Integrand size = 23, antiderivative size = 252 \[ \int \frac {(e \tan (c+d x))^m}{(a+a \sec (c+d x))^3} \, dx=\frac {3 e^5 (e \tan (c+d x))^{-5+m}}{a^3 d (5-m)}+\frac {e^5 \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2} (-5+m),\frac {1}{2} (-3+m),-\tan ^2(c+d x)\right ) (e \tan (c+d x))^{-5+m}}{a^3 d (5-m)}-\frac {3 e^5 \cos ^2(c+d x)^{\frac {1}{2} (-4+m)} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (-5+m),\frac {1}{2} (-4+m),\frac {1}{2} (-3+m),\sin ^2(c+d x)\right ) \sec (c+d x) (e \tan (c+d x))^{-5+m}}{a^3 d (5-m)}-\frac {e^5 \cos ^2(c+d x)^{\frac {1}{2} (-2+m)} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (-5+m),\frac {1}{2} (-2+m),\frac {1}{2} (-3+m),\sin ^2(c+d x)\right ) \sec ^3(c+d x) (e \tan (c+d x))^{-5+m}}{a^3 d (5-m)} \]
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Time = 0.41 (sec) , antiderivative size = 252, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3973, 3971, 3557, 371, 2697, 2687, 32} \[ \int \frac {(e \tan (c+d x))^m}{(a+a \sec (c+d x))^3} \, dx=\frac {e^5 (e \tan (c+d x))^{m-5} \operatorname {Hypergeometric2F1}\left (1,\frac {m-5}{2},\frac {m-3}{2},-\tan ^2(c+d x)\right )}{a^3 d (5-m)}-\frac {e^5 \sec ^3(c+d x) \cos ^2(c+d x)^{\frac {m-2}{2}} (e \tan (c+d x))^{m-5} \operatorname {Hypergeometric2F1}\left (\frac {m-5}{2},\frac {m-2}{2},\frac {m-3}{2},\sin ^2(c+d x)\right )}{a^3 d (5-m)}-\frac {3 e^5 \sec (c+d x) \cos ^2(c+d x)^{\frac {m-4}{2}} (e \tan (c+d x))^{m-5} \operatorname {Hypergeometric2F1}\left (\frac {m-5}{2},\frac {m-4}{2},\frac {m-3}{2},\sin ^2(c+d x)\right )}{a^3 d (5-m)}+\frac {3 e^5 (e \tan (c+d x))^{m-5}}{a^3 d (5-m)} \]
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Rule 32
Rule 371
Rule 2687
Rule 2697
Rule 3557
Rule 3971
Rule 3973
Rubi steps \begin{align*} \text {integral}& = \frac {e^6 \int (-a+a \sec (c+d x))^3 (e \tan (c+d x))^{-6+m} \, dx}{a^6} \\ & = \frac {e^6 \int \left (-a^3 (e \tan (c+d x))^{-6+m}+3 a^3 \sec (c+d x) (e \tan (c+d x))^{-6+m}-3 a^3 \sec ^2(c+d x) (e \tan (c+d x))^{-6+m}+a^3 \sec ^3(c+d x) (e \tan (c+d x))^{-6+m}\right ) \, dx}{a^6} \\ & = -\frac {e^6 \int (e \tan (c+d x))^{-6+m} \, dx}{a^3}+\frac {e^6 \int \sec ^3(c+d x) (e \tan (c+d x))^{-6+m} \, dx}{a^3}+\frac {\left (3 e^6\right ) \int \sec (c+d x) (e \tan (c+d x))^{-6+m} \, dx}{a^3}-\frac {\left (3 e^6\right ) \int \sec ^2(c+d x) (e \tan (c+d x))^{-6+m} \, dx}{a^3} \\ & = -\frac {3 e^5 \cos ^2(c+d x)^{\frac {1}{2} (-4+m)} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (-5+m),\frac {1}{2} (-4+m),\frac {1}{2} (-3+m),\sin ^2(c+d x)\right ) \sec (c+d x) (e \tan (c+d x))^{-5+m}}{a^3 d (5-m)}-\frac {e^5 \cos ^2(c+d x)^{\frac {1}{2} (-2+m)} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (-5+m),\frac {1}{2} (-2+m),\frac {1}{2} (-3+m),\sin ^2(c+d x)\right ) \sec ^3(c+d x) (e \tan (c+d x))^{-5+m}}{a^3 d (5-m)}-\frac {\left (3 e^6\right ) \text {Subst}\left (\int (e x)^{-6+m} \, dx,x,\tan (c+d x)\right )}{a^3 d}-\frac {e^7 \text {Subst}\left (\int \frac {x^{-6+m}}{e^2+x^2} \, dx,x,e \tan (c+d x)\right )}{a^3 d} \\ & = \frac {3 e^5 (e \tan (c+d x))^{-5+m}}{a^3 d (5-m)}+\frac {e^5 \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2} (-5+m),\frac {1}{2} (-3+m),-\tan ^2(c+d x)\right ) (e \tan (c+d x))^{-5+m}}{a^3 d (5-m)}-\frac {3 e^5 \cos ^2(c+d x)^{\frac {1}{2} (-4+m)} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (-5+m),\frac {1}{2} (-4+m),\frac {1}{2} (-3+m),\sin ^2(c+d x)\right ) \sec (c+d x) (e \tan (c+d x))^{-5+m}}{a^3 d (5-m)}-\frac {e^5 \cos ^2(c+d x)^{\frac {1}{2} (-2+m)} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (-5+m),\frac {1}{2} (-2+m),\frac {1}{2} (-3+m),\sin ^2(c+d x)\right ) \sec ^3(c+d x) (e \tan (c+d x))^{-5+m}}{a^3 d (5-m)} \\ \end{align*}
\[ \int \frac {(e \tan (c+d x))^m}{(a+a \sec (c+d x))^3} \, dx=\int \frac {(e \tan (c+d x))^m}{(a+a \sec (c+d x))^3} \, dx \]
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\[\int \frac {\left (e \tan \left (d x +c \right )\right )^{m}}{\left (a +a \sec \left (d x +c \right )\right )^{3}}d x\]
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\[ \int \frac {(e \tan (c+d x))^m}{(a+a \sec (c+d x))^3} \, dx=\int { \frac {\left (e \tan \left (d x + c\right )\right )^{m}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{3}} \,d x } \]
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\[ \int \frac {(e \tan (c+d x))^m}{(a+a \sec (c+d x))^3} \, dx=\frac {\int \frac {\left (e \tan {\left (c + d x \right )}\right )^{m}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx}{a^{3}} \]
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\[ \int \frac {(e \tan (c+d x))^m}{(a+a \sec (c+d x))^3} \, dx=\int { \frac {\left (e \tan \left (d x + c\right )\right )^{m}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{3}} \,d x } \]
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\[ \int \frac {(e \tan (c+d x))^m}{(a+a \sec (c+d x))^3} \, dx=\int { \frac {\left (e \tan \left (d x + c\right )\right )^{m}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{3}} \,d x } \]
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Timed out. \[ \int \frac {(e \tan (c+d x))^m}{(a+a \sec (c+d x))^3} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^3\,{\left (e\,\mathrm {tan}\left (c+d\,x\right )\right )}^m}{a^3\,{\left (\cos \left (c+d\,x\right )+1\right )}^3} \,d x \]
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