Integrand size = 23, antiderivative size = 126 \[ \int \frac {\sin ^2(c+d x)}{\sqrt [3]{a+a \sin (c+d x)}} \, dx=\frac {9 \cos (c+d x)}{10 d \sqrt [3]{a+a \sin (c+d x)}}-\frac {7 \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{6},\frac {3}{2},\frac {1}{2} (1-\sin (c+d x))\right )}{5\ 2^{5/6} d \sqrt [6]{1+\sin (c+d x)} \sqrt [3]{a+a \sin (c+d x)}}-\frac {3 \cos (c+d x) (a+a \sin (c+d x))^{2/3}}{5 a d} \]
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Time = 0.10 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {2838, 2830, 2731, 2730} \[ \int \frac {\sin ^2(c+d x)}{\sqrt [3]{a+a \sin (c+d x)}} \, dx=-\frac {7 \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{6},\frac {3}{2},\frac {1}{2} (1-\sin (c+d x))\right )}{5\ 2^{5/6} d \sqrt [6]{\sin (c+d x)+1} \sqrt [3]{a \sin (c+d x)+a}}-\frac {3 \cos (c+d x) (a \sin (c+d x)+a)^{2/3}}{5 a d}+\frac {9 \cos (c+d x)}{10 d \sqrt [3]{a \sin (c+d x)+a}} \]
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Rule 2730
Rule 2731
Rule 2830
Rule 2838
Rubi steps \begin{align*} \text {integral}& = -\frac {3 \cos (c+d x) (a+a \sin (c+d x))^{2/3}}{5 a d}+\frac {3 \int \frac {\frac {2 a}{3}-a \sin (c+d x)}{\sqrt [3]{a+a \sin (c+d x)}} \, dx}{5 a} \\ & = \frac {9 \cos (c+d x)}{10 d \sqrt [3]{a+a \sin (c+d x)}}-\frac {3 \cos (c+d x) (a+a \sin (c+d x))^{2/3}}{5 a d}+\frac {7}{10} \int \frac {1}{\sqrt [3]{a+a \sin (c+d x)}} \, dx \\ & = \frac {9 \cos (c+d x)}{10 d \sqrt [3]{a+a \sin (c+d x)}}-\frac {3 \cos (c+d x) (a+a \sin (c+d x))^{2/3}}{5 a d}+\frac {\left (7 \sqrt [3]{1+\sin (c+d x)}\right ) \int \frac {1}{\sqrt [3]{1+\sin (c+d x)}} \, dx}{10 \sqrt [3]{a+a \sin (c+d x)}} \\ & = \frac {9 \cos (c+d x)}{10 d \sqrt [3]{a+a \sin (c+d x)}}-\frac {7 \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{6},\frac {3}{2},\frac {1}{2} (1-\sin (c+d x))\right )}{5\ 2^{5/6} d \sqrt [6]{1+\sin (c+d x)} \sqrt [3]{a+a \sin (c+d x)}}-\frac {3 \cos (c+d x) (a+a \sin (c+d x))^{2/3}}{5 a d} \\ \end{align*}
Time = 0.57 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.75 \[ \int \frac {\sin ^2(c+d x)}{\sqrt [3]{a+a \sin (c+d x)}} \, dx=-\frac {3 \cos (c+d x) \left (-14 \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {7}{6},\sin ^2\left (\frac {1}{4} (2 c+\pi +2 d x)\right )\right )+\sqrt {2-2 \sin (c+d x)} (-1+2 \sin (c+d x))\right )}{10 d \sqrt {2-2 \sin (c+d x)} \sqrt [3]{a (1+\sin (c+d x))}} \]
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\[\int \frac {\sin ^{2}\left (d x +c \right )}{\left (a +a \sin \left (d x +c \right )\right )^{\frac {1}{3}}}d x\]
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\[ \int \frac {\sin ^2(c+d x)}{\sqrt [3]{a+a \sin (c+d x)}} \, dx=\int { \frac {\sin \left (d x + c\right )^{2}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {1}{3}}} \,d x } \]
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\[ \int \frac {\sin ^2(c+d x)}{\sqrt [3]{a+a \sin (c+d x)}} \, dx=\int \frac {\sin ^{2}{\left (c + d x \right )}}{\sqrt [3]{a \left (\sin {\left (c + d x \right )} + 1\right )}}\, dx \]
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\[ \int \frac {\sin ^2(c+d x)}{\sqrt [3]{a+a \sin (c+d x)}} \, dx=\int { \frac {\sin \left (d x + c\right )^{2}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {1}{3}}} \,d x } \]
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\[ \int \frac {\sin ^2(c+d x)}{\sqrt [3]{a+a \sin (c+d x)}} \, dx=\int { \frac {\sin \left (d x + c\right )^{2}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {1}{3}}} \,d x } \]
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Timed out. \[ \int \frac {\sin ^2(c+d x)}{\sqrt [3]{a+a \sin (c+d x)}} \, dx=\int \frac {{\sin \left (c+d\,x\right )}^2}{{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{1/3}} \,d x \]
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