Integrand size = 21, antiderivative size = 62 \[ \int \frac {1}{\left (a+b x^3\right )^{8/3} \left (c+d x^3\right )} \, dx=\frac {x \left (1+\frac {b x^3}{a}\right )^{2/3} \operatorname {AppellF1}\left (\frac {1}{3},\frac {8}{3},1,\frac {4}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{a^2 c \left (a+b x^3\right )^{2/3}} \]
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Time = 0.02 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {441, 440} \[ \int \frac {1}{\left (a+b x^3\right )^{8/3} \left (c+d x^3\right )} \, dx=\frac {x \left (\frac {b x^3}{a}+1\right )^{2/3} \operatorname {AppellF1}\left (\frac {1}{3},\frac {8}{3},1,\frac {4}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{a^2 c \left (a+b x^3\right )^{2/3}} \]
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Rule 440
Rule 441
Rubi steps \begin{align*} \text {integral}& = \frac {\left (1+\frac {b x^3}{a}\right )^{2/3} \int \frac {1}{\left (1+\frac {b x^3}{a}\right )^{8/3} \left (c+d x^3\right )} \, dx}{a^2 \left (a+b x^3\right )^{2/3}} \\ & = \frac {x \left (1+\frac {b x^3}{a}\right )^{2/3} F_1\left (\frac {1}{3};\frac {8}{3},1;\frac {4}{3};-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{a^2 c \left (a+b x^3\right )^{2/3}} \\ \end{align*}
Leaf count is larger than twice the leaf count of optimal. \(429\) vs. \(2(62)=124\).
Time = 10.78 (sec) , antiderivative size = 429, normalized size of antiderivative = 6.92 \[ \int \frac {1}{\left (a+b x^3\right )^{8/3} \left (c+d x^3\right )} \, dx=-\frac {x \left (\frac {b d (-4 b c+9 a d) x^3 \left (1+\frac {b x^3}{a}\right )^{2/3} \operatorname {AppellF1}\left (\frac {4}{3},\frac {2}{3},1,\frac {7}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{c}+\frac {4 \left (4 a c \left (10 a^3 d^2+4 b^3 c x^3 \left (2 c+d x^3\right )-a^2 b d \left (20 c+d x^3\right )+a b^2 \left (10 c^2-12 c d x^3-9 d^2 x^6\right )\right ) \operatorname {AppellF1}\left (\frac {1}{3},\frac {2}{3},1,\frac {4}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )+b x^3 \left (c+d x^3\right ) \left (11 a^2 d-4 b^2 c x^3+a b \left (-6 c+9 d x^3\right )\right ) \left (3 a d \operatorname {AppellF1}\left (\frac {4}{3},\frac {2}{3},2,\frac {7}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )+2 b c \operatorname {AppellF1}\left (\frac {4}{3},\frac {5}{3},1,\frac {7}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )\right )\right )}{\left (a+b x^3\right ) \left (c+d x^3\right ) \left (-4 a c \operatorname {AppellF1}\left (\frac {1}{3},\frac {2}{3},1,\frac {4}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )+x^3 \left (3 a d \operatorname {AppellF1}\left (\frac {4}{3},\frac {2}{3},2,\frac {7}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )+2 b c \operatorname {AppellF1}\left (\frac {4}{3},\frac {5}{3},1,\frac {7}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )\right )\right )}\right )}{40 a^2 (b c-a d)^2 \left (a+b x^3\right )^{2/3}} \]
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\[\int \frac {1}{\left (b \,x^{3}+a \right )^{\frac {8}{3}} \left (d \,x^{3}+c \right )}d x\]
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Timed out. \[ \int \frac {1}{\left (a+b x^3\right )^{8/3} \left (c+d x^3\right )} \, dx=\text {Timed out} \]
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\[ \int \frac {1}{\left (a+b x^3\right )^{8/3} \left (c+d x^3\right )} \, dx=\int \frac {1}{\left (a + b x^{3}\right )^{\frac {8}{3}} \left (c + d x^{3}\right )}\, dx \]
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\[ \int \frac {1}{\left (a+b x^3\right )^{8/3} \left (c+d x^3\right )} \, dx=\int { \frac {1}{{\left (b x^{3} + a\right )}^{\frac {8}{3}} {\left (d x^{3} + c\right )}} \,d x } \]
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\[ \int \frac {1}{\left (a+b x^3\right )^{8/3} \left (c+d x^3\right )} \, dx=\int { \frac {1}{{\left (b x^{3} + a\right )}^{\frac {8}{3}} {\left (d x^{3} + c\right )}} \,d x } \]
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Timed out. \[ \int \frac {1}{\left (a+b x^3\right )^{8/3} \left (c+d x^3\right )} \, dx=\int \frac {1}{{\left (b\,x^3+a\right )}^{8/3}\,\left (d\,x^3+c\right )} \,d x \]
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