Integrand size = 21, antiderivative size = 155 \[ \int \frac {\sin ^{\frac {5}{3}}(a+b x)}{\cos ^{\frac {5}{3}}(a+b x)} \, dx=-\frac {\sqrt {3} \arctan \left (\frac {1-\frac {2 \cos ^{\frac {2}{3}}(a+b x)}{\sin ^{\frac {2}{3}}(a+b x)}}{\sqrt {3}}\right )}{2 b}+\frac {\log \left (1+\frac {\cos ^{\frac {4}{3}}(a+b x)}{\sin ^{\frac {4}{3}}(a+b x)}-\frac {\cos ^{\frac {2}{3}}(a+b x)}{\sin ^{\frac {2}{3}}(a+b x)}\right )}{4 b}-\frac {\log \left (1+\frac {\cos ^{\frac {2}{3}}(a+b x)}{\sin ^{\frac {2}{3}}(a+b x)}\right )}{2 b}+\frac {3 \sin ^{\frac {2}{3}}(a+b x)}{2 b \cos ^{\frac {2}{3}}(a+b x)} \]
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Time = 0.11 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {2646, 2655, 281, 298, 31, 648, 632, 210, 642} \[ \int \frac {\sin ^{\frac {5}{3}}(a+b x)}{\cos ^{\frac {5}{3}}(a+b x)} \, dx=-\frac {\sqrt {3} \arctan \left (\frac {1-\frac {2 \cos ^{\frac {2}{3}}(a+b x)}{\sin ^{\frac {2}{3}}(a+b x)}}{\sqrt {3}}\right )}{2 b}+\frac {3 \sin ^{\frac {2}{3}}(a+b x)}{2 b \cos ^{\frac {2}{3}}(a+b x)}+\frac {\log \left (\frac {\cos ^{\frac {4}{3}}(a+b x)}{\sin ^{\frac {4}{3}}(a+b x)}-\frac {\cos ^{\frac {2}{3}}(a+b x)}{\sin ^{\frac {2}{3}}(a+b x)}+1\right )}{4 b}-\frac {\log \left (\frac {\cos ^{\frac {2}{3}}(a+b x)}{\sin ^{\frac {2}{3}}(a+b x)}+1\right )}{2 b} \]
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Rule 31
Rule 210
Rule 281
Rule 298
Rule 632
Rule 642
Rule 648
Rule 2646
Rule 2655
Rubi steps \begin{align*} \text {integral}& = \frac {3 \sin ^{\frac {2}{3}}(a+b x)}{2 b \cos ^{\frac {2}{3}}(a+b x)}-\int \frac {\sqrt [3]{\cos (a+b x)}}{\sqrt [3]{\sin (a+b x)}} \, dx \\ & = \frac {3 \sin ^{\frac {2}{3}}(a+b x)}{2 b \cos ^{\frac {2}{3}}(a+b x)}+\frac {3 \text {Subst}\left (\int \frac {x^3}{1+x^6} \, dx,x,\frac {\sqrt [3]{\cos (a+b x)}}{\sqrt [3]{\sin (a+b x)}}\right )}{b} \\ & = \frac {3 \sin ^{\frac {2}{3}}(a+b x)}{2 b \cos ^{\frac {2}{3}}(a+b x)}+\frac {3 \text {Subst}\left (\int \frac {x}{1+x^3} \, dx,x,\frac {\cos ^{\frac {2}{3}}(a+b x)}{\sin ^{\frac {2}{3}}(a+b x)}\right )}{2 b} \\ & = \frac {3 \sin ^{\frac {2}{3}}(a+b x)}{2 b \cos ^{\frac {2}{3}}(a+b x)}-\frac {\text {Subst}\left (\int \frac {1}{1+x} \, dx,x,\frac {\cos ^{\frac {2}{3}}(a+b x)}{\sin ^{\frac {2}{3}}(a+b x)}\right )}{2 b}+\frac {\text {Subst}\left (\int \frac {1+x}{1-x+x^2} \, dx,x,\frac {\cos ^{\frac {2}{3}}(a+b x)}{\sin ^{\frac {2}{3}}(a+b x)}\right )}{2 b} \\ & = -\frac {\log \left (1+\frac {\cos ^{\frac {2}{3}}(a+b x)}{\sin ^{\frac {2}{3}}(a+b x)}\right )}{2 b}+\frac {3 \sin ^{\frac {2}{3}}(a+b x)}{2 b \cos ^{\frac {2}{3}}(a+b x)}+\frac {\text {Subst}\left (\int \frac {-1+2 x}{1-x+x^2} \, dx,x,\frac {\cos ^{\frac {2}{3}}(a+b x)}{\sin ^{\frac {2}{3}}(a+b x)}\right )}{4 b}+\frac {3 \text {Subst}\left (\int \frac {1}{1-x+x^2} \, dx,x,\frac {\cos ^{\frac {2}{3}}(a+b x)}{\sin ^{\frac {2}{3}}(a+b x)}\right )}{4 b} \\ & = \frac {\log \left (1+\frac {\cos ^{\frac {4}{3}}(a+b x)}{\sin ^{\frac {4}{3}}(a+b x)}-\frac {\cos ^{\frac {2}{3}}(a+b x)}{\sin ^{\frac {2}{3}}(a+b x)}\right )}{4 b}-\frac {\log \left (1+\frac {\cos ^{\frac {2}{3}}(a+b x)}{\sin ^{\frac {2}{3}}(a+b x)}\right )}{2 b}+\frac {3 \sin ^{\frac {2}{3}}(a+b x)}{2 b \cos ^{\frac {2}{3}}(a+b x)}-\frac {3 \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,-1+\frac {2 \cos ^{\frac {2}{3}}(a+b x)}{\sin ^{\frac {2}{3}}(a+b x)}\right )}{2 b} \\ & = -\frac {\sqrt {3} \arctan \left (\frac {1-\frac {2 \cos ^{\frac {2}{3}}(a+b x)}{\sin ^{\frac {2}{3}}(a+b x)}}{\sqrt {3}}\right )}{2 b}+\frac {\log \left (1+\frac {\cos ^{\frac {4}{3}}(a+b x)}{\sin ^{\frac {4}{3}}(a+b x)}-\frac {\cos ^{\frac {2}{3}}(a+b x)}{\sin ^{\frac {2}{3}}(a+b x)}\right )}{4 b}-\frac {\log \left (1+\frac {\cos ^{\frac {2}{3}}(a+b x)}{\sin ^{\frac {2}{3}}(a+b x)}\right )}{2 b}+\frac {3 \sin ^{\frac {2}{3}}(a+b x)}{2 b \cos ^{\frac {2}{3}}(a+b x)} \\ \end{align*}
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.04 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.37 \[ \int \frac {\sin ^{\frac {5}{3}}(a+b x)}{\cos ^{\frac {5}{3}}(a+b x)} \, dx=\frac {3 \sqrt [3]{\cos ^2(a+b x)} \operatorname {Hypergeometric2F1}\left (\frac {4}{3},\frac {4}{3},\frac {7}{3},\sin ^2(a+b x)\right ) \sin ^{\frac {8}{3}}(a+b x)}{8 b \cos ^{\frac {2}{3}}(a+b x)} \]
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\[\int \frac {\sin ^{\frac {5}{3}}\left (b x +a \right )}{\cos \left (b x +a \right )^{\frac {5}{3}}}d x\]
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Time = 0.32 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.27 \[ \int \frac {\sin ^{\frac {5}{3}}(a+b x)}{\cos ^{\frac {5}{3}}(a+b x)} \, dx=\frac {2 \, \sqrt {3} \arctan \left (\frac {2 \, \sqrt {3} \cos \left (b x + a\right )^{\frac {2}{3}} \sin \left (b x + a\right )^{\frac {1}{3}} - \sqrt {3} \sin \left (b x + a\right )}{3 \, \sin \left (b x + a\right )}\right ) \cos \left (b x + a\right ) + \cos \left (b x + a\right ) \log \left (\frac {4 \, {\left (\cos \left (b x + a\right )^{2} - \cos \left (b x + a\right )^{\frac {4}{3}} \sin \left (b x + a\right )^{\frac {2}{3}} + \cos \left (b x + a\right )^{\frac {2}{3}} \sin \left (b x + a\right )^{\frac {4}{3}} - 1\right )}}{\cos \left (b x + a\right )^{2} - 1}\right ) - 2 \, \cos \left (b x + a\right ) \log \left (-\frac {2 \, {\left (\cos \left (b x + a\right )^{\frac {2}{3}} \sin \left (b x + a\right )^{\frac {1}{3}} + \sin \left (b x + a\right )\right )}}{\sin \left (b x + a\right )}\right ) + 6 \, \cos \left (b x + a\right )^{\frac {1}{3}} \sin \left (b x + a\right )^{\frac {2}{3}}}{4 \, b \cos \left (b x + a\right )} \]
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Timed out. \[ \int \frac {\sin ^{\frac {5}{3}}(a+b x)}{\cos ^{\frac {5}{3}}(a+b x)} \, dx=\text {Timed out} \]
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\[ \int \frac {\sin ^{\frac {5}{3}}(a+b x)}{\cos ^{\frac {5}{3}}(a+b x)} \, dx=\int { \frac {\sin \left (b x + a\right )^{\frac {5}{3}}}{\cos \left (b x + a\right )^{\frac {5}{3}}} \,d x } \]
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\[ \int \frac {\sin ^{\frac {5}{3}}(a+b x)}{\cos ^{\frac {5}{3}}(a+b x)} \, dx=\int { \frac {\sin \left (b x + a\right )^{\frac {5}{3}}}{\cos \left (b x + a\right )^{\frac {5}{3}}} \,d x } \]
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Time = 1.01 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.28 \[ \int \frac {\sin ^{\frac {5}{3}}(a+b x)}{\cos ^{\frac {5}{3}}(a+b x)} \, dx=\frac {3\,{\sin \left (a+b\,x\right )}^{8/3}\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{3},-\frac {1}{3};\ \frac {2}{3};\ {\cos \left (a+b\,x\right )}^2\right )}{2\,b\,{\cos \left (a+b\,x\right )}^{2/3}\,{\left ({\sin \left (a+b\,x\right )}^2\right )}^{4/3}} \]
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