Integrand size = 20, antiderivative size = 82 \[ \int \frac {1}{\sqrt {5+2 x+x^2} \left (-8+x^3\right )} \, dx=-\frac {\arctan \left (\frac {1+x}{\sqrt {3} \sqrt {5+2 x+x^2}}\right )}{4 \sqrt {3}}-\frac {\text {arctanh}\left (\frac {7+3 x}{\sqrt {13} \sqrt {5+2 x+x^2}}\right )}{12 \sqrt {13}}+\frac {1}{12} \text {arctanh}\left (\sqrt {5+2 x+x^2}\right ) \]
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Time = 0.10 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {2099, 738, 212, 1039, 996, 210, 1038} \[ \int \frac {1}{\sqrt {5+2 x+x^2} \left (-8+x^3\right )} \, dx=-\frac {\arctan \left (\frac {x+1}{\sqrt {3} \sqrt {x^2+2 x+5}}\right )}{4 \sqrt {3}}-\frac {\text {arctanh}\left (\frac {3 x+7}{\sqrt {13} \sqrt {x^2+2 x+5}}\right )}{12 \sqrt {13}}+\frac {1}{12} \text {arctanh}\left (\sqrt {x^2+2 x+5}\right ) \]
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Rule 210
Rule 212
Rule 738
Rule 996
Rule 1038
Rule 1039
Rule 2099
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {1}{12 (-2+x) \sqrt {5+2 x+x^2}}+\frac {-4-x}{12 \left (4+2 x+x^2\right ) \sqrt {5+2 x+x^2}}\right ) \, dx \\ & = \frac {1}{12} \int \frac {1}{(-2+x) \sqrt {5+2 x+x^2}} \, dx+\frac {1}{12} \int \frac {-4-x}{\left (4+2 x+x^2\right ) \sqrt {5+2 x+x^2}} \, dx \\ & = -\left (\frac {1}{24} \int \frac {2+2 x}{\left (4+2 x+x^2\right ) \sqrt {5+2 x+x^2}} \, dx\right )-\frac {1}{6} \text {Subst}\left (\int \frac {1}{52-x^2} \, dx,x,\frac {14+6 x}{\sqrt {5+2 x+x^2}}\right )-\frac {1}{4} \int \frac {1}{\left (4+2 x+x^2\right ) \sqrt {5+2 x+x^2}} \, dx \\ & = -\frac {\text {arctanh}\left (\frac {7+3 x}{\sqrt {13} \sqrt {5+2 x+x^2}}\right )}{12 \sqrt {13}}+\frac {1}{6} \text {Subst}\left (\int \frac {1}{2-2 x^2} \, dx,x,\sqrt {5+2 x+x^2}\right )+\text {Subst}\left (\int \frac {1}{-24-2 x^2} \, dx,x,\frac {2+2 x}{\sqrt {5+2 x+x^2}}\right ) \\ & = -\frac {\arctan \left (\frac {2+2 x}{2 \sqrt {3} \sqrt {5+2 x+x^2}}\right )}{4 \sqrt {3}}-\frac {\text {arctanh}\left (\frac {7+3 x}{\sqrt {13} \sqrt {5+2 x+x^2}}\right )}{12 \sqrt {13}}+\frac {1}{12} \text {arctanh}\left (\sqrt {5+2 x+x^2}\right ) \\ \end{align*}
Time = 0.37 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.10 \[ \int \frac {1}{\sqrt {5+2 x+x^2} \left (-8+x^3\right )} \, dx=\frac {1}{156} \left (13 \sqrt {3} \arctan \left (\frac {4+2 x+x^2-(1+x) \sqrt {5+2 x+x^2}}{\sqrt {3}}\right )+13 \text {arctanh}\left (\sqrt {5+2 x+x^2}\right )-2 \sqrt {13} \text {arctanh}\left (\frac {2-x+\sqrt {5+2 x+x^2}}{\sqrt {13}}\right )\right ) \]
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Time = 0.78 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.84
method | result | size |
default | \(-\frac {\sqrt {13}\, \operatorname {arctanh}\left (\frac {\left (14+6 x \right ) \sqrt {13}}{26 \sqrt {\left (-2+x \right )^{2}+1+6 x}}\right )}{156}+\frac {\operatorname {arctanh}\left (\sqrt {x^{2}+2 x +5}\right )}{12}-\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (2 x +2\right )}{6 \sqrt {x^{2}+2 x +5}}\right )}{12}\) | \(69\) |
trager | \(\operatorname {RootOf}\left (144 \textit {\_Z}^{2}-12 \textit {\_Z} +1\right ) \ln \left (\frac {2880 \operatorname {RootOf}\left (144 \textit {\_Z}^{2}-12 \textit {\_Z} +1\right )^{2} x -126 \operatorname {RootOf}\left (144 \textit {\_Z}^{2}-12 \textit {\_Z} +1\right ) \sqrt {x^{2}+2 x +5}-306 \operatorname {RootOf}\left (144 \textit {\_Z}^{2}-12 \textit {\_Z} +1\right ) x -570 \operatorname {RootOf}\left (144 \textit {\_Z}^{2}-12 \textit {\_Z} +1\right )+8 \sqrt {x^{2}+2 x +5}+7 x +19}{12 \operatorname {RootOf}\left (144 \textit {\_Z}^{2}-12 \textit {\_Z} +1\right ) x -x -2}\right )+\frac {\ln \left (\frac {5760 \operatorname {RootOf}\left (144 \textit {\_Z}^{2}-12 \textit {\_Z} +1\right )^{2} x +252 \operatorname {RootOf}\left (144 \textit {\_Z}^{2}-12 \textit {\_Z} +1\right ) \sqrt {x^{2}+2 x +5}-348 \operatorname {RootOf}\left (144 \textit {\_Z}^{2}-12 \textit {\_Z} +1\right ) x +1140 \operatorname {RootOf}\left (144 \textit {\_Z}^{2}-12 \textit {\_Z} +1\right )-5 \sqrt {x^{2}+2 x +5}+3 x -57}{6 \operatorname {RootOf}\left (144 \textit {\_Z}^{2}-12 \textit {\_Z} +1\right ) x +1}\right )}{12}-\ln \left (\frac {5760 \operatorname {RootOf}\left (144 \textit {\_Z}^{2}-12 \textit {\_Z} +1\right )^{2} x +252 \operatorname {RootOf}\left (144 \textit {\_Z}^{2}-12 \textit {\_Z} +1\right ) \sqrt {x^{2}+2 x +5}-348 \operatorname {RootOf}\left (144 \textit {\_Z}^{2}-12 \textit {\_Z} +1\right ) x +1140 \operatorname {RootOf}\left (144 \textit {\_Z}^{2}-12 \textit {\_Z} +1\right )-5 \sqrt {x^{2}+2 x +5}+3 x -57}{6 \operatorname {RootOf}\left (144 \textit {\_Z}^{2}-12 \textit {\_Z} +1\right ) x +1}\right ) \operatorname {RootOf}\left (144 \textit {\_Z}^{2}-12 \textit {\_Z} +1\right )+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}-13\right ) \ln \left (-\frac {-3 \operatorname {RootOf}\left (\textit {\_Z}^{2}-13\right ) x +13 \sqrt {x^{2}+2 x +5}-7 \operatorname {RootOf}\left (\textit {\_Z}^{2}-13\right )}{-2+x}\right )}{156}\) | \(387\) |
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Leaf count of result is larger than twice the leaf count of optimal. 151 vs. \(2 (64) = 128\).
Time = 0.25 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.84 \[ \int \frac {1}{\sqrt {5+2 x+x^2} \left (-8+x^3\right )} \, dx=\frac {1}{12} \, \sqrt {3} \arctan \left (-\frac {1}{3} \, \sqrt {3} {\left (x + 2\right )} + \frac {1}{3} \, \sqrt {3} \sqrt {x^{2} + 2 \, x + 5}\right ) - \frac {1}{12} \, \sqrt {3} \arctan \left (-\frac {1}{3} \, \sqrt {3} x + \frac {1}{3} \, \sqrt {3} \sqrt {x^{2} + 2 \, x + 5}\right ) + \frac {1}{156} \, \sqrt {13} \log \left (\frac {\sqrt {13} {\left (3 \, x + 7\right )} + \sqrt {x^{2} + 2 \, x + 5} {\left (3 \, \sqrt {13} - 13\right )} - 9 \, x - 21}{x - 2}\right ) - \frac {1}{24} \, \log \left (x^{2} - \sqrt {x^{2} + 2 \, x + 5} {\left (x + 2\right )} + 3 \, x + 6\right ) + \frac {1}{24} \, \log \left (x^{2} - \sqrt {x^{2} + 2 \, x + 5} x + x + 4\right ) \]
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\[ \int \frac {1}{\sqrt {5+2 x+x^2} \left (-8+x^3\right )} \, dx=\int \frac {1}{\left (x - 2\right ) \left (x^{2} + 2 x + 4\right ) \sqrt {x^{2} + 2 x + 5}}\, dx \]
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\[ \int \frac {1}{\sqrt {5+2 x+x^2} \left (-8+x^3\right )} \, dx=\int { \frac {1}{{\left (x^{3} - 8\right )} \sqrt {x^{2} + 2 \, x + 5}} \,d x } \]
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Leaf count of result is larger than twice the leaf count of optimal. 164 vs. \(2 (64) = 128\).
Time = 0.32 (sec) , antiderivative size = 164, normalized size of antiderivative = 2.00 \[ \int \frac {1}{\sqrt {5+2 x+x^2} \left (-8+x^3\right )} \, dx=\frac {1}{12} \, \sqrt {3} \arctan \left (-\frac {1}{3} \, \sqrt {3} {\left (x - \sqrt {x^{2} + 2 \, x + 5} + 2\right )}\right ) - \frac {1}{12} \, \sqrt {3} \arctan \left (-\frac {1}{3} \, \sqrt {3} {\left (x - \sqrt {x^{2} + 2 \, x + 5}\right )}\right ) + \frac {1}{156} \, \sqrt {13} \log \left (\frac {{\left | -2 \, x - 2 \, \sqrt {13} + 2 \, \sqrt {x^{2} + 2 \, x + 5} + 4 \right |}}{{\left | -2 \, x + 2 \, \sqrt {13} + 2 \, \sqrt {x^{2} + 2 \, x + 5} + 4 \right |}}\right ) - \frac {1}{24} \, \log \left ({\left (x - \sqrt {x^{2} + 2 \, x + 5}\right )}^{2} + 4 \, x - 4 \, \sqrt {x^{2} + 2 \, x + 5} + 7\right ) + \frac {1}{24} \, \log \left ({\left (x - \sqrt {x^{2} + 2 \, x + 5}\right )}^{2} + 3\right ) \]
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Timed out. \[ \int \frac {1}{\sqrt {5+2 x+x^2} \left (-8+x^3\right )} \, dx=\int \frac {1}{\left (x^3-8\right )\,\sqrt {x^2+2\,x+5}} \,d x \]
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