Integrand size = 24, antiderivative size = 164 \[ \int \frac {e^{n \arctan (a x)} x^2}{c+a^2 c x^2} \, dx=-\frac {(1+i n) (1-i a x)^{\frac {i n}{2}} (1+i a x)^{-\frac {i n}{2}}}{a^3 c n}+\frac {x (1-i a x)^{\frac {i n}{2}} (1+i a x)^{-\frac {i n}{2}}}{a^2 c}+\frac {i 2^{1-\frac {i n}{2}} (1-i a x)^{\frac {i n}{2}} \operatorname {Hypergeometric2F1}\left (\frac {i n}{2},\frac {i n}{2},1+\frac {i n}{2},\frac {1}{2} (1-i a x)\right )}{a^3 c} \]
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Time = 0.10 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {5190, 92, 80, 71} \[ \int \frac {e^{n \arctan (a x)} x^2}{c+a^2 c x^2} \, dx=\frac {i 2^{1-\frac {i n}{2}} (1-i a x)^{\frac {i n}{2}} \operatorname {Hypergeometric2F1}\left (\frac {i n}{2},\frac {i n}{2},\frac {i n}{2}+1,\frac {1}{2} (1-i a x)\right )}{a^3 c}-\frac {(1+i n) (1-i a x)^{\frac {i n}{2}} (1+i a x)^{-\frac {i n}{2}}}{a^3 c n}+\frac {x (1-i a x)^{\frac {i n}{2}} (1+i a x)^{-\frac {i n}{2}}}{a^2 c} \]
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Rule 71
Rule 80
Rule 92
Rule 5190
Rubi steps \begin{align*} \text {integral}& = \frac {\int x^2 (1-i a x)^{-1+\frac {i n}{2}} (1+i a x)^{-1-\frac {i n}{2}} \, dx}{c} \\ & = \frac {x (1-i a x)^{\frac {i n}{2}} (1+i a x)^{-\frac {i n}{2}}}{a^2 c}+\frac {\int (1-i a x)^{-1+\frac {i n}{2}} (1+i a x)^{-1-\frac {i n}{2}} (-1-a n x) \, dx}{a^2 c} \\ & = -\frac {(1+i n) (1-i a x)^{\frac {i n}{2}} (1+i a x)^{-\frac {i n}{2}}}{a^3 c n}+\frac {x (1-i a x)^{\frac {i n}{2}} (1+i a x)^{-\frac {i n}{2}}}{a^2 c}+\frac {(i n) \int (1-i a x)^{-1+\frac {i n}{2}} (1+i a x)^{-\frac {i n}{2}} \, dx}{a^2 c} \\ & = -\frac {(1+i n) (1-i a x)^{\frac {i n}{2}} (1+i a x)^{-\frac {i n}{2}}}{a^3 c n}+\frac {x (1-i a x)^{\frac {i n}{2}} (1+i a x)^{-\frac {i n}{2}}}{a^2 c}+\frac {i 2^{1-\frac {i n}{2}} (1-i a 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 x)\right )}{a^3 c} \\ \end{align*}
Time = 0.09 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.74 \[ \int \frac {e^{n \arctan (a x)} x^2}{c+a^2 c x^2} \, dx=\frac {(1-i a x)^{\frac {i n}{2}} (2+2 i a x)^{-\frac {i n}{2}} \left (2^{\frac {i n}{2}} (-1+n (-i+a x))+2 i n (1+i a x)^{\frac {i n}{2}} \operatorname {Hypergeometric2F1}\left (\frac {i n}{2},\frac {i n}{2},1+\frac {i n}{2},\frac {1}{2} (1-i a x)\right )\right )}{a^3 c n} \]
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\[\int \frac {{\mathrm e}^{n \arctan \left (a x \right )} x^{2}}{a^{2} c \,x^{2}+c}d x\]
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\[ \int \frac {e^{n \arctan (a x)} x^2}{c+a^2 c x^2} \, dx=\int { \frac {x^{2} e^{\left (n \arctan \left (a x\right )\right )}}{a^{2} c x^{2} + c} \,d x } \]
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\[ \int \frac {e^{n \arctan (a x)} x^2}{c+a^2 c x^2} \, dx=\frac {\int \frac {x^{2} e^{n \operatorname {atan}{\left (a x \right )}}}{a^{2} x^{2} + 1}\, dx}{c} \]
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\[ \int \frac {e^{n \arctan (a x)} x^2}{c+a^2 c x^2} \, dx=\int { \frac {x^{2} e^{\left (n \arctan \left (a x\right )\right )}}{a^{2} c x^{2} + c} \,d x } \]
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\[ \int \frac {e^{n \arctan (a x)} x^2}{c+a^2 c x^2} \, dx=\int { \frac {x^{2} e^{\left (n \arctan \left (a x\right )\right )}}{a^{2} c x^{2} + c} \,d x } \]
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Timed out. \[ \int \frac {e^{n \arctan (a x)} x^2}{c+a^2 c x^2} \, dx=\int \frac {x^2\,{\mathrm {e}}^{n\,\mathrm {atan}\left (a\,x\right )}}{c\,a^2\,x^2+c} \,d x \]
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