Integrand size = 26, antiderivative size = 147 \[ \int \frac {1}{\sqrt [3]{-2+b x^2} \left (-\frac {18 d}{b}+d x^2\right )} \, dx=\frac {\sqrt {b} \arctan \left (\frac {\sqrt [6]{2} \sqrt {3} \left (\sqrt [3]{2}+\sqrt [3]{-2+b x^2}\right )}{\sqrt {b} x}\right )}{4\ 2^{5/6} \sqrt {3} d}+\frac {\sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} x}{3 \sqrt {2}}\right )}{12\ 2^{5/6} d}-\frac {\sqrt {b} \text {arctanh}\left (\frac {\left (\sqrt [3]{2}+\sqrt [3]{-2+b x^2}\right )^2}{3 \sqrt [6]{2} \sqrt {b} x}\right )}{12\ 2^{5/6} d} \]
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Time = 0.02 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.038, Rules used = {404} \[ \int \frac {1}{\sqrt [3]{-2+b x^2} \left (-\frac {18 d}{b}+d x^2\right )} \, dx=\frac {\sqrt {b} \arctan \left (\frac {\sqrt [6]{2} \sqrt {3} \left (\sqrt [3]{b x^2-2}+\sqrt [3]{2}\right )}{\sqrt {b} x}\right )}{4\ 2^{5/6} \sqrt {3} d}-\frac {\sqrt {b} \text {arctanh}\left (\frac {\left (\sqrt [3]{b x^2-2}+\sqrt [3]{2}\right )^2}{3 \sqrt [6]{2} \sqrt {b} x}\right )}{12\ 2^{5/6} d}+\frac {\sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} x}{3 \sqrt {2}}\right )}{12\ 2^{5/6} d} \]
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Rule 404
Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt {b} \tan ^{-1}\left (\frac {\sqrt [6]{2} \sqrt {3} \left (\sqrt [3]{2}+\sqrt [3]{-2+b x^2}\right )}{\sqrt {b} x}\right )}{4\ 2^{5/6} \sqrt {3} d}+\frac {\sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} x}{3 \sqrt {2}}\right )}{12\ 2^{5/6} d}-\frac {\sqrt {b} \tanh ^{-1}\left (\frac {\left (\sqrt [3]{2}+\sqrt [3]{-2+b x^2}\right )^2}{3 \sqrt [6]{2} \sqrt {b} x}\right )}{12\ 2^{5/6} d} \\ \end{align*}
Result contains higher order function than in optimal. Order 6 vs. order 3 in optimal.
Time = 4.66 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.01 \[ \int \frac {1}{\sqrt [3]{-2+b x^2} \left (-\frac {18 d}{b}+d x^2\right )} \, dx=\frac {27 b x \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{3},1,\frac {3}{2},\frac {b x^2}{2},\frac {b x^2}{18}\right )}{d \left (-18+b x^2\right ) \sqrt [3]{-2+b x^2} \left (27 \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{3},1,\frac {3}{2},\frac {b x^2}{2},\frac {b x^2}{18}\right )+b x^2 \left (\operatorname {AppellF1}\left (\frac {3}{2},\frac {1}{3},2,\frac {5}{2},\frac {b x^2}{2},\frac {b x^2}{18}\right )+3 \operatorname {AppellF1}\left (\frac {3}{2},\frac {4}{3},1,\frac {5}{2},\frac {b x^2}{2},\frac {b x^2}{18}\right )\right )\right )} \]
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\[\int \frac {1}{\left (b \,x^{2}-2\right )^{\frac {1}{3}} \left (-\frac {18 d}{b}+d \,x^{2}\right )}d x\]
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Timed out. \[ \int \frac {1}{\sqrt [3]{-2+b x^2} \left (-\frac {18 d}{b}+d x^2\right )} \, dx=\text {Timed out} \]
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\[ \int \frac {1}{\sqrt [3]{-2+b x^2} \left (-\frac {18 d}{b}+d x^2\right )} \, dx=\frac {b \int \frac {1}{b x^{2} \sqrt [3]{b x^{2} - 2} - 18 \sqrt [3]{b x^{2} - 2}}\, dx}{d} \]
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\[ \int \frac {1}{\sqrt [3]{-2+b x^2} \left (-\frac {18 d}{b}+d x^2\right )} \, dx=\int { \frac {1}{{\left (b x^{2} - 2\right )}^{\frac {1}{3}} {\left (d x^{2} - \frac {18 \, d}{b}\right )}} \,d x } \]
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\[ \int \frac {1}{\sqrt [3]{-2+b x^2} \left (-\frac {18 d}{b}+d x^2\right )} \, dx=\int { \frac {1}{{\left (b x^{2} - 2\right )}^{\frac {1}{3}} {\left (d x^{2} - \frac {18 \, d}{b}\right )}} \,d x } \]
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Timed out. \[ \int \frac {1}{\sqrt [3]{-2+b x^2} \left (-\frac {18 d}{b}+d x^2\right )} \, dx=\int -\frac {1}{\left (\frac {18\,d}{b}-d\,x^2\right )\,{\left (b\,x^2-2\right )}^{1/3}} \,d x \]
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