Integrand size = 23, antiderivative size = 143 \[ \int \frac {1}{(1+x) \left (-2+2 x+x^2-x^4\right )^{3/2}} \, dx=-\frac {i \sqrt {-2+2 x+x^2-x^4} \left (-\frac {i \left (4-2 x-2 x^2+x^3\right )}{20 (-1+x)^2 \sqrt {2+2 x+x^2}}-\frac {1}{4} i \text {arctanh}\left (1+x-\sqrt {2+2 x+x^2}\right )-\frac {7 i \text {arctanh}\left (\frac {1}{\sqrt {5}}-\frac {x}{\sqrt {5}}+\frac {\sqrt {2+2 x+x^2}}{\sqrt {5}}\right )}{20 \sqrt {5}}\right )}{(-1+x) \sqrt {2+2 x+x^2}} \]
-I*(-x^4+x^2+2*x-2)^(1/2)*(-1/20*I*(x^3-2*x^2-2*x+4)/(-1+x)^2/(x^2+2*x+2)^ (1/2)-1/4*I*arctanh(1+x-(x^2+2*x+2)^(1/2))+7/100*I*arctanh(-1/5*5^(1/2)+1/ 5*x*5^(1/2)-1/5*(x^2+2*x+2)^(1/2)*5^(1/2))*5^(1/2))/(-1+x)/(x^2+2*x+2)^(1/ 2)
Time = 0.54 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.89 \[ \int \frac {1}{(1+x) \left (-2+2 x+x^2-x^4\right )^{3/2}} \, dx=\frac {5 \left (4-2 x-2 x^2+x^3\right )+25 (-1+x)^2 \sqrt {2+2 x+x^2} \text {arctanh}\left (1+x-\sqrt {2+2 x+x^2}\right )+7 \sqrt {5} (-1+x)^2 \sqrt {2+2 x+x^2} \text {arctanh}\left (\frac {1-x+\sqrt {2+2 x+x^2}}{\sqrt {5}}\right )}{100 (-1+x) \sqrt {-2+2 x+x^2-x^4}} \]
(5*(4 - 2*x - 2*x^2 + x^3) + 25*(-1 + x)^2*Sqrt[2 + 2*x + x^2]*ArcTanh[1 + x - Sqrt[2 + 2*x + x^2]] + 7*Sqrt[5]*(-1 + x)^2*Sqrt[2 + 2*x + x^2]*ArcTa nh[(1 - x + Sqrt[2 + 2*x + x^2])/Sqrt[5]])/(100*(-1 + x)*Sqrt[-2 + 2*x + x ^2 - x^4])
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.
| \(\displaystyle \int \frac {1}{(x+1) \left (-x^4+x^2+2 x-2\right )^{3/2}} \, dx\) |
| \(\Big \downarrow \) 7299 |
| \(\displaystyle \int \frac {1}{(x+1) \left (-x^4+x^2+2 x-2\right )^{3/2}}dx\) |
3.21.15.3.1 Defintions of rubi rules used
Time = 1.02 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.83
| method | result | size |
| risch | \(\frac {x^{3}-2 x^{2}-2 x +4}{20 \left (-1+x \right ) \sqrt {-\left (x^{2}+2 x +2\right ) \left (-1+x \right )^{2}}}-\frac {\left (\frac {\arctan \left (\frac {1}{\sqrt {-\left (1+x \right )^{2}-1}}\right )}{8}+\frac {7 \sqrt {5}\, \arctan \left (\frac {\left (-6-4 x \right ) \sqrt {5}}{10 \sqrt {-\left (-1+x \right )^{2}-1-4 x}}\right )}{200}\right ) \left (-1+x \right ) \sqrt {-x^{2}-2 x -2}}{\sqrt {-\left (x^{2}+2 x +2\right ) \left (-1+x \right )^{2}}}\) | \(118\) |
| trager | \(-\frac {\left (x^{3}-2 x^{2}-2 x +4\right ) \sqrt {-x^{4}+x^{2}+2 x -2}}{20 \left (-1+x \right )^{3} \left (x^{2}+2 x +2\right )}-\frac {7 \operatorname {RootOf}\left (\textit {\_Z}^{2}+5\right ) \ln \left (\frac {2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+5\right ) x^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{2}+5\right ) x +5 \sqrt {-x^{4}+x^{2}+2 x -2}-3 \operatorname {RootOf}\left (\textit {\_Z}^{2}+5\right )}{\left (-1+x \right )^{2}}\right )}{200}-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (-\frac {-\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x +\sqrt {-x^{4}+x^{2}+2 x -2}+\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right )}{\left (1+x \right ) \left (-1+x \right )}\right )}{8}\) | \(159\) |
| default | \(-\frac {\left (7 \sqrt {5}\, \sqrt {-x^{2}-2 x -2}\, \arctan \left (\frac {\left (3+2 x \right ) \sqrt {5}}{5 \sqrt {-x^{2}-2 x -2}}\right ) x^{2}-14 \sqrt {5}\, \sqrt {-x^{2}-2 x -2}\, \arctan \left (\frac {\left (3+2 x \right ) \sqrt {5}}{5 \sqrt {-x^{2}-2 x -2}}\right ) x -25 \sqrt {-x^{2}-2 x -2}\, \arctan \left (\frac {1}{\sqrt {-x^{2}-2 x -2}}\right ) x^{2}+7 \sqrt {5}\, \arctan \left (\frac {\left (3+2 x \right ) \sqrt {5}}{5 \sqrt {-x^{2}-2 x -2}}\right ) \sqrt {-x^{2}-2 x -2}+50 \sqrt {-x^{2}-2 x -2}\, \arctan \left (\frac {1}{\sqrt {-x^{2}-2 x -2}}\right ) x +10 x^{3}-25 \arctan \left (\frac {1}{\sqrt {-x^{2}-2 x -2}}\right ) \sqrt {-x^{2}-2 x -2}-20 x^{2}-20 x +40\right ) \left (-1+x \right ) \left (x^{2}+2 x +2\right )}{200 \left (-x^{4}+x^{2}+2 x -2\right )^{\frac {3}{2}}}\) | \(253\) |
1/20*(x^3-2*x^2-2*x+4)/(-1+x)/(-(x^2+2*x+2)*(-1+x)^2)^(1/2)-(1/8*arctan(1/ (-(1+x)^2-1)^(1/2))+7/200*5^(1/2)*arctan(1/10*(-6-4*x)*5^(1/2)/(-(-1+x)^2- 1-4*x)^(1/2)))*(-1+x)*(-x^2-2*x-2)^(1/2)/(-(x^2+2*x+2)*(-1+x)^2)^(1/2)
Time = 0.26 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.22 \[ \int \frac {1}{(1+x) \left (-2+2 x+x^2-x^4\right )^{3/2}} \, dx=-\frac {7 \, \sqrt {5} {\left (x^{5} - x^{4} - x^{3} - x^{2} + 4 \, x - 2\right )} \arctan \left (\frac {\sqrt {5} \sqrt {-x^{4} + x^{2} + 2 \, x - 2} {\left (2 \, x + 3\right )}}{5 \, {\left (x^{3} + x^{2} - 2\right )}}\right ) - 25 \, {\left (x^{5} - x^{4} - x^{3} - x^{2} + 4 \, x - 2\right )} \arctan \left (\frac {\sqrt {-x^{4} + x^{2} + 2 \, x - 2}}{x^{3} + x^{2} - 2}\right ) + 10 \, \sqrt {-x^{4} + x^{2} + 2 \, x - 2} {\left (x^{3} - 2 \, x^{2} - 2 \, x + 4\right )}}{200 \, {\left (x^{5} - x^{4} - x^{3} - x^{2} + 4 \, x - 2\right )}} \]
-1/200*(7*sqrt(5)*(x^5 - x^4 - x^3 - x^2 + 4*x - 2)*arctan(1/5*sqrt(5)*sqr t(-x^4 + x^2 + 2*x - 2)*(2*x + 3)/(x^3 + x^2 - 2)) - 25*(x^5 - x^4 - x^3 - x^2 + 4*x - 2)*arctan(sqrt(-x^4 + x^2 + 2*x - 2)/(x^3 + x^2 - 2)) + 10*sq rt(-x^4 + x^2 + 2*x - 2)*(x^3 - 2*x^2 - 2*x + 4))/(x^5 - x^4 - x^3 - x^2 + 4*x - 2)
\[ \int \frac {1}{(1+x) \left (-2+2 x+x^2-x^4\right )^{3/2}} \, dx=\int \frac {1}{\left (- \left (x - 1\right )^{2} \left (x^{2} + 2 x + 2\right )\right )^{\frac {3}{2}} \left (x + 1\right )}\, dx \]
\[ \int \frac {1}{(1+x) \left (-2+2 x+x^2-x^4\right )^{3/2}} \, dx=\int { \frac {1}{{\left (-x^{4} + x^{2} + 2 \, x - 2\right )}^{\frac {3}{2}} {\left (x + 1\right )}} \,d x } \]
\[ \int \frac {1}{(1+x) \left (-2+2 x+x^2-x^4\right )^{3/2}} \, dx=\int { \frac {1}{{\left (-x^{4} + x^{2} + 2 \, x - 2\right )}^{\frac {3}{2}} {\left (x + 1\right )}} \,d x } \]
Timed out. \[ \int \frac {1}{(1+x) \left (-2+2 x+x^2-x^4\right )^{3/2}} \, dx=\int \frac {1}{\left (x+1\right )\,{\left (-x^4+x^2+2\,x-2\right )}^{3/2}} \,d x \]