Integrand size = 21, antiderivative size = 230 \[ \int x^m \left (d+e x^2\right )^2 (a+b \arctan (c x)) \, dx=\frac {b e \left (e (3+m)-2 c^2 d (5+m)\right ) x^{2+m}}{c^3 (2+m) (3+m) (5+m)}-\frac {b e^2 x^{4+m}}{c (4+m) (5+m)}+\frac {d^2 x^{1+m} (a+b \arctan (c x))}{1+m}+\frac {2 d e x^{3+m} (a+b \arctan (c x))}{3+m}+\frac {e^2 x^{5+m} (a+b \arctan (c x))}{5+m}-\frac {b \left (e^2 \left (3+4 m+m^2\right )-2 c^2 d e \left (5+6 m+m^2\right )+c^4 d^2 \left (15+8 m+m^2\right )\right ) x^{2+m} \operatorname {Hypergeometric2F1}\left (1,\frac {2+m}{2},\frac {4+m}{2},-c^2 x^2\right )}{c^3 (1+m) (2+m) (3+m) (5+m)} \]
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Time = 0.20 (sec) , antiderivative size = 226, normalized size of antiderivative = 0.98, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {276, 5096, 1275, 371} \[ \int x^m \left (d+e x^2\right )^2 (a+b \arctan (c x)) \, dx=\frac {d^2 x^{m+1} (a+b \arctan (c x))}{m+1}+\frac {2 d e x^{m+3} (a+b \arctan (c x))}{m+3}+\frac {e^2 x^{m+5} (a+b \arctan (c x))}{m+5}-\frac {b e x^{m+2} \left (\frac {2 c^2 d}{m+3}-\frac {e}{m+5}\right )}{c^3 (m+2)}-\frac {b x^{m+2} \left (c^4 d^2 \left (m^2+8 m+15\right )-2 c^2 d e \left (m^2+6 m+5\right )+e^2 \left (m^2+4 m+3\right )\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {m+2}{2},\frac {m+4}{2},-c^2 x^2\right )}{c^3 (m+1) (m+2) (m+3) (m+5)}-\frac {b e^2 x^{m+4}}{c (m+4) (m+5)} \]
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Rule 276
Rule 371
Rule 1275
Rule 5096
Rubi steps \begin{align*} \text {integral}& = \frac {d^2 x^{1+m} (a+b \arctan (c x))}{1+m}+\frac {2 d e x^{3+m} (a+b \arctan (c x))}{3+m}+\frac {e^2 x^{5+m} (a+b \arctan (c x))}{5+m}-(b c) \int \frac {x^{1+m} \left (\frac {d^2}{1+m}+\frac {2 d e x^2}{3+m}+\frac {e^2 x^4}{5+m}\right )}{1+c^2 x^2} \, dx \\ & = \frac {d^2 x^{1+m} (a+b \arctan (c x))}{1+m}+\frac {2 d e x^{3+m} (a+b \arctan (c x))}{3+m}+\frac {e^2 x^{5+m} (a+b \arctan (c x))}{5+m}-(b c) \int \left (\frac {e \left (\frac {2 c^2 d}{3+m}-\frac {e}{5+m}\right ) x^{1+m}}{c^4}+\frac {e^2 x^{3+m}}{c^2 (5+m)}+\frac {\left (15 c^4 d^2-10 c^2 d e+3 e^2+8 c^4 d^2 m-12 c^2 d e m+4 e^2 m+c^4 d^2 m^2-2 c^2 d e m^2+e^2 m^2\right ) x^{1+m}}{c^4 (1+m) (3+m) (5+m) \left (1+c^2 x^2\right )}\right ) \, dx \\ & = -\frac {b e \left (\frac {2 c^2 d}{3+m}-\frac {e}{5+m}\right ) x^{2+m}}{c^3 (2+m)}-\frac {b e^2 x^{4+m}}{c (4+m) (5+m)}+\frac {d^2 x^{1+m} (a+b \arctan (c x))}{1+m}+\frac {2 d e x^{3+m} (a+b \arctan (c x))}{3+m}+\frac {e^2 x^{5+m} (a+b \arctan (c x))}{5+m}-\frac {\left (b \left (e^2 \left (3+4 m+m^2\right )-2 c^2 d e \left (5+6 m+m^2\right )+c^4 d^2 \left (15+8 m+m^2\right )\right )\right ) \int \frac {x^{1+m}}{1+c^2 x^2} \, dx}{c^3 (1+m) (3+m) (5+m)} \\ & = -\frac {b e \left (\frac {2 c^2 d}{3+m}-\frac {e}{5+m}\right ) x^{2+m}}{c^3 (2+m)}-\frac {b e^2 x^{4+m}}{c (4+m) (5+m)}+\frac {d^2 x^{1+m} (a+b \arctan (c x))}{1+m}+\frac {2 d e x^{3+m} (a+b \arctan (c x))}{3+m}+\frac {e^2 x^{5+m} (a+b \arctan (c x))}{5+m}-\frac {b \left (e^2 \left (3+4 m+m^2\right )-2 c^2 d e \left (5+6 m+m^2\right )+c^4 d^2 \left (15+8 m+m^2\right )\right ) x^{2+m} \operatorname {Hypergeometric2F1}\left (1,\frac {2+m}{2},\frac {4+m}{2},-c^2 x^2\right )}{c^3 (1+m) (2+m) (3+m) (5+m)} \\ \end{align*}
Time = 0.15 (sec) , antiderivative size = 193, normalized size of antiderivative = 0.84 \[ \int x^m \left (d+e x^2\right )^2 (a+b \arctan (c x)) \, dx=x^{1+m} \left (\frac {d^2 (a+b \arctan (c x))}{1+m}+\frac {2 d e x^2 (a+b \arctan (c x))}{3+m}+\frac {e^2 x^4 (a+b \arctan (c x))}{5+m}-\frac {b c d^2 x \operatorname {Hypergeometric2F1}\left (1,\frac {2+m}{2},\frac {4+m}{2},-c^2 x^2\right )}{2+3 m+m^2}-\frac {2 b c d e x^3 \operatorname {Hypergeometric2F1}\left (1,\frac {4+m}{2},\frac {6+m}{2},-c^2 x^2\right )}{12+7 m+m^2}-\frac {b c e^2 x^5 \operatorname {Hypergeometric2F1}\left (1,\frac {6+m}{2},\frac {8+m}{2},-c^2 x^2\right )}{(5+m) (6+m)}\right ) \]
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\[\int x^{m} \left (e \,x^{2}+d \right )^{2} \left (a +b \arctan \left (c x \right )\right )d x\]
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\[ \int x^m \left (d+e x^2\right )^2 (a+b \arctan (c x)) \, dx=\int { {\left (e x^{2} + d\right )}^{2} {\left (b \arctan \left (c x\right ) + a\right )} x^{m} \,d x } \]
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\[ \int x^m \left (d+e x^2\right )^2 (a+b \arctan (c x)) \, dx=\int x^{m} \left (a + b \operatorname {atan}{\left (c x \right )}\right ) \left (d + e x^{2}\right )^{2}\, dx \]
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\[ \int x^m \left (d+e x^2\right )^2 (a+b \arctan (c x)) \, dx=\int { {\left (e x^{2} + d\right )}^{2} {\left (b \arctan \left (c x\right ) + a\right )} x^{m} \,d x } \]
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\[ \int x^m \left (d+e x^2\right )^2 (a+b \arctan (c x)) \, dx=\int { {\left (e x^{2} + d\right )}^{2} {\left (b \arctan \left (c x\right ) + a\right )} x^{m} \,d x } \]
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Timed out. \[ \int x^m \left (d+e x^2\right )^2 (a+b \arctan (c x)) \, dx=\int x^m\,\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )\,{\left (e\,x^2+d\right )}^2 \,d x \]
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