Integrand size = 27, antiderivative size = 131 \[ \int \frac {1+x}{(-1+x) (1+2 x) \sqrt [3]{-1+3 x^2}} \, dx=\frac {\arctan \left (\frac {\sqrt {3} \sqrt [3]{-1+3 x^2}}{2 \sqrt [3]{2} x+\sqrt [3]{-1+3 x^2}}\right )}{\sqrt [3]{2} \sqrt {3}}+\frac {\log \left (-\sqrt [3]{2} x+\sqrt [3]{-1+3 x^2}\right )}{3 \sqrt [3]{2}}-\frac {\log \left (2^{2/3} x^2+\sqrt [3]{2} x \sqrt [3]{-1+3 x^2}+\left (-1+3 x^2\right )^{2/3}\right )}{6 \sqrt [3]{2}} \]
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Result contains higher order function than in optimal. Order 6 vs. order 3 in optimal.
Time = 0.38 (sec) , antiderivative size = 265, normalized size of antiderivative = 2.02, number of steps used = 20, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.370, Rules used = {6874, 771, 441, 440, 455, 57, 631, 210, 31, 58} \[ \int \frac {1+x}{(-1+x) (1+2 x) \sqrt [3]{-1+3 x^2}} \, dx=-\frac {2 x \sqrt [3]{1-3 x^2} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{3},1,\frac {3}{2},3 x^2,x^2\right )}{3 \sqrt [3]{3 x^2-1}}-\frac {x \sqrt [3]{1-3 x^2} \operatorname {AppellF1}\left (\frac {1}{2},1,\frac {1}{3},\frac {3}{2},4 x^2,3 x^2\right )}{3 \sqrt [3]{3 x^2-1}}+\frac {\arctan \left (\frac {1-2\ 2^{2/3} \sqrt [3]{3 x^2-1}}{\sqrt {3}}\right )}{2 \sqrt [3]{2} \sqrt {3}}+\frac {\arctan \left (\frac {2^{2/3} \sqrt [3]{3 x^2-1}+1}{\sqrt {3}}\right )}{\sqrt [3]{2} \sqrt {3}}-\frac {\log \left (1-4 x^2\right )}{12 \sqrt [3]{2}}-\frac {\log \left (1-x^2\right )}{6 \sqrt [3]{2}}+\frac {\log \left (\sqrt [3]{2}-\sqrt [3]{3 x^2-1}\right )}{2 \sqrt [3]{2}}+\frac {\log \left (2^{2/3} \sqrt [3]{3 x^2-1}+1\right )}{4 \sqrt [3]{2}} \]
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Rule 31
Rule 57
Rule 58
Rule 210
Rule 440
Rule 441
Rule 455
Rule 631
Rule 771
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {2}{3 (-1+x) \sqrt [3]{-1+3 x^2}}-\frac {1}{3 (1+2 x) \sqrt [3]{-1+3 x^2}}\right ) \, dx \\ & = -\left (\frac {1}{3} \int \frac {1}{(1+2 x) \sqrt [3]{-1+3 x^2}} \, dx\right )+\frac {2}{3} \int \frac {1}{(-1+x) \sqrt [3]{-1+3 x^2}} \, dx \\ & = -\left (\frac {1}{3} \int \left (\frac {1}{\left (1-4 x^2\right ) \sqrt [3]{-1+3 x^2}}+\frac {2 x}{\sqrt [3]{-1+3 x^2} \left (-1+4 x^2\right )}\right ) \, dx\right )+\frac {2}{3} \int \left (\frac {1}{\left (-1+x^2\right ) \sqrt [3]{-1+3 x^2}}+\frac {x}{\left (-1+x^2\right ) \sqrt [3]{-1+3 x^2}}\right ) \, dx \\ & = -\left (\frac {1}{3} \int \frac {1}{\left (1-4 x^2\right ) \sqrt [3]{-1+3 x^2}} \, dx\right )+\frac {2}{3} \int \frac {1}{\left (-1+x^2\right ) \sqrt [3]{-1+3 x^2}} \, dx+\frac {2}{3} \int \frac {x}{\left (-1+x^2\right ) \sqrt [3]{-1+3 x^2}} \, dx-\frac {2}{3} \int \frac {x}{\sqrt [3]{-1+3 x^2} \left (-1+4 x^2\right )} \, dx \\ & = \frac {1}{3} \text {Subst}\left (\int \frac {1}{(-1+x) \sqrt [3]{-1+3 x}} \, dx,x,x^2\right )-\frac {1}{3} \text {Subst}\left (\int \frac {1}{\sqrt [3]{-1+3 x} (-1+4 x)} \, dx,x,x^2\right )-\frac {\sqrt [3]{1-3 x^2} \int \frac {1}{\left (1-4 x^2\right ) \sqrt [3]{1-3 x^2}} \, dx}{3 \sqrt [3]{-1+3 x^2}}+\frac {\left (2 \sqrt [3]{1-3 x^2}\right ) \int \frac {1}{\sqrt [3]{1-3 x^2} \left (-1+x^2\right )} \, dx}{3 \sqrt [3]{-1+3 x^2}} \\ & = -\frac {2 x \sqrt [3]{1-3 x^2} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{3},1,\frac {3}{2},3 x^2,x^2\right )}{3 \sqrt [3]{-1+3 x^2}}-\frac {x \sqrt [3]{1-3 x^2} \operatorname {AppellF1}\left (\frac {1}{2},1,\frac {1}{3},\frac {3}{2},4 x^2,3 x^2\right )}{3 \sqrt [3]{-1+3 x^2}}-\frac {\log \left (1-4 x^2\right )}{12 \sqrt [3]{2}}-\frac {\log \left (1-x^2\right )}{6 \sqrt [3]{2}}-\frac {1}{8} \text {Subst}\left (\int \frac {1}{\frac {1}{2 \sqrt [3]{2}}-\frac {x}{2^{2/3}}+x^2} \, dx,x,\sqrt [3]{-1+3 x^2}\right )+\frac {1}{2} \text {Subst}\left (\int \frac {1}{2^{2/3}+\sqrt [3]{2} x+x^2} \, dx,x,\sqrt [3]{-1+3 x^2}\right )+\frac {\text {Subst}\left (\int \frac {1}{\frac {1}{2^{2/3}}+x} \, dx,x,\sqrt [3]{-1+3 x^2}\right )}{4 \sqrt [3]{2}}-\frac {\text {Subst}\left (\int \frac {1}{\sqrt [3]{2}-x} \, dx,x,\sqrt [3]{-1+3 x^2}\right )}{2 \sqrt [3]{2}} \\ & = -\frac {2 x \sqrt [3]{1-3 x^2} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{3},1,\frac {3}{2},3 x^2,x^2\right )}{3 \sqrt [3]{-1+3 x^2}}-\frac {x \sqrt [3]{1-3 x^2} \operatorname {AppellF1}\left (\frac {1}{2},1,\frac {1}{3},\frac {3}{2},4 x^2,3 x^2\right )}{3 \sqrt [3]{-1+3 x^2}}-\frac {\log \left (1-4 x^2\right )}{12 \sqrt [3]{2}}-\frac {\log \left (1-x^2\right )}{6 \sqrt [3]{2}}+\frac {\log \left (\sqrt [3]{2}-\sqrt [3]{-1+3 x^2}\right )}{2 \sqrt [3]{2}}+\frac {\log \left (1+2^{2/3} \sqrt [3]{-1+3 x^2}\right )}{4 \sqrt [3]{2}}-\frac {\text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1-2\ 2^{2/3} \sqrt [3]{-1+3 x^2}\right )}{2 \sqrt [3]{2}}-\frac {\text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+2^{2/3} \sqrt [3]{-1+3 x^2}\right )}{\sqrt [3]{2}} \\ & = -\frac {2 x \sqrt [3]{1-3 x^2} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{3},1,\frac {3}{2},3 x^2,x^2\right )}{3 \sqrt [3]{-1+3 x^2}}-\frac {x \sqrt [3]{1-3 x^2} \operatorname {AppellF1}\left (\frac {1}{2},1,\frac {1}{3},\frac {3}{2},4 x^2,3 x^2\right )}{3 \sqrt [3]{-1+3 x^2}}+\frac {\arctan \left (\frac {1-2\ 2^{2/3} \sqrt [3]{-1+3 x^2}}{\sqrt {3}}\right )}{2 \sqrt [3]{2} \sqrt {3}}+\frac {\arctan \left (\frac {1+2^{2/3} \sqrt [3]{-1+3 x^2}}{\sqrt {3}}\right )}{\sqrt [3]{2} \sqrt {3}}-\frac {\log \left (1-4 x^2\right )}{12 \sqrt [3]{2}}-\frac {\log \left (1-x^2\right )}{6 \sqrt [3]{2}}+\frac {\log \left (\sqrt [3]{2}-\sqrt [3]{-1+3 x^2}\right )}{2 \sqrt [3]{2}}+\frac {\log \left (1+2^{2/3} \sqrt [3]{-1+3 x^2}\right )}{4 \sqrt [3]{2}} \\ \end{align*}
Time = 0.17 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.89 \[ \int \frac {1+x}{(-1+x) (1+2 x) \sqrt [3]{-1+3 x^2}} \, dx=\frac {2 \sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{-1+3 x^2}}{2 \sqrt [3]{2} x+\sqrt [3]{-1+3 x^2}}\right )+2 \log \left (-\sqrt [3]{2} x+\sqrt [3]{-1+3 x^2}\right )-\log \left (2^{2/3} x^2+\left (-1+3 x^2\right )^{2/3}+x \sqrt [3]{-2+6 x^2}\right )}{6 \sqrt [3]{2}} \]
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Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 19.30 (sec) , antiderivative size = 663, normalized size of antiderivative = 5.06
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Exception generated. \[ \int \frac {1+x}{(-1+x) (1+2 x) \sqrt [3]{-1+3 x^2}} \, dx=\text {Exception raised: TypeError} \]
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\[ \int \frac {1+x}{(-1+x) (1+2 x) \sqrt [3]{-1+3 x^2}} \, dx=\int \frac {x + 1}{\left (x - 1\right ) \left (2 x + 1\right ) \sqrt [3]{3 x^{2} - 1}}\, dx \]
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\[ \int \frac {1+x}{(-1+x) (1+2 x) \sqrt [3]{-1+3 x^2}} \, dx=\int { \frac {x + 1}{{\left (3 \, x^{2} - 1\right )}^{\frac {1}{3}} {\left (2 \, x + 1\right )} {\left (x - 1\right )}} \,d x } \]
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\[ \int \frac {1+x}{(-1+x) (1+2 x) \sqrt [3]{-1+3 x^2}} \, dx=\int { \frac {x + 1}{{\left (3 \, x^{2} - 1\right )}^{\frac {1}{3}} {\left (2 \, x + 1\right )} {\left (x - 1\right )}} \,d x } \]
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Timed out. \[ \int \frac {1+x}{(-1+x) (1+2 x) \sqrt [3]{-1+3 x^2}} \, dx=\int \frac {x+1}{\left (2\,x+1\right )\,{\left (3\,x^2-1\right )}^{1/3}\,\left (x-1\right )} \,d x \]
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