Integrand size = 23, antiderivative size = 155 \[ \int \frac {\left (a+b x^3\right )^{5/4}}{\left (c+d x^3\right )^{31/12}} \, dx=\frac {4 x \left (a+b x^3\right )^{5/4}}{19 c \left (c+d x^3\right )^{19/12}}+\frac {60 a x \sqrt [4]{a+b x^3}}{133 c^2 \left (c+d x^3\right )^{7/12}}+\frac {45 a^2 x \left (\frac {c \left (a+b x^3\right )}{a \left (c+d x^3\right )}\right )^{3/4} \left (c+d x^3\right )^{5/12} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {3}{4},\frac {4}{3},-\frac {(b c-a d) x^3}{a \left (c+d x^3\right )}\right )}{133 c^3 \left (a+b x^3\right )^{3/4}} \]
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Time = 0.04 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {386, 388} \[ \int \frac {\left (a+b x^3\right )^{5/4}}{\left (c+d x^3\right )^{31/12}} \, dx=\frac {45 a^2 x \left (c+d x^3\right )^{5/12} \left (\frac {c \left (a+b x^3\right )}{a \left (c+d x^3\right )}\right )^{3/4} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {3}{4},\frac {4}{3},-\frac {(b c-a d) x^3}{a \left (d x^3+c\right )}\right )}{133 c^3 \left (a+b x^3\right )^{3/4}}+\frac {60 a x \sqrt [4]{a+b x^3}}{133 c^2 \left (c+d x^3\right )^{7/12}}+\frac {4 x \left (a+b x^3\right )^{5/4}}{19 c \left (c+d x^3\right )^{19/12}} \]
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Rule 386
Rule 388
Rubi steps \begin{align*} \text {integral}& = \frac {4 x \left (a+b x^3\right )^{5/4}}{19 c \left (c+d x^3\right )^{19/12}}+\frac {(15 a) \int \frac {\sqrt [4]{a+b x^3}}{\left (c+d x^3\right )^{19/12}} \, dx}{19 c} \\ & = \frac {4 x \left (a+b x^3\right )^{5/4}}{19 c \left (c+d x^3\right )^{19/12}}+\frac {60 a x \sqrt [4]{a+b x^3}}{133 c^2 \left (c+d x^3\right )^{7/12}}+\frac {\left (45 a^2\right ) \int \frac {1}{\left (a+b x^3\right )^{3/4} \left (c+d x^3\right )^{7/12}} \, dx}{133 c^2} \\ & = \frac {4 x \left (a+b x^3\right )^{5/4}}{19 c \left (c+d x^3\right )^{19/12}}+\frac {60 a x \sqrt [4]{a+b x^3}}{133 c^2 \left (c+d x^3\right )^{7/12}}+\frac {45 a^2 x \left (\frac {c \left (a+b x^3\right )}{a \left (c+d x^3\right )}\right )^{3/4} \left (c+d x^3\right )^{5/12} \, _2F_1\left (\frac {1}{3},\frac {3}{4};\frac {4}{3};-\frac {(b c-a d) x^3}{a \left (c+d x^3\right )}\right )}{133 c^3 \left (a+b x^3\right )^{3/4}} \\ \end{align*}
Time = 5.46 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.58 \[ \int \frac {\left (a+b x^3\right )^{5/4}}{\left (c+d x^3\right )^{31/12}} \, dx=\frac {a x \sqrt [4]{a+b x^3} \sqrt [4]{1+\frac {d x^3}{c}} \operatorname {Hypergeometric2F1}\left (-\frac {5}{4},\frac {1}{3},\frac {4}{3},\frac {(-b c+a d) x^3}{a \left (c+d x^3\right )}\right )}{c^2 \sqrt [4]{1+\frac {b x^3}{a}} \left (c+d x^3\right )^{7/12}} \]
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\[\int \frac {\left (b \,x^{3}+a \right )^{\frac {5}{4}}}{\left (d \,x^{3}+c \right )^{\frac {31}{12}}}d x\]
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\[ \int \frac {\left (a+b x^3\right )^{5/4}}{\left (c+d x^3\right )^{31/12}} \, dx=\int { \frac {{\left (b x^{3} + a\right )}^{\frac {5}{4}}}{{\left (d x^{3} + c\right )}^{\frac {31}{12}}} \,d x } \]
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Timed out. \[ \int \frac {\left (a+b x^3\right )^{5/4}}{\left (c+d x^3\right )^{31/12}} \, dx=\text {Timed out} \]
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\[ \int \frac {\left (a+b x^3\right )^{5/4}}{\left (c+d x^3\right )^{31/12}} \, dx=\int { \frac {{\left (b x^{3} + a\right )}^{\frac {5}{4}}}{{\left (d x^{3} + c\right )}^{\frac {31}{12}}} \,d x } \]
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\[ \int \frac {\left (a+b x^3\right )^{5/4}}{\left (c+d x^3\right )^{31/12}} \, dx=\int { \frac {{\left (b x^{3} + a\right )}^{\frac {5}{4}}}{{\left (d x^{3} + c\right )}^{\frac {31}{12}}} \,d x } \]
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Timed out. \[ \int \frac {\left (a+b x^3\right )^{5/4}}{\left (c+d x^3\right )^{31/12}} \, dx=\int \frac {{\left (b\,x^3+a\right )}^{5/4}}{{\left (d\,x^3+c\right )}^{31/12}} \,d x \]
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