Integrand size = 35, antiderivative size = 91 \[ \int \frac {5+x+3 x^2+2 x^3}{x^3 \left (2+x+3 x^2+x^3+2 x^4\right )} \, dx=-\frac {5}{4 x^2}+\frac {3}{4 x}+\frac {1}{24} \sqrt {\frac {5}{3}} \arctan \left (\frac {1-4 x}{\sqrt {15}}\right )+\frac {8 \arctan \left (\frac {1+2 x}{\sqrt {3}}\right )}{3 \sqrt {3}}-\frac {15 \log (x)}{8}+\frac {2}{3} \log \left (1+x+x^2\right )+\frac {13}{48} \log \left (2-x+2 x^2\right ) \]
[Out]
Time = 0.10 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {2112, 814, 648, 632, 210, 642} \[ \int \frac {5+x+3 x^2+2 x^3}{x^3 \left (2+x+3 x^2+x^3+2 x^4\right )} \, dx=\frac {1}{24} \sqrt {\frac {5}{3}} \arctan \left (\frac {1-4 x}{\sqrt {15}}\right )+\frac {8 \arctan \left (\frac {2 x+1}{\sqrt {3}}\right )}{3 \sqrt {3}}-\frac {5}{4 x^2}+\frac {2}{3} \log \left (x^2+x+1\right )+\frac {13}{48} \log \left (2 x^2-x+2\right )+\frac {3}{4 x}-\frac {15 \log (x)}{8} \]
[In]
[Out]
Rule 210
Rule 632
Rule 642
Rule 648
Rule 814
Rule 2112
Rubi steps \begin{align*} \text {integral}& = -\left (\frac {1}{3} \int \frac {-6+4 x}{x^3 \left (4-2 x+4 x^2\right )} \, dx\right )+\frac {1}{3} \int \frac {24+16 x}{x^3 \left (4+4 x+4 x^2\right )} \, dx \\ & = \frac {1}{3} \int \left (\frac {6}{x^3}-\frac {2}{x^2}-\frac {4}{x}+\frac {2 (3+2 x)}{1+x+x^2}\right ) \, dx-\frac {1}{3} \int \left (-\frac {3}{2 x^3}+\frac {1}{4 x^2}+\frac {13}{8 x}+\frac {9-26 x}{8 \left (2-x+2 x^2\right )}\right ) \, dx \\ & = -\frac {5}{4 x^2}+\frac {3}{4 x}-\frac {15 \log (x)}{8}-\frac {1}{24} \int \frac {9-26 x}{2-x+2 x^2} \, dx+\frac {2}{3} \int \frac {3+2 x}{1+x+x^2} \, dx \\ & = -\frac {5}{4 x^2}+\frac {3}{4 x}-\frac {15 \log (x)}{8}-\frac {5}{48} \int \frac {1}{2-x+2 x^2} \, dx+\frac {13}{48} \int \frac {-1+4 x}{2-x+2 x^2} \, dx+\frac {2}{3} \int \frac {1+2 x}{1+x+x^2} \, dx+\frac {4}{3} \int \frac {1}{1+x+x^2} \, dx \\ & = -\frac {5}{4 x^2}+\frac {3}{4 x}-\frac {15 \log (x)}{8}+\frac {2}{3} \log \left (1+x+x^2\right )+\frac {13}{48} \log \left (2-x+2 x^2\right )+\frac {5}{24} \text {Subst}\left (\int \frac {1}{-15-x^2} \, dx,x,-1+4 x\right )-\frac {8}{3} \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+2 x\right ) \\ & = -\frac {5}{4 x^2}+\frac {3}{4 x}+\frac {1}{24} \sqrt {\frac {5}{3}} \tan ^{-1}\left (\frac {1-4 x}{\sqrt {15}}\right )+\frac {8 \tan ^{-1}\left (\frac {1+2 x}{\sqrt {3}}\right )}{3 \sqrt {3}}-\frac {15 \log (x)}{8}+\frac {2}{3} \log \left (1+x+x^2\right )+\frac {13}{48} \log \left (2-x+2 x^2\right ) \\ \end{align*}
Time = 0.04 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.90 \[ \int \frac {5+x+3 x^2+2 x^3}{x^3 \left (2+x+3 x^2+x^3+2 x^4\right )} \, dx=\frac {1}{144} \left (128 \sqrt {3} \arctan \left (\frac {1+2 x}{\sqrt {3}}\right )-2 \sqrt {15} \arctan \left (\frac {-1+4 x}{\sqrt {15}}\right )+3 \left (-\frac {60}{x^2}+\frac {36}{x}-90 \log (x)+32 \log \left (1+x+x^2\right )+13 \log \left (2-x+2 x^2\right )\right )\right ) \]
[In]
[Out]
Time = 0.08 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.77
method | result | size |
default | \(-\frac {5}{4 x^{2}}+\frac {3}{4 x}-\frac {15 \ln \left (x \right )}{8}+\frac {2 \ln \left (x^{2}+x +1\right )}{3}+\frac {8 \arctan \left (\frac {\left (1+2 x \right ) \sqrt {3}}{3}\right ) \sqrt {3}}{9}+\frac {13 \ln \left (2 x^{2}-x +2\right )}{48}-\frac {\sqrt {15}\, \arctan \left (\frac {\left (-1+4 x \right ) \sqrt {15}}{15}\right )}{72}\) | \(70\) |
risch | \(\frac {\frac {3 x}{4}-\frac {5}{4}}{x^{2}}-\frac {\sqrt {15}\, \arctan \left (\frac {\left (-1+4 x \right ) \sqrt {15}}{15}\right )}{72}+\frac {13 \ln \left (16 x^{2}-8 x +16\right )}{48}+\frac {2 \ln \left (4 x^{2}+4 x +4\right )}{3}+\frac {8 \arctan \left (\frac {\left (1+2 x \right ) \sqrt {3}}{3}\right ) \sqrt {3}}{9}-\frac {15 \ln \left (x \right )}{8}\) | \(73\) |
[In]
[Out]
none
Time = 0.27 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.98 \[ \int \frac {5+x+3 x^2+2 x^3}{x^3 \left (2+x+3 x^2+x^3+2 x^4\right )} \, dx=-\frac {2 \, \sqrt {5} \sqrt {3} x^{2} \arctan \left (\frac {1}{15} \, \sqrt {5} \sqrt {3} {\left (4 \, x - 1\right )}\right ) - 128 \, \sqrt {3} x^{2} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) - 39 \, x^{2} \log \left (2 \, x^{2} - x + 2\right ) - 96 \, x^{2} \log \left (x^{2} + x + 1\right ) + 270 \, x^{2} \log \left (x\right ) - 108 \, x + 180}{144 \, x^{2}} \]
[In]
[Out]
Time = 0.16 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.03 \[ \int \frac {5+x+3 x^2+2 x^3}{x^3 \left (2+x+3 x^2+x^3+2 x^4\right )} \, dx=- \frac {15 \log {\left (x \right )}}{8} + \frac {13 \log {\left (x^{2} - \frac {x}{2} + 1 \right )}}{48} + \frac {2 \log {\left (x^{2} + x + 1 \right )}}{3} - \frac {\sqrt {15} \operatorname {atan}{\left (\frac {4 \sqrt {15} x}{15} - \frac {\sqrt {15}}{15} \right )}}{72} + \frac {8 \sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} x}{3} + \frac {\sqrt {3}}{3} \right )}}{9} + \frac {3 x - 5}{4 x^{2}} \]
[In]
[Out]
none
Time = 0.27 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.76 \[ \int \frac {5+x+3 x^2+2 x^3}{x^3 \left (2+x+3 x^2+x^3+2 x^4\right )} \, dx=-\frac {1}{72} \, \sqrt {15} \arctan \left (\frac {1}{15} \, \sqrt {15} {\left (4 \, x - 1\right )}\right ) + \frac {8}{9} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) + \frac {3 \, x - 5}{4 \, x^{2}} + \frac {13}{48} \, \log \left (2 \, x^{2} - x + 2\right ) + \frac {2}{3} \, \log \left (x^{2} + x + 1\right ) - \frac {15}{8} \, \log \left (x\right ) \]
[In]
[Out]
none
Time = 0.28 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.77 \[ \int \frac {5+x+3 x^2+2 x^3}{x^3 \left (2+x+3 x^2+x^3+2 x^4\right )} \, dx=-\frac {1}{72} \, \sqrt {15} \arctan \left (\frac {1}{15} \, \sqrt {15} {\left (4 \, x - 1\right )}\right ) + \frac {8}{9} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) + \frac {3 \, x - 5}{4 \, x^{2}} + \frac {13}{48} \, \log \left (2 \, x^{2} - x + 2\right ) + \frac {2}{3} \, \log \left (x^{2} + x + 1\right ) - \frac {15}{8} \, \log \left ({\left | x \right |}\right ) \]
[In]
[Out]
Time = 0.09 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.01 \[ \int \frac {5+x+3 x^2+2 x^3}{x^3 \left (2+x+3 x^2+x^3+2 x^4\right )} \, dx=\frac {\frac {3\,x}{4}-\frac {5}{4}}{x^2}-\frac {15\,\ln \left (x\right )}{8}-\ln \left (x+\frac {1}{2}-\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (-\frac {2}{3}+\frac {\sqrt {3}\,4{}\mathrm {i}}{9}\right )+\ln \left (x+\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {2}{3}+\frac {\sqrt {3}\,4{}\mathrm {i}}{9}\right )+\ln \left (x-\frac {1}{4}-\frac {\sqrt {15}\,1{}\mathrm {i}}{4}\right )\,\left (\frac {13}{48}+\frac {\sqrt {15}\,1{}\mathrm {i}}{144}\right )-\ln \left (x-\frac {1}{4}+\frac {\sqrt {15}\,1{}\mathrm {i}}{4}\right )\,\left (-\frac {13}{48}+\frac {\sqrt {15}\,1{}\mathrm {i}}{144}\right ) \]
[In]
[Out]