Integrand size = 11, antiderivative size = 71 \[ \int e^{i n \arctan (a x)} \, dx=\frac {i 2^{1+\frac {n}{2}} (1-i a x)^{1-\frac {n}{2}} \operatorname {Hypergeometric2F1}\left (1-\frac {n}{2},-\frac {n}{2},2-\frac {n}{2},\frac {1}{2} (1-i a x)\right )}{a (2-n)} \]
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Time = 0.01 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {5169, 71} \[ \int e^{i n \arctan (a x)} \, dx=\frac {i 2^{\frac {n}{2}+1} (1-i a x)^{1-\frac {n}{2}} \operatorname {Hypergeometric2F1}\left (1-\frac {n}{2},-\frac {n}{2},2-\frac {n}{2},\frac {1}{2} (1-i a x)\right )}{a (2-n)} \]
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Rule 71
Rule 5169
Rubi steps \begin{align*} \text {integral}& = \int (1-i a x)^{-n/2} (1+i a x)^{n/2} \, dx \\ & = \frac {i 2^{1+\frac {n}{2}} (1-i a x)^{1-\frac {n}{2}} \operatorname {Hypergeometric2F1}\left (1-\frac {n}{2},-\frac {n}{2},2-\frac {n}{2},\frac {1}{2} (1-i a x)\right )}{a (2-n)} \\ \end{align*}
Time = 0.03 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.75 \[ \int e^{i n \arctan (a x)} \, dx=-\frac {4 i e^{i (2+n) \arctan (a x)} \operatorname {Hypergeometric2F1}\left (2,1+\frac {n}{2},2+\frac {n}{2},-e^{2 i \arctan (a x)}\right )}{a (2+n)} \]
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\[\int {\mathrm e}^{i n \arctan \left (a x \right )}d x\]
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\[ \int e^{i n \arctan (a x)} \, dx=\int { e^{\left (i \, n \arctan \left (a x\right )\right )} \,d x } \]
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\[ \int e^{i n \arctan (a x)} \, dx=\int e^{i n \operatorname {atan}{\left (a x \right )}}\, dx \]
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\[ \int e^{i n \arctan (a x)} \, dx=\int { e^{\left (i \, n \arctan \left (a x\right )\right )} \,d x } \]
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\[ \int e^{i n \arctan (a x)} \, dx=\int { e^{\left (i \, n \arctan \left (a x\right )\right )} \,d x } \]
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Timed out. \[ \int e^{i n \arctan (a x)} \, dx=\int {\mathrm {e}}^{n\,\mathrm {atan}\left (a\,x\right )\,1{}\mathrm {i}} \,d x \]
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