Integrand size = 30, antiderivative size = 32 \[ \int \frac {-1-2 x+2 x^2}{\left (-1+3 x+x^2\right ) \sqrt {x+x^4}} \, dx=\sqrt {2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {x+x^4}}{1-x+x^2}\right ) \]
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\[ \int \frac {-1-2 x+2 x^2}{\left (-1+3 x+x^2\right ) \sqrt {x+x^4}} \, dx=\int \frac {-1-2 x+2 x^2}{\left (-1+3 x+x^2\right ) \sqrt {x+x^4}} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \frac {\left (\sqrt {x} \sqrt {1+x^3}\right ) \int \frac {-1-2 x+2 x^2}{\sqrt {x} \left (-1+3 x+x^2\right ) \sqrt {1+x^3}} \, dx}{\sqrt {x+x^4}} \\ & = \frac {\left (\sqrt {x} \sqrt {1+x^3}\right ) \int \left (\frac {2}{\sqrt {x} \sqrt {1+x^3}}+\frac {1-8 x}{\sqrt {x} \left (-1+3 x+x^2\right ) \sqrt {1+x^3}}\right ) \, dx}{\sqrt {x+x^4}} \\ & = \frac {\left (\sqrt {x} \sqrt {1+x^3}\right ) \int \frac {1-8 x}{\sqrt {x} \left (-1+3 x+x^2\right ) \sqrt {1+x^3}} \, dx}{\sqrt {x+x^4}}+\frac {\left (2 \sqrt {x} \sqrt {1+x^3}\right ) \int \frac {1}{\sqrt {x} \sqrt {1+x^3}} \, dx}{\sqrt {x+x^4}} \\ & = \frac {\left (\sqrt {x} \sqrt {1+x^3}\right ) \int \left (\frac {-8+2 \sqrt {13}}{\sqrt {x} \left (3-\sqrt {13}+2 x\right ) \sqrt {1+x^3}}+\frac {-8-2 \sqrt {13}}{\sqrt {x} \left (3+\sqrt {13}+2 x\right ) \sqrt {1+x^3}}\right ) \, dx}{\sqrt {x+x^4}}+\frac {\left (4 \sqrt {x} \sqrt {1+x^3}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1+x^6}} \, dx,x,\sqrt {x}\right )}{\sqrt {x+x^4}} \\ & = \frac {2 x (1+x) \sqrt {\frac {1-x+x^2}{\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 )}{\sqrt [4]{3} \sqrt {\frac {x (1+x)}{\left (1+\left (1+\sqrt {3}\right ) x\right )^2}} \sqrt {x+x^4}}-\frac {\left (2 \left (4-\sqrt {13}\right ) \sqrt {x} \sqrt {1+x^3}\right ) \int \frac {1}{\sqrt {x} \left (3-\sqrt {13}+2 x\right ) \sqrt {1+x^3}} \, dx}{\sqrt {x+x^4}}-\frac {\left (2 \left (4+\sqrt {13}\right ) \sqrt {x} \sqrt {1+x^3}\right ) \int \frac {1}{\sqrt {x} \left (3+\sqrt {13}+2 x\right ) \sqrt {1+x^3}} \, dx}{\sqrt {x+x^4}} \\ & = \frac {2 x (1+x) \sqrt {\frac {1-x+x^2}{\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 )}{\sqrt [4]{3} \sqrt {\frac {x (1+x)}{\left (1+\left (1+\sqrt {3}\right ) x\right )^2}} \sqrt {x+x^4}}-\frac {\left (4 \left (4-\sqrt {13}\right ) \sqrt {x} \sqrt {1+x^3}\right ) \text {Subst}\left (\int \frac {1}{\left (3-\sqrt {13}+2 x^2\right ) \sqrt {1+x^6}} \, dx,x,\sqrt {x}\right )}{\sqrt {x+x^4}}-\frac {\left (4 \left (4+\sqrt {13}\right ) \sqrt {x} \sqrt {1+x^3}\right ) \text {Subst}\left (\int \frac {1}{\left (3+\sqrt {13}+2 x^2\right ) \sqrt {1+x^6}} \, dx,x,\sqrt {x}\right )}{\sqrt {x+x^4}} \\ & = \frac {2 x (1+x) \sqrt {\frac {1-x+x^2}{\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 )}{\sqrt [4]{3} \sqrt {\frac {x (1+x)}{\left (1+\left (1+\sqrt {3}\right ) x\right )^2}} \sqrt {x+x^4}}-\frac {\left (4 \left (4-\sqrt {13}\right ) \sqrt {x} \sqrt {1+x^3}\right ) \text {Subst}\left (\int \left (\frac {\sqrt {-3+\sqrt {13}}}{2 \left (3-\sqrt {13}\right ) \left (\sqrt {-3+\sqrt {13}}-\sqrt {2} x\right ) \sqrt {1+x^6}}+\frac {\sqrt {-3+\sqrt {13}}}{2 \left (3-\sqrt {13}\right ) \left (\sqrt {-3+\sqrt {13}}+\sqrt {2} x\right ) \sqrt {1+x^6}}\right ) \, dx,x,\sqrt {x}\right )}{\sqrt {x+x^4}}-\frac {\left (4 \left (4+\sqrt {13}\right ) \sqrt {x} \sqrt {1+x^3}\right ) \text {Subst}\left (\int \left (\frac {i}{2 \sqrt {3+\sqrt {13}} \left (i \sqrt {3+\sqrt {13}}-\sqrt {2} x\right ) \sqrt {1+x^6}}+\frac {i}{2 \sqrt {3+\sqrt {13}} \left (i \sqrt {3+\sqrt {13}}+\sqrt {2} x\right ) \sqrt {1+x^6}}\right ) \, dx,x,\sqrt {x}\right )}{\sqrt {x+x^4}} \\ & = \frac {2 x (1+x) \sqrt {\frac {1-x+x^2}{\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 )}{\sqrt [4]{3} \sqrt {\frac {x (1+x)}{\left (1+\left (1+\sqrt {3}\right ) x\right )^2}} \sqrt {x+x^4}}+\frac {\left (2 \left (4-\sqrt {13}\right ) \sqrt {x} \sqrt {1+x^3}\right ) \text {Subst}\left (\int \frac {1}{\left (\sqrt {-3+\sqrt {13}}-\sqrt {2} x\right ) \sqrt {1+x^6}} \, dx,x,\sqrt {x}\right )}{\sqrt {-3+\sqrt {13}} \sqrt {x+x^4}}+\frac {\left (2 \left (4-\sqrt {13}\right ) \sqrt {x} \sqrt {1+x^3}\right ) \text {Subst}\left (\int \frac {1}{\left (\sqrt {-3+\sqrt {13}}+\sqrt {2} x\right ) \sqrt {1+x^6}} \, dx,x,\sqrt {x}\right )}{\sqrt {-3+\sqrt {13}} \sqrt {x+x^4}}-\frac {\left (2 i \left (4+\sqrt {13}\right ) \sqrt {x} \sqrt {1+x^3}\right ) \text {Subst}\left (\int \frac {1}{\left (i \sqrt {3+\sqrt {13}}-\sqrt {2} x\right ) \sqrt {1+x^6}} \, dx,x,\sqrt {x}\right )}{\sqrt {3+\sqrt {13}} \sqrt {x+x^4}}-\frac {\left (2 i \left (4+\sqrt {13}\right ) \sqrt {x} \sqrt {1+x^3}\right ) \text {Subst}\left (\int \frac {1}{\left (i \sqrt {3+\sqrt {13}}+\sqrt {2} x\right ) \sqrt {1+x^6}} \, dx,x,\sqrt {x}\right )}{\sqrt {3+\sqrt {13}} \sqrt {x+x^4}} \\ \end{align*}
Time = 24.35 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.72 \[ \int \frac {-1-2 x+2 x^2}{\left (-1+3 x+x^2\right ) \sqrt {x+x^4}} \, dx=\frac {\sqrt {2} \sqrt {1+\frac {1}{x^3}} x^2 \text {arctanh}\left (\frac {\sqrt {2} \sqrt {1+\frac {1}{x^3}}}{1+\frac {1}{x^2}-\frac {1}{x}}\right )}{\sqrt {x+x^4}} \]
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Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 3.84 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.75
method | result | size |
trager | \(\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right ) \ln \left (\frac {3 \operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right ) x^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right ) x +\operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right )+4 \sqrt {x^{4}+x}}{x^{2}+3 x -1}\right )}{2}\) | \(56\) |
default | \(\text {Expression too large to display}\) | \(13798\) |
elliptic | \(\text {Expression too large to display}\) | \(15100\) |
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Leaf count of result is larger than twice the leaf count of optimal. 68 vs. \(2 (26) = 52\).
Time = 0.27 (sec) , antiderivative size = 68, normalized size of antiderivative = 2.12 \[ \int \frac {-1-2 x+2 x^2}{\left (-1+3 x+x^2\right ) \sqrt {x+x^4}} \, dx=\frac {1}{4} \, \sqrt {2} \log \left (-\frac {17 \, x^{4} + 6 \, x^{3} + 4 \, \sqrt {2} \sqrt {x^{4} + x} {\left (3 \, x^{2} + x + 1\right )} + 7 \, x^{2} + 10 \, x + 1}{x^{4} + 6 \, x^{3} + 7 \, x^{2} - 6 \, x + 1}\right ) \]
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\[ \int \frac {-1-2 x+2 x^2}{\left (-1+3 x+x^2\right ) \sqrt {x+x^4}} \, dx=\int \frac {2 x^{2} - 2 x - 1}{\sqrt {x \left (x + 1\right ) \left (x^{2} - x + 1\right )} \left (x^{2} + 3 x - 1\right )}\, dx \]
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\[ \int \frac {-1-2 x+2 x^2}{\left (-1+3 x+x^2\right ) \sqrt {x+x^4}} \, dx=\int { \frac {2 \, x^{2} - 2 \, x - 1}{\sqrt {x^{4} + x} {\left (x^{2} + 3 \, x - 1\right )}} \,d x } \]
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\[ \int \frac {-1-2 x+2 x^2}{\left (-1+3 x+x^2\right ) \sqrt {x+x^4}} \, dx=\int { \frac {2 \, x^{2} - 2 \, x - 1}{\sqrt {x^{4} + x} {\left (x^{2} + 3 \, x - 1\right )}} \,d x } \]
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Timed out. \[ \int \frac {-1-2 x+2 x^2}{\left (-1+3 x+x^2\right ) \sqrt {x+x^4}} \, dx=\int -\frac {-2\,x^2+2\,x+1}{\sqrt {x^4+x}\,\left (x^2+3\,x-1\right )} \,d x \]
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