Integrand size = 34, antiderivative size = 32 \[ \int \frac {-1+2 x+2 x^2}{\left (1-x+3 x^2\right ) \sqrt {-x+x^4}} \, dx=\sqrt {2} \arctan \left (\frac {\sqrt {2} \sqrt {-x+x^4}}{1+x+x^2}\right ) \]
Time = 24.43 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.75 \[ \int \frac {-1+2 x+2 x^2}{\left (1-x+3 x^2\right ) \sqrt {-x+x^4}} \, dx=-\frac {\sqrt {2} \sqrt {-1+\frac {1}{x^3}} x^2 \text {arctanh}\left (\frac {\sqrt {2} \sqrt {-1+\frac {1}{x^3}}}{1+\frac {1}{x^2}+\frac {1}{x}}\right )}{\sqrt {x \left (-1+x^3\right )}} \]
-((Sqrt[2]*Sqrt[-1 + x^(-3)]*x^2*ArcTanh[(Sqrt[2]*Sqrt[-1 + x^(-3)])/(1 + x^(-2) + x^(-1))])/Sqrt[x*(-1 + x^3)])
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.
| \(\displaystyle \int \frac {2 x^2+2 x-1}{\left (3 x^2-x+1\right ) \sqrt {x^4-x}} \, dx\) |
| \(\Big \downarrow \) 2467 |
| \(\displaystyle \frac {\sqrt {x} \sqrt {x^3-1} \int -\frac {-2 x^2-2 x+1}{\sqrt {x} \left (3 x^2-x+1\right ) \sqrt {x^3-1}}dx}{\sqrt {x^4-x}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {\sqrt {x} \sqrt {x^3-1} \int \frac {-2 x^2-2 x+1}{\sqrt {x} \left (3 x^2-x+1\right ) \sqrt {x^3-1}}dx}{\sqrt {x^4-x}}\) |
| \(\Big \downarrow \) 2035 |
| \(\displaystyle -\frac {2 \sqrt {x} \sqrt {x^3-1} \int \frac {-2 x^2-2 x+1}{\left (3 x^2-x+1\right ) \sqrt {x^3-1}}d\sqrt {x}}{\sqrt {x^4-x}}\) |
| \(\Big \downarrow \) 7279 |
| \(\displaystyle -\frac {2 \sqrt {x} \sqrt {x^3-1} \int \left (\frac {5-8 x}{3 \left (3 x^2-x+1\right ) \sqrt {x^3-1}}-\frac {2}{3 \sqrt {x^3-1}}\right )d\sqrt {x}}{\sqrt {x^4-x}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {2 \sqrt {x} \sqrt {x^3-1} \left (\frac {\left (4-i \sqrt {11}\right ) \int \frac {1}{\left (\sqrt {1-i \sqrt {11}}-\sqrt {6} \sqrt {x}\right ) \sqrt {x^3-1}}d\sqrt {x}}{3 \sqrt {1-i \sqrt {11}}}+\frac {\left (4+i \sqrt {11}\right ) \int \frac {1}{\left (\sqrt {1+i \sqrt {11}}-\sqrt {6} \sqrt {x}\right ) \sqrt {x^3-1}}d\sqrt {x}}{3 \sqrt {1+i \sqrt {11}}}+\frac {\left (4-i \sqrt {11}\right ) \int \frac {1}{\left (\sqrt {6} \sqrt {x}+\sqrt {1-i \sqrt {11}}\right ) \sqrt {x^3-1}}d\sqrt {x}}{3 \sqrt {1-i \sqrt {11}}}+\frac {\left (4+i \sqrt {11}\right ) \int \frac {1}{\left (\sqrt {6} \sqrt {x}+\sqrt {1+i \sqrt {11}}\right ) \sqrt {x^3-1}}d\sqrt {x}}{3 \sqrt {1+i \sqrt {11}}}-\frac {(1-x) \sqrt {x} \sqrt {\frac {x^2+x+1}{\left (1-\left (1+\sqrt {3}\right ) x\right )^2}} \operatorname {EllipticF}\left (\arccos \left (\frac {1-\left (1-\sqrt {3}\right ) x}{1-\left (1+\sqrt {3}\right ) x}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{3 \sqrt [4]{3} \sqrt {-\frac {(1-x) x}{\left (1-\left (1+\sqrt {3}\right ) x\right )^2}} \sqrt {x^3-1}}\right )}{\sqrt {x^4-x}}\) |
3.4.88.3.1 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]
Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[ {v = RationalFunctionExpand[u/(a + b*x^n + c*x^(2*n)), x]}, Int[v, x] /; Su mQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 5.26 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.97
| method | result | size |
| trager | \(\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+2\right ) \ln \left (-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+2\right ) x^{2}-3 \operatorname {RootOf}\left (\textit {\_Z}^{2}+2\right ) x -\operatorname {RootOf}\left (\textit {\_Z}^{2}+2\right )-4 \sqrt {x^{4}-x}}{3 x^{2}-x +1}\right )}{2}\) | \(63\) |
| default | \(\text {Expression too large to display}\) | \(828\) |
| elliptic | \(\text {Expression too large to display}\) | \(828\) |
1/2*RootOf(_Z^2+2)*ln(-(RootOf(_Z^2+2)*x^2-3*RootOf(_Z^2+2)*x-RootOf(_Z^2+ 2)-4*(x^4-x)^(1/2))/(3*x^2-x+1))
Time = 0.28 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.88 \[ \int \frac {-1+2 x+2 x^2}{\left (1-x+3 x^2\right ) \sqrt {-x+x^4}} \, dx=\frac {1}{2} \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (x^{2} - 3 \, x - 1\right )}}{4 \, \sqrt {x^{4} - x}}\right ) \]
\[ \int \frac {-1+2 x+2 x^2}{\left (1-x+3 x^2\right ) \sqrt {-x+x^4}} \, dx=\int \frac {2 x^{2} + 2 x - 1}{\sqrt {x \left (x - 1\right ) \left (x^{2} + x + 1\right )} \left (3 x^{2} - x + 1\right )}\, dx \]
\[ \int \frac {-1+2 x+2 x^2}{\left (1-x+3 x^2\right ) \sqrt {-x+x^4}} \, dx=\int { \frac {2 \, x^{2} + 2 \, x - 1}{\sqrt {x^{4} - x} {\left (3 \, x^{2} - x + 1\right )}} \,d x } \]
\[ \int \frac {-1+2 x+2 x^2}{\left (1-x+3 x^2\right ) \sqrt {-x+x^4}} \, dx=\int { \frac {2 \, x^{2} + 2 \, x - 1}{\sqrt {x^{4} - x} {\left (3 \, x^{2} - x + 1\right )}} \,d x } \]
Timed out. \[ \int \frac {-1+2 x+2 x^2}{\left (1-x+3 x^2\right ) \sqrt {-x+x^4}} \, dx=\int \frac {2\,x^2+2\,x-1}{\sqrt {x^4-x}\,\left (3\,x^2-x+1\right )} \,d x \]