Integrand size = 23, antiderivative size = 130 \[ \int \frac {(e \tan (c+d x))^m}{a+a \sec (c+d x)} \, dx=\frac {e \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2} (-1+m),\frac {1+m}{2},-\tan ^2(c+d x)\right ) (e \tan (c+d x))^{-1+m}}{a d (1-m)}-\frac {e \cos ^2(c+d x)^{m/2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (-1+m),\frac {m}{2},\frac {1+m}{2},\sin ^2(c+d x)\right ) \sec (c+d x) (e \tan (c+d x))^{-1+m}}{a d (1-m)} \]
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Time = 0.19 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3973, 3969, 3557, 371, 2697} \[ \int \frac {(e \tan (c+d x))^m}{a+a \sec (c+d x)} \, dx=\frac {e (e \tan (c+d x))^{m-1} \operatorname {Hypergeometric2F1}\left (1,\frac {m-1}{2},\frac {m+1}{2},-\tan ^2(c+d x)\right )}{a d (1-m)}-\frac {e \sec (c+d x) \cos ^2(c+d x)^{m/2} (e \tan (c+d x))^{m-1} \operatorname {Hypergeometric2F1}\left (\frac {m-1}{2},\frac {m}{2},\frac {m+1}{2},\sin ^2(c+d x)\right )}{a d (1-m)} \]
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Rule 371
Rule 2697
Rule 3557
Rule 3969
Rule 3973
Rubi steps \begin{align*} \text {integral}& = \frac {e^2 \int (-a+a \sec (c+d x)) (e \tan (c+d x))^{-2+m} \, dx}{a^2} \\ & = -\frac {e^2 \int (e \tan (c+d x))^{-2+m} \, dx}{a}+\frac {e^2 \int \sec (c+d x) (e \tan (c+d x))^{-2+m} \, dx}{a} \\ & = -\frac {e \cos ^2(c+d x)^{m/2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (-1+m),\frac {m}{2},\frac {1+m}{2},\sin ^2(c+d x)\right ) \sec (c+d x) (e \tan (c+d x))^{-1+m}}{a d (1-m)}-\frac {e^3 \text {Subst}\left (\int \frac {x^{-2+m}}{e^2+x^2} \, dx,x,e \tan (c+d x)\right )}{a d} \\ & = \frac {e \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2} (-1+m),\frac {1+m}{2},-\tan ^2(c+d x)\right ) (e \tan (c+d x))^{-1+m}}{a d (1-m)}-\frac {e \cos ^2(c+d x)^{m/2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (-1+m),\frac {m}{2},\frac {1+m}{2},\sin ^2(c+d x)\right ) \sec (c+d x) (e \tan (c+d x))^{-1+m}}{a d (1-m)} \\ \end{align*}
\[ \int \frac {(e \tan (c+d x))^m}{a+a \sec (c+d x)} \, dx=\int \frac {(e \tan (c+d x))^m}{a+a \sec (c+d x)} \, dx \]
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\[\int \frac {\left (e \tan \left (d x +c \right )\right )^{m}}{a +a \sec \left (d x +c \right )}d x\]
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\[ \int \frac {(e \tan (c+d x))^m}{a+a \sec (c+d x)} \, dx=\int { \frac {\left (e \tan \left (d x + c\right )\right )^{m}}{a \sec \left (d x + c\right ) + a} \,d x } \]
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\[ \int \frac {(e \tan (c+d x))^m}{a+a \sec (c+d x)} \, dx=\frac {\int \frac {\left (e \tan {\left (c + d x \right )}\right )^{m}}{\sec {\left (c + d x \right )} + 1}\, dx}{a} \]
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\[ \int \frac {(e \tan (c+d x))^m}{a+a \sec (c+d x)} \, dx=\int { \frac {\left (e \tan \left (d x + c\right )\right )^{m}}{a \sec \left (d x + c\right ) + a} \,d x } \]
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\[ \int \frac {(e \tan (c+d x))^m}{a+a \sec (c+d x)} \, dx=\int { \frac {\left (e \tan \left (d x + c\right )\right )^{m}}{a \sec \left (d x + c\right ) + a} \,d x } \]
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Timed out. \[ \int \frac {(e \tan (c+d x))^m}{a+a \sec (c+d x)} \, dx=\int \frac {\cos \left (c+d\,x\right )\,{\left (e\,\mathrm {tan}\left (c+d\,x\right )\right )}^m}{a\,\left (\cos \left (c+d\,x\right )+1\right )} \,d x \]
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