Integrand size = 19, antiderivative size = 122 \[ \int x^m \left (d+e x^2\right ) (a+b \arctan (c x)) \, dx=-\frac {b e x^{2+m}}{c \left (6+5 m+m^2\right )}+\frac {d x^{1+m} (a+b \arctan (c x))}{1+m}+\frac {e x^{3+m} (a+b \arctan (c x))}{3+m}-\frac {b \left (\frac {c^2 d}{1+m}-\frac {e}{3+m}\right ) x^{2+m} \operatorname {Hypergeometric2F1}\left (1,\frac {2+m}{2},\frac {4+m}{2},-c^2 x^2\right )}{c (2+m)} \]
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Time = 0.09 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {14, 5096, 470, 371} \[ \int x^m \left (d+e x^2\right ) (a+b \arctan (c x)) \, dx=\frac {d x^{m+1} (a+b \arctan (c x))}{m+1}+\frac {e x^{m+3} (a+b \arctan (c x))}{m+3}-\frac {b x^{m+2} \left (\frac {c^2 d}{m+1}-\frac {e}{m+3}\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {m+2}{2},\frac {m+4}{2},-c^2 x^2\right )}{c (m+2)}-\frac {b e x^{m+2}}{c \left (m^2+5 m+6\right )} \]
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Rule 14
Rule 371
Rule 470
Rule 5096
Rubi steps \begin{align*} \text {integral}& = \frac {d x^{1+m} (a+b \arctan (c x))}{1+m}+\frac {e x^{3+m} (a+b \arctan (c x))}{3+m}-(b c) \int \frac {x^{1+m} \left (\frac {d}{1+m}+\frac {e x^2}{3+m}\right )}{1+c^2 x^2} \, dx \\ & = -\frac {b e x^{2+m}}{c \left (6+5 m+m^2\right )}+\frac {d x^{1+m} (a+b \arctan (c x))}{1+m}+\frac {e x^{3+m} (a+b \arctan (c x))}{3+m}+\left (b c \left (-\frac {d}{1+m}+\frac {e}{c^2 (3+m)}\right )\right ) \int \frac {x^{1+m}}{1+c^2 x^2} \, dx \\ & = -\frac {b e x^{2+m}}{c \left (6+5 m+m^2\right )}+\frac {d x^{1+m} (a+b \arctan (c x))}{1+m}+\frac {e x^{3+m} (a+b \arctan (c x))}{3+m}-\frac {b c \left (\frac {d}{1+m}-\frac {e}{c^2 (3+m)}\right ) x^{2+m} \operatorname {Hypergeometric2F1}\left (1,\frac {2+m}{2},\frac {4+m}{2},-c^2 x^2\right )}{2+m} \\ \end{align*}
Time = 0.13 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.98 \[ \int x^m \left (d+e x^2\right ) (a+b \arctan (c x)) \, dx=x^{1+m} \left (-\frac {b c d x \operatorname {Hypergeometric2F1}\left (1,\frac {2+m}{2},\frac {4+m}{2},-c^2 x^2\right )}{2+3 m+m^2}+\frac {\frac {\left (d (3+m)+e (1+m) x^2\right ) (a+b \arctan (c x))}{1+m}-\frac {b c e x^3 \operatorname {Hypergeometric2F1}\left (1,\frac {4+m}{2},\frac {6+m}{2},-c^2 x^2\right )}{4+m}}{3+m}\right ) \]
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\[\int x^{m} \left (e \,x^{2}+d \right ) \left (a +b \arctan \left (c x \right )\right )d x\]
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\[ \int x^m \left (d+e x^2\right ) (a+b \arctan (c x)) \, dx=\int { {\left (e x^{2} + d\right )} {\left (b \arctan \left (c x\right ) + a\right )} x^{m} \,d x } \]
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\[ \int x^m \left (d+e x^2\right ) (a+b \arctan (c x)) \, dx=\int x^{m} \left (a + b \operatorname {atan}{\left (c x \right )}\right ) \left (d + e x^{2}\right )\, dx \]
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\[ \int x^m \left (d+e x^2\right ) (a+b \arctan (c x)) \, dx=\int { {\left (e x^{2} + d\right )} {\left (b \arctan \left (c x\right ) + a\right )} x^{m} \,d x } \]
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\[ \int x^m \left (d+e x^2\right ) (a+b \arctan (c x)) \, dx=\int { {\left (e x^{2} + d\right )} {\left (b \arctan \left (c x\right ) + a\right )} x^{m} \,d x } \]
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Timed out. \[ \int x^m \left (d+e x^2\right ) (a+b \arctan (c x)) \, dx=\int x^m\,\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )\,\left (e\,x^2+d\right ) \,d x \]
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