Integrand size = 22, antiderivative size = 204 \[ \int \frac {1}{\sqrt [3]{c-3 d x^2} \left (c+d x^2\right )} \, dx=\frac {\arctan \left (\frac {\sqrt {c}}{\sqrt {d} x}\right )}{2\ 2^{2/3} c^{5/6} \sqrt {d}}+\frac {\arctan \left (\frac {\sqrt [6]{c} \left (\sqrt [3]{c}-\sqrt [3]{2} \sqrt [3]{c-3 d x^2}\right )}{\sqrt {d} x}\right )}{2\ 2^{2/3} c^{5/6} \sqrt {d}}-\frac {\text {arctanh}\left (\frac {\sqrt {3} \sqrt {d} x}{\sqrt {c}}\right )}{2\ 2^{2/3} \sqrt {3} c^{5/6} \sqrt {d}}+\frac {\sqrt {3} \text {arctanh}\left (\frac {\sqrt {3} \sqrt {d} x}{\sqrt [6]{c} \left (\sqrt [3]{c}+\sqrt [3]{2} \sqrt [3]{c-3 d x^2}\right )}\right )}{2\ 2^{2/3} c^{5/6} \sqrt {d}} \]
[Out]
Time = 0.02 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.045, Rules used = {402} \[ \int \frac {1}{\sqrt [3]{c-3 d x^2} \left (c+d x^2\right )} \, dx=\frac {\arctan \left (\frac {\sqrt [6]{c} \left (\sqrt [3]{c}-\sqrt [3]{2} \sqrt [3]{c-3 d x^2}\right )}{\sqrt {d} x}\right )}{2\ 2^{2/3} c^{5/6} \sqrt {d}}+\frac {\arctan \left (\frac {\sqrt {c}}{\sqrt {d} x}\right )}{2\ 2^{2/3} c^{5/6} \sqrt {d}}+\frac {\sqrt {3} \text {arctanh}\left (\frac {\sqrt {3} \sqrt {d} x}{\sqrt [6]{c} \left (\sqrt [3]{2} \sqrt [3]{c-3 d x^2}+\sqrt [3]{c}\right )}\right )}{2\ 2^{2/3} c^{5/6} \sqrt {d}}-\frac {\text {arctanh}\left (\frac {\sqrt {3} \sqrt {d} x}{\sqrt {c}}\right )}{2\ 2^{2/3} \sqrt {3} c^{5/6} \sqrt {d}} \]
[In]
[Out]
Rule 402
Rubi steps \begin{align*} \text {integral}& = \frac {\tan ^{-1}\left (\frac {\sqrt {c}}{\sqrt {d} x}\right )}{2\ 2^{2/3} c^{5/6} \sqrt {d}}+\frac {\tan ^{-1}\left (\frac {\sqrt [6]{c} \left (\sqrt [3]{c}-\sqrt [3]{2} \sqrt [3]{c-3 d x^2}\right )}{\sqrt {d} x}\right )}{2\ 2^{2/3} c^{5/6} \sqrt {d}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {3} \sqrt {d} x}{\sqrt {c}}\right )}{2\ 2^{2/3} \sqrt {3} c^{5/6} \sqrt {d}}+\frac {\sqrt {3} \tanh ^{-1}\left (\frac {\sqrt {3} \sqrt {d} x}{\sqrt [6]{c} \left (\sqrt [3]{c}+\sqrt [3]{2} \sqrt [3]{c-3 d x^2}\right )}\right )}{2\ 2^{2/3} c^{5/6} \sqrt {d}} \\ \end{align*}
Result contains higher order function than in optimal. Order 6 vs. order 3 in optimal.
Time = 5.07 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.76 \[ \int \frac {1}{\sqrt [3]{c-3 d x^2} \left (c+d x^2\right )} \, dx=\frac {3 c x \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{3},1,\frac {3}{2},\frac {3 d x^2}{c},-\frac {d x^2}{c}\right )}{\sqrt [3]{c-3 d x^2} \left (c+d x^2\right ) \left (3 c \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{3},1,\frac {3}{2},\frac {3 d x^2}{c},-\frac {d x^2}{c}\right )+2 d x^2 \left (-\operatorname {AppellF1}\left (\frac {3}{2},\frac {1}{3},2,\frac {5}{2},\frac {3 d x^2}{c},-\frac {d x^2}{c}\right )+\operatorname {AppellF1}\left (\frac {3}{2},\frac {4}{3},1,\frac {5}{2},\frac {3 d x^2}{c},-\frac {d x^2}{c}\right )\right )\right )} \]
[In]
[Out]
\[\int \frac {1}{\left (-3 d \,x^{2}+c \right )^{\frac {1}{3}} \left (d \,x^{2}+c \right )}d x\]
[In]
[Out]
Timed out. \[ \int \frac {1}{\sqrt [3]{c-3 d x^2} \left (c+d x^2\right )} \, dx=\text {Timed out} \]
[In]
[Out]
\[ \int \frac {1}{\sqrt [3]{c-3 d x^2} \left (c+d x^2\right )} \, dx=\int \frac {1}{\sqrt [3]{c - 3 d x^{2}} \left (c + d x^{2}\right )}\, dx \]
[In]
[Out]
\[ \int \frac {1}{\sqrt [3]{c-3 d x^2} \left (c+d x^2\right )} \, dx=\int { \frac {1}{{\left (d x^{2} + c\right )} {\left (-3 \, d x^{2} + c\right )}^{\frac {1}{3}}} \,d x } \]
[In]
[Out]
\[ \int \frac {1}{\sqrt [3]{c-3 d x^2} \left (c+d x^2\right )} \, dx=\int { \frac {1}{{\left (d x^{2} + c\right )} {\left (-3 \, d x^{2} + c\right )}^{\frac {1}{3}}} \,d x } \]
[In]
[Out]
Timed out. \[ \int \frac {1}{\sqrt [3]{c-3 d x^2} \left (c+d x^2\right )} \, dx=\int \frac {1}{\left (d\,x^2+c\right )\,{\left (c-3\,d\,x^2\right )}^{1/3}} \,d x \]
[In]
[Out]