Integrand size = 26, antiderivative size = 161 \[ \int \frac {x^m (a+b \text {arcsinh}(c x))}{\sqrt {d+c^2 d x^2}} \, dx=\frac {x^{1+m} \sqrt {1+c^2 x^2} (a+b \text {arcsinh}(c x)) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},-c^2 x^2\right )}{(1+m) \sqrt {d+c^2 d x^2}}-\frac {b c x^{2+m} \sqrt {1+c^2 x^2} \, _3F_2\left (1,1+\frac {m}{2},1+\frac {m}{2};\frac {3}{2}+\frac {m}{2},2+\frac {m}{2};-c^2 x^2\right )}{\left (2+3 m+m^2\right ) \sqrt {d+c^2 d x^2}} \]
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Time = 0.07 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.038, Rules used = {5817} \[ \int \frac {x^m (a+b \text {arcsinh}(c x))}{\sqrt {d+c^2 d x^2}} \, dx=\frac {\sqrt {c^2 x^2+1} x^{m+1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+1}{2},\frac {m+3}{2},-c^2 x^2\right ) (a+b \text {arcsinh}(c x))}{(m+1) \sqrt {c^2 d x^2+d}}-\frac {b c \sqrt {c^2 x^2+1} x^{m+2} \, _3F_2\left (1,\frac {m}{2}+1,\frac {m}{2}+1;\frac {m}{2}+\frac {3}{2},\frac {m}{2}+2;-c^2 x^2\right )}{\left (m^2+3 m+2\right ) \sqrt {c^2 d x^2+d}} \]
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Rule 5817
Rubi steps \begin{align*} \text {integral}& = \frac {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 )}{(1+m) \sqrt {d+c^2 d x^2}}-\frac {b c 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 )}{\left (2+3 m+m^2\right ) \sqrt {d+c^2 d x^2}} \\ \end{align*}
Time = 0.04 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.80 \[ \int \frac {x^m (a+b \text {arcsinh}(c x))}{\sqrt {d+c^2 d x^2}} \, dx=\frac {x^{1+m} \sqrt {1+c^2 x^2} \left ((2+m) (a+b \text {arcsinh}(c x)) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},-c^2 x^2\right )-b c x \, _3F_2\left (1,1+\frac {m}{2},1+\frac {m}{2};\frac {3}{2}+\frac {m}{2},2+\frac {m}{2};-c^2 x^2\right )\right )}{(1+m) (2+m) \sqrt {d+c^2 d x^2}} \]
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\[\int \frac {x^{m} \left (a +b \,\operatorname {arcsinh}\left (c x \right )\right )}{\sqrt {c^{2} d \,x^{2}+d}}d x\]
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\[ \int \frac {x^m (a+b \text {arcsinh}(c x))}{\sqrt {d+c^2 d x^2}} \, dx=\int { \frac {{\left (b \operatorname {arsinh}\left (c x\right ) + a\right )} x^{m}}{\sqrt {c^{2} d x^{2} + d}} \,d x } \]
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\[ \int \frac {x^m (a+b \text {arcsinh}(c x))}{\sqrt {d+c^2 d x^2}} \, dx=\int \frac {x^{m} \left (a + b \operatorname {asinh}{\left (c x \right )}\right )}{\sqrt {d \left (c^{2} x^{2} + 1\right )}}\, dx \]
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\[ \int \frac {x^m (a+b \text {arcsinh}(c x))}{\sqrt {d+c^2 d x^2}} \, dx=\int { \frac {{\left (b \operatorname {arsinh}\left (c x\right ) + a\right )} x^{m}}{\sqrt {c^{2} d x^{2} + d}} \,d x } \]
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\[ \int \frac {x^m (a+b \text {arcsinh}(c x))}{\sqrt {d+c^2 d x^2}} \, dx=\int { \frac {{\left (b \operatorname {arsinh}\left (c x\right ) + a\right )} x^{m}}{\sqrt {c^{2} d x^{2} + d}} \,d x } \]
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Timed out. \[ \int \frac {x^m (a+b \text {arcsinh}(c x))}{\sqrt {d+c^2 d x^2}} \, dx=\int \frac {x^m\,\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}{\sqrt {d\,c^2\,x^2+d}} \,d x \]
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