Integrand size = 27, antiderivative size = 27 \[ \int \frac {x^m (a+b \arcsin (c x))^2}{\left (d-c^2 d x^2\right )^3} \, dx=-\frac {b c x^{2+m} (a+b \arcsin (c x))}{6 d^3 \left (1-c^2 x^2\right )^{3/2}}-\frac {b c (1-m) x^{2+m} (a+b \arcsin (c x))}{6 d^3 \sqrt {1-c^2 x^2}}-\frac {b c (3-m) x^{2+m} (a+b \arcsin (c x))}{4 d^3 \sqrt {1-c^2 x^2}}+\frac {x^{1+m} (a+b \arcsin (c x))^2}{4 d^3 \left (1-c^2 x^2\right )^2}+\frac {(3-m) x^{1+m} (a+b \arcsin (c x))^2}{8 d^3 \left (1-c^2 x^2\right )}+\frac {b c (1-m) (1+m) x^{2+m} (a+b \arcsin (c x)) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+m}{2},\frac {4+m}{2},c^2 x^2\right )}{6 d^3 (2+m)}+\frac {b c (3-m) (1+m) x^{2+m} (a+b \arcsin (c x)) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+m}{2},\frac {4+m}{2},c^2 x^2\right )}{4 d^3 (2+m)}+\frac {b^2 c^2 (1-m) x^{3+m} \operatorname {Hypergeometric2F1}\left (1,\frac {3+m}{2},\frac {5+m}{2},c^2 x^2\right )}{6 d^3 (3+m)}+\frac {b^2 c^2 (3-m) x^{3+m} \operatorname {Hypergeometric2F1}\left (1,\frac {3+m}{2},\frac {5+m}{2},c^2 x^2\right )}{4 d^3 (3+m)}+\frac {b^2 c^2 x^{3+m} \operatorname {Hypergeometric2F1}\left (2,\frac {3+m}{2},\frac {5+m}{2},c^2 x^2\right )}{6 d^3 (3+m)}-\frac {b^2 c^2 (1-m) (1+m) x^{3+m} \, _3F_2\left (1,\frac {3}{2}+\frac {m}{2},\frac {3}{2}+\frac {m}{2};2+\frac {m}{2},\frac {5}{2}+\frac {m}{2};c^2 x^2\right )}{6 d^3 \left (6+5 m+m^2\right )}-\frac {b^2 c^2 (3-m) (1+m) x^{3+m} \, _3F_2\left (1,\frac {3}{2}+\frac {m}{2},\frac {3}{2}+\frac {m}{2};2+\frac {m}{2},\frac {5}{2}+\frac {m}{2};c^2 x^2\right )}{4 d^3 \left (6+5 m+m^2\right )}+\frac {(1-m) (3-m) \text {Int}\left (\frac {x^m (a+b \arcsin (c x))^2}{d-c^2 d x^2},x\right )}{8 d^2} \]
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Not integrable
Time = 0.62 (sec) , antiderivative size = 27, 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 = {} \[ \int \frac {x^m (a+b \arcsin (c x))^2}{\left (d-c^2 d x^2\right )^3} \, dx=\int \frac {x^m (a+b \arcsin (c x))^2}{\left (d-c^2 d x^2\right )^3} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \frac {x^{1+m} (a+b \arcsin (c x))^2}{4 d^3 \left (1-c^2 x^2\right )^2}-\frac {(b c) \int \frac {x^{1+m} (a+b \arcsin (c x))}{\left (1-c^2 x^2\right )^{5/2}} \, dx}{2 d^3}+\frac {(3-m) \int \frac {x^m (a+b \arcsin (c x))^2}{\left (d-c^2 d x^2\right )^2} \, dx}{4 d} \\ & = -\frac {b c x^{2+m} (a+b \arcsin (c x))}{6 d^3 \left (1-c^2 x^2\right )^{3/2}}+\frac {x^{1+m} (a+b \arcsin (c x))^2}{4 d^3 \left (1-c^2 x^2\right )^2}+\frac {(3-m) x^{1+m} (a+b \arcsin (c x))^2}{8 d^3 \left (1-c^2 x^2\right )}+\frac {\left (b^2 c^2\right ) \int \frac {x^{2+m}}{\left (1-c^2 x^2\right )^2} \, dx}{6 d^3}-\frac {(b c (1-m)) \int \frac {x^{1+m} (a+b \arcsin (c x))}{\left (1-c^2 x^2\right )^{3/2}} \, dx}{6 d^3}-\frac {(b c (3-m)) \int \frac {x^{1+m} (a+b \arcsin (c x))}{\left (1-c^2 x^2\right )^{3/2}} \, dx}{4 d^3}+\frac {((1-m) (3-m)) \int \frac {x^m (a+b \arcsin (c x))^2}{d-c^2 d x^2} \, dx}{8 d^2} \\ & = -\frac {b c x^{2+m} (a+b \arcsin (c x))}{6 d^3 \left (1-c^2 x^2\right )^{3/2}}-\frac {b c (1-m) x^{2+m} (a+b \arcsin (c x))}{6 d^3 \sqrt {1-c^2 x^2}}-\frac {b c (3-m) x^{2+m} (a+b \arcsin (c x))}{4 d^3 \sqrt {1-c^2 x^2}}+\frac {x^{1+m} (a+b \arcsin (c x))^2}{4 d^3 \left (1-c^2 x^2\right )^2}+\frac {(3-m) x^{1+m} (a+b \arcsin (c x))^2}{8 d^3 \left (1-c^2 x^2\right )}+\frac {b^2 c^2 x^{3+m} \operatorname {Hypergeometric2F1}\left (2,\frac {3+m}{2},\frac {5+m}{2},c^2 x^2\right )}{6 d^3 (3+m)}+\frac {\left (b^2 c^2 (1-m)\right ) \int \frac {x^{2+m}}{1-c^2 x^2} \, dx}{6 d^3}+\frac {\left (b^2 c^2 (3-m)\right ) \int \frac {x^{2+m}}{1-c^2 x^2} \, dx}{4 d^3}+\frac {((1-m) (3-m)) \int \frac {x^m (a+b \arcsin (c x))^2}{d-c^2 d x^2} \, dx}{8 d^2}+\frac {(b c (1-m) (1+m)) \int \frac {x^{1+m} (a+b \arcsin (c x))}{\sqrt {1-c^2 x^2}} \, dx}{6 d^3}+\frac {(b c (3-m) (1+m)) \int \frac {x^{1+m} (a+b \arcsin (c x))}{\sqrt {1-c^2 x^2}} \, dx}{4 d^3} \\ & = -\frac {b c x^{2+m} (a+b \arcsin (c x))}{6 d^3 \left (1-c^2 x^2\right )^{3/2}}-\frac {b c (1-m) x^{2+m} (a+b \arcsin (c x))}{6 d^3 \sqrt {1-c^2 x^2}}-\frac {b c (3-m) x^{2+m} (a+b \arcsin (c x))}{4 d^3 \sqrt {1-c^2 x^2}}+\frac {x^{1+m} (a+b \arcsin (c x))^2}{4 d^3 \left (1-c^2 x^2\right )^2}+\frac {(3-m) x^{1+m} (a+b \arcsin (c x))^2}{8 d^3 \left (1-c^2 x^2\right )}+\frac {b c (1-m) (1+m) x^{2+m} (a+b \arcsin (c x)) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+m}{2},\frac {4+m}{2},c^2 x^2\right )}{6 d^3 (2+m)}+\frac {b c (3-m) (1+m) x^{2+m} (a+b \arcsin (c x)) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+m}{2},\frac {4+m}{2},c^2 x^2\right )}{4 d^3 (2+m)}+\frac {b^2 c^2 (1-m) x^{3+m} \operatorname {Hypergeometric2F1}\left (1,\frac {3+m}{2},\frac {5+m}{2},c^2 x^2\right )}{6 d^3 (3+m)}+\frac {b^2 c^2 (3-m) x^{3+m} \operatorname {Hypergeometric2F1}\left (1,\frac {3+m}{2},\frac {5+m}{2},c^2 x^2\right )}{4 d^3 (3+m)}+\frac {b^2 c^2 x^{3+m} \operatorname {Hypergeometric2F1}\left (2,\frac {3+m}{2},\frac {5+m}{2},c^2 x^2\right )}{6 d^3 (3+m)}-\frac {b^2 c^2 (1-m) (1+m) x^{3+m} \, _3F_2\left (1,\frac {3}{2}+\frac {m}{2},\frac {3}{2}+\frac {m}{2};2+\frac {m}{2},\frac {5}{2}+\frac {m}{2};c^2 x^2\right )}{6 d^3 \left (6+5 m+m^2\right )}-\frac {b^2 c^2 (3-m) (1+m) x^{3+m} \, _3F_2\left (1,\frac {3}{2}+\frac {m}{2},\frac {3}{2}+\frac {m}{2};2+\frac {m}{2},\frac {5}{2}+\frac {m}{2};c^2 x^2\right )}{4 d^3 \left (6+5 m+m^2\right )}+\frac {((1-m) (3-m)) \int \frac {x^m (a+b \arcsin (c x))^2}{d-c^2 d x^2} \, dx}{8 d^2} \\ \end{align*}
Not integrable
Time = 9.81 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.07 \[ \int \frac {x^m (a+b \arcsin (c x))^2}{\left (d-c^2 d x^2\right )^3} \, dx=\int \frac {x^m (a+b \arcsin (c x))^2}{\left (d-c^2 d x^2\right )^3} \, dx \]
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Not integrable
Time = 0.32 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00
\[\int \frac {x^{m} \left (a +b \arcsin \left (c x \right )\right )^{2}}{\left (-c^{2} d \,x^{2}+d \right )^{3}}d x\]
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Not integrable
Time = 0.25 (sec) , antiderivative size = 69, normalized size of antiderivative = 2.56 \[ \int \frac {x^m (a+b \arcsin (c x))^2}{\left (d-c^2 d x^2\right )^3} \, dx=\int { -\frac {{\left (b \arcsin \left (c x\right ) + a\right )}^{2} x^{m}}{{\left (c^{2} d x^{2} - d\right )}^{3}} \,d x } \]
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Not integrable
Time = 117.24 (sec) , antiderivative size = 119, normalized size of antiderivative = 4.41 \[ \int \frac {x^m (a+b \arcsin (c x))^2}{\left (d-c^2 d x^2\right )^3} \, dx=- \frac {\int \frac {a^{2} x^{m}}{c^{6} x^{6} - 3 c^{4} x^{4} + 3 c^{2} x^{2} - 1}\, dx + \int \frac {b^{2} x^{m} \operatorname {asin}^{2}{\left (c x \right )}}{c^{6} x^{6} - 3 c^{4} x^{4} + 3 c^{2} x^{2} - 1}\, dx + \int \frac {2 a b x^{m} \operatorname {asin}{\left (c x \right )}}{c^{6} x^{6} - 3 c^{4} x^{4} + 3 c^{2} x^{2} - 1}\, dx}{d^{3}} \]
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Not integrable
Time = 0.90 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.19 \[ \int \frac {x^m (a+b \arcsin (c x))^2}{\left (d-c^2 d x^2\right )^3} \, dx=\int { -\frac {{\left (b \arcsin \left (c x\right ) + a\right )}^{2} x^{m}}{{\left (c^{2} d x^{2} - d\right )}^{3}} \,d x } \]
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Not integrable
Time = 0.88 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.15 \[ \int \frac {x^m (a+b \arcsin (c x))^2}{\left (d-c^2 d x^2\right )^3} \, dx=\int { -\frac {{\left (b \arcsin \left (c x\right ) + a\right )}^{2} x^{m}}{{\left (c^{2} d x^{2} - d\right )}^{3}} \,d x } \]
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Not integrable
Time = 0.17 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.07 \[ \int \frac {x^m (a+b \arcsin (c x))^2}{\left (d-c^2 d x^2\right )^3} \, dx=\int \frac {x^m\,{\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}^2}{{\left (d-c^2\,d\,x^2\right )}^3} \,d x \]
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