Integrand size = 15, antiderivative size = 171 \[ \int \frac {e^{i n \arctan (a x)}}{x^4} \, dx=-\frac {(1-i a x)^{1-\frac {n}{2}} (1+i a x)^{\frac {2+n}{2}}}{3 x^3}-\frac {i a n (1-i a x)^{1-\frac {n}{2}} (1+i a x)^{\frac {2+n}{2}}}{6 x^2}+\frac {2 i a^3 \left (2+n^2\right ) (1-i a x)^{1-\frac {n}{2}} (1+i a x)^{\frac {1}{2} (-2+n)} \operatorname {Hypergeometric2F1}\left (2,1-\frac {n}{2},2-\frac {n}{2},\frac {1-i a x}{1+i a x}\right )}{3 (2-n)} \]
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Time = 0.05 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {5170, 105, 156, 12, 133} \[ \int \frac {e^{i n \arctan (a x)}}{x^4} \, dx=\frac {2 i a^3 \left (n^2+2\right ) (1+i a x)^{\frac {n-2}{2}} (1-i a x)^{1-\frac {n}{2}} \operatorname {Hypergeometric2F1}\left (2,1-\frac {n}{2},2-\frac {n}{2},\frac {1-i a x}{i a x+1}\right )}{3 (2-n)}-\frac {(1+i a x)^{\frac {n+2}{2}} (1-i a x)^{1-\frac {n}{2}}}{3 x^3}-\frac {i a n (1+i a x)^{\frac {n+2}{2}} (1-i a x)^{1-\frac {n}{2}}}{6 x^2} \]
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Rule 12
Rule 105
Rule 133
Rule 156
Rule 5170
Rubi steps \begin{align*} \text {integral}& = \int \frac {(1-i a x)^{-n/2} (1+i a x)^{n/2}}{x^4} \, dx \\ & = -\frac {(1-i a x)^{1-\frac {n}{2}} (1+i a x)^{\frac {2+n}{2}}}{3 x^3}-\frac {1}{3} \int \frac {(1-i a x)^{-n/2} (1+i a x)^{n/2} \left (-i a n+a^2 x\right )}{x^3} \, dx \\ & = -\frac {(1-i a x)^{1-\frac {n}{2}} (1+i a x)^{\frac {2+n}{2}}}{3 x^3}-\frac {i a n (1-i a x)^{1-\frac {n}{2}} (1+i a x)^{\frac {2+n}{2}}}{6 x^2}-\frac {1}{6} \int \frac {a^2 \left (2+n^2\right ) (1-i a x)^{-n/2} (1+i a x)^{n/2}}{x^2} \, dx \\ & = -\frac {(1-i a x)^{1-\frac {n}{2}} (1+i a x)^{\frac {2+n}{2}}}{3 x^3}-\frac {i a n (1-i a x)^{1-\frac {n}{2}} (1+i a x)^{\frac {2+n}{2}}}{6 x^2}-\frac {1}{6} \left (a^2 \left (2+n^2\right )\right ) \int \frac {(1-i a x)^{-n/2} (1+i a x)^{n/2}}{x^2} \, dx \\ & = -\frac {(1-i a x)^{1-\frac {n}{2}} (1+i a x)^{\frac {2+n}{2}}}{3 x^3}-\frac {i a n (1-i a x)^{1-\frac {n}{2}} (1+i a x)^{\frac {2+n}{2}}}{6 x^2}+\frac {2 i a^3 \left (2+n^2\right ) (1-i a x)^{1-\frac {n}{2}} (1+i a x)^{\frac {1}{2} (-2+n)} \operatorname {Hypergeometric2F1}\left (2,1-\frac {n}{2},2-\frac {n}{2},\frac {1-i a x}{1+i a x}\right )}{3 (2-n)} \\ \end{align*}
Time = 0.06 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.70 \[ \int \frac {e^{i n \arctan (a x)}}{x^4} \, dx=-\frac {(1-i a x)^{-n/2} (1+i a x)^{\frac {1}{2} (-2+n)} (i+a x) \left (-\left ((-2+n) (-i+a x)^2 (-2 i+a n x)\right )+4 a^3 \left (2+n^2\right ) x^3 \operatorname {Hypergeometric2F1}\left (2,1-\frac {n}{2},2-\frac {n}{2},\frac {i+a x}{i-a x}\right )\right )}{6 (-2+n) x^3} \]
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\[\int \frac {{\mathrm e}^{i n \arctan \left (a x \right )}}{x^{4}}d x\]
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\[ \int \frac {e^{i n \arctan (a x)}}{x^4} \, dx=\int { \frac {e^{\left (i \, n \arctan \left (a x\right )\right )}}{x^{4}} \,d x } \]
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\[ \int \frac {e^{i n \arctan (a x)}}{x^4} \, dx=\int \frac {e^{i n \operatorname {atan}{\left (a x \right )}}}{x^{4}}\, dx \]
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\[ \int \frac {e^{i n \arctan (a x)}}{x^4} \, dx=\int { \frac {e^{\left (i \, n \arctan \left (a x\right )\right )}}{x^{4}} \,d x } \]
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\[ \int \frac {e^{i n \arctan (a x)}}{x^4} \, dx=\int { \frac {e^{\left (i \, n \arctan \left (a x\right )\right )}}{x^{4}} \,d x } \]
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Timed out. \[ \int \frac {e^{i n \arctan (a x)}}{x^4} \, dx=\int \frac {{\mathrm {e}}^{n\,\mathrm {atan}\left (a\,x\right )\,1{}\mathrm {i}}}{x^4} \,d x \]
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