Integrand size = 8, antiderivative size = 46 \[ \int e^{2 \arctan (a x)} \, dx=\frac {(1+i) 2^{-1-i} (1-i a x)^{1+i} \operatorname {Hypergeometric2F1}\left (i,1+i,2+i,\frac {1}{2} (1-i a x)\right )}{a} \]
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Time = 0.01 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {5169, 71} \[ \int e^{2 \arctan (a x)} \, dx=\frac {(1+i) 2^{-1-i} (1-i a x)^{1+i} \operatorname {Hypergeometric2F1}\left (i,1+i,2+i,\frac {1}{2} (1-i a x)\right )}{a} \]
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Rule 71
Rule 5169
Rubi steps \begin{align*} \text {integral}& = \int (1-i a x)^i (1+i a x)^{-i} \, dx \\ & = \frac {(1+i) 2^{-1-i} (1-i a x)^{1+i} \operatorname {Hypergeometric2F1}\left (i,1+i,2+i,\frac {1}{2} (1-i a x)\right )}{a} \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.80 \[ \int e^{2 \arctan (a x)} \, dx=\frac {(1-i) e^{(2+2 i) \arctan (a x)} \operatorname {Hypergeometric2F1}\left (1-i,2,2-i,-e^{2 i \arctan (a x)}\right )}{a} \]
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\[\int {\mathrm e}^{2 \arctan \left (a x \right )}d x\]
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\[ \int e^{2 \arctan (a x)} \, dx=\int { e^{\left (2 \, \arctan \left (a x\right )\right )} \,d x } \]
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\[ \int e^{2 \arctan (a x)} \, dx=\int e^{2 \operatorname {atan}{\left (a x \right )}}\, dx \]
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\[ \int e^{2 \arctan (a x)} \, dx=\int { e^{\left (2 \, \arctan \left (a x\right )\right )} \,d x } \]
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\[ \int e^{2 \arctan (a x)} \, dx=\int { e^{\left (2 \, \arctan \left (a x\right )\right )} \,d x } \]
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Timed out. \[ \int e^{2 \arctan (a x)} \, dx=\int {\mathrm {e}}^{2\,\mathrm {atan}\left (a\,x\right )} \,d x \]
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