Integrand size = 39, antiderivative size = 87 \[ \int \frac {(3+4 x) \sqrt {x+2 x^2-2 x^4}}{(1+2 x) \left (1+2 x+x^3\right )} \, dx=2 \sqrt {2} \arctan \left (\frac {\sqrt {2} x \sqrt {x+2 x^2-2 x^4}}{-1-2 x+2 x^3}\right )-2 \sqrt {3} \arctan \left (\frac {\sqrt {3} x \sqrt {x+2 x^2-2 x^4}}{-1-2 x+2 x^3}\right ) \]
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\[ \int \frac {(3+4 x) \sqrt {x+2 x^2-2 x^4}}{(1+2 x) \left (1+2 x+x^3\right )} \, dx=\int \frac {(3+4 x) \sqrt {x+2 x^2-2 x^4}}{(1+2 x) \left (1+2 x+x^3\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt {x+2 x^2-2 x^4} \int \frac {\sqrt {x} (3+4 x) \sqrt {1+2 x-2 x^3}}{(1+2 x) \left (1+2 x+x^3\right )} \, dx}{\sqrt {x} \sqrt {1+2 x-2 x^3}} \\ & = \frac {\left (2 \sqrt {x+2 x^2-2 x^4}\right ) \text {Subst}\left (\int \frac {x^2 \left (3+4 x^2\right ) \sqrt {1+2 x^2-2 x^6}}{\left (1+2 x^2\right ) \left (1+2 x^2+x^6\right )} \, dx,x,\sqrt {x}\right )}{\sqrt {x} \sqrt {1+2 x-2 x^3}} \\ & = \frac {\left (2 \sqrt {x+2 x^2-2 x^4}\right ) \text {Subst}\left (\int \left (\frac {4 \sqrt {1+2 x^2-2 x^6}}{1+2 x^2}+\frac {\left (-4+3 x^2-2 x^4\right ) \sqrt {1+2 x^2-2 x^6}}{1+2 x^2+x^6}\right ) \, dx,x,\sqrt {x}\right )}{\sqrt {x} \sqrt {1+2 x-2 x^3}} \\ & = \frac {\left (2 \sqrt {x+2 x^2-2 x^4}\right ) \text {Subst}\left (\int \frac {\left (-4+3 x^2-2 x^4\right ) \sqrt {1+2 x^2-2 x^6}}{1+2 x^2+x^6} \, dx,x,\sqrt {x}\right )}{\sqrt {x} \sqrt {1+2 x-2 x^3}}+\frac {\left (8 \sqrt {x+2 x^2-2 x^4}\right ) \text {Subst}\left (\int \frac {\sqrt {1+2 x^2-2 x^6}}{1+2 x^2} \, dx,x,\sqrt {x}\right )}{\sqrt {x} \sqrt {1+2 x-2 x^3}} \\ & = \frac {\left (2 \sqrt {x+2 x^2-2 x^4}\right ) \text {Subst}\left (\int \left (-\frac {4 \sqrt {1+2 x^2-2 x^6}}{1+2 x^2+x^6}+\frac {3 x^2 \sqrt {1+2 x^2-2 x^6}}{1+2 x^2+x^6}-\frac {2 x^4 \sqrt {1+2 x^2-2 x^6}}{1+2 x^2+x^6}\right ) \, dx,x,\sqrt {x}\right )}{\sqrt {x} \sqrt {1+2 x-2 x^3}}+\frac {\left (8 \sqrt {x+2 x^2-2 x^4}\right ) \text {Subst}\left (\int \left (\frac {i \sqrt {1+2 x^2-2 x^6}}{2 \left (i-\sqrt {2} x\right )}+\frac {i \sqrt {1+2 x^2-2 x^6}}{2 \left (i+\sqrt {2} x\right )}\right ) \, dx,x,\sqrt {x}\right )}{\sqrt {x} \sqrt {1+2 x-2 x^3}} \\ & = \frac {\left (4 i \sqrt {x+2 x^2-2 x^4}\right ) \text {Subst}\left (\int \frac {\sqrt {1+2 x^2-2 x^6}}{i-\sqrt {2} x} \, dx,x,\sqrt {x}\right )}{\sqrt {x} \sqrt {1+2 x-2 x^3}}+\frac {\left (4 i \sqrt {x+2 x^2-2 x^4}\right ) \text {Subst}\left (\int \frac {\sqrt {1+2 x^2-2 x^6}}{i+\sqrt {2} x} \, dx,x,\sqrt {x}\right )}{\sqrt {x} \sqrt {1+2 x-2 x^3}}-\frac {\left (4 \sqrt {x+2 x^2-2 x^4}\right ) \text {Subst}\left (\int \frac {x^4 \sqrt {1+2 x^2-2 x^6}}{1+2 x^2+x^6} \, dx,x,\sqrt {x}\right )}{\sqrt {x} \sqrt {1+2 x-2 x^3}}+\frac {\left (6 \sqrt {x+2 x^2-2 x^4}\right ) \text {Subst}\left (\int \frac {x^2 \sqrt {1+2 x^2-2 x^6}}{1+2 x^2+x^6} \, dx,x,\sqrt {x}\right )}{\sqrt {x} \sqrt {1+2 x-2 x^3}}-\frac {\left (8 \sqrt {x+2 x^2-2 x^4}\right ) \text {Subst}\left (\int \frac {\sqrt {1+2 x^2-2 x^6}}{1+2 x^2+x^6} \, dx,x,\sqrt {x}\right )}{\sqrt {x} \sqrt {1+2 x-2 x^3}} \\ \end{align*}
Time = 2.18 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.13 \[ \int \frac {(3+4 x) \sqrt {x+2 x^2-2 x^4}}{(1+2 x) \left (1+2 x+x^3\right )} \, dx=\frac {2 \sqrt {x+2 x^2-2 x^4} \left (\sqrt {2} \text {arctanh}\left (\frac {x^{3/2}}{\sqrt {-\frac {1}{2}-x+x^3}}\right )-\sqrt {3} \text {arctanh}\left (\frac {\sqrt {3} x^{3/2}}{\sqrt {-1-2 x+2 x^3}}\right )\right )}{\sqrt {x} \sqrt {-1-2 x+2 x^3}} \]
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Time = 19.15 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.67
method | result | size |
default | \(2 \sqrt {2}\, \arctan \left (\frac {\sqrt {-2 x^{4}+2 x^{2}+x}\, \sqrt {2}}{2 x^{2}}\right )-2 \sqrt {3}\, \arctan \left (\frac {\sqrt {-2 x^{4}+2 x^{2}+x}\, \sqrt {3}}{3 x^{2}}\right )\) | \(58\) |
pseudoelliptic | \(2 \sqrt {2}\, \arctan \left (\frac {\sqrt {-2 x^{4}+2 x^{2}+x}\, \sqrt {2}}{2 x^{2}}\right )-2 \sqrt {3}\, \arctan \left (\frac {\sqrt {-2 x^{4}+2 x^{2}+x}\, \sqrt {3}}{3 x^{2}}\right )\) | \(58\) |
trager | \(\operatorname {RootOf}\left (\textit {\_Z}^{2}+2\right ) \ln \left (-\frac {4 \operatorname {RootOf}\left (\textit {\_Z}^{2}+2\right ) x^{3}+4 \sqrt {-2 x^{4}+2 x^{2}+x}\, x -2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+2\right ) x -\operatorname {RootOf}\left (\textit {\_Z}^{2}+2\right )}{2 x +1}\right )+\operatorname {RootOf}\left (\textit {\_Z}^{2}+3\right ) \ln \left (-\frac {-5 \operatorname {RootOf}\left (\textit {\_Z}^{2}+3\right ) x^{3}+6 \sqrt {-2 x^{4}+2 x^{2}+x}\, x +2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+3\right ) x +\operatorname {RootOf}\left (\textit {\_Z}^{2}+3\right )}{x^{3}+2 x +1}\right )\) | \(129\) |
elliptic | \(\text {Expression too large to display}\) | \(413672\) |
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Leaf count of result is larger than twice the leaf count of optimal. 168 vs. \(2 (75) = 150\).
Time = 0.35 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.93 \[ \int \frac {(3+4 x) \sqrt {x+2 x^2-2 x^4}}{(1+2 x) \left (1+2 x+x^3\right )} \, dx=\frac {2}{5} \, \sqrt {2} \arctan \left (\frac {2 \, \sqrt {2} \sqrt {-2 \, x^{4} + 2 \, x^{2} + x} {\left (4 \, x^{3} - 4 \, x^{2} - x + 1\right )}}{16 \, x^{5} - 16 \, x^{4} - 12 \, x^{3} + 8 \, x^{2} + 4 \, x - 1}\right ) - \frac {1}{5} \, \sqrt {2} \arctan \left (\frac {2 \, \sqrt {2} \sqrt {-2 \, x^{4} + 2 \, x^{2} + x} {\left (4 \, x^{2} + 5 \, x + 2\right )}}{32 \, x^{5} + 80 \, x^{4} + 84 \, x^{3} + 40 \, x^{2} + 6 \, x - 1}\right ) - \sqrt {3} \arctan \left (\frac {2 \, \sqrt {3} \sqrt {-2 \, x^{4} + 2 \, x^{2} + x} x}{5 \, x^{3} - 2 \, x - 1}\right ) \]
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Timed out. \[ \int \frac {(3+4 x) \sqrt {x+2 x^2-2 x^4}}{(1+2 x) \left (1+2 x+x^3\right )} \, dx=\text {Timed out} \]
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\[ \int \frac {(3+4 x) \sqrt {x+2 x^2-2 x^4}}{(1+2 x) \left (1+2 x+x^3\right )} \, dx=\int { \frac {\sqrt {-2 \, x^{4} + 2 \, x^{2} + x} {\left (4 \, x + 3\right )}}{{\left (x^{3} + 2 \, x + 1\right )} {\left (2 \, x + 1\right )}} \,d x } \]
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\[ \int \frac {(3+4 x) \sqrt {x+2 x^2-2 x^4}}{(1+2 x) \left (1+2 x+x^3\right )} \, dx=\int { \frac {\sqrt {-2 \, x^{4} + 2 \, x^{2} + x} {\left (4 \, x + 3\right )}}{{\left (x^{3} + 2 \, x + 1\right )} {\left (2 \, x + 1\right )}} \,d x } \]
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Timed out. \[ \int \frac {(3+4 x) \sqrt {x+2 x^2-2 x^4}}{(1+2 x) \left (1+2 x+x^3\right )} \, dx=\int \frac {\left (4\,x+3\right )\,\sqrt {-2\,x^4+2\,x^2+x}}{\left (2\,x+1\right )\,\left (x^3+2\,x+1\right )} \,d x \]
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