Integrand size = 25, antiderivative size = 143 \[ \int \frac {\cot ^3(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=\frac {(2 a+5 b) \text {arctanh}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a}}\right )}{2 a^{7/2} f}-\frac {2 a+5 b}{6 a^2 f \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac {\csc ^2(e+f x)}{2 a f \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac {2 a+5 b}{2 a^3 f \sqrt {a+b \sin ^2(e+f x)}} \]
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Time = 0.17 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3273, 79, 53, 65, 214} \[ \int \frac {\cot ^3(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=\frac {(2 a+5 b) \text {arctanh}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a}}\right )}{2 a^{7/2} f}-\frac {2 a+5 b}{2 a^3 f \sqrt {a+b \sin ^2(e+f x)}}-\frac {2 a+5 b}{6 a^2 f \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac {\csc ^2(e+f x)}{2 a f \left (a+b \sin ^2(e+f x)\right )^{3/2}} \]
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Rule 53
Rule 65
Rule 79
Rule 214
Rule 3273
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {1-x}{x^2 (a+b x)^{5/2}} \, dx,x,\sin ^2(e+f x)\right )}{2 f} \\ & = -\frac {\csc ^2(e+f x)}{2 a f \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac {(2 a+5 b) \text {Subst}\left (\int \frac {1}{x (a+b x)^{5/2}} \, dx,x,\sin ^2(e+f x)\right )}{4 a f} \\ & = -\frac {2 a+5 b}{6 a^2 f \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac {\csc ^2(e+f x)}{2 a f \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac {(2 a+5 b) \text {Subst}\left (\int \frac {1}{x (a+b x)^{3/2}} \, dx,x,\sin ^2(e+f x)\right )}{4 a^2 f} \\ & = -\frac {2 a+5 b}{6 a^2 f \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac {\csc ^2(e+f x)}{2 a f \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac {2 a+5 b}{2 a^3 f \sqrt {a+b \sin ^2(e+f x)}}-\frac {(2 a+5 b) \text {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,\sin ^2(e+f x)\right )}{4 a^3 f} \\ & = -\frac {2 a+5 b}{6 a^2 f \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac {\csc ^2(e+f x)}{2 a f \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac {2 a+5 b}{2 a^3 f \sqrt {a+b \sin ^2(e+f x)}}-\frac {(2 a+5 b) \text {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b \sin ^2(e+f x)}\right )}{2 a^3 b f} \\ & = \frac {(2 a+5 b) \text {arctanh}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a}}\right )}{2 a^{7/2} f}-\frac {2 a+5 b}{6 a^2 f \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac {\csc ^2(e+f x)}{2 a f \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac {2 a+5 b}{2 a^3 f \sqrt {a+b \sin ^2(e+f x)}} \\ \end{align*}
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.27 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.48 \[ \int \frac {\cot ^3(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=-\frac {3 a \csc ^2(e+f x)+(2 a+5 b) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},1+\frac {b \sin ^2(e+f x)}{a}\right )}{6 a^2 f \left (a+b \sin ^2(e+f x)\right )^{3/2}} \]
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Leaf count of result is larger than twice the leaf count of optimal. \(1037\) vs. \(2(123)=246\).
Time = 2.25 (sec) , antiderivative size = 1038, normalized size of antiderivative = 7.26
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Leaf count of result is larger than twice the leaf count of optimal. 320 vs. \(2 (123) = 246\).
Time = 0.38 (sec) , antiderivative size = 666, normalized size of antiderivative = 4.66 \[ \int \frac {\cot ^3(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=\left [\frac {3 \, {\left ({\left (2 \, a b^{2} + 5 \, b^{3}\right )} \cos \left (f x + e\right )^{6} - {\left (4 \, a^{2} b + 16 \, a b^{2} + 15 \, b^{3}\right )} \cos \left (f x + e\right )^{4} - 2 \, a^{3} - 9 \, a^{2} b - 12 \, a b^{2} - 5 \, b^{3} + {\left (2 \, a^{3} + 13 \, a^{2} b + 26 \, a b^{2} + 15 \, b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {a} \log \left (\frac {2 \, {\left (b \cos \left (f x + e\right )^{2} - 2 \, \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {a} - 2 \, a - b\right )}}{\cos \left (f x + e\right )^{2} - 1}\right ) + 2 \, {\left (3 \, {\left (2 \, a^{2} b + 5 \, a b^{2}\right )} \cos \left (f x + e\right )^{4} + 11 \, a^{3} + 26 \, a^{2} b + 15 \, a b^{2} - 2 \, {\left (4 \, a^{3} + 16 \, a^{2} b + 15 \, a b^{2}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{12 \, {\left (a^{4} b^{2} f \cos \left (f x + e\right )^{6} - {\left (2 \, a^{5} b + 3 \, a^{4} b^{2}\right )} f \cos \left (f x + e\right )^{4} + {\left (a^{6} + 4 \, a^{5} b + 3 \, a^{4} b^{2}\right )} f \cos \left (f x + e\right )^{2} - {\left (a^{6} + 2 \, a^{5} b + a^{4} b^{2}\right )} f\right )}}, -\frac {3 \, {\left ({\left (2 \, a b^{2} + 5 \, b^{3}\right )} \cos \left (f x + e\right )^{6} - {\left (4 \, a^{2} b + 16 \, a b^{2} + 15 \, b^{3}\right )} \cos \left (f x + e\right )^{4} - 2 \, a^{3} - 9 \, a^{2} b - 12 \, a b^{2} - 5 \, b^{3} + {\left (2 \, a^{3} + 13 \, a^{2} b + 26 \, a b^{2} + 15 \, b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {-a}}{a}\right ) - {\left (3 \, {\left (2 \, a^{2} b + 5 \, a b^{2}\right )} \cos \left (f x + e\right )^{4} + 11 \, a^{3} + 26 \, a^{2} b + 15 \, a b^{2} - 2 \, {\left (4 \, a^{3} + 16 \, a^{2} b + 15 \, a b^{2}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{6 \, {\left (a^{4} b^{2} f \cos \left (f x + e\right )^{6} - {\left (2 \, a^{5} b + 3 \, a^{4} b^{2}\right )} f \cos \left (f x + e\right )^{4} + {\left (a^{6} + 4 \, a^{5} b + 3 \, a^{4} b^{2}\right )} f \cos \left (f x + e\right )^{2} - {\left (a^{6} + 2 \, a^{5} b + a^{4} b^{2}\right )} f\right )}}\right ] \]
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\[ \int \frac {\cot ^3(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=\int \frac {\cot ^{3}{\left (e + f x \right )}}{\left (a + b \sin ^{2}{\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \]
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Time = 0.26 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.09 \[ \int \frac {\cot ^3(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=\frac {\frac {6 \, \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | \sin \left (f x + e\right ) \right |}}\right )}{a^{\frac {5}{2}}} + \frac {15 \, b \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | \sin \left (f x + e\right ) \right |}}\right )}{a^{\frac {7}{2}}} - \frac {6}{\sqrt {b \sin \left (f x + e\right )^{2} + a} a^{2}} - \frac {2}{{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} a} - \frac {15 \, b}{\sqrt {b \sin \left (f x + e\right )^{2} + a} a^{3}} - \frac {5 \, b}{{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} a^{2}} - \frac {3}{{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} a \sin \left (f x + e\right )^{2}}}{6 \, f} \]
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\[ \int \frac {\cot ^3(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=\int { \frac {\cot \left (f x + e\right )^{3}}{{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {5}{2}}} \,d x } \]
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Timed out. \[ \int \frac {\cot ^3(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=\int \frac {{\mathrm {cot}\left (e+f\,x\right )}^3}{{\left (b\,{\sin \left (e+f\,x\right )}^2+a\right )}^{5/2}} \,d x \]
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