Integrand size = 39, antiderivative size = 100 \[ \int \frac {\left (-1+3 x^4\right ) \sqrt {1+x+2 x^4+x^5+x^8}}{x^2 \left (4+x+4 x^4\right )} \, dx=\frac {\sqrt {1+x+2 x^4+x^5+x^8}}{4 x}-\frac {1}{8} \sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt {1+x+2 x^4+x^5+x^8}}{1+x+x^4}\right )+\frac {1}{8} \text {arctanh}\left (\frac {\sqrt {1+x+2 x^4+x^5+x^8}}{1+x+x^4}\right ) \]
1/4*(x^8+x^5+2*x^4+x+1)^(1/2)/x-1/8*3^(1/2)*arctan(3^(1/2)*(x^8+x^5+2*x^4+ x+1)^(1/2)/(x^4+x+1))+1/8*arctanh((x^8+x^5+2*x^4+x+1)^(1/2)/(x^4+x+1))
Time = 0.23 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.00 \[ \int \frac {\left (-1+3 x^4\right ) \sqrt {1+x+2 x^4+x^5+x^8}}{x^2 \left (4+x+4 x^4\right )} \, dx=\frac {\sqrt {1+x+2 x^4+x^5+x^8}}{4 x}-\frac {1}{8} \sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt {1+x+2 x^4+x^5+x^8}}{1+x+x^4}\right )+\frac {1}{8} \text {arctanh}\left (\frac {\sqrt {1+x+2 x^4+x^5+x^8}}{1+x+x^4}\right ) \]
Sqrt[1 + x + 2*x^4 + x^5 + x^8]/(4*x) - (Sqrt[3]*ArcTan[(Sqrt[3]*Sqrt[1 + x + 2*x^4 + x^5 + x^8])/(1 + x + x^4)])/8 + ArcTanh[Sqrt[1 + x + 2*x^4 + x ^5 + x^8]/(1 + x + x^4)]/8
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.
\(\displaystyle \int \frac {\left (3 x^4-1\right ) \sqrt {x^8+x^5+2 x^4+x+1}}{x^2 \left (4 x^4+x+4\right )} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {\sqrt {x^8+x^5+2 x^4+x+1}}{16 x}-\frac {\sqrt {x^8+x^5+2 x^4+x+1}}{4 x^2}+\frac {\sqrt {x^8+x^5+2 x^4+x+1} \left (-4 x^3+64 x^2-1\right )}{16 \left (4 x^4+x+4\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{16} \int \frac {\sqrt {x^8+x^5+2 x^4+x+1}}{x}dx+\frac {1}{16} \int \frac {\sqrt {x^8+x^5+2 x^4+x+1}}{-4 x^4-x-4}dx-\frac {1}{4} \int \frac {x^3 \sqrt {x^8+x^5+2 x^4+x+1}}{4 x^4+x+4}dx-\frac {1}{4} \int \frac {\sqrt {x^8+x^5+2 x^4+x+1}}{x^2}dx+4 \int \frac {x^2 \sqrt {x^8+x^5+2 x^4+x+1}}{4 x^4+x+4}dx\) |
3.15.2.3.1 Defintions of rubi rules used
Time = 1.08 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.96
method | result | size |
pseudoelliptic | \(\frac {-\sqrt {3}\, \arctan \left (\frac {\left (2 x^{4}-x +2\right ) \sqrt {3}}{6 \sqrt {x^{8}+x^{5}+2 x^{4}+x +1}}\right ) x +\ln \left (\frac {2 x^{4}+2 \sqrt {x^{8}+x^{5}+2 x^{4}+x +1}+x +2}{x}\right ) x +4 \sqrt {x^{8}+x^{5}+2 x^{4}+x +1}}{16 x}\) | \(96\) |
risch | \(\frac {\sqrt {x^{8}+x^{5}+2 x^{4}+x +1}}{4 x}+\frac {\ln \left (-\frac {2 x^{4}+2 \sqrt {x^{8}+x^{5}+2 x^{4}+x +1}+x +2}{x}\right )}{16}-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+3\right ) \ln \left (\frac {2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+3\right ) x^{4}-\operatorname {RootOf}\left (\textit {\_Z}^{2}+3\right ) x +2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+3\right )-6 \sqrt {x^{8}+x^{5}+2 x^{4}+x +1}}{4 x^{4}+x +4}\right )}{16}\) | \(124\) |
trager | \(\frac {\sqrt {x^{8}+x^{5}+2 x^{4}+x +1}}{4 x}-\frac {\ln \left (-\frac {-2 x^{4}+2 \sqrt {x^{8}+x^{5}+2 x^{4}+x +1}-x -2}{x}\right )}{16}+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+3\right ) \ln \left (-\frac {2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+3\right ) x^{4}-\operatorname {RootOf}\left (\textit {\_Z}^{2}+3\right ) x +2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+3\right )+6 \sqrt {x^{8}+x^{5}+2 x^{4}+x +1}}{4 x^{4}+x +4}\right )}{16}\) | \(127\) |
1/16*(-3^(1/2)*arctan(1/6*(2*x^4-x+2)*3^(1/2)/(x^8+x^5+2*x^4+x+1)^(1/2))*x +ln((2*x^4+2*(x^8+x^5+2*x^4+x+1)^(1/2)+x+2)/x)*x+4*(x^8+x^5+2*x^4+x+1)^(1/ 2))/x
Time = 1.17 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.94 \[ \int \frac {\left (-1+3 x^4\right ) \sqrt {1+x+2 x^4+x^5+x^8}}{x^2 \left (4+x+4 x^4\right )} \, dx=-\frac {\sqrt {3} x \arctan \left (\frac {\sqrt {3} {\left (2 \, x^{4} - x + 2\right )}}{6 \, \sqrt {x^{8} + x^{5} + 2 \, x^{4} + x + 1}}\right ) + x \log \left (\frac {2 \, x^{4} + x - 2 \, \sqrt {x^{8} + x^{5} + 2 \, x^{4} + x + 1} + 2}{x}\right ) - 4 \, \sqrt {x^{8} + x^{5} + 2 \, x^{4} + x + 1}}{16 \, x} \]
-1/16*(sqrt(3)*x*arctan(1/6*sqrt(3)*(2*x^4 - x + 2)/sqrt(x^8 + x^5 + 2*x^4 + x + 1)) + x*log((2*x^4 + x - 2*sqrt(x^8 + x^5 + 2*x^4 + x + 1) + 2)/x) - 4*sqrt(x^8 + x^5 + 2*x^4 + x + 1))/x
Timed out. \[ \int \frac {\left (-1+3 x^4\right ) \sqrt {1+x+2 x^4+x^5+x^8}}{x^2 \left (4+x+4 x^4\right )} \, dx=\text {Timed out} \]
\[ \int \frac {\left (-1+3 x^4\right ) \sqrt {1+x+2 x^4+x^5+x^8}}{x^2 \left (4+x+4 x^4\right )} \, dx=\int { \frac {\sqrt {x^{8} + x^{5} + 2 \, x^{4} + x + 1} {\left (3 \, x^{4} - 1\right )}}{{\left (4 \, x^{4} + x + 4\right )} x^{2}} \,d x } \]
\[ \int \frac {\left (-1+3 x^4\right ) \sqrt {1+x+2 x^4+x^5+x^8}}{x^2 \left (4+x+4 x^4\right )} \, dx=\int { \frac {\sqrt {x^{8} + x^{5} + 2 \, x^{4} + x + 1} {\left (3 \, x^{4} - 1\right )}}{{\left (4 \, x^{4} + x + 4\right )} x^{2}} \,d x } \]
Timed out. \[ \int \frac {\left (-1+3 x^4\right ) \sqrt {1+x+2 x^4+x^5+x^8}}{x^2 \left (4+x+4 x^4\right )} \, dx=\int \frac {\left (3\,x^4-1\right )\,\sqrt {x^8+x^5+2\,x^4+x+1}}{x^2\,\left (4\,x^4+x+4\right )} \,d x \]