Integrand size = 34, antiderivative size = 81 \[ \int \frac {(-3+2 x) \sqrt {-2 x+2 x^2+3 x^4}}{\left (-2+2 x+x^3\right )^2} \, dx=\frac {x \sqrt {-2 x+2 x^2+3 x^4}}{2 \left (-2+2 x+x^3\right )}+\frac {\text {arctanh}\left (\frac {\sqrt {2} x \sqrt {-2 x+2 x^2+3 x^4}}{-2+2 x+3 x^3}\right )}{2 \sqrt {2}} \]
x*(3*x^4+2*x^2-2*x)^(1/2)/(2*x^3+4*x-4)+1/4*arctanh(2^(1/2)*x*(3*x^4+2*x^2 -2*x)^(1/2)/(3*x^3+2*x-2))*2^(1/2)
Time = 3.36 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.98 \[ \int \frac {(-3+2 x) \sqrt {-2 x+2 x^2+3 x^4}}{\left (-2+2 x+x^3\right )^2} \, dx=\frac {\sqrt {x \left (-2+2 x+3 x^3\right )} \left (\frac {2 x^{3/2}}{-2+2 x+x^3}+\frac {\text {arctanh}\left (\frac {x^{3/2}}{\sqrt {-1+x+\frac {3 x^3}{2}}}\right )}{\sqrt {-1+x+\frac {3 x^3}{2}}}\right )}{4 \sqrt {x}} \]
(Sqrt[x*(-2 + 2*x + 3*x^3)]*((2*x^(3/2))/(-2 + 2*x + x^3) + ArcTanh[x^(3/2 )/Sqrt[-1 + x + (3*x^3)/2]]/Sqrt[-1 + x + (3*x^3)/2]))/(4*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.
\(\displaystyle \int \frac {(2 x-3) \sqrt {3 x^4+2 x^2-2 x}}{\left (x^3+2 x-2\right )^2} \, dx\) |
\(\Big \downarrow \) 2467 |
\(\displaystyle \frac {\sqrt {3 x^4+2 x^2-2 x} \int -\frac {(3-2 x) \sqrt {x} \sqrt {3 x^3+2 x-2}}{\left (-x^3-2 x+2\right )^2}dx}{\sqrt {x} \sqrt {3 x^3+2 x-2}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\sqrt {3 x^4+2 x^2-2 x} \int \frac {(3-2 x) \sqrt {x} \sqrt {3 x^3+2 x-2}}{\left (-x^3-2 x+2\right )^2}dx}{\sqrt {x} \sqrt {3 x^3+2 x-2}}\) |
\(\Big \downarrow \) 2035 |
\(\displaystyle -\frac {2 \sqrt {3 x^4+2 x^2-2 x} \int \frac {(3-2 x) x \sqrt {3 x^3+2 x-2}}{\left (-x^3-2 x+2\right )^2}d\sqrt {x}}{\sqrt {x} \sqrt {3 x^3+2 x-2}}\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {2 \sqrt {3 x^4+2 x^2-2 x} \int \left (\frac {3 x \sqrt {3 x^3+2 x-2}}{\left (x^3+2 x-2\right )^2}-\frac {2 x^2 \sqrt {3 x^3+2 x-2}}{\left (x^3+2 x-2\right )^2}\right )d\sqrt {x}}{\sqrt {x} \sqrt {3 x^3+2 x-2}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {2 \sqrt {3 x^4+2 x^2-2 x} \left (3 \int \frac {x \sqrt {3 x^3+2 x-2}}{\left (x^3+2 x-2\right )^2}d\sqrt {x}-2 \int \frac {x^2 \sqrt {3 x^3+2 x-2}}{\left (x^3+2 x-2\right )^2}d\sqrt {x}\right )}{\sqrt {x} \sqrt {3 x^3+2 x-2}}\) |
3.11.83.3.1 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 = 4.85 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.88
method | result | size |
default | \(\frac {\sqrt {2}\, \left (x^{3}+2 x -2\right ) \operatorname {arctanh}\left (\frac {\sqrt {3 x^{4}+2 x^{2}-2 x}\, \sqrt {2}}{2 x^{2}}\right )+2 x \sqrt {3 x^{4}+2 x^{2}-2 x}}{4 x^{3}+8 x -8}\) | \(71\) |
risch | \(\frac {x^{2} \left (3 x^{3}+2 x -2\right )}{2 \left (x^{3}+2 x -2\right ) \sqrt {x \left (3 x^{3}+2 x -2\right )}}+\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {3 x^{4}+2 x^{2}-2 x}\, \sqrt {2}}{2 x^{2}}\right )}{4}\) | \(71\) |
pseudoelliptic | \(\frac {\sqrt {2}\, \left (x^{3}+2 x -2\right ) \operatorname {arctanh}\left (\frac {\sqrt {3 x^{4}+2 x^{2}-2 x}\, \sqrt {2}}{2 x^{2}}\right )+2 x \sqrt {3 x^{4}+2 x^{2}-2 x}}{4 x^{3}+8 x -8}\) | \(71\) |
trager | \(\frac {x \sqrt {3 x^{4}+2 x^{2}-2 x}}{2 x^{3}+4 x -4}+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right ) \ln \left (-\frac {5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right ) x^{3}+2 \operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right ) x +4 x \sqrt {3 x^{4}+2 x^{2}-2 x}-2 \operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right )}{x^{3}+2 x -2}\right )}{8}\) | \(100\) |
elliptic | \(\text {Expression too large to display}\) | \(1689\) |
(2^(1/2)*(x^3+2*x-2)*arctanh(1/2*(3*x^4+2*x^2-2*x)^(1/2)/x^2*2^(1/2))+2*x* (3*x^4+2*x^2-2*x)^(1/2))/(4*x^3+8*x-8)
Time = 0.30 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.63 \[ \int \frac {(-3+2 x) \sqrt {-2 x+2 x^2+3 x^4}}{\left (-2+2 x+x^3\right )^2} \, dx=\frac {\sqrt {2} {\left (x^{3} + 2 \, x - 2\right )} \log \left (-\frac {49 \, x^{6} + 36 \, x^{4} - 36 \, x^{3} + 4 \, \sqrt {2} {\left (5 \, x^{4} + 2 \, x^{2} - 2 \, x\right )} \sqrt {3 \, x^{4} + 2 \, x^{2} - 2 \, x} + 4 \, x^{2} - 8 \, x + 4}{x^{6} + 4 \, x^{4} - 4 \, x^{3} + 4 \, x^{2} - 8 \, x + 4}\right ) + 8 \, \sqrt {3 \, x^{4} + 2 \, x^{2} - 2 \, x} x}{16 \, {\left (x^{3} + 2 \, x - 2\right )}} \]
1/16*(sqrt(2)*(x^3 + 2*x - 2)*log(-(49*x^6 + 36*x^4 - 36*x^3 + 4*sqrt(2)*( 5*x^4 + 2*x^2 - 2*x)*sqrt(3*x^4 + 2*x^2 - 2*x) + 4*x^2 - 8*x + 4)/(x^6 + 4 *x^4 - 4*x^3 + 4*x^2 - 8*x + 4)) + 8*sqrt(3*x^4 + 2*x^2 - 2*x)*x)/(x^3 + 2 *x - 2)
\[ \int \frac {(-3+2 x) \sqrt {-2 x+2 x^2+3 x^4}}{\left (-2+2 x+x^3\right )^2} \, dx=\int \frac {\sqrt {x \left (3 x^{3} + 2 x - 2\right )} \left (2 x - 3\right )}{\left (x^{3} + 2 x - 2\right )^{2}}\, dx \]
\[ \int \frac {(-3+2 x) \sqrt {-2 x+2 x^2+3 x^4}}{\left (-2+2 x+x^3\right )^2} \, dx=\int { \frac {\sqrt {3 \, x^{4} + 2 \, x^{2} - 2 \, x} {\left (2 \, x - 3\right )}}{{\left (x^{3} + 2 \, x - 2\right )}^{2}} \,d x } \]
\[ \int \frac {(-3+2 x) \sqrt {-2 x+2 x^2+3 x^4}}{\left (-2+2 x+x^3\right )^2} \, dx=\int { \frac {\sqrt {3 \, x^{4} + 2 \, x^{2} - 2 \, x} {\left (2 \, x - 3\right )}}{{\left (x^{3} + 2 \, x - 2\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {(-3+2 x) \sqrt {-2 x+2 x^2+3 x^4}}{\left (-2+2 x+x^3\right )^2} \, dx=\int \frac {\left (2\,x-3\right )\,\sqrt {3\,x^4+2\,x^2-2\,x}}{{\left (x^3+2\,x-2\right )}^2} \,d x \]