Integrand size = 25, antiderivative size = 69 \[ \int \frac {1}{\sqrt {d \sec (e+f x)} (b \tan (e+f x))^{5/2}} \, dx=-\frac {2}{3 b f \sqrt {d \sec (e+f x)} (b \tan (e+f x))^{3/2}}-\frac {8 \sqrt {b \tan (e+f x)}}{3 b^3 f \sqrt {d \sec (e+f x)}} \]
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Time = 0.11 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {2689, 2685} \[ \int \frac {1}{\sqrt {d \sec (e+f x)} (b \tan (e+f x))^{5/2}} \, dx=-\frac {8 \sqrt {b \tan (e+f x)}}{3 b^3 f \sqrt {d \sec (e+f x)}}-\frac {2}{3 b f (b \tan (e+f x))^{3/2} \sqrt {d \sec (e+f x)}} \]
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Rule 2685
Rule 2689
Rubi steps \begin{align*} \text {integral}& = -\frac {2}{3 b f \sqrt {d \sec (e+f x)} (b \tan (e+f x))^{3/2}}-\frac {4 \int \frac {1}{\sqrt {d \sec (e+f x)} \sqrt {b \tan (e+f x)}} \, dx}{3 b^2} \\ & = -\frac {2}{3 b f \sqrt {d \sec (e+f x)} (b \tan (e+f x))^{3/2}}-\frac {8 \sqrt {b \tan (e+f x)}}{3 b^3 f \sqrt {d \sec (e+f x)}} \\ \end{align*}
Time = 0.79 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.64 \[ \int \frac {1}{\sqrt {d \sec (e+f x)} (b \tan (e+f x))^{5/2}} \, dx=-\frac {2 \left (3+\csc ^2(e+f x)\right ) \sqrt {b \tan (e+f x)}}{3 b^3 f \sqrt {d \sec (e+f x)}} \]
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Time = 1.36 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.75
method | result | size |
default | \(\frac {2 \cot \left (f x +e \right )-\frac {8 \sec \left (f x +e \right ) \csc \left (f x +e \right )}{3}}{f \sqrt {d \sec \left (f x +e \right )}\, \sqrt {b \tan \left (f x +e \right )}\, b^{2}}\) | \(52\) |
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Time = 0.27 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.09 \[ \int \frac {1}{\sqrt {d \sec (e+f x)} (b \tan (e+f x))^{5/2}} \, dx=-\frac {2 \, {\left (3 \, \cos \left (f x + e\right )^{3} - 4 \, \cos \left (f x + e\right )\right )} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt {\frac {d}{\cos \left (f x + e\right )}}}{3 \, {\left (b^{3} d f \cos \left (f x + e\right )^{2} - b^{3} d f\right )}} \]
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Time = 56.80 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.33 \[ \int \frac {1}{\sqrt {d \sec (e+f x)} (b \tan (e+f x))^{5/2}} \, dx=\begin {cases} - \frac {8 \tan ^{3}{\left (e + f x \right )}}{3 f \left (b \tan {\left (e + f x \right )}\right )^{\frac {5}{2}} \sqrt {d \sec {\left (e + f x \right )}}} - \frac {2 \tan {\left (e + f x \right )}}{3 f \left (b \tan {\left (e + f x \right )}\right )^{\frac {5}{2}} \sqrt {d \sec {\left (e + f x \right )}}} & \text {for}\: f \neq 0 \\\frac {x}{\left (b \tan {\left (e \right )}\right )^{\frac {5}{2}} \sqrt {d \sec {\left (e \right )}}} & \text {otherwise} \end {cases} \]
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\[ \int \frac {1}{\sqrt {d \sec (e+f x)} (b \tan (e+f x))^{5/2}} \, dx=\int { \frac {1}{\sqrt {d \sec \left (f x + e\right )} \left (b \tan \left (f x + e\right )\right )^{\frac {5}{2}}} \,d x } \]
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\[ \int \frac {1}{\sqrt {d \sec (e+f x)} (b \tan (e+f x))^{5/2}} \, dx=\int { \frac {1}{\sqrt {d \sec \left (f x + e\right )} \left (b \tan \left (f x + e\right )\right )^{\frac {5}{2}}} \,d x } \]
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Time = 4.31 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.17 \[ \int \frac {1}{\sqrt {d \sec (e+f x)} (b \tan (e+f x))^{5/2}} \, dx=\frac {\left (\frac {13\,\sin \left (e+f\,x\right )}{3}-\sin \left (3\,e+3\,f\,x\right )\right )\,\sqrt {\frac {d}{\cos \left (e+f\,x\right )}}}{b^2\,d\,f\,\left (\cos \left (2\,e+2\,f\,x\right )-1\right )\,\sqrt {\frac {b\,\sin \left (2\,e+2\,f\,x\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}} \]
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