Integrand size = 26, antiderivative size = 158 \[ \int \frac {(d \tan (e+f x))^n}{a-i a \tan (e+f x)} \, dx=\frac {(1-n) \operatorname {Hypergeometric2F1}\left (1,\frac {1+n}{2},\frac {3+n}{2},-\tan ^2(e+f x)\right ) (d \tan (e+f x))^{1+n}}{2 a d f (1+n)}-\frac {i n \operatorname {Hypergeometric2F1}\left (1,\frac {2+n}{2},\frac {4+n}{2},-\tan ^2(e+f x)\right ) (d \tan (e+f x))^{2+n}}{2 a d^2 f (2+n)}+\frac {(d \tan (e+f x))^{1+n}}{2 d f (a-i a \tan (e+f x))} \]
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Time = 0.20 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {3633, 3619, 3557, 371} \[ \int \frac {(d \tan (e+f x))^n}{a-i a \tan (e+f x)} \, dx=-\frac {i n (d \tan (e+f x))^{n+2} \operatorname {Hypergeometric2F1}\left (1,\frac {n+2}{2},\frac {n+4}{2},-\tan ^2(e+f x)\right )}{2 a d^2 f (n+2)}+\frac {(1-n) (d \tan (e+f x))^{n+1} \operatorname {Hypergeometric2F1}\left (1,\frac {n+1}{2},\frac {n+3}{2},-\tan ^2(e+f x)\right )}{2 a d f (n+1)}+\frac {(d \tan (e+f x))^{n+1}}{2 d f (a-i a \tan (e+f x))} \]
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Rule 371
Rule 3557
Rule 3619
Rule 3633
Rubi steps \begin{align*} \text {integral}& = \frac {(d \tan (e+f x))^{1+n}}{2 d f (a-i a \tan (e+f x))}-\frac {\int (d \tan (e+f x))^n (-a d (1-n)+i a d n \tan (e+f x)) \, dx}{2 a^2 d} \\ & = \frac {(d \tan (e+f x))^{1+n}}{2 d f (a-i a \tan (e+f x))}+\frac {(1-n) \int (d \tan (e+f x))^n \, dx}{2 a}-\frac {(i n) \int (d \tan (e+f x))^{1+n} \, dx}{2 a d} \\ & = \frac {(d \tan (e+f x))^{1+n}}{2 d f (a-i a \tan (e+f x))}+\frac {(d (1-n)) \text {Subst}\left (\int \frac {x^n}{d^2+x^2} \, dx,x,d \tan (e+f x)\right )}{2 a f}-\frac {(i n) \text {Subst}\left (\int \frac {x^{1+n}}{d^2+x^2} \, dx,x,d \tan (e+f x)\right )}{2 a f} \\ & = \frac {(1-n) \operatorname {Hypergeometric2F1}\left (1,\frac {1+n}{2},\frac {3+n}{2},-\tan ^2(e+f x)\right ) (d \tan (e+f x))^{1+n}}{2 a d f (1+n)}-\frac {i n \operatorname {Hypergeometric2F1}\left (1,\frac {2+n}{2},\frac {4+n}{2},-\tan ^2(e+f x)\right ) (d \tan (e+f x))^{2+n}}{2 a d^2 f (2+n)}+\frac {(d \tan (e+f x))^{1+n}}{2 d f (a-i a \tan (e+f x))} \\ \end{align*}
Time = 0.86 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.78 \[ \int \frac {(d \tan (e+f x))^n}{a-i a \tan (e+f x)} \, dx=\frac {\tan (e+f x) (d \tan (e+f x))^n \left (-\frac {(-1+n) \operatorname {Hypergeometric2F1}\left (1,\frac {1+n}{2},\frac {3+n}{2},-\tan ^2(e+f x)\right )}{a (1+n)}-\frac {i n \operatorname {Hypergeometric2F1}\left (1,\frac {2+n}{2},\frac {4+n}{2},-\tan ^2(e+f x)\right ) \tan (e+f x)}{a (2+n)}+\frac {1}{a-i a \tan (e+f x)}\right )}{2 f} \]
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\[\int \frac {\left (d \tan \left (f x +e \right )\right )^{n}}{a -i a \tan \left (f x +e \right )}d x\]
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\[ \int \frac {(d \tan (e+f x))^n}{a-i a \tan (e+f x)} \, dx=\int { \frac {\left (d \tan \left (f x + e\right )\right )^{n}}{-i \, a \tan \left (f x + e\right ) + a} \,d x } \]
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\[ \int \frac {(d \tan (e+f x))^n}{a-i a \tan (e+f x)} \, dx=\frac {i \int \frac {\left (d \tan {\left (e + f x \right )}\right )^{n}}{\tan {\left (e + f x \right )} + i}\, dx}{a} \]
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Exception generated. \[ \int \frac {(d \tan (e+f x))^n}{a-i a \tan (e+f x)} \, dx=\text {Exception raised: RuntimeError} \]
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\[ \int \frac {(d \tan (e+f x))^n}{a-i a \tan (e+f x)} \, dx=\int { \frac {\left (d \tan \left (f x + e\right )\right )^{n}}{-i \, a \tan \left (f x + e\right ) + a} \,d x } \]
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Timed out. \[ \int \frac {(d \tan (e+f x))^n}{a-i a \tan (e+f x)} \, dx=\int \frac {{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^n}{a-a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}} \,d x \]
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