Integrand size = 23, antiderivative size = 147 \[ \int (d \sec (e+f x))^m (a+b \tan (e+f x))^2 \, dx=\frac {a b (2+m) (d \sec (e+f x))^m}{f m (1+m)}+\frac {d \left (b^2-a^2 (1+m)\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1-m}{2},\frac {3-m}{2},\cos ^2(e+f x)\right ) (d \sec (e+f x))^{-1+m} \sin (e+f x)}{f (1-m) (1+m) \sqrt {\sin ^2(e+f x)}}+\frac {b (d \sec (e+f x))^m (a+b \tan (e+f x))}{f (1+m)} \]
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Time = 0.24 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3589, 3567, 3857, 2722} \[ \int (d \sec (e+f x))^m (a+b \tan (e+f x))^2 \, dx=\frac {d \left (b^2-a^2 (m+1)\right ) \sin (e+f x) (d \sec (e+f x))^{m-1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1-m}{2},\frac {3-m}{2},\cos ^2(e+f x)\right )}{f (1-m) (m+1) \sqrt {\sin ^2(e+f x)}}+\frac {a b (m+2) (d \sec (e+f x))^m}{f m (m+1)}+\frac {b (a+b \tan (e+f x)) (d \sec (e+f x))^m}{f (m+1)} \]
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Rule 2722
Rule 3567
Rule 3589
Rule 3857
Rubi steps \begin{align*} \text {integral}& = \frac {b (d \sec (e+f x))^m (a+b \tan (e+f x))}{f (1+m)}+\frac {\int (d \sec (e+f x))^m \left (-b^2+a^2 (1+m)+a b (2+m) \tan (e+f x)\right ) \, dx}{1+m} \\ & = \frac {a b (2+m) (d \sec (e+f x))^m}{f m (1+m)}+\frac {b (d \sec (e+f x))^m (a+b \tan (e+f x))}{f (1+m)}+\left (a^2-\frac {b^2}{1+m}\right ) \int (d \sec (e+f x))^m \, dx \\ & = \frac {a b (2+m) (d \sec (e+f x))^m}{f m (1+m)}+\frac {b (d \sec (e+f x))^m (a+b \tan (e+f x))}{f (1+m)}+\left (\left (a^2-\frac {b^2}{1+m}\right ) \left (\frac {\cos (e+f x)}{d}\right )^m (d \sec (e+f x))^m\right ) \int \left (\frac {\cos (e+f x)}{d}\right )^{-m} \, dx \\ & = \frac {a b (2+m) (d \sec (e+f x))^m}{f m (1+m)}-\frac {\left (a^2-\frac {b^2}{1+m}\right ) \cos (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1-m}{2},\frac {3-m}{2},\cos ^2(e+f x)\right ) (d \sec (e+f x))^m \sin (e+f x)}{f (1-m) \sqrt {\sin ^2(e+f x)}}+\frac {b (d \sec (e+f x))^m (a+b \tan (e+f x))}{f (1+m)} \\ \end{align*}
Time = 0.82 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.81 \[ \int (d \sec (e+f x))^m (a+b \tan (e+f x))^2 \, dx=\frac {(d \sec (e+f x))^m \left (b^2 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {m}{2},\frac {2+m}{2},\sec ^2(e+f x)\right ) \tan (e+f x)+a \left (-a \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m}{2},\frac {2+m}{2},\sec ^2(e+f x)\right ) \tan (e+f x)+2 b \sqrt {-\tan ^2(e+f x)}\right )\right )}{f m \sqrt {-\tan ^2(e+f x)}} \]
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\[\int \left (d \sec \left (f x +e \right )\right )^{m} \left (a +b \tan \left (f x +e \right )\right )^{2}d x\]
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\[ \int (d \sec (e+f x))^m (a+b \tan (e+f x))^2 \, dx=\int { {\left (b \tan \left (f x + e\right ) + a\right )}^{2} \left (d \sec \left (f x + e\right )\right )^{m} \,d x } \]
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\[ \int (d \sec (e+f x))^m (a+b \tan (e+f x))^2 \, dx=\int \left (d \sec {\left (e + f x \right )}\right )^{m} \left (a + b \tan {\left (e + f x \right )}\right )^{2}\, dx \]
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\[ \int (d \sec (e+f x))^m (a+b \tan (e+f x))^2 \, dx=\int { {\left (b \tan \left (f x + e\right ) + a\right )}^{2} \left (d \sec \left (f x + e\right )\right )^{m} \,d x } \]
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\[ \int (d \sec (e+f x))^m (a+b \tan (e+f x))^2 \, dx=\int { {\left (b \tan \left (f x + e\right ) + a\right )}^{2} \left (d \sec \left (f x + e\right )\right )^{m} \,d x } \]
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Timed out. \[ \int (d \sec (e+f x))^m (a+b \tan (e+f x))^2 \, dx=\int {\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^m\,{\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^2 \,d x \]
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