Integrand size = 21, antiderivative size = 43 \[ \int \frac {\sin ^3(e+f x)}{(b \sec (e+f x))^{3/2}} \, dx=\frac {2 b^3}{9 f (b \sec (e+f x))^{9/2}}-\frac {2 b}{5 f (b \sec (e+f x))^{5/2}} \]
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Time = 0.03 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2702, 14} \[ \int \frac {\sin ^3(e+f x)}{(b \sec (e+f x))^{3/2}} \, dx=\frac {2 b^3}{9 f (b \sec (e+f x))^{9/2}}-\frac {2 b}{5 f (b \sec (e+f x))^{5/2}} \]
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Rule 14
Rule 2702
Rubi steps \begin{align*} \text {integral}& = \frac {b^3 \text {Subst}\left (\int \frac {-1+\frac {x^2}{b^2}}{x^{11/2}} \, dx,x,b \sec (e+f x)\right )}{f} \\ & = \frac {b^3 \text {Subst}\left (\int \left (-\frac {1}{x^{11/2}}+\frac {1}{b^2 x^{7/2}}\right ) \, dx,x,b \sec (e+f x)\right )}{f} \\ & = \frac {2 b^3}{9 f (b \sec (e+f x))^{9/2}}-\frac {2 b}{5 f (b \sec (e+f x))^{5/2}} \\ \end{align*}
Time = 0.27 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.74 \[ \int \frac {\sin ^3(e+f x)}{(b \sec (e+f x))^{3/2}} \, dx=\frac {b (-13+5 \cos (2 (e+f x)))}{45 f (b \sec (e+f x))^{5/2}} \]
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Time = 0.18 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.93
method | result | size |
default | \(\frac {\frac {2 \left (\cos ^{4}\left (f x +e \right )\right )}{9}-\frac {2 \left (\cos ^{2}\left (f x +e \right )\right )}{5}}{f b \sqrt {b \sec \left (f x +e \right )}}\) | \(40\) |
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Time = 0.33 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.95 \[ \int \frac {\sin ^3(e+f x)}{(b \sec (e+f x))^{3/2}} \, dx=\frac {2 \, {\left (5 \, \cos \left (f x + e\right )^{5} - 9 \, \cos \left (f x + e\right )^{3}\right )} \sqrt {\frac {b}{\cos \left (f x + e\right )}}}{45 \, b^{2} f} \]
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Timed out. \[ \int \frac {\sin ^3(e+f x)}{(b \sec (e+f x))^{3/2}} \, dx=\text {Timed out} \]
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Time = 0.21 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.86 \[ \int \frac {\sin ^3(e+f x)}{(b \sec (e+f x))^{3/2}} \, dx=\frac {2 \, {\left (5 \, b^{2} - \frac {9 \, b^{2}}{\cos \left (f x + e\right )^{2}}\right )} b}{45 \, f \left (\frac {b}{\cos \left (f x + e\right )}\right )^{\frac {9}{2}}} \]
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Time = 0.32 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.49 \[ \int \frac {\sin ^3(e+f x)}{(b \sec (e+f x))^{3/2}} \, dx=\frac {2 \, {\left (5 \, \sqrt {b \cos \left (f x + e\right )} b^{4} \cos \left (f x + e\right )^{4} - 9 \, \sqrt {b \cos \left (f x + e\right )} b^{4} \cos \left (f x + e\right )^{2}\right )}}{45 \, b^{6} f \mathrm {sgn}\left (\cos \left (f x + e\right )\right )} \]
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Timed out. \[ \int \frac {\sin ^3(e+f x)}{(b \sec (e+f x))^{3/2}} \, dx=\int \frac {{\sin \left (e+f\,x\right )}^3}{{\left (\frac {b}{\cos \left (e+f\,x\right )}\right )}^{3/2}} \,d x \]
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