Integrand size = 25, antiderivative size = 25 \[ \int \frac {x^m (a+b \arcsin (c x))}{\left (d-c^2 d x^2\right )^3} \, dx=\frac {x^{1+m} (a+b \arcsin (c x))}{4 d^3 \left (1-c^2 x^2\right )^2}+\frac {(3-m) x^{1+m} (a+b \arcsin (c x))}{8 d^3 \left (1-c^2 x^2\right )}-\frac {b c (3-m) x^{2+m} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {2+m}{2},\frac {4+m}{2},c^2 x^2\right )}{8 d^3 (2+m)}-\frac {b c x^{2+m} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},\frac {2+m}{2},\frac {4+m}{2},c^2 x^2\right )}{4 d^3 (2+m)}+\frac {(1-m) (3-m) \text {Int}\left (\frac {x^m (a+b \arcsin (c x))}{d-c^2 d x^2},x\right )}{8 d^2} \]
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Not integrable
Time = 0.15 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 0, number of rules used = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \[ \int \frac {x^m (a+b \arcsin (c x))}{\left (d-c^2 d x^2\right )^3} \, dx=\int \frac {x^m (a+b \arcsin (c x))}{\left (d-c^2 d x^2\right )^3} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \frac {x^{1+m} (a+b \arcsin (c x))}{4 d^3 \left (1-c^2 x^2\right )^2}-\frac {(b c) \int \frac {x^{1+m}}{\left (1-c^2 x^2\right )^{5/2}} \, dx}{4 d^3}+\frac {(3-m) \int \frac {x^m (a+b \arcsin (c x))}{\left (d-c^2 d x^2\right )^2} \, dx}{4 d} \\ & = \frac {x^{1+m} (a+b \arcsin (c x))}{4 d^3 \left (1-c^2 x^2\right )^2}+\frac {(3-m) x^{1+m} (a+b \arcsin (c x))}{8 d^3 \left (1-c^2 x^2\right )}-\frac {b c x^{2+m} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},\frac {2+m}{2},\frac {4+m}{2},c^2 x^2\right )}{4 d^3 (2+m)}-\frac {(b c (3-m)) \int \frac {x^{1+m}}{\left (1-c^2 x^2\right )^{3/2}} \, dx}{8 d^3}+\frac {((1-m) (3-m)) \int \frac {x^m (a+b \arcsin (c x))}{d-c^2 d x^2} \, dx}{8 d^2} \\ & = \frac {x^{1+m} (a+b \arcsin (c x))}{4 d^3 \left (1-c^2 x^2\right )^2}+\frac {(3-m) x^{1+m} (a+b \arcsin (c x))}{8 d^3 \left (1-c^2 x^2\right )}-\frac {b c (3-m) x^{2+m} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {2+m}{2},\frac {4+m}{2},c^2 x^2\right )}{8 d^3 (2+m)}-\frac {b c x^{2+m} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},\frac {2+m}{2},\frac {4+m}{2},c^2 x^2\right )}{4 d^3 (2+m)}+\frac {((1-m) (3-m)) \int \frac {x^m (a+b \arcsin (c x))}{d-c^2 d x^2} \, dx}{8 d^2} \\ \end{align*}
Not integrable
Time = 6.48 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08 \[ \int \frac {x^m (a+b \arcsin (c x))}{\left (d-c^2 d x^2\right )^3} \, dx=\int \frac {x^m (a+b \arcsin (c x))}{\left (d-c^2 d x^2\right )^3} \, dx \]
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Not integrable
Time = 0.31 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00
\[\int \frac {x^{m} \left (a +b \arcsin \left (c x \right )\right )}{\left (-c^{2} d \,x^{2}+d \right )^{3}}d x\]
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Not integrable
Time = 0.26 (sec) , antiderivative size = 55, normalized size of antiderivative = 2.20 \[ \int \frac {x^m (a+b \arcsin (c x))}{\left (d-c^2 d x^2\right )^3} \, dx=\int { -\frac {{\left (b \arcsin \left (c x\right ) + a\right )} x^{m}}{{\left (c^{2} d x^{2} - d\right )}^{3}} \,d x } \]
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Not integrable
Time = 117.00 (sec) , antiderivative size = 73, normalized size of antiderivative = 2.92 \[ \int \frac {x^m (a+b \arcsin (c x))}{\left (d-c^2 d x^2\right )^3} \, dx=- \frac {\int \frac {a x^{m}}{c^{6} x^{6} - 3 c^{4} x^{4} + 3 c^{2} x^{2} - 1}\, dx + \int \frac {b x^{m} \operatorname {asin}{\left (c x \right )}}{c^{6} x^{6} - 3 c^{4} x^{4} + 3 c^{2} x^{2} - 1}\, dx}{d^{3}} \]
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Not integrable
Time = 0.58 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.20 \[ \int \frac {x^m (a+b \arcsin (c x))}{\left (d-c^2 d x^2\right )^3} \, dx=\int { -\frac {{\left (b \arcsin \left (c x\right ) + a\right )} x^{m}}{{\left (c^{2} d x^{2} - d\right )}^{3}} \,d x } \]
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Not integrable
Time = 0.63 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.16 \[ \int \frac {x^m (a+b \arcsin (c x))}{\left (d-c^2 d x^2\right )^3} \, dx=\int { -\frac {{\left (b \arcsin \left (c x\right ) + a\right )} x^{m}}{{\left (c^{2} d x^{2} - d\right )}^{3}} \,d x } \]
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Not integrable
Time = 0.34 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08 \[ \int \frac {x^m (a+b \arcsin (c x))}{\left (d-c^2 d x^2\right )^3} \, dx=\int \frac {x^m\,\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}{{\left (d-c^2\,d\,x^2\right )}^3} \,d x \]
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