Integrand size = 21, antiderivative size = 161 \[ \int (f x)^m \left (d+e x^2\right ) (a+b \arcsin (c x)) \, dx=\frac {b e (f x)^{2+m} \sqrt {1-c^2 x^2}}{c f^2 (3+m)^2}+\frac {d (f x)^{1+m} (a+b \arcsin (c x))}{f (1+m)}+\frac {e (f x)^{3+m} (a+b \arcsin (c x))}{f^3 (3+m)}-\frac {b \left (e (1+m) (2+m)+c^2 d (3+m)^2\right ) (f x)^{2+m} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+m}{2},\frac {4+m}{2},c^2 x^2\right )}{c f^2 (1+m) (2+m) (3+m)^2} \]
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Time = 0.12 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.92, number of steps used = 4, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {14, 4815, 12, 470, 371} \[ \int (f x)^m \left (d+e x^2\right ) (a+b \arcsin (c x)) \, dx=\frac {d (f x)^{m+1} (a+b \arcsin (c x))}{f (m+1)}+\frac {e (f x)^{m+3} (a+b \arcsin (c x))}{f^3 (m+3)}-\frac {b c (f x)^{m+2} \left (\frac {e}{c^2 (m+3)^2}+\frac {d}{m^2+3 m+2}\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+2}{2},\frac {m+4}{2},c^2 x^2\right )}{f^2}+\frac {b e \sqrt {1-c^2 x^2} (f x)^{m+2}}{c f^2 (m+3)^2} \]
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Rule 12
Rule 14
Rule 371
Rule 470
Rule 4815
Rubi steps \begin{align*} \text {integral}& = \frac {d (f x)^{1+m} (a+b \arcsin (c x))}{f (1+m)}+\frac {e (f x)^{3+m} (a+b \arcsin (c x))}{f^3 (3+m)}-(b c) \int \frac {(f x)^{1+m} \left (d (3+m)+e (1+m) x^2\right )}{f (1+m) (3+m) \sqrt {1-c^2 x^2}} \, dx \\ & = \frac {d (f x)^{1+m} (a+b \arcsin (c x))}{f (1+m)}+\frac {e (f x)^{3+m} (a+b \arcsin (c x))}{f^3 (3+m)}-\frac {(b c) \int \frac {(f x)^{1+m} \left (d (3+m)+e (1+m) x^2\right )}{\sqrt {1-c^2 x^2}} \, dx}{f \left (3+4 m+m^2\right )} \\ & = \frac {b e (f x)^{2+m} \sqrt {1-c^2 x^2}}{c f^2 (3+m)^2}+\frac {d (f x)^{1+m} (a+b \arcsin (c x))}{f (1+m)}+\frac {e (f x)^{3+m} (a+b \arcsin (c x))}{f^3 (3+m)}-\frac {\left (b c \left (\frac {e (1+m) (2+m)}{c^2 (3+m)}+d (3+m)\right )\right ) \int \frac {(f x)^{1+m}}{\sqrt {1-c^2 x^2}} \, dx}{f \left (3+4 m+m^2\right )} \\ & = \frac {b e (f x)^{2+m} \sqrt {1-c^2 x^2}}{c f^2 (3+m)^2}+\frac {d (f x)^{1+m} (a+b \arcsin (c x))}{f (1+m)}+\frac {e (f x)^{3+m} (a+b \arcsin (c x))}{f^3 (3+m)}-\frac {b c \left (\frac {e (1+m) (2+m)}{c^2 (3+m)}+d (3+m)\right ) (f x)^{2+m} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+m}{2},\frac {4+m}{2},c^2 x^2\right )}{f^2 (2+m) \left (3+4 m+m^2\right )} \\ \end{align*}
Time = 0.13 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.76 \[ \int (f x)^m \left (d+e x^2\right ) (a+b \arcsin (c x)) \, dx=x (f x)^m \left (-\frac {b c d x \operatorname {Hypergeometric2F1}\left (\frac {1}{2},1+\frac {m}{2},2+\frac {m}{2},c^2 x^2\right )}{2+3 m+m^2}+\frac {\frac {\left (d (3+m)+e (1+m) x^2\right ) (a+b \arcsin (c x))}{1+m}-\frac {b c e x^3 \operatorname {Hypergeometric2F1}\left (\frac {1}{2},2+\frac {m}{2},3+\frac {m}{2},c^2 x^2\right )}{4+m}}{3+m}\right ) \]
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\[\int \left (f x \right )^{m} \left (e \,x^{2}+d \right ) \left (a +b \arcsin \left (c x \right )\right )d x\]
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\[ \int (f x)^m \left (d+e x^2\right ) (a+b \arcsin (c x)) \, dx=\int { {\left (e x^{2} + d\right )} {\left (b \arcsin \left (c x\right ) + a\right )} \left (f x\right )^{m} \,d x } \]
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\[ \int (f x)^m \left (d+e x^2\right ) (a+b \arcsin (c x)) \, dx=\int \left (f x\right )^{m} \left (a + b \operatorname {asin}{\left (c x \right )}\right ) \left (d + e x^{2}\right )\, dx \]
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\[ \int (f x)^m \left (d+e x^2\right ) (a+b \arcsin (c x)) \, dx=\int { {\left (e x^{2} + d\right )} {\left (b \arcsin \left (c x\right ) + a\right )} \left (f x\right )^{m} \,d x } \]
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\[ \int (f x)^m \left (d+e x^2\right ) (a+b \arcsin (c x)) \, dx=\int { {\left (e x^{2} + d\right )} {\left (b \arcsin \left (c x\right ) + a\right )} \left (f x\right )^{m} \,d x } \]
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Timed out. \[ \int (f x)^m \left (d+e x^2\right ) (a+b \arcsin (c x)) \, dx=\int \left (a+b\,\mathrm {asin}\left (c\,x\right )\right )\,{\left (f\,x\right )}^m\,\left (e\,x^2+d\right ) \,d x \]
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