Integrand size = 25, antiderivative size = 66 \[ \int \frac {\sqrt {1+6 x^2+x^4}}{(-1+x) (1+x)^3} \, dx=\frac {\sqrt {1+6 x^2+x^4}}{4 (1+x)^2}-\frac {\text {arctanh}\left (\frac {2 \sqrt {2} x}{1-2 x+x^2+\sqrt {1+6 x^2+x^4}}\right )}{2 \sqrt {2}} \]
1/4*(x^4+6*x^2+1)^(1/2)/(1+x)^2-1/4*arctanh(2*2^(1/2)*x/(1-2*x+x^2+(x^4+6* x^2+1)^(1/2)))*2^(1/2)
Time = 0.56 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.00 \[ \int \frac {\sqrt {1+6 x^2+x^4}}{(-1+x) (1+x)^3} \, dx=\frac {\sqrt {1+6 x^2+x^4}}{4 (1+x)^2}-\frac {\text {arctanh}\left (\frac {2 \sqrt {2} x}{1-2 x+x^2+\sqrt {1+6 x^2+x^4}}\right )}{2 \sqrt {2}} \]
Sqrt[1 + 6*x^2 + x^4]/(4*(1 + x)^2) - ArcTanh[(2*Sqrt[2]*x)/(1 - 2*x + x^2 + Sqrt[1 + 6*x^2 + x^4])]/(2*Sqrt[2])
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.
\(\displaystyle \int \frac {\sqrt {x^4+6 x^2+1}}{(x-1) (x+1)^3} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {\sqrt {x^4+6 x^2+1}}{4 \left (x^2-1\right )}-\frac {\sqrt {x^4+6 x^2+1}}{4 (x+1)^2}-\frac {\sqrt {x^4+6 x^2+1}}{2 (x+1)^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {1}{2} \int \frac {\sqrt {x^4+6 x^2+1}}{(x+1)^3}dx-\frac {1}{4} \int \frac {\sqrt {x^4+6 x^2+1}}{(x+1)^2}dx+\frac {3 \sqrt {\frac {\left (3-2 \sqrt {2}\right ) x^2+1}{\left (3+2 \sqrt {2}\right ) x^2+1}} \left (\left (3+2 \sqrt {2}\right ) x^2+1\right ) \operatorname {EllipticF}\left (\arctan \left (\sqrt {3+2 \sqrt {2}} x\right ),-4 \left (4-3 \sqrt {2}\right )\right )}{4 \sqrt {3+2 \sqrt {2}} \sqrt {x^4+6 x^2+1}}-\frac {\sqrt {3+2 \sqrt {2}} \sqrt {\frac {\left (3-2 \sqrt {2}\right ) x^2+1}{\left (3+2 \sqrt {2}\right ) x^2+1}} \left (\left (3+2 \sqrt {2}\right ) x^2+1\right ) E\left (\arctan \left (\sqrt {3+2 \sqrt {2}} x\right )|-4 \left (4-3 \sqrt {2}\right )\right )}{4 \sqrt {x^4+6 x^2+1}}-\frac {\text {arctanh}\left (\frac {2 \sqrt {2} x}{\sqrt {x^4+6 x^2+1}}\right )}{2 \sqrt {2}}+\frac {x \left (x^2+2 \sqrt {2}+3\right )}{4 \sqrt {x^4+6 x^2+1}}\) |
3.9.68.3.1 Defintions of rubi rules used
Time = 2.77 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.74
method | result | size |
risch | \(\frac {\sqrt {x^{4}+6 x^{2}+1}}{4 \left (1+x \right )^{2}}-\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\left (1+x \right )^{2} \sqrt {2}}{2 \sqrt {x^{4}+6 x^{2}+1}}\right )}{8}\) | \(49\) |
default | \(\frac {-\left (1+x \right )^{2} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\left (1+x \right )^{2} \sqrt {2}}{2 \sqrt {x^{4}+6 x^{2}+1}}\right )+2 \sqrt {x^{4}+6 x^{2}+1}}{8 \left (1+x \right )^{2}}\) | \(56\) |
pseudoelliptic | \(\frac {-\left (1+x \right )^{2} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\left (1+x \right )^{2} \sqrt {2}}{2 \sqrt {x^{4}+6 x^{2}+1}}\right )+2 \sqrt {x^{4}+6 x^{2}+1}}{8 \left (1+x \right )^{2}}\) | \(56\) |
trager | \(\frac {\sqrt {x^{4}+6 x^{2}+1}}{4 \left (1+x \right )^{2}}+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right ) \ln \left (-\frac {-\operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right ) x^{2}-2 \operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right ) x +2 \sqrt {x^{4}+6 x^{2}+1}-\operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right )}{\left (x -1\right )^{2}}\right )}{8}\) | \(80\) |
elliptic | \(\frac {{\left (\left (x^{2}-1\right )^{2}+8 x^{2}\right )}^{\frac {3}{2}}}{16 \left (x^{2}-1\right )^{2}}-\frac {{\left (\left (x^{2}-1\right )^{2}+8 x^{2}\right )}^{\frac {3}{2}}}{32 \left (x^{2}-1\right )}+\frac {\sqrt {\left (x^{2}-1\right )^{2}+8 x^{2}}}{16}-\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\left (8 x^{2}+8\right ) \sqrt {2}}{8 \sqrt {\left (x^{2}-1\right )^{2}+8 x^{2}}}\right )}{8}+\frac {\left (2 x^{2}+6\right ) \sqrt {\left (x^{2}-1\right )^{2}+8 x^{2}}}{64}+\frac {\left (-\frac {1}{4 \left (\frac {\sqrt {x^{4}+6 x^{2}+1}\, \sqrt {2}}{2 x}-2\right )}+\frac {\ln \left (\frac {\sqrt {x^{4}+6 x^{2}+1}\, \sqrt {2}}{2 x}-2\right )}{8}-\frac {1}{4 \left (\frac {\sqrt {x^{4}+6 x^{2}+1}\, \sqrt {2}}{2 x}+2\right )}-\frac {\ln \left (\frac {\sqrt {x^{4}+6 x^{2}+1}\, \sqrt {2}}{2 x}+2\right )}{8}\right ) \sqrt {2}}{2}\) | \(232\) |
1/4*(x^4+6*x^2+1)^(1/2)/(1+x)^2-1/8*2^(1/2)*arctanh(1/2*(1+x)^2*2^(1/2)/(x ^4+6*x^2+1)^(1/2))
Time = 0.29 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.61 \[ \int \frac {\sqrt {1+6 x^2+x^4}}{(-1+x) (1+x)^3} \, dx=\frac {\sqrt {2} {\left (x^{2} + 2 \, x + 1\right )} \log \left (\frac {3 \, x^{4} + 4 \, x^{3} - 2 \, \sqrt {2} \sqrt {x^{4} + 6 \, x^{2} + 1} {\left (x^{2} + 2 \, x + 1\right )} + 18 \, x^{2} + 4 \, x + 3}{x^{4} - 4 \, x^{3} + 6 \, x^{2} - 4 \, x + 1}\right ) + 4 \, \sqrt {x^{4} + 6 \, x^{2} + 1}}{16 \, {\left (x^{2} + 2 \, x + 1\right )}} \]
1/16*(sqrt(2)*(x^2 + 2*x + 1)*log((3*x^4 + 4*x^3 - 2*sqrt(2)*sqrt(x^4 + 6* x^2 + 1)*(x^2 + 2*x + 1) + 18*x^2 + 4*x + 3)/(x^4 - 4*x^3 + 6*x^2 - 4*x + 1)) + 4*sqrt(x^4 + 6*x^2 + 1))/(x^2 + 2*x + 1)
\[ \int \frac {\sqrt {1+6 x^2+x^4}}{(-1+x) (1+x)^3} \, dx=\int \frac {\sqrt {x^{4} + 6 x^{2} + 1}}{\left (x - 1\right ) \left (x + 1\right )^{3}}\, dx \]
\[ \int \frac {\sqrt {1+6 x^2+x^4}}{(-1+x) (1+x)^3} \, dx=\int { \frac {\sqrt {x^{4} + 6 \, x^{2} + 1}}{{\left (x + 1\right )}^{3} {\left (x - 1\right )}} \,d x } \]
\[ \int \frac {\sqrt {1+6 x^2+x^4}}{(-1+x) (1+x)^3} \, dx=\int { \frac {\sqrt {x^{4} + 6 \, x^{2} + 1}}{{\left (x + 1\right )}^{3} {\left (x - 1\right )}} \,d x } \]
Timed out. \[ \int \frac {\sqrt {1+6 x^2+x^4}}{(-1+x) (1+x)^3} \, dx=\int \frac {\sqrt {x^4+6\,x^2+1}}{\left (x-1\right )\,{\left (x+1\right )}^3} \,d x \]