Integrand size = 24, antiderivative size = 217 \[ \int x^m \left (d+c^2 d x^2\right )^2 (a+b \text {arcsinh}(c x)) \, dx=-\frac {b c d^2 \left (38+13 m+m^2\right ) x^{2+m} \sqrt {1+c^2 x^2}}{(3+m)^2 (5+m)^2}-\frac {b c^3 d^2 x^{4+m} \sqrt {1+c^2 x^2}}{(5+m)^2}+\frac {d^2 x^{1+m} (a+b \text {arcsinh}(c x))}{1+m}+\frac {2 c^2 d^2 x^{3+m} (a+b \text {arcsinh}(c x))}{3+m}+\frac {c^4 d^2 x^{5+m} (a+b \text {arcsinh}(c x))}{5+m}-\frac {b c d^2 \left (149+100 m+15 m^2\right ) x^{2+m} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+m}{2},\frac {4+m}{2},-c^2 x^2\right )}{(1+m) (2+m) (3+m)^2 (5+m)^2} \]
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Time = 0.20 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {276, 5803, 12, 1281, 470, 371} \[ \int x^m \left (d+c^2 d x^2\right )^2 (a+b \text {arcsinh}(c x)) \, dx=\frac {c^4 d^2 x^{m+5} (a+b \text {arcsinh}(c x))}{m+5}+\frac {2 c^2 d^2 x^{m+3} (a+b \text {arcsinh}(c x))}{m+3}+\frac {d^2 x^{m+1} (a+b \text {arcsinh}(c x))}{m+1}-\frac {b c d^2 \left (15 m^2+100 m+149\right ) x^{m+2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+2}{2},\frac {m+4}{2},-c^2 x^2\right )}{(m+1) (m+2) (m+3)^2 (m+5)^2}-\frac {b c d^2 \left (m^2+13 m+38\right ) \sqrt {c^2 x^2+1} x^{m+2}}{(m+3)^2 (m+5)^2}-\frac {b c^3 d^2 \sqrt {c^2 x^2+1} x^{m+4}}{(m+5)^2} \]
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Rule 12
Rule 276
Rule 371
Rule 470
Rule 1281
Rule 5803
Rubi steps \begin{align*} \text {integral}& = \frac {d^2 x^{1+m} (a+b \text {arcsinh}(c x))}{1+m}+\frac {2 c^2 d^2 x^{3+m} (a+b \text {arcsinh}(c x))}{3+m}+\frac {c^4 d^2 x^{5+m} (a+b \text {arcsinh}(c x))}{5+m}-(b c) \int \frac {d^2 x^{1+m} \left (\frac {1}{1+m}+\frac {2 c^2 x^2}{3+m}+\frac {c^4 x^4}{5+m}\right )}{\sqrt {1+c^2 x^2}} \, dx \\ & = \frac {d^2 x^{1+m} (a+b \text {arcsinh}(c x))}{1+m}+\frac {2 c^2 d^2 x^{3+m} (a+b \text {arcsinh}(c x))}{3+m}+\frac {c^4 d^2 x^{5+m} (a+b \text {arcsinh}(c x))}{5+m}-\left (b c d^2\right ) \int \frac {x^{1+m} \left (\frac {1}{1+m}+\frac {2 c^2 x^2}{3+m}+\frac {c^4 x^4}{5+m}\right )}{\sqrt {1+c^2 x^2}} \, dx \\ & = -\frac {b c^3 d^2 x^{4+m} \sqrt {1+c^2 x^2}}{(5+m)^2}+\frac {d^2 x^{1+m} (a+b \text {arcsinh}(c x))}{1+m}+\frac {2 c^2 d^2 x^{3+m} (a+b \text {arcsinh}(c x))}{3+m}+\frac {c^4 d^2 x^{5+m} (a+b \text {arcsinh}(c x))}{5+m}-\frac {\left (b d^2\right ) \int \frac {x^{1+m} \left (\frac {c^2 (5+m)}{1+m}+\frac {c^4 \left (38+13 m+m^2\right ) x^2}{(3+m) (5+m)}\right )}{\sqrt {1+c^2 x^2}} \, dx}{c (5+m)} \\ & = -\frac {b c d^2 \left (38+13 m+m^2\right ) x^{2+m} \sqrt {1+c^2 x^2}}{(3+m)^2 (5+m)^2}-\frac {b c^3 d^2 x^{4+m} \sqrt {1+c^2 x^2}}{(5+m)^2}+\frac {d^2 x^{1+m} (a+b \text {arcsinh}(c x))}{1+m}+\frac {2 c^2 d^2 x^{3+m} (a+b \text {arcsinh}(c x))}{3+m}+\frac {c^4 d^2 x^{5+m} (a+b \text {arcsinh}(c x))}{5+m}-\frac {\left (b c d^2 \left (149+100 m+15 m^2\right )\right ) \int \frac {x^{1+m}}{\sqrt {1+c^2 x^2}} \, dx}{(1+m) (3+m)^2 (5+m)^2} \\ & = -\frac {b c d^2 \left (38+13 m+m^2\right ) x^{2+m} \sqrt {1+c^2 x^2}}{(3+m)^2 (5+m)^2}-\frac {b c^3 d^2 x^{4+m} \sqrt {1+c^2 x^2}}{(5+m)^2}+\frac {d^2 x^{1+m} (a+b \text {arcsinh}(c x))}{1+m}+\frac {2 c^2 d^2 x^{3+m} (a+b \text {arcsinh}(c x))}{3+m}+\frac {c^4 d^2 x^{5+m} (a+b \text {arcsinh}(c x))}{5+m}-\frac {b c d^2 \left (149+100 m+15 m^2\right ) x^{2+m} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+m}{2},\frac {4+m}{2},-c^2 x^2\right )}{(1+m) (2+m) (3+m)^2 (5+m)^2} \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 188, normalized size of antiderivative = 0.87 \[ \int x^m \left (d+c^2 d x^2\right )^2 (a+b \text {arcsinh}(c x)) \, dx=\frac {x^{1+m} \left (\left (d+c^2 d x^2\right )^2 (a+b \text {arcsinh}(c x))-\frac {b c d^2 x \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1+\frac {m}{2},2+\frac {m}{2},-c^2 x^2\right )}{2+m}+\frac {4 d^2 \left ((2+m) \left (3+m+c^2 x^2+c^2 m x^2\right ) (a+b \text {arcsinh}(c x))-b c (1+m) x \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1+\frac {m}{2},2+\frac {m}{2},-c^2 x^2\right )-2 b c x \operatorname {Hypergeometric2F1}\left (\frac {1}{2},1+\frac {m}{2},2+\frac {m}{2},-c^2 x^2\right )\right )}{(1+m) (2+m) (3+m)}\right )}{5+m} \]
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\[\int x^{m} \left (c^{2} d \,x^{2}+d \right )^{2} \left (a +b \,\operatorname {arcsinh}\left (c x \right )\right )d x\]
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\[ \int x^m \left (d+c^2 d x^2\right )^2 (a+b \text {arcsinh}(c x)) \, dx=\int { {\left (c^{2} d x^{2} + d\right )}^{2} {\left (b \operatorname {arsinh}\left (c x\right ) + a\right )} x^{m} \,d x } \]
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\[ \int x^m \left (d+c^2 d x^2\right )^2 (a+b \text {arcsinh}(c x)) \, dx=d^{2} \left (\int a x^{m}\, dx + \int b x^{m} \operatorname {asinh}{\left (c x \right )}\, dx + \int 2 a c^{2} x^{2} x^{m}\, dx + \int a c^{4} x^{4} x^{m}\, dx + \int 2 b c^{2} x^{2} x^{m} \operatorname {asinh}{\left (c x \right )}\, dx + \int b c^{4} x^{4} x^{m} \operatorname {asinh}{\left (c x \right )}\, dx\right ) \]
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\[ \int x^m \left (d+c^2 d x^2\right )^2 (a+b \text {arcsinh}(c x)) \, dx=\int { {\left (c^{2} d x^{2} + d\right )}^{2} {\left (b \operatorname {arsinh}\left (c x\right ) + a\right )} x^{m} \,d x } \]
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Exception generated. \[ \int x^m \left (d+c^2 d x^2\right )^2 (a+b \text {arcsinh}(c x)) \, dx=\text {Exception raised: TypeError} \]
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Timed out. \[ \int x^m \left (d+c^2 d x^2\right )^2 (a+b \text {arcsinh}(c x)) \, dx=\int x^m\,\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )\,{\left (d\,c^2\,x^2+d\right )}^2 \,d x \]
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