Integrand size = 14, antiderivative size = 207 \[ \int \frac {e^{n \arctan (a+b x)}}{x^3} \, dx=-\frac {(1-i a-i b x)^{1+\frac {i n}{2}} (1+i a+i b x)^{1-\frac {i n}{2}}}{2 \left (1+a^2\right ) x^2}-\frac {2 b^2 (2 a-n) (1-i a-i b x)^{1+\frac {i n}{2}} (1+i a+i b x)^{-1-\frac {i n}{2}} \operatorname {Hypergeometric2F1}\left (2,1+\frac {i n}{2},2+\frac {i n}{2},\frac {(i-a) (1-i a-i b x)}{(i+a) (1+i a+i b x)}\right )}{(i-a) (i+a)^3 (2 i-n)} \]
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Time = 0.08 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {5203, 98, 133} \[ \int \frac {e^{n \arctan (a+b x)}}{x^3} \, dx=-\frac {(-i a-i b x+1)^{1+\frac {i n}{2}} (i a+i b x+1)^{1-\frac {i n}{2}}}{2 \left (a^2+1\right ) x^2}-\frac {2 b^2 (2 a-n) (-i a-i b x+1)^{1+\frac {i n}{2}} (i a+i b x+1)^{-1-\frac {i n}{2}} \operatorname {Hypergeometric2F1}\left (2,\frac {i n}{2}+1,\frac {i n}{2}+2,\frac {(i-a) (-i a-i b x+1)}{(a+i) (i a+i b x+1)}\right )}{(-a+i) (a+i)^3 (-n+2 i)} \]
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Rule 98
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^3} \, dx \\ & = -\frac {(1-i a-i b x)^{1+\frac {i n}{2}} (1+i a+i b x)^{1-\frac {i n}{2}}}{2 \left (1+a^2\right ) x^2}-\frac {(b (2 a-n)) \int \frac {(1-i a-i b x)^{\frac {i n}{2}} (1+i a+i b x)^{-\frac {i n}{2}}}{x^2} \, dx}{2 \left (1+a^2\right )} \\ & = -\frac {(1-i a-i b x)^{1+\frac {i n}{2}} (1+i a+i b x)^{1-\frac {i n}{2}}}{2 \left (1+a^2\right ) x^2}+\frac {2 b^2 (2 a-n) (1-i a-i b x)^{1+\frac {i n}{2}} (1+i a+i b x)^{-1-\frac {i n}{2}} \operatorname {Hypergeometric2F1}\left (2,1+\frac {i n}{2},2+\frac {i n}{2},\frac {(i-a) (1-i a-i b x)}{(i+a) (1+i a+i b x)}\right )}{(i+a)^2 \left (1+a^2\right ) (2 i-n)} \\ \end{align*}
Time = 0.05 (sec) , antiderivative size = 173, normalized size of antiderivative = 0.84 \[ \int \frac {e^{n \arctan (a+b x)}}{x^3} \, dx=-\frac {i (1+i a+i b x)^{-\frac {i n}{2}} (-i (i+a+b x))^{1+\frac {i n}{2}} \left ((i+a)^2 (-2 i+n) (-i+a+b x)^2+4 b^2 (-2 a+n) x^2 \operatorname {Hypergeometric2F1}\left (2,1+\frac {i n}{2},2+\frac {i n}{2},\frac {1+a^2-i b x+a b x}{1+a^2+i b x+a b x}\right )\right )}{2 (-i+a) (i+a)^3 (-2 i+n) x^2 (-i+a+b x)} \]
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\[\int \frac {{\mathrm e}^{n \arctan \left (b x +a \right )}}{x^{3}}d x\]
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\[ \int \frac {e^{n \arctan (a+b x)}}{x^3} \, dx=\int { \frac {e^{\left (n \arctan \left (b x + a\right )\right )}}{x^{3}} \,d x } \]
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\[ \int \frac {e^{n \arctan (a+b x)}}{x^3} \, dx=\int \frac {e^{n \operatorname {atan}{\left (a + b x \right )}}}{x^{3}}\, dx \]
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\[ \int \frac {e^{n \arctan (a+b x)}}{x^3} \, dx=\int { \frac {e^{\left (n \arctan \left (b x + a\right )\right )}}{x^{3}} \,d x } \]
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Timed out. \[ \int \frac {e^{n \arctan (a+b x)}}{x^3} \, dx=\text {Timed out} \]
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Timed out. \[ \int \frac {e^{n \arctan (a+b x)}}{x^3} \, dx=\int \frac {{\mathrm {e}}^{n\,\mathrm {atan}\left (a+b\,x\right )}}{x^3} \,d x \]
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