Integrand size = 14, antiderivative size = 191 \[ \int \frac {e^{n \arctan (a+b x)}}{x} \, dx=\frac {2 i (1-i a-i b x)^{\frac {i n}{2}} (1+i a+i b x)^{-\frac {i n}{2}} \operatorname {Hypergeometric2F1}\left (1,\frac {i n}{2},1+\frac {i n}{2},\frac {(i-a) (1-i a-i b x)}{(i+a) (1+i a+i b x)}\right )}{n}-\frac {i 2^{1-\frac {i n}{2}} (1-i a-i b x)^{\frac {i n}{2}} \operatorname {Hypergeometric2F1}\left (\frac {i n}{2},\frac {i n}{2},1+\frac {i n}{2},\frac {1}{2} (1-i a-i b x)\right )}{n} \]
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Time = 0.06 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {5203, 132, 71, 12, 133} \[ \int \frac {e^{n \arctan (a+b x)}}{x} \, dx=\frac {2 i (-i a-i b x+1)^{\frac {i n}{2}} (i a+i b x+1)^{-\frac {i n}{2}} \operatorname {Hypergeometric2F1}\left (1,\frac {i n}{2},\frac {i n}{2}+1,\frac {(i-a) (-i a-i b x+1)}{(a+i) (i a+i b x+1)}\right )}{n}-\frac {i 2^{1-\frac {i n}{2}} (-i a-i b x+1)^{\frac {i n}{2}} \operatorname {Hypergeometric2F1}\left (\frac {i n}{2},\frac {i n}{2},\frac {i n}{2}+1,\frac {1}{2} (-i a-i b x+1)\right )}{n} \]
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Rule 12
Rule 71
Rule 132
Rule 133
Rule 5203
Rubi steps \begin{align*} \text {integral}& = \int \frac {(1-i a-i b x)^{\frac {i n}{2}} (1+i a+i b x)^{-\frac {i n}{2}}}{x} \, dx \\ & = -\left ((i b) \int (1-i a-i b x)^{-1+\frac {i n}{2}} (1+i a+i b x)^{-\frac {i n}{2}} \, dx\right )+\int \frac {(1-i a) (1-i a-i b x)^{-1+\frac {i n}{2}} (1+i a+i b x)^{-\frac {i n}{2}}}{x} \, dx \\ & = -\frac {i 2^{1-\frac {i n}{2}} (1-i a-i b x)^{\frac {i n}{2}} \operatorname {Hypergeometric2F1}\left (\frac {i n}{2},\frac {i n}{2},1+\frac {i n}{2},\frac {1}{2} (1-i a-i b x)\right )}{n}+(1-i a) \int \frac {(1-i a-i b x)^{-1+\frac {i n}{2}} (1+i a+i b x)^{-\frac {i n}{2}}}{x} \, dx \\ & = \frac {2 i (1-i a-i b x)^{\frac {i n}{2}} (1+i a+i b x)^{-\frac {i n}{2}} \operatorname {Hypergeometric2F1}\left (1,\frac {i n}{2},1+\frac {i n}{2},\frac {(i-a) (1-i a-i b x)}{(i+a) (1+i a+i b x)}\right )}{n}-\frac {i 2^{1-\frac {i n}{2}} (1-i a-i b x)^{\frac {i n}{2}} \operatorname {Hypergeometric2F1}\left (\frac {i n}{2},\frac {i n}{2},1+\frac {i n}{2},\frac {1}{2} (1-i a-i b x)\right )}{n} \\ \end{align*}
Time = 0.03 (sec) , antiderivative size = 170, normalized size of antiderivative = 0.89 \[ \int \frac {e^{n \arctan (a+b x)}}{x} \, dx=\frac {2 i (1+i a+i b x)^{-\frac {i n}{2}} (-i (i+a+b x))^{\frac {i n}{2}} \left (\operatorname {Hypergeometric2F1}\left (1,\frac {i n}{2},1+\frac {i n}{2},\frac {1+a^2-i b x+a b x}{1+a^2+i b x+a b x}\right )-2^{-\frac {i n}{2}} (1+i a+i b x)^{\frac {i n}{2}} \operatorname {Hypergeometric2F1}\left (\frac {i n}{2},\frac {i n}{2},1+\frac {i n}{2},-\frac {1}{2} i (i+a+b x)\right )\right )}{n} \]
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\[\int \frac {{\mathrm e}^{n \arctan \left (b x +a \right )}}{x}d x\]
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\[ \int \frac {e^{n \arctan (a+b x)}}{x} \, dx=\int { \frac {e^{\left (n \arctan \left (b x + a\right )\right )}}{x} \,d x } \]
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\[ \int \frac {e^{n \arctan (a+b x)}}{x} \, dx=\int \frac {e^{n \operatorname {atan}{\left (a + b x \right )}}}{x}\, dx \]
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\[ \int \frac {e^{n \arctan (a+b x)}}{x} \, dx=\int { \frac {e^{\left (n \arctan \left (b x + a\right )\right )}}{x} \,d x } \]
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\[ \int \frac {e^{n \arctan (a+b x)}}{x} \, dx=\int { \frac {e^{\left (n \arctan \left (b x + a\right )\right )}}{x} \,d x } \]
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Timed out. \[ \int \frac {e^{n \arctan (a+b x)}}{x} \, dx=\int \frac {{\mathrm {e}}^{n\,\mathrm {atan}\left (a+b\,x\right )}}{x} \,d x \]
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