Integrand size = 25, antiderivative size = 129 \[ \int x^{-3-2 p} \left (d+e x^2\right )^p (a+b \arctan (c x)) \, dx=-\frac {b c x^{-1-2 p} \left (d+e x^2\right )^p \left (1+\frac {e x^2}{d}\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2} (-1-2 p),1,-1-p,\frac {1}{2} (1-2 p),-c^2 x^2,-\frac {e x^2}{d}\right )}{2 \left (1+3 p+2 p^2\right )}-\frac {x^{-2 (1+p)} \left (d+e x^2\right )^{1+p} (a+b \arctan (c x))}{2 d (1+p)} \]
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Time = 0.11 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {270, 5096, 12, 525, 524} \[ \int x^{-3-2 p} \left (d+e x^2\right )^p (a+b \arctan (c x)) \, dx=-\frac {x^{-2 (p+1)} \left (d+e x^2\right )^{p+1} (a+b \arctan (c x))}{2 d (p+1)}-\frac {b c x^{-2 p-1} \left (d+e x^2\right )^p \left (\frac {e x^2}{d}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2} (-2 p-1),1,-p-1,\frac {1}{2} (1-2 p),-c^2 x^2,-\frac {e x^2}{d}\right )}{2 \left (2 p^2+3 p+1\right )} \]
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Rule 12
Rule 270
Rule 524
Rule 525
Rule 5096
Rubi steps \begin{align*} \text {integral}& = -\frac {x^{-2 (1+p)} \left (d+e x^2\right )^{1+p} (a+b \arctan (c x))}{2 d (1+p)}-(b c) \int -\frac {x^{-2 (1+p)} \left (d+e x^2\right )^{1+p}}{2 d (1+p) \left (1+c^2 x^2\right )} \, dx \\ & = -\frac {x^{-2 (1+p)} \left (d+e x^2\right )^{1+p} (a+b \arctan (c x))}{2 d (1+p)}+\frac {(b c) \int \frac {x^{-2 (1+p)} \left (d+e x^2\right )^{1+p}}{1+c^2 x^2} \, dx}{2 d (1+p)} \\ & = -\frac {x^{-2 (1+p)} \left (d+e x^2\right )^{1+p} (a+b \arctan (c x))}{2 d (1+p)}+\frac {\left (b c \left (d+e x^2\right )^p \left (1+\frac {e x^2}{d}\right )^{-p}\right ) \int \frac {x^{-2 (1+p)} \left (1+\frac {e x^2}{d}\right )^{1+p}}{1+c^2 x^2} \, dx}{2 (1+p)} \\ & = -\frac {b c x^{-1-2 p} \left (d+e x^2\right )^p \left (1+\frac {e x^2}{d}\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2} (-1-2 p),1,-1-p,\frac {1}{2} (1-2 p),-c^2 x^2,-\frac {e x^2}{d}\right )}{2 \left (1+3 p+2 p^2\right )}-\frac {x^{-2 (1+p)} \left (d+e x^2\right )^{1+p} (a+b \arctan (c x))}{2 d (1+p)} \\ \end{align*}
Time = 0.47 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.29 \[ \int x^{-3-2 p} \left (d+e x^2\right )^p (a+b \arctan (c x)) \, dx=-\frac {x^{-2 (1+p)} \left (d+e x^2\right )^p \left (1+\frac {e x^2}{d}\right )^{-p} \left (b \left (c^2 d-e\right ) x \operatorname {AppellF1}\left (-\frac {1}{2}-p,-p,1,\frac {1}{2}-p,-\frac {e x^2}{d},-c^2 x^2\right )+c (1+2 p) \left (d+e x^2\right ) \left (1+\frac {e x^2}{d}\right )^p (a+b \arctan (c x))+b e x \operatorname {Hypergeometric2F1}\left (-\frac {1}{2}-p,-p,\frac {1}{2}-p,-\frac {e x^2}{d}\right )\right )}{2 c d (1+p) (1+2 p)} \]
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\[\int x^{-3-2 p} \left (e \,x^{2}+d \right )^{p} \left (a +b \arctan \left (c x \right )\right )d x\]
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\[ \int x^{-3-2 p} \left (d+e x^2\right )^p (a+b \arctan (c x)) \, dx=\int { {\left (b \arctan \left (c x\right ) + a\right )} {\left (e x^{2} + d\right )}^{p} x^{-2 \, p - 3} \,d x } \]
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Timed out. \[ \int x^{-3-2 p} \left (d+e x^2\right )^p (a+b \arctan (c x)) \, dx=\text {Timed out} \]
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\[ \int x^{-3-2 p} \left (d+e x^2\right )^p (a+b \arctan (c x)) \, dx=\int { {\left (b \arctan \left (c x\right ) + a\right )} {\left (e x^{2} + d\right )}^{p} x^{-2 \, p - 3} \,d x } \]
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\[ \int x^{-3-2 p} \left (d+e x^2\right )^p (a+b \arctan (c x)) \, dx=\int { {\left (b \arctan \left (c x\right ) + a\right )} {\left (e x^{2} + d\right )}^{p} x^{-2 \, p - 3} \,d x } \]
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Timed out. \[ \int x^{-3-2 p} \left (d+e x^2\right )^p (a+b \arctan (c x)) \, dx=\int \frac {\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )\,{\left (e\,x^2+d\right )}^p}{x^{2\,p+3}} \,d x \]
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