Integrand size = 21, antiderivative size = 155 \[ \int \frac {\cos ^{\frac {5}{3}}(a+b x)}{\sin ^{\frac {5}{3}}(a+b x)} \, dx=\frac {\sqrt {3} \arctan \left (\frac {1-\frac {2 \sin ^{\frac {2}{3}}(a+b x)}{\cos ^{\frac {2}{3}}(a+b x)}}{\sqrt {3}}\right )}{2 b}+\frac {\log \left (1+\frac {\sin ^{\frac {2}{3}}(a+b x)}{\cos ^{\frac {2}{3}}(a+b x)}\right )}{2 b}-\frac {\log \left (1-\frac {\sin ^{\frac {2}{3}}(a+b x)}{\cos ^{\frac {2}{3}}(a+b x)}+\frac {\sin ^{\frac {4}{3}}(a+b x)}{\cos ^{\frac {4}{3}}(a+b x)}\right )}{4 b}-\frac {3 \cos ^{\frac {2}{3}}(a+b x)}{2 b \sin ^{\frac {2}{3}}(a+b x)} \]
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Time = 0.07 (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 = {2647, 2654, 281, 298, 31, 648, 632, 210, 642} \[ \int \frac {\cos ^{\frac {5}{3}}(a+b x)}{\sin ^{\frac {5}{3}}(a+b x)} \, dx=\frac {\sqrt {3} \arctan \left (\frac {1-\frac {2 \sin ^{\frac {2}{3}}(a+b x)}{\cos ^{\frac {2}{3}}(a+b x)}}{\sqrt {3}}\right )}{2 b}-\frac {3 \cos ^{\frac {2}{3}}(a+b x)}{2 b \sin ^{\frac {2}{3}}(a+b x)}+\frac {\log \left (\frac {\sin ^{\frac {2}{3}}(a+b x)}{\cos ^{\frac {2}{3}}(a+b x)}+1\right )}{2 b}-\frac {\log \left (\frac {\sin ^{\frac {4}{3}}(a+b x)}{\cos ^{\frac {4}{3}}(a+b x)}-\frac {\sin ^{\frac {2}{3}}(a+b x)}{\cos ^{\frac {2}{3}}(a+b x)}+1\right )}{4 b} \]
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Rule 31
Rule 210
Rule 281
Rule 298
Rule 632
Rule 642
Rule 648
Rule 2647
Rule 2654
Rubi steps \begin{align*} \text {integral}& = -\frac {3 \cos ^{\frac {2}{3}}(a+b x)}{2 b \sin ^{\frac {2}{3}}(a+b x)}-\int \frac {\sqrt [3]{\sin (a+b x)}}{\sqrt [3]{\cos (a+b x)}} \, dx \\ & = -\frac {3 \cos ^{\frac {2}{3}}(a+b x)}{2 b \sin ^{\frac {2}{3}}(a+b x)}-\frac {3 \text {Subst}\left (\int \frac {x^3}{1+x^6} \, dx,x,\frac {\sqrt [3]{\sin (a+b x)}}{\sqrt [3]{\cos (a+b x)}}\right )}{b} \\ & = -\frac {3 \cos ^{\frac {2}{3}}(a+b x)}{2 b \sin ^{\frac {2}{3}}(a+b x)}-\frac {3 \text {Subst}\left (\int \frac {x}{1+x^3} \, dx,x,\frac {\sin ^{\frac {2}{3}}(a+b x)}{\cos ^{\frac {2}{3}}(a+b x)}\right )}{2 b} \\ & = -\frac {3 \cos ^{\frac {2}{3}}(a+b x)}{2 b \sin ^{\frac {2}{3}}(a+b x)}+\frac {\text {Subst}\left (\int \frac {1}{1+x} \, dx,x,\frac {\sin ^{\frac {2}{3}}(a+b x)}{\cos ^{\frac {2}{3}}(a+b x)}\right )}{2 b}-\frac {\text {Subst}\left (\int \frac {1+x}{1-x+x^2} \, dx,x,\frac {\sin ^{\frac {2}{3}}(a+b x)}{\cos ^{\frac {2}{3}}(a+b x)}\right )}{2 b} \\ & = \frac {\log \left (1+\frac {\sin ^{\frac {2}{3}}(a+b x)}{\cos ^{\frac {2}{3}}(a+b x)}\right )}{2 b}-\frac {3 \cos ^{\frac {2}{3}}(a+b x)}{2 b \sin ^{\frac {2}{3}}(a+b x)}-\frac {\text {Subst}\left (\int \frac {-1+2 x}{1-x+x^2} \, dx,x,\frac {\sin ^{\frac {2}{3}}(a+b x)}{\cos ^{\frac {2}{3}}(a+b x)}\right )}{4 b}-\frac {3 \text {Subst}\left (\int \frac {1}{1-x+x^2} \, dx,x,\frac {\sin ^{\frac {2}{3}}(a+b x)}{\cos ^{\frac {2}{3}}(a+b x)}\right )}{4 b} \\ & = \frac {\log \left (1+\frac {\sin ^{\frac {2}{3}}(a+b x)}{\cos ^{\frac {2}{3}}(a+b x)}\right )}{2 b}-\frac {\log \left (1-\frac {\sin ^{\frac {2}{3}}(a+b x)}{\cos ^{\frac {2}{3}}(a+b x)}+\frac {\sin ^{\frac {4}{3}}(a+b x)}{\cos ^{\frac {4}{3}}(a+b x)}\right )}{4 b}-\frac {3 \cos ^{\frac {2}{3}}(a+b x)}{2 b \sin ^{\frac {2}{3}}(a+b x)}+\frac {3 \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,-1+\frac {2 \sin ^{\frac {2}{3}}(a+b x)}{\cos ^{\frac {2}{3}}(a+b x)}\right )}{2 b} \\ & = \frac {\sqrt {3} \arctan \left (\frac {1-\frac {2 \sin ^{\frac {2}{3}}(a+b x)}{\cos ^{\frac {2}{3}}(a+b x)}}{\sqrt {3}}\right )}{2 b}+\frac {\log \left (1+\frac {\sin ^{\frac {2}{3}}(a+b x)}{\cos ^{\frac {2}{3}}(a+b x)}\right )}{2 b}-\frac {\log \left (1-\frac {\sin ^{\frac {2}{3}}(a+b x)}{\cos ^{\frac {2}{3}}(a+b x)}+\frac {\sin ^{\frac {4}{3}}(a+b x)}{\cos ^{\frac {4}{3}}(a+b x)}\right )}{4 b}-\frac {3 \cos ^{\frac {2}{3}}(a+b x)}{2 b \sin ^{\frac {2}{3}}(a+b x)} \\ \end{align*}
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.02 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.37 \[ \int \frac {\cos ^{\frac {5}{3}}(a+b x)}{\sin ^{\frac {5}{3}}(a+b x)} \, dx=-\frac {3 \cos ^2(a+b x)^{2/3} \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},-\frac {1}{3},\frac {2}{3},\sin ^2(a+b x)\right )}{2 b \cos ^{\frac {4}{3}}(a+b x) \sin ^{\frac {2}{3}}(a+b x)} \]
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\[\int \frac {\cos ^{\frac {5}{3}}\left (b x +a \right )}{\sin \left (b x +a \right )^{\frac {5}{3}}}d x\]
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none
Time = 0.32 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.22 \[ \int \frac {\cos ^{\frac {5}{3}}(a+b x)}{\sin ^{\frac {5}{3}}(a+b x)} \, dx=-\frac {2 \, \sqrt {3} \arctan \left (-\frac {\sqrt {3} \cos \left (b x + a\right ) - 2 \, \sqrt {3} \cos \left (b x + a\right )^{\frac {1}{3}} \sin \left (b x + a\right )^{\frac {2}{3}}}{3 \, \cos \left (b x + a\right )}\right ) \sin \left (b x + a\right ) - 2 \, \log \left (\frac {\cos \left (b x + a\right )^{\frac {1}{3}} \sin \left (b x + a\right )^{\frac {2}{3}} + \cos \left (b x + a\right )}{\cos \left (b x + a\right )}\right ) \sin \left (b x + a\right ) + \log \left (\frac {\cos \left (b x + a\right )^{2} - \cos \left (b x + a\right )^{\frac {4}{3}} \sin \left (b x + a\right )^{\frac {2}{3}} + \cos \left (b x + a\right )^{\frac {2}{3}} \sin \left (b x + a\right )^{\frac {4}{3}}}{\cos \left (b x + a\right )^{2}}\right ) \sin \left (b x + a\right ) + 6 \, \cos \left (b x + a\right )^{\frac {2}{3}} \sin \left (b x + a\right )^{\frac {1}{3}}}{4 \, b \sin \left (b x + a\right )} \]
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Timed out. \[ \int \frac {\cos ^{\frac {5}{3}}(a+b x)}{\sin ^{\frac {5}{3}}(a+b x)} \, dx=\text {Timed out} \]
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\[ \int \frac {\cos ^{\frac {5}{3}}(a+b x)}{\sin ^{\frac {5}{3}}(a+b x)} \, dx=\int { \frac {\cos \left (b x + a\right )^{\frac {5}{3}}}{\sin \left (b x + a\right )^{\frac {5}{3}}} \,d x } \]
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\[ \int \frac {\cos ^{\frac {5}{3}}(a+b x)}{\sin ^{\frac {5}{3}}(a+b x)} \, dx=\int { \frac {\cos \left (b x + a\right )^{\frac {5}{3}}}{\sin \left (b x + a\right )^{\frac {5}{3}}} \,d x } \]
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Time = 1.03 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.28 \[ \int \frac {\cos ^{\frac {5}{3}}(a+b x)}{\sin ^{\frac {5}{3}}(a+b x)} \, dx=-\frac {3\,{\cos \left (a+b\,x\right )}^{8/3}\,{\left ({\sin \left (a+b\,x\right )}^2\right )}^{1/3}\,{{}}_2{\mathrm {F}}_1\left (\frac {4}{3},\frac {4}{3};\ \frac {7}{3};\ {\cos \left (a+b\,x\right )}^2\right )}{8\,b\,{\sin \left (a+b\,x\right )}^{2/3}} \]
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