Integrand size = 25, antiderivative size = 285 \[ \int x^{-5-2 p} \left (d+e x^2\right )^p (a+b \arctan (c x)) \, dx=-\frac {b \left (e+c^2 d (1+p)\right ) x^{-3-2 p} \left (d+e x^2\right )^p \left (1+\frac {e x^2}{d}\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2} (-3-2 p),1,-1-p,\frac {1}{2} (-1-2 p),-c^2 x^2,-\frac {e x^2}{d}\right )}{2 c d (1+p) (2+p) (3+2 p)}+\frac {e x^{-2 (1+p)} \left (d+e x^2\right )^{1+p} (a+b \arctan (c x))}{2 d^2 (1+p) (2+p)}-\frac {x^{-2 (2+p)} \left (d+e x^2\right )^{1+p} (a+b \arctan (c x))}{2 d (2+p)}+\frac {b e x^{-3-2 p} \left (d+e x^2\right )^p \left (1+\frac {e x^2}{d}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (-3-2 p),-1-p,\frac {1}{2} (-1-2 p),-\frac {e x^2}{d}\right )}{2 c d \left (6+13 p+9 p^2+2 p^3\right )} \]
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Time = 0.25 (sec) , antiderivative size = 285, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {277, 270, 5096, 12, 598, 372, 371, 525, 524} \[ \int x^{-5-2 p} \left (d+e x^2\right )^p (a+b \arctan (c x)) \, dx=\frac {e x^{-2 (p+1)} \left (d+e x^2\right )^{p+1} (a+b \arctan (c x))}{2 d^2 (p+1) (p+2)}-\frac {x^{-2 (p+2)} \left (d+e x^2\right )^{p+1} (a+b \arctan (c x))}{2 d (p+2)}-\frac {b x^{-2 p-3} \left (c^2 d (p+1)+e\right ) \left (d+e x^2\right )^p \left (\frac {e x^2}{d}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2} (-2 p-3),1,-p-1,\frac {1}{2} (-2 p-1),-c^2 x^2,-\frac {e x^2}{d}\right )}{2 c d (p+1) (p+2) (2 p+3)}+\frac {b e x^{-2 p-3} \left (d+e x^2\right )^p \left (\frac {e x^2}{d}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (-2 p-3),-p-1,\frac {1}{2} (-2 p-1),-\frac {e x^2}{d}\right )}{2 c d \left (2 p^3+9 p^2+13 p+6\right )} \]
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Rule 12
Rule 270
Rule 277
Rule 371
Rule 372
Rule 524
Rule 525
Rule 598
Rule 5096
Rubi steps \begin{align*} \text {integral}& = \frac {e x^{-2 (1+p)} \left (d+e x^2\right )^{1+p} (a+b \arctan (c x))}{2 d^2 (1+p) (2+p)}-\frac {x^{-2 (2+p)} \left (d+e x^2\right )^{1+p} (a+b \arctan (c x))}{2 d (2+p)}-(b c) \int \frac {x^{-2 (2+p)} \left (d+e x^2\right )^{1+p} \left (-d-d p+e x^2\right )}{2 d^2 (1+p) (2+p) \left (1+c^2 x^2\right )} \, dx \\ & = \frac {e x^{-2 (1+p)} \left (d+e x^2\right )^{1+p} (a+b \arctan (c x))}{2 d^2 (1+p) (2+p)}-\frac {x^{-2 (2+p)} \left (d+e x^2\right )^{1+p} (a+b \arctan (c x))}{2 d (2+p)}-\frac {(b c) \int \frac {x^{-2 (2+p)} \left (d+e x^2\right )^{1+p} \left (-d-d p+e x^2\right )}{1+c^2 x^2} \, dx}{2 d^2 (1+p) (2+p)} \\ & = \frac {e x^{-2 (1+p)} \left (d+e x^2\right )^{1+p} (a+b \arctan (c x))}{2 d^2 (1+p) (2+p)}-\frac {x^{-2 (2+p)} \left (d+e x^2\right )^{1+p} (a+b \arctan (c x))}{2 d (2+p)}-\frac {(b c) \int \left (\frac {e x^{-2 (2+p)} \left (d+e x^2\right )^{1+p}}{c^2}+\frac {\left (-e+c^2 (-d-d p)\right ) x^{-2 (2+p)} \left (d+e x^2\right )^{1+p}}{c^2 \left (1+c^2 x^2\right )}\right ) \, dx}{2 d^2 (1+p) (2+p)} \\ & = \frac {e x^{-2 (1+p)} \left (d+e x^2\right )^{1+p} (a+b \arctan (c x))}{2 d^2 (1+p) (2+p)}-\frac {x^{-2 (2+p)} \left (d+e x^2\right )^{1+p} (a+b \arctan (c x))}{2 d (2+p)}-\frac {(b e) \int x^{-2 (2+p)} \left (d+e x^2\right )^{1+p} \, dx}{2 c d^2 (1+p) (2+p)}+\frac {\left (b \left (e+c^2 d (1+p)\right )\right ) \int \frac {x^{-2 (2+p)} \left (d+e x^2\right )^{1+p}}{1+c^2 x^2} \, dx}{2 c d^2 (1+p) (2+p)} \\ & = \frac {e x^{-2 (1+p)} \left (d+e x^2\right )^{1+p} (a+b \arctan (c x))}{2 d^2 (1+p) (2+p)}-\frac {x^{-2 (2+p)} \left (d+e x^2\right )^{1+p} (a+b \arctan (c x))}{2 d (2+p)}-\frac {\left (b e \left (d+e x^2\right )^p \left (1+\frac {e x^2}{d}\right )^{-p}\right ) \int x^{-2 (2+p)} \left (1+\frac {e x^2}{d}\right )^{1+p} \, dx}{2 c d (1+p) (2+p)}+\frac {\left (b \left (e+c^2 d (1+p)\right ) \left (d+e x^2\right )^p \left (1+\frac {e x^2}{d}\right )^{-p}\right ) \int \frac {x^{-2 (2+p)} \left (1+\frac {e x^2}{d}\right )^{1+p}}{1+c^2 x^2} \, dx}{2 c d (1+p) (2+p)} \\ & = -\frac {b \left (e+c^2 d (1+p)\right ) x^{-3-2 p} \left (d+e x^2\right )^p \left (1+\frac {e x^2}{d}\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2} (-3-2 p),1,-1-p,\frac {1}{2} (-1-2 p),-c^2 x^2,-\frac {e x^2}{d}\right )}{2 c d (1+p) (2+p) (3+2 p)}+\frac {e x^{-2 (1+p)} \left (d+e x^2\right )^{1+p} (a+b \arctan (c x))}{2 d^2 (1+p) (2+p)}-\frac {x^{-2 (2+p)} \left (d+e x^2\right )^{1+p} (a+b \arctan (c x))}{2 d (2+p)}+\frac {b e x^{-3-2 p} \left (d+e x^2\right )^p \left (1+\frac {e x^2}{d}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (-3-2 p),-1-p,\frac {1}{2} (-1-2 p),-\frac {e x^2}{d}\right )}{2 c d \left (6+13 p+9 p^2+2 p^3\right )} \\ \end{align*}
\[ \int x^{-5-2 p} \left (d+e x^2\right )^p (a+b \arctan (c x)) \, dx=\int x^{-5-2 p} \left (d+e x^2\right )^p (a+b \arctan (c x)) \, dx \]
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\[\int x^{-5-2 p} \left (e \,x^{2}+d \right )^{p} \left (a +b \arctan \left (c x \right )\right )d x\]
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\[ \int x^{-5-2 p} \left (d+e x^2\right )^p (a+b \arctan (c x)) \, dx=\int { {\left (b \arctan \left (c x\right ) + a\right )} {\left (e x^{2} + d\right )}^{p} x^{-2 \, p - 5} \,d x } \]
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Timed out. \[ \int x^{-5-2 p} \left (d+e x^2\right )^p (a+b \arctan (c x)) \, dx=\text {Timed out} \]
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\[ \int x^{-5-2 p} \left (d+e x^2\right )^p (a+b \arctan (c x)) \, dx=\int { {\left (b \arctan \left (c x\right ) + a\right )} {\left (e x^{2} + d\right )}^{p} x^{-2 \, p - 5} \,d x } \]
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\[ \int x^{-5-2 p} \left (d+e x^2\right )^p (a+b \arctan (c x)) \, dx=\int { {\left (b \arctan \left (c x\right ) + a\right )} {\left (e x^{2} + d\right )}^{p} x^{-2 \, p - 5} \,d x } \]
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Timed out. \[ \int x^{-5-2 p} \left (d+e x^2\right )^p (a+b \arctan (c x)) \, dx=\int \frac {\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )\,{\left (e\,x^2+d\right )}^p}{x^{2\,p+5}} \,d x \]
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