Integrand size = 31, antiderivative size = 70 \[ \int \frac {\left (3+2 x^5\right ) \sqrt {x-2 x^4-x^6}}{\left (-1+x^5\right )^2} \, dx=-\frac {x \sqrt {x-2 x^4-x^6}}{-1+x^5}-\frac {\arctan \left (\frac {\sqrt {2} x \sqrt {x-2 x^4-x^6}}{-1+2 x^3+x^5}\right )}{\sqrt {2}} \]
-x*(-x^6-2*x^4+x)^(1/2)/(x^5-1)-1/2*arctan(2^(1/2)*x*(-x^6-2*x^4+x)^(1/2)/ (x^5+2*x^3-1))*2^(1/2)
Time = 4.30 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.24 \[ \int \frac {\left (3+2 x^5\right ) \sqrt {x-2 x^4-x^6}}{\left (-1+x^5\right )^2} \, dx=\frac {\sqrt {x-2 x^4-x^6} \left (-\frac {2 x^{3/2}}{-1+x^5}-\frac {\sqrt {2} \text {arctanh}\left (\frac {\sqrt {2} x^{3/2}}{\sqrt {-1+2 x^3+x^5}}\right )}{\sqrt {-1+2 x^3+x^5}}\right )}{2 \sqrt {x}} \]
(Sqrt[x - 2*x^4 - x^6]*((-2*x^(3/2))/(-1 + x^5) - (Sqrt[2]*ArcTanh[(Sqrt[2 ]*x^(3/2))/Sqrt[-1 + 2*x^3 + x^5]])/Sqrt[-1 + 2*x^3 + x^5]))/(2*Sqrt[x])
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.
| \(\displaystyle \int \frac {\left (2 x^5+3\right ) \sqrt {-x^6-2 x^4+x}}{\left (x^5-1\right )^2} \, dx\) |
| \(\Big \downarrow \) 2467 |
| \(\displaystyle \frac {\sqrt {-x^6-2 x^4+x} \int \frac {\sqrt {x} \sqrt {-x^5-2 x^3+1} \left (2 x^5+3\right )}{\left (1-x^5\right )^2}dx}{\sqrt {x} \sqrt {-x^5-2 x^3+1}}\) |
| \(\Big \downarrow \) 2035 |
| \(\displaystyle \frac {2 \sqrt {-x^6-2 x^4+x} \int \frac {x \sqrt {-x^5-2 x^3+1} \left (2 x^5+3\right )}{\left (1-x^5\right )^2}d\sqrt {x}}{\sqrt {x} \sqrt {-x^5-2 x^3+1}}\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \frac {2 \sqrt {-x^6-2 x^4+x} \int \left (\frac {\sqrt {-x^5-2 x^3+1} \left (-x-3 \sqrt {x}-1\right )}{4 \left (x^2+x^{3/2}+x+\sqrt {x}+1\right )^2}-\frac {3 \sqrt {-x^5-2 x^3+1}}{10 (x-1)}+\frac {\left (-3 x^{3/2}+5 x-\sqrt {x}+1\right ) \sqrt {-x^5-2 x^3+1}}{20 \left (x^2-x^{3/2}+x-\sqrt {x}+1\right )}+\frac {\left (3 x^{3/2}+5 x+\sqrt {x}+1\right ) \sqrt {-x^5-2 x^3+1}}{20 \left (x^2+x^{3/2}+x+\sqrt {x}+1\right )}+\frac {\sqrt {-x^5-2 x^3+1}}{20 \left (\sqrt {x}-1\right )^2}+\frac {\sqrt {-x^5-2 x^3+1}}{20 \left (\sqrt {x}+1\right )^2}+\frac {\left (-x+3 \sqrt {x}-1\right ) \sqrt {-x^5-2 x^3+1}}{4 \left (x^2-x^{3/2}+x-\sqrt {x}+1\right )^2}\right )d\sqrt {x}}{\sqrt {x} \sqrt {-x^5-2 x^3+1}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {2 \sqrt {-x^6-2 x^4+x} \left (\frac {3}{20} \int \frac {\sqrt {-x^5-2 x^3+1}}{1-\sqrt {x}}d\sqrt {x}+\frac {1}{20} \int \frac {\sqrt {-x^5-2 x^3+1}}{\left (\sqrt {x}-1\right )^2}d\sqrt {x}+\frac {1}{20} \int \frac {\sqrt {-x^5-2 x^3+1}}{\left (\sqrt {x}+1\right )^2}d\sqrt {x}+\frac {3}{20} \int \frac {\sqrt {-x^5-2 x^3+1}}{\sqrt {x}+1}d\sqrt {x}-\frac {1}{4} \int \frac {\sqrt {-x^5-2 x^3+1}}{\left (x^2-x^{3/2}+x-\sqrt {x}+1\right )^2}d\sqrt {x}+\frac {3}{4} \int \frac {\sqrt {x} \sqrt {-x^5-2 x^3+1}}{\left (x^2-x^{3/2}+x-\sqrt {x}+1\right )^2}d\sqrt {x}-\frac {1}{4} \int \frac {x \sqrt {-x^5-2 x^3+1}}{\left (x^2-x^{3/2}+x-\sqrt {x}+1\right )^2}d\sqrt {x}+\frac {1}{20} \int \frac {\sqrt {-x^5-2 x^3+1}}{x^2-x^{3/2}+x-\sqrt {x}+1}d\sqrt {x}-\frac {1}{20} \int \frac {\sqrt {x} \sqrt {-x^5-2 x^3+1}}{x^2-x^{3/2}+x-\sqrt {x}+1}d\sqrt {x}+\frac {1}{4} \int \frac {x \sqrt {-x^5-2 x^3+1}}{x^2-x^{3/2}+x-\sqrt {x}+1}d\sqrt {x}-\frac {3}{20} \int \frac {x^{3/2} \sqrt {-x^5-2 x^3+1}}{x^2-x^{3/2}+x-\sqrt {x}+1}d\sqrt {x}-\frac {1}{4} \int \frac {\sqrt {-x^5-2 x^3+1}}{\left (x^2+x^{3/2}+x+\sqrt {x}+1\right )^2}d\sqrt {x}-\frac {3}{4} \int \frac {\sqrt {x} \sqrt {-x^5-2 x^3+1}}{\left (x^2+x^{3/2}+x+\sqrt {x}+1\right )^2}d\sqrt {x}-\frac {1}{4} \int \frac {x \sqrt {-x^5-2 x^3+1}}{\left (x^2+x^{3/2}+x+\sqrt {x}+1\right )^2}d\sqrt {x}+\frac {1}{20} \int \frac {\sqrt {-x^5-2 x^3+1}}{x^2+x^{3/2}+x+\sqrt {x}+1}d\sqrt {x}+\frac {1}{20} \int \frac {\sqrt {x} \sqrt {-x^5-2 x^3+1}}{x^2+x^{3/2}+x+\sqrt {x}+1}d\sqrt {x}+\frac {1}{4} \int \frac {x \sqrt {-x^5-2 x^3+1}}{x^2+x^{3/2}+x+\sqrt {x}+1}d\sqrt {x}+\frac {3}{20} \int \frac {x^{3/2} \sqrt {-x^5-2 x^3+1}}{x^2+x^{3/2}+x+\sqrt {x}+1}d\sqrt {x}\right )}{\sqrt {x} \sqrt {-x^5-2 x^3+1}}\) |
3.10.23.3.1 Defintions of rubi rules used
Int[(Fx_)*(x_)^(m_), x_Symbol] :> With[{k = Denominator[m]}, Simp[k Subst [Int[x^(k*(m + 1) - 1)*SubstPower[Fx, x, k], x], x, x^(1/k)], x]] /; Fracti onQ[m] && AlgebraicFunctionQ[Fx, x]
Int[(Fx_.)*(Px_)^(p_), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Simp[Px^F racPart[p]/(x^(r*FracPart[p])*ExpandToSum[Px/x^r, x]^FracPart[p]) Int[x^( p*r)*ExpandToSum[Px/x^r, x]^p*Fx, x], x] /; IGtQ[r, 0]] /; FreeQ[p, x] && P olyQ[Px, x] && !IntegerQ[p] && !MonomialQ[Px, x] && !PolyQ[Fx, x]
Time = 4.60 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.93
| method | result | size |
| pseudoelliptic | \(\frac {\left (-x^{5}+1\right ) \sqrt {2}\, \arctan \left (\frac {\sqrt {-x \left (x^{5}+2 x^{3}-1\right )}\, \sqrt {2}}{2 x^{2}}\right )-2 \sqrt {-x \left (x^{5}+2 x^{3}-1\right )}\, x}{2 x^{5}-2}\) | \(65\) |
| trager | \(-\frac {x \sqrt {-x^{6}-2 x^{4}+x}}{x^{5}-1}+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+2\right ) \ln \left (-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+2\right ) x^{5}+4 \operatorname {RootOf}\left (\textit {\_Z}^{2}+2\right ) x^{3}-4 \sqrt {-x^{6}-2 x^{4}+x}\, x -\operatorname {RootOf}\left (\textit {\_Z}^{2}+2\right )}{\left (x -1\right ) \left (x^{4}+x^{3}+x^{2}+x +1\right )}\right )}{4}\) | \(103\) |
| risch | \(\frac {x^{2} \left (x^{5}+2 x^{3}-1\right )}{\left (x^{5}-1\right ) \sqrt {-x \left (x^{5}+2 x^{3}-1\right )}}-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+2\right ) \ln \left (\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+2\right ) x^{5}+4 \operatorname {RootOf}\left (\textit {\_Z}^{2}+2\right ) x^{3}+4 \sqrt {-x^{6}-2 x^{4}+x}\, x -\operatorname {RootOf}\left (\textit {\_Z}^{2}+2\right )}{\left (x -1\right ) \left (x^{4}+x^{3}+x^{2}+x +1\right )}\right )}{4}\) | \(114\) |
((-x^5+1)*2^(1/2)*arctan(1/2*(-x*(x^5+2*x^3-1))^(1/2)/x^2*2^(1/2))-2*(-x*( x^5+2*x^3-1))^(1/2)*x)/(2*x^5-2)
Time = 0.28 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.99 \[ \int \frac {\left (3+2 x^5\right ) \sqrt {x-2 x^4-x^6}}{\left (-1+x^5\right )^2} \, dx=-\frac {\sqrt {2} {\left (x^{5} - 1\right )} \arctan \left (\frac {2 \, \sqrt {2} \sqrt {-x^{6} - 2 \, x^{4} + x} x}{x^{5} + 4 \, x^{3} - 1}\right ) + 4 \, \sqrt {-x^{6} - 2 \, x^{4} + x} x}{4 \, {\left (x^{5} - 1\right )}} \]
-1/4*(sqrt(2)*(x^5 - 1)*arctan(2*sqrt(2)*sqrt(-x^6 - 2*x^4 + x)*x/(x^5 + 4 *x^3 - 1)) + 4*sqrt(-x^6 - 2*x^4 + x)*x)/(x^5 - 1)
Timed out. \[ \int \frac {\left (3+2 x^5\right ) \sqrt {x-2 x^4-x^6}}{\left (-1+x^5\right )^2} \, dx=\text {Timed out} \]
\[ \int \frac {\left (3+2 x^5\right ) \sqrt {x-2 x^4-x^6}}{\left (-1+x^5\right )^2} \, dx=\int { \frac {\sqrt {-x^{6} - 2 \, x^{4} + x} {\left (2 \, x^{5} + 3\right )}}{{\left (x^{5} - 1\right )}^{2}} \,d x } \]
\[ \int \frac {\left (3+2 x^5\right ) \sqrt {x-2 x^4-x^6}}{\left (-1+x^5\right )^2} \, dx=\int { \frac {\sqrt {-x^{6} - 2 \, x^{4} + x} {\left (2 \, x^{5} + 3\right )}}{{\left (x^{5} - 1\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {\left (3+2 x^5\right ) \sqrt {x-2 x^4-x^6}}{\left (-1+x^5\right )^2} \, dx=\int \frac {\left (2\,x^5+3\right )\,\sqrt {-x^6-2\,x^4+x}}{{\left (x^5-1\right )}^2} \,d x \]