16} - \frac {\log {\left (x^{2} + x + 1 \right )}}{2} + \frac {7 \operatorname {atan}{\left (x \right )}}{16} - \frac {\sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} x}{3} + \frac {\sqrt {3}}{3} \right )}}{3} + \frac {9 x^{3} - 6 x^{2} + 11 x - 4}{16 x^{4} + 32 x^{2} + 16} \]
-log(x) + log(x - 1)/8 + 15*log(x**2 + 1)/16 - log(x**2 + x + 1)/2 + 7*ata n(x)/16 - sqrt(3)*atan(2*sqrt(3)*x/3 + sqrt(3)/3)/3 + (9*x**3 - 6*x**2 + 1 1*x - 4)/(16*x**4 + 32*x**2 + 16)
Time = 0.33 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.75 \[ \int \frac {1+x^2+x^3}{(-1+x) x \left (1+x^2\right )^3 \left (1+x+x^2\right )} \, dx=-\frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) + \frac {9 \, x^{3} - 6 \, x^{2} + 11 \, x - 4}{16 \, {\left (x^{4} + 2 \, x^{2} + 1\right )}} + \frac {7}{16} \, \arctan \left (x\right ) - \frac {1}{2} \, \log \left (x^{2} + x + 1\right ) + \frac {15}{16} \, \log \left (x^{2} + 1\right ) + \frac {1}{8} \, \log \left (x - 1\right ) - \log \left (x\right ) \]
-1/3*sqrt(3)*arctan(1/3*sqrt(3)*(2*x + 1)) + 1/16*(9*x^3 - 6*x^2 + 11*x - 4)/(x^4 + 2*x^2 + 1) + 7/16*arctan(x) - 1/2*log(x^2 + x + 1) + 15/16*log(x ^2 + 1) + 1/8*log(x - 1) - log(x)
Time = 0.29 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.72 \[ \int \frac {1+x^2+x^3}{(-1+x) x \left (1+x^2\right )^3 \left (1+x+x^2\right )} \, dx=-\frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) + \frac {9 \, x^{3} - 6 \, x^{2} + 11 \, x - 4}{16 \, {\left (x^{2} + 1\right )}^{2}} + \frac {7}{16} \, \arctan \left (x\right ) - \frac {1}{2} \, \log \left (x^{2} + x + 1\right ) + \frac {15}{16} \, \log \left (x^{2} + 1\right ) + \frac {1}{8} \, \log \left ({\left | x - 1 \right |}\right ) - \log \left ({\left | x \right |}\right ) \]
-1/3*sqrt(3)*arctan(1/3*sqrt(3)*(2*x + 1)) + 1/16*(9*x^3 - 6*x^2 + 11*x - 4)/(x^2 + 1)^2 + 7/16*arctan(x) - 1/2*log(x^2 + x + 1) + 15/16*log(x^2 + 1 ) + 1/8*log(abs(x - 1)) - log(abs(x))
Time = 0.35 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.93 \[ \int \frac {1+x^2+x^3}{(-1+x) x \left (1+x^2\right )^3 \left (1+x+x^2\right )} \, dx=\frac {\ln \left (x-1\right )}{8}-\ln \left (x\right )+\ln \left (x+\frac {1}{2}-\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )-\ln \left (x+\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )+\frac {\frac {9\,x^3}{16}-\frac {3\,x^2}{8}+\frac {11\,x}{16}-\frac {1}{4}}{x^4+2\,x^2+1}+\ln \left (x-\mathrm {i}\right )\,\left (\frac {15}{16}-\frac {7}{32}{}\mathrm {i}\right )+\ln \left (x+1{}\mathrm {i}\right )\,\left (\frac {15}{16}+\frac {7}{32}{}\mathrm {i}\right ) \]