Integrand size = 34, antiderivative size = 81 \[ \int \frac {(-3+2 x) \sqrt {-2 x+2 x^2+3 x^4}}{\left (-2+2 x+x^3\right )^2} \, dx=\frac {x \sqrt {-2 x+2 x^2+3 x^4}}{2 \left (-2+2 x+x^3\right )}+\frac {\text {arctanh}\left (\frac {\sqrt {2} x \sqrt {-2 x+2 x^2+3 x^4}}{-2+2 x+3 x^3}\right )}{2 \sqrt {2}} \]
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\[ \int \frac {(-3+2 x) \sqrt {-2 x+2 x^2+3 x^4}}{\left (-2+2 x+x^3\right )^2} \, dx=\int \frac {(-3+2 x) \sqrt {-2 x+2 x^2+3 x^4}}{\left (-2+2 x+x^3\right )^2} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt {-2 x+2 x^2+3 x^4} \int \frac {\sqrt {x} (-3+2 x) \sqrt {-2+2 x+3 x^3}}{\left (-2+2 x+x^3\right )^2} \, dx}{\sqrt {x} \sqrt {-2+2 x+3 x^3}} \\ & = \frac {\left (2 \sqrt {-2 x+2 x^2+3 x^4}\right ) \text {Subst}\left (\int \frac {x^2 \left (-3+2 x^2\right ) \sqrt {-2+2 x^2+3 x^6}}{\left (-2+2 x^2+x^6\right )^2} \, dx,x,\sqrt {x}\right )}{\sqrt {x} \sqrt {-2+2 x+3 x^3}} \\ & = \frac {\left (2 \sqrt {-2 x+2 x^2+3 x^4}\right ) \text {Subst}\left (\int \left (-\frac {3 x^2 \sqrt {-2+2 x^2+3 x^6}}{\left (-2+2 x^2+x^6\right )^2}+\frac {2 x^4 \sqrt {-2+2 x^2+3 x^6}}{\left (-2+2 x^2+x^6\right )^2}\right ) \, dx,x,\sqrt {x}\right )}{\sqrt {x} \sqrt {-2+2 x+3 x^3}} \\ & = \frac {\left (4 \sqrt {-2 x+2 x^2+3 x^4}\right ) \text {Subst}\left (\int \frac {x^4 \sqrt {-2+2 x^2+3 x^6}}{\left (-2+2 x^2+x^6\right )^2} \, dx,x,\sqrt {x}\right )}{\sqrt {x} \sqrt {-2+2 x+3 x^3}}-\frac {\left (6 \sqrt {-2 x+2 x^2+3 x^4}\right ) \text {Subst}\left (\int \frac {x^2 \sqrt {-2+2 x^2+3 x^6}}{\left (-2+2 x^2+x^6\right )^2} \, dx,x,\sqrt {x}\right )}{\sqrt {x} \sqrt {-2+2 x+3 x^3}} \\ \end{align*}
Time = 3.36 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.98 \[ \int \frac {(-3+2 x) \sqrt {-2 x+2 x^2+3 x^4}}{\left (-2+2 x+x^3\right )^2} \, dx=\frac {\sqrt {x \left (-2+2 x+3 x^3\right )} \left (\frac {2 x^{3/2}}{-2+2 x+x^3}+\frac {\text {arctanh}\left (\frac {x^{3/2}}{\sqrt {-1+x+\frac {3 x^3}{2}}}\right )}{\sqrt {-1+x+\frac {3 x^3}{2}}}\right )}{4 \sqrt {x}} \]
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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 | \(\text {Expression too large to display}\) | \(1689\) |
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Time = 0.30 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.63 \[ \int \frac {(-3+2 x) \sqrt {-2 x+2 x^2+3 x^4}}{\left (-2+2 x+x^3\right )^2} \, dx=\frac {\sqrt {2} {\left (x^{3} + 2 \, x - 2\right )} \log \left (-\frac {49 \, x^{6} + 36 \, x^{4} - 36 \, x^{3} + 4 \, \sqrt {2} {\left (5 \, x^{4} + 2 \, x^{2} - 2 \, x\right )} \sqrt {3 \, x^{4} + 2 \, x^{2} - 2 \, x} + 4 \, x^{2} - 8 \, x + 4}{x^{6} + 4 \, x^{4} - 4 \, x^{3} + 4 \, x^{2} - 8 \, x + 4}\right ) + 8 \, \sqrt {3 \, x^{4} + 2 \, x^{2} - 2 \, x} x}{16 \, {\left (x^{3} + 2 \, x - 2\right )}} \]
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\[ \int \frac {(-3+2 x) \sqrt {-2 x+2 x^2+3 x^4}}{\left (-2+2 x+x^3\right )^2} \, dx=\int \frac {\sqrt {x \left (3 x^{3} + 2 x - 2\right )} \left (2 x - 3\right )}{\left (x^{3} + 2 x - 2\right )^{2}}\, dx \]
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\[ \int \frac {(-3+2 x) \sqrt {-2 x+2 x^2+3 x^4}}{\left (-2+2 x+x^3\right )^2} \, dx=\int { \frac {\sqrt {3 \, x^{4} + 2 \, x^{2} - 2 \, x} {\left (2 \, x - 3\right )}}{{\left (x^{3} + 2 \, x - 2\right )}^{2}} \,d x } \]
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\[ \int \frac {(-3+2 x) \sqrt {-2 x+2 x^2+3 x^4}}{\left (-2+2 x+x^3\right )^2} \, dx=\int { \frac {\sqrt {3 \, x^{4} + 2 \, x^{2} - 2 \, x} {\left (2 \, x - 3\right )}}{{\left (x^{3} + 2 \, x - 2\right )}^{2}} \,d x } \]
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Timed out. \[ \int \frac {(-3+2 x) \sqrt {-2 x+2 x^2+3 x^4}}{\left (-2+2 x+x^3\right )^2} \, dx=\int \frac {\left (2\,x-3\right )\,\sqrt {3\,x^4+2\,x^2-2\,x}}{{\left (x^3+2\,x-2\right )}^2} \,d x \]
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