Integrand size = 23, antiderivative size = 153 \[ \int \frac {\left (c+d x^3\right )^{17/12}}{\left (a+b x^3\right )^{11/4}} \, dx=\frac {68 c x \left (c+d x^3\right )^{5/12}}{189 a^2 \left (a+b x^3\right )^{3/4}}+\frac {4 x \left (c+d x^3\right )^{17/12}}{21 a \left (a+b x^3\right )^{7/4}}+\frac {85 c x \left (\frac {c \left (a+b x^3\right )}{a \left (c+d x^3\right )}\right )^{3/4} \left (c+d x^3\right )^{5/12} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {3}{4},\frac {4}{3},-\frac {(b c-a d) x^3}{a \left (c+d x^3\right )}\right )}{189 a^2 \left (a+b x^3\right )^{3/4}} \]
[Out]
Time = 0.05 (sec) , antiderivative size = 153, 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} \[ \int \frac {\left (c+d x^3\right )^{17/12}}{\left (a+b x^3\right )^{11/4}} \, dx=\frac {85 c x \left (c+d x^3\right )^{5/12} \left (\frac {c \left (a+b x^3\right )}{a \left (c+d x^3\right )}\right )^{3/4} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {3}{4},\frac {4}{3},-\frac {(b c-a d) x^3}{a \left (d x^3+c\right )}\right )}{189 a^2 \left (a+b x^3\right )^{3/4}}+\frac {68 c x \left (c+d x^3\right )^{5/12}}{189 a^2 \left (a+b x^3\right )^{3/4}}+\frac {4 x \left (c+d x^3\right )^{17/12}}{21 a \left (a+b x^3\right )^{7/4}} \]
[In]
[Out]
Rule 386
Rule 388
Rubi steps \begin{align*} \text {integral}& = \frac {4 x \left (c+d x^3\right )^{17/12}}{21 a \left (a+b x^3\right )^{7/4}}+\frac {(17 c) \int \frac {\left (c+d x^3\right )^{5/12}}{\left (a+b x^3\right )^{7/4}} \, dx}{21 a} \\ & = \frac {68 c x \left (c+d x^3\right )^{5/12}}{189 a^2 \left (a+b x^3\right )^{3/4}}+\frac {4 x \left (c+d x^3\right )^{17/12}}{21 a \left (a+b x^3\right )^{7/4}}+\frac {\left (85 c^2\right ) \int \frac {1}{\left (a+b x^3\right )^{3/4} \left (c+d x^3\right )^{7/12}} \, dx}{189 a^2} \\ & = \frac {68 c x \left (c+d x^3\right )^{5/12}}{189 a^2 \left (a+b x^3\right )^{3/4}}+\frac {4 x \left (c+d x^3\right )^{17/12}}{21 a \left (a+b x^3\right )^{7/4}}+\frac {85 c 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 )}{189 a^2 \left (a+b x^3\right )^{3/4}} \\ \end{align*}
Time = 5.70 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.59 \[ \int \frac {\left (c+d x^3\right )^{17/12}}{\left (a+b x^3\right )^{11/4}} \, dx=\frac {c x \left (1+\frac {b x^3}{a}\right )^{3/4} \left (c+d x^3\right )^{5/12} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {11}{4},\frac {4}{3},\frac {(-b c+a d) x^3}{a \left (c+d x^3\right )}\right )}{a^2 \left (a+b x^3\right )^{3/4} \left (1+\frac {d x^3}{c}\right )^{3/4}} \]
[In]
[Out]
\[\int \frac {\left (d \,x^{3}+c \right )^{\frac {17}{12}}}{\left (b \,x^{3}+a \right )^{\frac {11}{4}}}d x\]
[In]
[Out]
\[ \int \frac {\left (c+d x^3\right )^{17/12}}{\left (a+b x^3\right )^{11/4}} \, dx=\int { \frac {{\left (d x^{3} + c\right )}^{\frac {17}{12}}}{{\left (b x^{3} + a\right )}^{\frac {11}{4}}} \,d x } \]
[In]
[Out]
Timed out. \[ \int \frac {\left (c+d x^3\right )^{17/12}}{\left (a+b x^3\right )^{11/4}} \, dx=\text {Timed out} \]
[In]
[Out]
\[ \int \frac {\left (c+d x^3\right )^{17/12}}{\left (a+b x^3\right )^{11/4}} \, dx=\int { \frac {{\left (d x^{3} + c\right )}^{\frac {17}{12}}}{{\left (b x^{3} + a\right )}^{\frac {11}{4}}} \,d x } \]
[In]
[Out]
\[ \int \frac {\left (c+d x^3\right )^{17/12}}{\left (a+b x^3\right )^{11/4}} \, dx=\int { \frac {{\left (d x^{3} + c\right )}^{\frac {17}{12}}}{{\left (b x^{3} + a\right )}^{\frac {11}{4}}} \,d x } \]
[In]
[Out]
Timed out. \[ \int \frac {\left (c+d x^3\right )^{17/12}}{\left (a+b x^3\right )^{11/4}} \, dx=\int \frac {{\left (d\,x^3+c\right )}^{17/12}}{{\left (b\,x^3+a\right )}^{11/4}} \,d x \]
[In]
[Out]