Integrand size = 28, antiderivative size = 106 \[ \int \frac {(e \sec (c+d x))^m}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {i 2^{\frac {1}{2} (-1+m)} \operatorname {Hypergeometric2F1}\left (\frac {3-m}{2},\frac {m}{2},\frac {2+m}{2},\frac {1}{2} (1-i \tan (c+d x))\right ) (e \sec (c+d x))^m (1+i \tan (c+d x))^{\frac {1-m}{2}}}{d m \sqrt {a+i a \tan (c+d x)}} \]
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Time = 0.25 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3586, 3604, 72, 71} \[ \int \frac {(e \sec (c+d x))^m}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {i 2^{\frac {m-1}{2}} (1+i \tan (c+d x))^{\frac {1-m}{2}} (e \sec (c+d x))^m \operatorname {Hypergeometric2F1}\left (\frac {3-m}{2},\frac {m}{2},\frac {m+2}{2},\frac {1}{2} (1-i \tan (c+d x))\right )}{d m \sqrt {a+i a \tan (c+d x)}} \]
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Rule 71
Rule 72
Rule 3586
Rule 3604
Rubi steps \begin{align*} \text {integral}& = \left ((e \sec (c+d x))^m (a-i a \tan (c+d x))^{-m/2} (a+i a \tan (c+d x))^{-m/2}\right ) \int (a-i a \tan (c+d x))^{m/2} (a+i a \tan (c+d x))^{-\frac {1}{2}+\frac {m}{2}} \, dx \\ & = \frac {\left (a^2 (e \sec (c+d x))^m (a-i a \tan (c+d x))^{-m/2} (a+i a \tan (c+d x))^{-m/2}\right ) \text {Subst}\left (\int (a-i a x)^{-1+\frac {m}{2}} (a+i a x)^{-\frac {3}{2}+\frac {m}{2}} \, dx,x,\tan (c+d x)\right )}{d} \\ & = \frac {\left (2^{-\frac {3}{2}+\frac {m}{2}} a (e \sec (c+d x))^m (a-i a \tan (c+d x))^{-m/2} \left (\frac {a+i a \tan (c+d x)}{a}\right )^{\frac {1}{2}-\frac {m}{2}}\right ) \text {Subst}\left (\int \left (\frac {1}{2}+\frac {i x}{2}\right )^{-\frac {3}{2}+\frac {m}{2}} (a-i a x)^{-1+\frac {m}{2}} \, dx,x,\tan (c+d x)\right )}{d \sqrt {a+i a \tan (c+d x)}} \\ & = \frac {i 2^{\frac {1}{2} (-1+m)} \operatorname {Hypergeometric2F1}\left (\frac {3-m}{2},\frac {m}{2},\frac {2+m}{2},\frac {1}{2} (1-i \tan (c+d x))\right ) (e \sec (c+d x))^m (1+i \tan (c+d x))^{\frac {1-m}{2}}}{d m \sqrt {a+i a \tan (c+d x)}} \\ \end{align*}
Time = 2.49 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.64 \[ \int \frac {(e \sec (c+d x))^m}{\sqrt {a+i a \tan (c+d x)}} \, dx=-\frac {i 2^{-\frac {1}{2}+m} \left (\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^{-\frac {1}{2}+m} \left (1+e^{2 i (c+d x)}\right )^{-\frac {1}{2}+m} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (-1+m),-\frac {1}{2}+m,\frac {1+m}{2},-e^{2 i (c+d x)}\right ) \sec ^{\frac {1}{2}-m}(c+d x) (e \sec (c+d x))^m \sqrt {\cos (d x)+i \sin (d x)}}{d \sqrt {e^{i d x}} (-1+m) \sqrt {a+i a \tan (c+d x)}} \]
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\[\int \frac {\left (e \sec \left (d x +c \right )\right )^{m}}{\sqrt {a +i a \tan \left (d x +c \right )}}d x\]
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\[ \int \frac {(e \sec (c+d x))^m}{\sqrt {a+i a \tan (c+d x)}} \, dx=\int { \frac {\left (e \sec \left (d x + c\right )\right )^{m}}{\sqrt {i \, a \tan \left (d x + c\right ) + a}} \,d x } \]
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\[ \int \frac {(e \sec (c+d x))^m}{\sqrt {a+i a \tan (c+d x)}} \, dx=\int \frac {\left (e \sec {\left (c + d x \right )}\right )^{m}}{\sqrt {i a \left (\tan {\left (c + d x \right )} - i\right )}}\, dx \]
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\[ \int \frac {(e \sec (c+d x))^m}{\sqrt {a+i a \tan (c+d x)}} \, dx=\int { \frac {\left (e \sec \left (d x + c\right )\right )^{m}}{\sqrt {i \, a \tan \left (d x + c\right ) + a}} \,d x } \]
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\[ \int \frac {(e \sec (c+d x))^m}{\sqrt {a+i a \tan (c+d x)}} \, dx=\int { \frac {\left (e \sec \left (d x + c\right )\right )^{m}}{\sqrt {i \, a \tan \left (d x + c\right ) + a}} \,d x } \]
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Timed out. \[ \int \frac {(e \sec (c+d x))^m}{\sqrt {a+i a \tan (c+d x)}} \, dx=\int \frac {{\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^m}{\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}} \,d x \]
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