Integrand size = 25, antiderivative size = 199 \[ \int \frac {\sin ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=-\frac {\left (15 a^2+10 a b-b^2\right ) \cos (e+f x)}{15 (a-b)^3 f \sqrt {a-b+b \sec ^2(e+f x)}}+\frac {2 (5 a-2 b) \cos ^3(e+f x)}{15 (a-b)^2 f \sqrt {a-b+b \sec ^2(e+f x)}}-\frac {\cos ^5(e+f x)}{5 (a-b) f \sqrt {a-b+b \sec ^2(e+f x)}}-\frac {2 b \left (15 a^2+10 a b-b^2\right ) \sec (e+f x)}{15 (a-b)^4 f \sqrt {a-b+b \sec ^2(e+f x)}} \]
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Time = 0.22 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3745, 473, 464, 277, 197} \[ \int \frac {\sin ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=-\frac {2 b \left (15 a^2+10 a b-b^2\right ) \sec (e+f x)}{15 f (a-b)^4 \sqrt {a+b \sec ^2(e+f x)-b}}-\frac {\left (15 a^2+10 a b-b^2\right ) \cos (e+f x)}{15 f (a-b)^3 \sqrt {a+b \sec ^2(e+f x)-b}}-\frac {\cos ^5(e+f x)}{5 f (a-b) \sqrt {a+b \sec ^2(e+f x)-b}}+\frac {2 (5 a-2 b) \cos ^3(e+f x)}{15 f (a-b)^2 \sqrt {a+b \sec ^2(e+f x)-b}} \]
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Rule 197
Rule 277
Rule 464
Rule 473
Rule 3745
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {\left (-1+x^2\right )^2}{x^6 \left (a-b+b x^2\right )^{3/2}} \, dx,x,\sec (e+f x)\right )}{f} \\ & = -\frac {\cos ^5(e+f x)}{5 (a-b) f \sqrt {a-b+b \sec ^2(e+f x)}}+\frac {\text {Subst}\left (\int \frac {-2 (5 a-2 b)+5 (a-b) x^2}{x^4 \left (a-b+b x^2\right )^{3/2}} \, dx,x,\sec (e+f x)\right )}{5 (a-b) f} \\ & = \frac {2 (5 a-2 b) \cos ^3(e+f x)}{15 (a-b)^2 f \sqrt {a-b+b \sec ^2(e+f x)}}-\frac {\cos ^5(e+f x)}{5 (a-b) f \sqrt {a-b+b \sec ^2(e+f x)}}+\frac {\left (15 a^2+10 a b-b^2\right ) \text {Subst}\left (\int \frac {1}{x^2 \left (a-b+b x^2\right )^{3/2}} \, dx,x,\sec (e+f x)\right )}{15 (a-b)^2 f} \\ & = -\frac {\left (15 a^2+10 a b-b^2\right ) \cos (e+f x)}{15 (a-b)^3 f \sqrt {a-b+b \sec ^2(e+f x)}}+\frac {2 (5 a-2 b) \cos ^3(e+f x)}{15 (a-b)^2 f \sqrt {a-b+b \sec ^2(e+f x)}}-\frac {\cos ^5(e+f x)}{5 (a-b) f \sqrt {a-b+b \sec ^2(e+f x)}}-\frac {\left (2 b \left (15 a^2+10 a b-b^2\right )\right ) \text {Subst}\left (\int \frac {1}{\left (a-b+b x^2\right )^{3/2}} \, dx,x,\sec (e+f x)\right )}{15 (a-b)^3 f} \\ & = -\frac {\left (15 a^2+10 a b-b^2\right ) \cos (e+f x)}{15 (a-b)^3 f \sqrt {a-b+b \sec ^2(e+f x)}}+\frac {2 (5 a-2 b) \cos ^3(e+f x)}{15 (a-b)^2 f \sqrt {a-b+b \sec ^2(e+f x)}}-\frac {\cos ^5(e+f x)}{5 (a-b) f \sqrt {a-b+b \sec ^2(e+f x)}}-\frac {2 b \left (15 a^2+10 a b-b^2\right ) \sec (e+f x)}{15 (a-b)^4 f \sqrt {a-b+b \sec ^2(e+f x)}} \\ \end{align*}
Time = 3.63 (sec) , antiderivative size = 186, normalized size of antiderivative = 0.93 \[ \int \frac {\sin ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=-\frac {\left (150 a^3+1078 a^2 b+338 a b^2-30 b^3+\left (125 a^3+169 a^2 b-329 a b^2+35 b^3\right ) \cos (2 (e+f x))-2 (a-b)^2 (11 a+b) \cos (4 (e+f x))+3 a^3 \cos (6 (e+f x))-9 a^2 b \cos (6 (e+f x))+9 a b^2 \cos (6 (e+f x))-3 b^3 \cos (6 (e+f x))\right ) \sec (e+f x)}{240 \sqrt {2} (a-b)^4 f \sqrt {(a+b+(a-b) \cos (2 (e+f x))) \sec ^2(e+f x)}} \]
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Time = 3.00 (sec) , antiderivative size = 273, normalized size of antiderivative = 1.37
method | result | size |
default | \(\frac {a^{6} \left (a -b \right )^{2} \left (3 \sin \left (f x +e \right )^{4} \cos \left (f x +e \right )^{2} b^{3}+9 \sin \left (f x +e \right )^{2} \cos \left (f x +e \right )^{4} a \,b^{2}-3 a^{3} \cos \left (f x +e \right )^{6}+9 a^{2} b \cos \left (f x +e \right )^{6}+2 \sin \left (f x +e \right )^{4} b^{3}-9 \sin \left (f x +e \right )^{2} \cos \left (f x +e \right )^{2} a \,b^{2}+10 a^{3} \cos \left (f x +e \right )^{4}-24 a^{2} b \cos \left (f x +e \right )^{4}-20 \sin \left (f x +e \right )^{2} a \,b^{2}-15 a^{3} \cos \left (f x +e \right )^{2}+5 a^{2} b \cos \left (f x +e \right )^{2}-30 a^{2} b \right ) \left (a \cos \left (f x +e \right )^{2}+b \sin \left (f x +e \right )^{2}\right ) \sec \left (f x +e \right )^{3}}{15 f \left (\sqrt {-b \left (a -b \right )}+a -b \right )^{6} \left (\sqrt {-b \left (a -b \right )}-a +b \right )^{6} \left (a +b \tan \left (f x +e \right )^{2}\right )^{\frac {3}{2}}}\) | \(273\) |
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Time = 0.40 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.17 \[ \int \frac {\sin ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=-\frac {{\left (3 \, {\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} \cos \left (f x + e\right )^{7} - 2 \, {\left (5 \, a^{3} - 12 \, a^{2} b + 9 \, a b^{2} - 2 \, b^{3}\right )} \cos \left (f x + e\right )^{5} + {\left (15 \, a^{3} - 5 \, a^{2} b - 11 \, a b^{2} + b^{3}\right )} \cos \left (f x + e\right )^{3} + 2 \, {\left (15 \, a^{2} b + 10 \, a b^{2} - b^{3}\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{15 \, {\left ({\left (a^{5} - 5 \, a^{4} b + 10 \, a^{3} b^{2} - 10 \, a^{2} b^{3} + 5 \, a b^{4} - b^{5}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{4} b - 4 \, a^{3} b^{2} + 6 \, a^{2} b^{3} - 4 \, a b^{4} + b^{5}\right )} f\right )}} \]
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Timed out. \[ \int \frac {\sin ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=\text {Timed out} \]
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Leaf count of result is larger than twice the leaf count of optimal. 389 vs. \(2 (183) = 366\).
Time = 0.26 (sec) , antiderivative size = 389, normalized size of antiderivative = 1.95 \[ \int \frac {\sin ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=-\frac {\frac {15 \, b^{3}}{{\left (a^{4} - 4 \, a^{3} b + 6 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}\right )} \sqrt {a - b + \frac {b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )} + \frac {15 \, \sqrt {a - b + \frac {b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{a^{2} - 2 \, a b + b^{2}} + \frac {3 \, {\left ({\left (a - b + \frac {b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac {5}{2}} \cos \left (f x + e\right )^{5} - 5 \, {\left (a - b + \frac {b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac {3}{2}} b \cos \left (f x + e\right )^{3} + 15 \, \sqrt {a - b + \frac {b}{\cos \left (f x + e\right )^{2}}} b^{2} \cos \left (f x + e\right )\right )}}{a^{4} - 4 \, a^{3} b + 6 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}} - \frac {10 \, {\left ({\left (a - b + \frac {b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac {3}{2}} \cos \left (f x + e\right )^{3} - 6 \, \sqrt {a - b + \frac {b}{\cos \left (f x + e\right )^{2}}} b \cos \left (f x + e\right )\right )}}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} + \frac {30 \, b^{2}}{{\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} \sqrt {a - b + \frac {b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )} + \frac {15 \, b}{{\left (a^{2} - 2 \, a b + b^{2}\right )} \sqrt {a - b + \frac {b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}}{15 \, f} \]
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\[ \int \frac {\sin ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=\int { \frac {\sin \left (f x + e\right )^{5}}{{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}}} \,d x } \]
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Timed out. \[ \int \frac {\sin ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=\int \frac {{\sin \left (e+f\,x\right )}^5}{{\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a\right )}^{3/2}} \,d x \]
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