Integrand size = 24, antiderivative size = 202 \[ \int \frac {e^{n \arctan (a x)} x}{\sqrt {c+a^2 c x^2}} \, dx=\frac {(1-i a x)^{\frac {1}{2} (1+i n)} (1+i a x)^{\frac {1}{2} (1-i n)} \sqrt {1+a^2 x^2}}{a^2 (1-i n) \sqrt {c+a^2 c x^2}}-\frac {i 2^{\frac {3}{2}-\frac {i n}{2}} n (1-i a x)^{\frac {1}{2} (1+i n)} \sqrt {1+a^2 x^2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (-1+i n),\frac {1}{2} (1+i n),\frac {1}{2} (3+i n),\frac {1}{2} (1-i a x)\right )}{a^2 \left (1+n^2\right ) \sqrt {c+a^2 c x^2}} \]
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Time = 0.13 (sec) , antiderivative size = 202, 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 = {5193, 5190, 80, 71} \[ \int \frac {e^{n \arctan (a x)} x}{\sqrt {c+a^2 c x^2}} \, dx=\frac {\sqrt {a^2 x^2+1} (1-i a x)^{\frac {1}{2} (1+i n)} (1+i a x)^{\frac {1}{2} (1-i n)}}{a^2 (1-i n) \sqrt {a^2 c x^2+c}}-\frac {i 2^{\frac {3}{2}-\frac {i n}{2}} n \sqrt {a^2 x^2+1} (1-i a x)^{\frac {1}{2} (1+i n)} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (i n-1),\frac {1}{2} (i n+1),\frac {1}{2} (i n+3),\frac {1}{2} (1-i a x)\right )}{a^2 \left (n^2+1\right ) \sqrt {a^2 c x^2+c}} \]
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Rule 71
Rule 80
Rule 5190
Rule 5193
Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt {1+a^2 x^2} \int \frac {e^{n \arctan (a x)} x}{\sqrt {1+a^2 x^2}} \, dx}{\sqrt {c+a^2 c x^2}} \\ & = \frac {\sqrt {1+a^2 x^2} \int x (1-i a x)^{-\frac {1}{2}+\frac {i n}{2}} (1+i a x)^{-\frac {1}{2}-\frac {i n}{2}} \, dx}{\sqrt {c+a^2 c x^2}} \\ & = \frac {(1-i a x)^{\frac {1}{2} (1+i n)} (1+i a x)^{\frac {1}{2} (1-i n)} \sqrt {1+a^2 x^2}}{a^2 (1-i n) \sqrt {c+a^2 c x^2}}-\frac {\left (n \sqrt {1+a^2 x^2}\right ) \int (1-i a x)^{-\frac {1}{2}+\frac {i n}{2}} (1+i a x)^{-\frac {1}{2} i (i+n)} \, dx}{a (1-i n) \sqrt {c+a^2 c x^2}} \\ & = \frac {(1-i a x)^{\frac {1}{2} (1+i n)} (1+i a x)^{\frac {1}{2} (1-i n)} \sqrt {1+a^2 x^2}}{a^2 (1-i n) \sqrt {c+a^2 c x^2}}-\frac {i 2^{\frac {3}{2}-\frac {i n}{2}} n (1-i a x)^{\frac {1}{2} (1+i n)} \sqrt {1+a^2 x^2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (-1+i n),\frac {1}{2} (1+i n),\frac {1}{2} (3+i n),\frac {1}{2} (1-i a x)\right )}{a^2 \left (1+n^2\right ) \sqrt {c+a^2 c x^2}} \\ \end{align*}
Time = 0.11 (sec) , antiderivative size = 176, normalized size of antiderivative = 0.87 \[ \int \frac {e^{n \arctan (a x)} x}{\sqrt {c+a^2 c x^2}} \, dx=\frac {(1-i a x)^{\frac {1}{2}+\frac {i n}{2}} (2+2 i a x)^{-\frac {i n}{2}} \sqrt {1+a^2 x^2} \left (2^{\frac {i n}{2}} (1+i n) \sqrt {1+i a x}-2 i \sqrt {2} n (1+i a x)^{\frac {i n}{2}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (1+i n),\frac {1}{2} i (i+n),\frac {1}{2} (3+i n),\frac {1}{2} (1-i a x)\right )\right )}{a^2 \left (1+n^2\right ) \sqrt {c+a^2 c x^2}} \]
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\[\int \frac {{\mathrm e}^{n \arctan \left (a x \right )} x}{\sqrt {a^{2} c \,x^{2}+c}}d x\]
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\[ \int \frac {e^{n \arctan (a x)} x}{\sqrt {c+a^2 c x^2}} \, dx=\int { \frac {x e^{\left (n \arctan \left (a x\right )\right )}}{\sqrt {a^{2} c x^{2} + c}} \,d x } \]
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\[ \int \frac {e^{n \arctan (a x)} x}{\sqrt {c+a^2 c x^2}} \, dx=\int \frac {x e^{n \operatorname {atan}{\left (a x \right )}}}{\sqrt {c \left (a^{2} x^{2} + 1\right )}}\, dx \]
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\[ \int \frac {e^{n \arctan (a x)} x}{\sqrt {c+a^2 c x^2}} \, dx=\int { \frac {x e^{\left (n \arctan \left (a x\right )\right )}}{\sqrt {a^{2} c x^{2} + c}} \,d x } \]
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\[ \int \frac {e^{n \arctan (a x)} x}{\sqrt {c+a^2 c x^2}} \, dx=\int { \frac {x e^{\left (n \arctan \left (a x\right )\right )}}{\sqrt {a^{2} c x^{2} + c}} \,d x } \]
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Timed out. \[ \int \frac {e^{n \arctan (a x)} x}{\sqrt {c+a^2 c x^2}} \, dx=\int \frac {x\,{\mathrm {e}}^{n\,\mathrm {atan}\left (a\,x\right )}}{\sqrt {c\,a^2\,x^2+c}} \,d x \]
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