Integrand size = 28, antiderivative size = 264 \[ \int \frac {(a+i a \tan (e+f x))^m}{(c+d \tan (e+f x))^3} \, dx=-\frac {\operatorname {Hypergeometric2F1}\left (1,m,1+m,\frac {1}{2} (1+i \tan (e+f x))\right ) (a+i a \tan (e+f x))^m}{2 (i c+d)^3 f m}-\frac {d \left (2 i c d (3-m) m+c^2 \left (6-5 m+m^2\right )-d^2 \left (2-m+m^2\right )\right ) \operatorname {Hypergeometric2F1}\left (1,m,1+m,-\frac {d (1+i \tan (e+f x))}{i c-d}\right ) (a+i a \tan (e+f x))^m}{2 \left (c^2+d^2\right )^3 f m}-\frac {d (a+i a \tan (e+f x))^m}{2 \left (c^2+d^2\right ) f (c+d \tan (e+f x))^2}-\frac {d (c (4-m)+i d m) (a+i a \tan (e+f x))^m}{2 \left (c^2+d^2\right )^2 f (c+d \tan (e+f x))} \]
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Time = 1.01 (sec) , antiderivative size = 264, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {3642, 3679, 3681, 3562, 70, 3680} \[ \int \frac {(a+i a \tan (e+f x))^m}{(c+d \tan (e+f x))^3} \, dx=-\frac {d \left (c^2 \left (m^2-5 m+6\right )+2 i c d (3-m) m-d^2 \left (m^2-m+2\right )\right ) (a+i a \tan (e+f x))^m \operatorname {Hypergeometric2F1}\left (1,m,m+1,-\frac {d (i \tan (e+f x)+1)}{i c-d}\right )}{2 f m \left (c^2+d^2\right )^3}-\frac {d (c (4-m)+i d m) (a+i a \tan (e+f x))^m}{2 f \left (c^2+d^2\right )^2 (c+d \tan (e+f x))}-\frac {d (a+i a \tan (e+f x))^m}{2 f \left (c^2+d^2\right ) (c+d \tan (e+f x))^2}-\frac {(a+i a \tan (e+f x))^m \operatorname {Hypergeometric2F1}\left (1,m,m+1,\frac {1}{2} (i \tan (e+f x)+1)\right )}{2 f m (d+i c)^3} \]
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Rule 70
Rule 3562
Rule 3642
Rule 3679
Rule 3680
Rule 3681
Rubi steps \begin{align*} \text {integral}& = -\frac {d (a+i a \tan (e+f x))^m}{2 \left (c^2+d^2\right ) f (c+d \tan (e+f x))^2}+\frac {\int \frac {(a+i a \tan (e+f x))^m (a (2 c+i d m)-a d (2-m) \tan (e+f x))}{(c+d \tan (e+f x))^2} \, dx}{2 a \left (c^2+d^2\right )} \\ & = -\frac {d (a+i a \tan (e+f x))^m}{2 \left (c^2+d^2\right ) f (c+d \tan (e+f x))^2}-\frac {d (c (4-m)+i d m) (a+i a \tan (e+f x))^m}{2 \left (c^2+d^2\right )^2 f (c+d \tan (e+f x))}+\frac {\int \frac {(a+i a \tan (e+f x))^m \left (a^2 \left (2 c^2+i c d (5-m) m-d^2 \left (2-m+m^2\right )\right )-a^2 d (1-m) (c (4-m)+i d m) \tan (e+f x)\right )}{c+d \tan (e+f x)} \, dx}{2 a^2 \left (c^2+d^2\right )^2} \\ & = -\frac {d (a+i a \tan (e+f x))^m}{2 \left (c^2+d^2\right ) f (c+d \tan (e+f x))^2}-\frac {d (c (4-m)+i d m) (a+i a \tan (e+f x))^m}{2 \left (c^2+d^2\right )^2 f (c+d \tan (e+f x))}+\frac {\int (a+i a \tan (e+f x))^m \, dx}{(c-i d)^3}-\frac {\left (d \left (2 i c d (3-m) m+c^2 \left (6-5 m+m^2\right )-d^2 \left (2-m+m^2\right )\right )\right ) \int \frac {(a-i a \tan (e+f x)) (a+i a \tan (e+f x))^m}{c+d \tan (e+f x)} \, dx}{2 a (c+i d)^2 (i c+d)^3} \\ & = -\frac {d (a+i a \tan (e+f x))^m}{2 \left (c^2+d^2\right ) f (c+d \tan (e+f x))^2}-\frac {d (c (4-m)+i d m) (a+i a \tan (e+f x))^m}{2 \left (c^2+d^2\right )^2 f (c+d \tan (e+f x))}-\frac {a \text {Subst}\left (\int \frac {(a+x)^{-1+m}}{a-x} \, dx,x,i a \tan (e+f x)\right )}{(i c+d)^3 f}-\frac {\left (a d \left (2 i c d (3-m) m+c^2 \left (6-5 m+m^2\right )-d^2 \left (2-m+m^2\right )\right )\right ) \text {Subst}\left (\int \frac {(a+i a x)^{-1+m}}{c+d x} \, dx,x,\tan (e+f x)\right )}{2 (c+i d)^2 (i c+d)^3 f} \\ & = -\frac {\operatorname {Hypergeometric2F1}\left (1,m,1+m,\frac {1}{2} (1+i \tan (e+f x))\right ) (a+i a \tan (e+f x))^m}{2 (i c+d)^3 f m}-\frac {d \left (2 i c d (3-m) m+c^2 \left (6-5 m+m^2\right )-d^2 \left (2-m+m^2\right )\right ) \operatorname {Hypergeometric2F1}\left (1,m,1+m,-\frac {d (1+i \tan (e+f x))}{i c-d}\right ) (a+i a \tan (e+f x))^m}{2 \left (c^2+d^2\right )^3 f m}-\frac {d (a+i a \tan (e+f x))^m}{2 \left (c^2+d^2\right ) f (c+d \tan (e+f x))^2}-\frac {d (c (4-m)+i d m) (a+i a \tan (e+f x))^m}{2 \left (c^2+d^2\right )^2 f (c+d \tan (e+f x))} \\ \end{align*}
Time = 3.44 (sec) , antiderivative size = 215, normalized size of antiderivative = 0.81 \[ \int \frac {(a+i a \tan (e+f x))^m}{(c+d \tan (e+f x))^3} \, dx=\frac {(a+i a \tan (e+f x))^m \left (-2 d+\frac {2 (c+d \tan (e+f x)) \left ((c-i d) (c+i d) d m (c (-4+m)-i d m)-\left (i (c+i d)^3 \operatorname {Hypergeometric2F1}\left (1,m,1+m,\frac {1}{2} (1+i \tan (e+f x))\right )+d \left (-2 i c d (-3+m) m+d^2 \left (-2+m-m^2\right )+c^2 \left (6-5 m+m^2\right )\right ) \operatorname {Hypergeometric2F1}\left (1,m,1+m,\frac {d (1+i \tan (e+f x))}{-i c+d}\right )\right ) (c+d \tan (e+f x))\right )}{\left (c^2+d^2\right )^2 m}\right )}{4 \left (c^2+d^2\right ) f (c+d \tan (e+f x))^2} \]
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\[\int \frac {\left (a +i a \tan \left (f x +e \right )\right )^{m}}{\left (c +d \tan \left (f x +e \right )\right )^{3}}d x\]
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\[ \int \frac {(a+i a \tan (e+f x))^m}{(c+d \tan (e+f x))^3} \, dx=\int { \frac {{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{m}}{{\left (d \tan \left (f x + e\right ) + c\right )}^{3}} \,d x } \]
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\[ \int \frac {(a+i a \tan (e+f x))^m}{(c+d \tan (e+f x))^3} \, dx=\int \frac {\left (i a \left (\tan {\left (e + f x \right )} - i\right )\right )^{m}}{\left (c + d \tan {\left (e + f x \right )}\right )^{3}}\, dx \]
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Timed out. \[ \int \frac {(a+i a \tan (e+f x))^m}{(c+d \tan (e+f x))^3} \, dx=\text {Timed out} \]
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\[ \int \frac {(a+i a \tan (e+f x))^m}{(c+d \tan (e+f x))^3} \, dx=\int { \frac {{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{m}}{{\left (d \tan \left (f x + e\right ) + c\right )}^{3}} \,d x } \]
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Timed out. \[ \int \frac {(a+i a \tan (e+f x))^m}{(c+d \tan (e+f x))^3} \, dx=\int \frac {{\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^m}{{\left (c+d\,\mathrm {tan}\left (e+f\,x\right )\right )}^3} \,d x \]
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