Integrand size = 24, antiderivative size = 131 \[ \int \frac {e^{n \arctan (a x)} x^3}{c+a^2 c x^2} \, dx=\frac {e^{n \arctan (a x)} \left (2 i+n-i n^2\right )}{2 a^4 c n}-\frac {e^{n \arctan (a x)} n x}{2 a^3 c}+\frac {e^{n \arctan (a x)} x^2}{2 a^2 c}+\frac {i e^{n \arctan (a x)} \left (-2+n^2\right ) \operatorname {Hypergeometric2F1}\left (1,-\frac {i n}{2},1-\frac {i n}{2},-e^{2 i \arctan (a x)}\right )}{a^4 c n} \]
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Time = 0.17 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.57, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {5190, 102, 148, 71} \[ \int \frac {e^{n \arctan (a x)} x^3}{c+a^2 c x^2} \, dx=\frac {2^{-1-\frac {i n}{2}} \left (2-n^2\right ) (1-i a x)^{1+\frac {i n}{2}} \operatorname {Hypergeometric2F1}\left (\frac {i n}{2}+1,\frac {i n}{2}+1,\frac {i n}{2}+2,\frac {1}{2} (1-i a x)\right )}{a^4 c (2+i n)}+\frac {i (1+i a x)^{-\frac {i n}{2}} \left (i a n^2 x-n^2-i n+2\right ) (1-i a x)^{\frac {i n}{2}}}{2 a^4 c n}+\frac {x^2 (1+i a x)^{-\frac {i n}{2}} (1-i a x)^{\frac {i n}{2}}}{2 a^2 c} \]
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Rule 71
Rule 102
Rule 148
Rule 5190
Rubi steps \begin{align*} \text {integral}& = \frac {\int x^3 (1-i a x)^{-1+\frac {i n}{2}} (1+i a x)^{-1-\frac {i n}{2}} \, dx}{c} \\ & = \frac {x^2 (1-i a x)^{\frac {i n}{2}} (1+i a x)^{-\frac {i n}{2}}}{2 a^2 c}+\frac {\int x (1-i a x)^{-1+\frac {i n}{2}} (1+i a x)^{-1-\frac {i n}{2}} (-2-a n x) \, dx}{2 a^2 c} \\ & = \frac {x^2 (1-i a x)^{\frac {i n}{2}} (1+i a x)^{-\frac {i n}{2}}}{2 a^2 c}+\frac {i (1-i a x)^{\frac {i n}{2}} (1+i a x)^{-\frac {i n}{2}} \left (2-i n-n^2+i a n^2 x\right )}{2 a^4 c n}-\frac {\left (i \left (2-n^2\right )\right ) \int (1-i a x)^{\frac {i n}{2}} (1+i a x)^{-1-\frac {i n}{2}} \, dx}{2 a^3 c} \\ & = \frac {x^2 (1-i a x)^{\frac {i n}{2}} (1+i a x)^{-\frac {i n}{2}}}{2 a^2 c}+\frac {i (1-i a x)^{\frac {i n}{2}} (1+i a x)^{-\frac {i n}{2}} \left (2-i n-n^2+i a n^2 x\right )}{2 a^4 c n}+\frac {2^{-1-\frac {i n}{2}} \left (2-n^2\right ) (1-i a x)^{1+\frac {i n}{2}} \operatorname {Hypergeometric2F1}\left (1+\frac {i n}{2},1+\frac {i n}{2},2+\frac {i n}{2},\frac {1}{2} (1-i a x)\right )}{a^4 c (2+i n)} \\ \end{align*}
Time = 0.09 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.08 \[ \int \frac {e^{n \arctan (a x)} x^3}{c+a^2 c x^2} \, dx=\frac {(1-i a x)^{\frac {i n}{2}} \left (\frac {(1+i a x)^{-\frac {i n}{2}} \left (2 i+n+a^2 n x^2-n^2 (i+a x)\right )}{n}+\frac {2^{-\frac {i n}{2}} \left (-2+n^2\right ) (i+a x) \operatorname {Hypergeometric2F1}\left (1+\frac {i n}{2},1+\frac {i n}{2},2+\frac {i n}{2},\frac {1}{2} (1-i a x)\right )}{-2 i+n}\right )}{2 a^4 c} \]
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\[\int \frac {{\mathrm e}^{n \arctan \left (a x \right )} x^{3}}{a^{2} c \,x^{2}+c}d x\]
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\[ \int \frac {e^{n \arctan (a x)} x^3}{c+a^2 c x^2} \, dx=\int { \frac {x^{3} e^{\left (n \arctan \left (a x\right )\right )}}{a^{2} c x^{2} + c} \,d x } \]
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\[ \int \frac {e^{n \arctan (a x)} x^3}{c+a^2 c x^2} \, dx=\frac {\int \frac {x^{3} e^{n \operatorname {atan}{\left (a x \right )}}}{a^{2} x^{2} + 1}\, dx}{c} \]
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\[ \int \frac {e^{n \arctan (a x)} x^3}{c+a^2 c x^2} \, dx=\int { \frac {x^{3} e^{\left (n \arctan \left (a x\right )\right )}}{a^{2} c x^{2} + c} \,d x } \]
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\[ \int \frac {e^{n \arctan (a x)} x^3}{c+a^2 c x^2} \, dx=\int { \frac {x^{3} e^{\left (n \arctan \left (a x\right )\right )}}{a^{2} c x^{2} + c} \,d x } \]
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Timed out. \[ \int \frac {e^{n \arctan (a x)} x^3}{c+a^2 c x^2} \, dx=\int \frac {x^3\,{\mathrm {e}}^{n\,\mathrm {atan}\left (a\,x\right )}}{c\,a^2\,x^2+c} \,d x \]
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