Integrand size = 26, antiderivative size = 402 \[ \int \frac {x^m (a+b \text {arcsinh}(c x))}{\left (d+c^2 d x^2\right )^{5/2}} \, dx=\frac {x^{1+m} (a+b \text {arcsinh}(c x))}{3 d \left (d+c^2 d x^2\right )^{3/2}}+\frac {(2-m) x^{1+m} (a+b \text {arcsinh}(c x))}{3 d^2 \sqrt {d+c^2 d x^2}}-\frac {(2-m) m x^{1+m} \sqrt {1+c^2 x^2} (a+b \text {arcsinh}(c x)) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},-c^2 x^2\right )}{3 d^2 (1+m) \sqrt {d+c^2 d x^2}}-\frac {b c (2-m) x^{2+m} \sqrt {1+c^2 x^2} \operatorname {Hypergeometric2F1}\left (1,\frac {2+m}{2},\frac {4+m}{2},-c^2 x^2\right )}{3 d^2 (2+m) \sqrt {d+c^2 d x^2}}-\frac {b c x^{2+m} \sqrt {1+c^2 x^2} \operatorname {Hypergeometric2F1}\left (2,\frac {2+m}{2},\frac {4+m}{2},-c^2 x^2\right )}{3 d^2 (2+m) \sqrt {d+c^2 d x^2}}+\frac {b c (2-m) m x^{2+m} \sqrt {1+c^2 x^2} \, _3F_2\left (1,1+\frac {m}{2},1+\frac {m}{2};\frac {3}{2}+\frac {m}{2},2+\frac {m}{2};-c^2 x^2\right )}{3 d^2 \left (2+3 m+m^2\right ) \sqrt {d+c^2 d x^2}} \]
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Time = 0.26 (sec) , antiderivative size = 402, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {5811, 5817, 371} \[ \int \frac {x^m (a+b \text {arcsinh}(c x))}{\left (d+c^2 d x^2\right )^{5/2}} \, dx=\frac {b c (2-m) m \sqrt {c^2 x^2+1} x^{m+2} \, _3F_2\left (1,\frac {m}{2}+1,\frac {m}{2}+1;\frac {m}{2}+\frac {3}{2},\frac {m}{2}+2;-c^2 x^2\right )}{3 d^2 \left (m^2+3 m+2\right ) \sqrt {c^2 d x^2+d}}-\frac {(2-m) m \sqrt {c^2 x^2+1} x^{m+1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+1}{2},\frac {m+3}{2},-c^2 x^2\right ) (a+b \text {arcsinh}(c x))}{3 d^2 (m+1) \sqrt {c^2 d x^2+d}}+\frac {(2-m) x^{m+1} (a+b \text {arcsinh}(c x))}{3 d^2 \sqrt {c^2 d x^2+d}}+\frac {x^{m+1} (a+b \text {arcsinh}(c x))}{3 d \left (c^2 d x^2+d\right )^{3/2}}-\frac {b c (2-m) \sqrt {c^2 x^2+1} x^{m+2} \operatorname {Hypergeometric2F1}\left (1,\frac {m+2}{2},\frac {m+4}{2},-c^2 x^2\right )}{3 d^2 (m+2) \sqrt {c^2 d x^2+d}}-\frac {b c \sqrt {c^2 x^2+1} x^{m+2} \operatorname {Hypergeometric2F1}\left (2,\frac {m+2}{2},\frac {m+4}{2},-c^2 x^2\right )}{3 d^2 (m+2) \sqrt {c^2 d x^2+d}} \]
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Rule 371
Rule 5811
Rule 5817
Rubi steps \begin{align*} \text {integral}& = \frac {x^{1+m} (a+b \text {arcsinh}(c x))}{3 d \left (d+c^2 d x^2\right )^{3/2}}+\frac {(2-m) \int \frac {x^m (a+b \text {arcsinh}(c x))}{\left (d+c^2 d x^2\right )^{3/2}} \, dx}{3 d}-\frac {\left (b c \sqrt {1+c^2 x^2}\right ) \int \frac {x^{1+m}}{\left (1+c^2 x^2\right )^2} \, dx}{3 d^2 \sqrt {d+c^2 d x^2}} \\ & = \frac {x^{1+m} (a+b \text {arcsinh}(c x))}{3 d \left (d+c^2 d x^2\right )^{3/2}}+\frac {(2-m) x^{1+m} (a+b \text {arcsinh}(c x))}{3 d^2 \sqrt {d+c^2 d x^2}}-\frac {b c x^{2+m} \sqrt {1+c^2 x^2} \operatorname {Hypergeometric2F1}\left (2,\frac {2+m}{2},\frac {4+m}{2},-c^2 x^2\right )}{3 d^2 (2+m) \sqrt {d+c^2 d x^2}}-\frac {((2-m) m) \int \frac {x^m (a+b \text {arcsinh}(c x))}{\sqrt {d+c^2 d x^2}} \, dx}{3 d^2}-\frac {\left (b c (2-m) \sqrt {1+c^2 x^2}\right ) \int \frac {x^{1+m}}{1+c^2 x^2} \, dx}{3 d^2 \sqrt {d+c^2 d x^2}} \\ & = \frac {x^{1+m} (a+b \text {arcsinh}(c x))}{3 d \left (d+c^2 d x^2\right )^{3/2}}+\frac {(2-m) x^{1+m} (a+b \text {arcsinh}(c x))}{3 d^2 \sqrt {d+c^2 d x^2}}-\frac {(2-m) m x^{1+m} \sqrt {1+c^2 x^2} (a+b \text {arcsinh}(c x)) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},-c^2 x^2\right )}{3 d^2 (1+m) \sqrt {d+c^2 d x^2}}-\frac {b c (2-m) x^{2+m} \sqrt {1+c^2 x^2} \operatorname {Hypergeometric2F1}\left (1,\frac {2+m}{2},\frac {4+m}{2},-c^2 x^2\right )}{3 d^2 (2+m) \sqrt {d+c^2 d x^2}}-\frac {b c x^{2+m} \sqrt {1+c^2 x^2} \operatorname {Hypergeometric2F1}\left (2,\frac {2+m}{2},\frac {4+m}{2},-c^2 x^2\right )}{3 d^2 (2+m) \sqrt {d+c^2 d x^2}}+\frac {b c (2-m) m x^{2+m} \sqrt {1+c^2 x^2} \, _3F_2\left (1,1+\frac {m}{2},1+\frac {m}{2};\frac {3}{2}+\frac {m}{2},2+\frac {m}{2};-c^2 x^2\right )}{3 d^2 \left (2+3 m+m^2\right ) \sqrt {d+c^2 d x^2}} \\ \end{align*}
Time = 0.27 (sec) , antiderivative size = 286, normalized size of antiderivative = 0.71 \[ \int \frac {x^m (a+b \text {arcsinh}(c x))}{\left (d+c^2 d x^2\right )^{5/2}} \, dx=\frac {x^{1+m} \left ((1+m) (2+m) (a+b \text {arcsinh}(c x))-b c (1+m) x \left (1+c^2 x^2\right )^{3/2} \operatorname {Hypergeometric2F1}\left (2,1+\frac {m}{2},2+\frac {m}{2},-c^2 x^2\right )+(2-m) \left (1+c^2 x^2\right ) \left ((1+m) (2+m) (a+b \text {arcsinh}(c x))-b c (1+m) x \sqrt {1+c^2 x^2} \operatorname {Hypergeometric2F1}\left (1,1+\frac {m}{2},2+\frac {m}{2},-c^2 x^2\right )-m \sqrt {1+c^2 x^2} \left ((2+m) (a+b \text {arcsinh}(c x)) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},-c^2 x^2\right )-b c x \, _3F_2\left (1,1+\frac {m}{2},1+\frac {m}{2};\frac {3}{2}+\frac {m}{2},2+\frac {m}{2};-c^2 x^2\right )\right )\right )\right )}{3 d^2 (1+m) (2+m) \left (1+c^2 x^2\right ) \sqrt {d+c^2 d x^2}} \]
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\[\int \frac {x^{m} \left (a +b \,\operatorname {arcsinh}\left (c x \right )\right )}{\left (c^{2} d \,x^{2}+d \right )^{\frac {5}{2}}}d x\]
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\[ \int \frac {x^m (a+b \text {arcsinh}(c x))}{\left (d+c^2 d x^2\right )^{5/2}} \, dx=\int { \frac {{\left (b \operatorname {arsinh}\left (c x\right ) + a\right )} x^{m}}{{\left (c^{2} d x^{2} + d\right )}^{\frac {5}{2}}} \,d x } \]
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\[ \int \frac {x^m (a+b \text {arcsinh}(c x))}{\left (d+c^2 d x^2\right )^{5/2}} \, dx=\int \frac {x^{m} \left (a + b \operatorname {asinh}{\left (c x \right )}\right )}{\left (d \left (c^{2} x^{2} + 1\right )\right )^{\frac {5}{2}}}\, dx \]
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\[ \int \frac {x^m (a+b \text {arcsinh}(c x))}{\left (d+c^2 d x^2\right )^{5/2}} \, dx=\int { \frac {{\left (b \operatorname {arsinh}\left (c x\right ) + a\right )} x^{m}}{{\left (c^{2} d x^{2} + d\right )}^{\frac {5}{2}}} \,d x } \]
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\[ \int \frac {x^m (a+b \text {arcsinh}(c x))}{\left (d+c^2 d x^2\right )^{5/2}} \, dx=\int { \frac {{\left (b \operatorname {arsinh}\left (c x\right ) + a\right )} x^{m}}{{\left (c^{2} d x^{2} + d\right )}^{\frac {5}{2}}} \,d x } \]
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Timed out. \[ \int \frac {x^m (a+b \text {arcsinh}(c x))}{\left (d+c^2 d x^2\right )^{5/2}} \, dx=\int \frac {x^m\,\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}{{\left (d\,c^2\,x^2+d\right )}^{5/2}} \,d x \]
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