Integrand size = 17, antiderivative size = 75 \[ \int \frac {1}{\sqrt [3]{9+3 x-5 x^2+x^3}} \, dx=\sqrt {3} \arctan \left (\frac {1+\frac {2 (-3+x)}{\sqrt [3]{9+3 x-5 x^2+x^3}}}{\sqrt {3}}\right )-\frac {1}{2} \log (1+x)-\frac {3}{2} \log \left (1-\frac {-3+x}{\sqrt [3]{9+3 x-5 x^2+x^3}}\right ) \]
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Leaf count is larger than twice the leaf count of optimal. \(188\) vs. \(2(75)=150\).
Time = 0.09 (sec) , antiderivative size = 188, normalized size of antiderivative = 2.51, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {2092, 2089, 62} \[ \int \frac {1}{\sqrt [3]{9+3 x-5 x^2+x^3}} \, dx=-\frac {(9-3 x)^{2/3} \sqrt [3]{x+1} \arctan \left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{x+1}}{\sqrt [6]{3} \sqrt [3]{9-3 x}}\right )}{\sqrt [6]{3} \sqrt [3]{x^3-5 x^2+3 x+9}}-\frac {(9-3 x)^{2/3} \sqrt [3]{x+1} \log \left (-\frac {32}{3} (x-3)\right )}{2\ 3^{2/3} \sqrt [3]{x^3-5 x^2+3 x+9}}-\frac {\sqrt [3]{3} (9-3 x)^{2/3} \sqrt [3]{x+1} \log \left (\frac {\sqrt [3]{3} \sqrt [3]{x+1}}{\sqrt [3]{9-3 x}}+1\right )}{2 \sqrt [3]{x^3-5 x^2+3 x+9}} \]
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Rule 62
Rule 2089
Rule 2092
Rubi steps \begin{align*} \text {integral}& = \text {Subst}\left (\int \frac {1}{\sqrt [3]{\frac {128}{27}-\frac {16 x}{3}+x^3}} \, dx,x,-\frac {5}{3}+x\right ) \\ & = \frac {\left (16\ 2^{2/3} (3-x)^{2/3} \sqrt [3]{1+x}\right ) \text {Subst}\left (\int \frac {1}{\left (\frac {128}{9}-\frac {32 x}{3}\right )^{2/3} \sqrt [3]{\frac {128}{9}+\frac {16 x}{3}}} \, dx,x,-\frac {5}{3}+x\right )}{3 \sqrt [3]{9+3 x-5 x^2+x^3}} \\ & = -\frac {\sqrt {3} (3-x)^{2/3} \sqrt [3]{1+x} \arctan \left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{1+x}}{\sqrt {3} \sqrt [3]{3-x}}\right )}{\sqrt [3]{9+3 x-5 x^2+x^3}}-\frac {(3-x)^{2/3} \sqrt [3]{1+x} \log (3-x)}{2 \sqrt [3]{9+3 x-5 x^2+x^3}}-\frac {3 (3-x)^{2/3} \sqrt [3]{1+x} \log \left (\frac {\sqrt [3]{3-x}+\sqrt [3]{1+x}}{\sqrt [3]{3-x}}\right )}{2 \sqrt [3]{9+3 x-5 x^2+x^3}} \\ \end{align*}
Time = 0.13 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.64 \[ \int \frac {1}{\sqrt [3]{9+3 x-5 x^2+x^3}} \, dx=\frac {(-3+x)^{2/3} \sqrt [3]{1+x} \left (-2 \sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{1+x}}{2 \sqrt [3]{-3+x}+\sqrt [3]{1+x}}\right )-2 \log \left (\sqrt [3]{-3+x}-\sqrt [3]{1+x}\right )+\log \left ((-3+x)^{2/3}+\sqrt [3]{-3+x} \sqrt [3]{1+x}+(1+x)^{2/3}\right )\right )}{2 \sqrt [3]{(-3+x)^2 (1+x)}} \]
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Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.57 (sec) , antiderivative size = 446, normalized size of antiderivative = 5.95
method | result | size |
trager | \(-\ln \left (-\frac {-16 \operatorname {RootOf}\left (\textit {\_Z}^{2}-3 \textit {\_Z} +9\right )^{2} x^{2}+27 \operatorname {RootOf}\left (\textit {\_Z}^{2}-3 \textit {\_Z} +9\right ) \left (x^{3}-5 x^{2}+3 x +9\right )^{\frac {2}{3}}+45 \operatorname {RootOf}\left (\textit {\_Z}^{2}-3 \textit {\_Z} +9\right ) \left (x^{3}-5 x^{2}+3 x +9\right )^{\frac {1}{3}} x +48 \operatorname {RootOf}\left (\textit {\_Z}^{2}-3 \textit {\_Z} +9\right )^{2} x -24 \operatorname {RootOf}\left (\textit {\_Z}^{2}-3 \textit {\_Z} +9\right ) x^{2}-216 \left (x^{3}-5 x^{2}+3 x +9\right )^{\frac {2}{3}}-135 \operatorname {RootOf}\left (\textit {\_Z}^{2}-3 \textit {\_Z} +9\right ) \left (x^{3}-5 x^{2}+3 x +9\right )^{\frac {1}{3}}+81 \left (x^{3}-5 x^{2}+3 x +9\right )^{\frac {1}{3}} x +156 \operatorname {RootOf}\left (\textit {\_Z}^{2}-3 \textit {\_Z} +9\right ) x -9 x^{2}-243 \left (x^{3}-5 x^{2}+3 x +9\right )^{\frac {1}{3}}-252 \operatorname {RootOf}\left (\textit {\_Z}^{2}-3 \textit {\_Z} +9\right )+90 x -189}{-3+x}\right )+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}-3 \textit {\_Z} +9\right ) \ln \left (\frac {4 \operatorname {RootOf}\left (\textit {\_Z}^{2}-3 \textit {\_Z} +9\right )^{2} x^{2}+27 \operatorname {RootOf}\left (\textit {\_Z}^{2}-3 \textit {\_Z} +9\right ) \left (x^{3}-5 x^{2}+3 x +9\right )^{\frac {2}{3}}-72 \operatorname {RootOf}\left (\textit {\_Z}^{2}-3 \textit {\_Z} +9\right ) \left (x^{3}-5 x^{2}+3 x +9\right )^{\frac {1}{3}} x -12 \operatorname {RootOf}\left (\textit {\_Z}^{2}-3 \textit {\_Z} +9\right )^{2} x +33 \operatorname {RootOf}\left (\textit {\_Z}^{2}-3 \textit {\_Z} +9\right ) x^{2}+135 \left (x^{3}-5 x^{2}+3 x +9\right )^{\frac {2}{3}}+216 \operatorname {RootOf}\left (\textit {\_Z}^{2}-3 \textit {\_Z} +9\right ) \left (x^{3}-5 x^{2}+3 x +9\right )^{\frac {1}{3}}+81 \left (x^{3}-5 x^{2}+3 x +9\right )^{\frac {1}{3}} x -78 \operatorname {RootOf}\left (\textit {\_Z}^{2}-3 \textit {\_Z} +9\right ) x -180 x^{2}-243 \left (x^{3}-5 x^{2}+3 x +9\right )^{\frac {1}{3}}-63 \operatorname {RootOf}\left (\textit {\_Z}^{2}-3 \textit {\_Z} +9\right )+792 x -756}{-3+x}\right )}{3}\) | \(446\) |
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none
Time = 0.24 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.71 \[ \int \frac {1}{\sqrt [3]{9+3 x-5 x^2+x^3}} \, dx=-\sqrt {3} \arctan \left (\frac {\sqrt {3} {\left (x - 3\right )} + 2 \, \sqrt {3} {\left (x^{3} - 5 \, x^{2} + 3 \, x + 9\right )}^{\frac {1}{3}}}{3 \, {\left (x - 3\right )}}\right ) + \frac {1}{2} \, \log \left (\frac {x^{2} + {\left (x^{3} - 5 \, x^{2} + 3 \, x + 9\right )}^{\frac {1}{3}} {\left (x - 3\right )} - 6 \, x + {\left (x^{3} - 5 \, x^{2} + 3 \, x + 9\right )}^{\frac {2}{3}} + 9}{x^{2} - 6 \, x + 9}\right ) - \log \left (-\frac {x - {\left (x^{3} - 5 \, x^{2} + 3 \, x + 9\right )}^{\frac {1}{3}} - 3}{x - 3}\right ) \]
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\[ \int \frac {1}{\sqrt [3]{9+3 x-5 x^2+x^3}} \, dx=\int \frac {1}{\sqrt [3]{x^{3} - 5 x^{2} + 3 x + 9}}\, dx \]
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\[ \int \frac {1}{\sqrt [3]{9+3 x-5 x^2+x^3}} \, dx=\int { \frac {1}{{\left (x^{3} - 5 \, x^{2} + 3 \, x + 9\right )}^{\frac {1}{3}}} \,d x } \]
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\[ \int \frac {1}{\sqrt [3]{9+3 x-5 x^2+x^3}} \, dx=\int { \frac {1}{{\left (x^{3} - 5 \, x^{2} + 3 \, x + 9\right )}^{\frac {1}{3}}} \,d x } \]
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Timed out. \[ \int \frac {1}{\sqrt [3]{9+3 x-5 x^2+x^3}} \, dx=\int \frac {1}{{\left (x^3-5\,x^2+3\,x+9\right )}^{1/3}} \,d x \]
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