Integrand size = 23, antiderivative size = 187 \[ \int (c e+d e x)^m (a+b \text {arcsinh}(c+d x))^2 \, dx=\frac {(e (c+d x))^{1+m} (a+b \text {arcsinh}(c+d x))^2}{d e (1+m)}-\frac {2 b (e (c+d x))^{2+m} (a+b \text {arcsinh}(c+d x)) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+m}{2},\frac {4+m}{2},-(c+d x)^2\right )}{d e^2 (1+m) (2+m)}+\frac {2 b^2 (e (c+d x))^{3+m} \, _3F_2\left (1,\frac {3}{2}+\frac {m}{2},\frac {3}{2}+\frac {m}{2};2+\frac {m}{2},\frac {5}{2}+\frac {m}{2};-(c+d x)^2\right )}{d e^3 (1+m) (2+m) (3+m)} \]
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Time = 0.15 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {5859, 5776, 5817} \[ \int (c e+d e x)^m (a+b \text {arcsinh}(c+d x))^2 \, dx=\frac {2 b^2 (e (c+d x))^{m+3} \, _3F_2\left (1,\frac {m}{2}+\frac {3}{2},\frac {m}{2}+\frac {3}{2};\frac {m}{2}+2,\frac {m}{2}+\frac {5}{2};-(c+d x)^2\right )}{d e^3 (m+1) (m+2) (m+3)}-\frac {2 b (e (c+d x))^{m+2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+2}{2},\frac {m+4}{2},-(c+d x)^2\right ) (a+b \text {arcsinh}(c+d x))}{d e^2 (m+1) (m+2)}+\frac {(e (c+d x))^{m+1} (a+b \text {arcsinh}(c+d x))^2}{d e (m+1)} \]
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Rule 5776
Rule 5817
Rule 5859
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int (e x)^m (a+b \text {arcsinh}(x))^2 \, dx,x,c+d x\right )}{d} \\ & = \frac {(e (c+d x))^{1+m} (a+b \text {arcsinh}(c+d x))^2}{d e (1+m)}-\frac {(2 b) \text {Subst}\left (\int \frac {(e x)^{1+m} (a+b \text {arcsinh}(x))}{\sqrt {1+x^2}} \, dx,x,c+d x\right )}{d e (1+m)} \\ & = \frac {(e (c+d x))^{1+m} (a+b \text {arcsinh}(c+d x))^2}{d e (1+m)}-\frac {2 b (e (c+d x))^{2+m} (a+b \text {arcsinh}(c+d x)) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+m}{2},\frac {4+m}{2},-(c+d x)^2\right )}{d e^2 (1+m) (2+m)}+\frac {2 b^2 (e (c+d 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+d x)^2\right )}{d e^3 (1+m) (2+m) (3+m)} \\ \end{align*}
Time = 0.07 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.83 \[ \int (c e+d e x)^m (a+b \text {arcsinh}(c+d x))^2 \, dx=\frac {(c+d x) (e (c+d x))^m \left ((a+b \text {arcsinh}(c+d x))^2-\frac {2 b (c+d x) (a+b \text {arcsinh}(c+d x)) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+m}{2},\frac {4+m}{2},-(c+d x)^2\right )}{2+m}+\frac {2 b^2 (c+d x)^2 \, _3F_2\left (1,\frac {3}{2}+\frac {m}{2},\frac {3}{2}+\frac {m}{2};2+\frac {m}{2},\frac {5}{2}+\frac {m}{2};-(c+d x)^2\right )}{(2+m) (3+m)}\right )}{d (1+m)} \]
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\[\int \left (d e x +c e \right )^{m} \left (a +b \,\operatorname {arcsinh}\left (d x +c \right )\right )^{2}d x\]
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\[ \int (c e+d e x)^m (a+b \text {arcsinh}(c+d x))^2 \, dx=\int { {\left (b \operatorname {arsinh}\left (d x + c\right ) + a\right )}^{2} {\left (d e x + c e\right )}^{m} \,d x } \]
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\[ \int (c e+d e x)^m (a+b \text {arcsinh}(c+d x))^2 \, dx=\int \left (e \left (c + d x\right )\right )^{m} \left (a + b \operatorname {asinh}{\left (c + d x \right )}\right )^{2}\, dx \]
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\[ \int (c e+d e x)^m (a+b \text {arcsinh}(c+d x))^2 \, dx=\int { {\left (b \operatorname {arsinh}\left (d x + c\right ) + a\right )}^{2} {\left (d e x + c e\right )}^{m} \,d x } \]
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\[ \int (c e+d e x)^m (a+b \text {arcsinh}(c+d x))^2 \, dx=\int { {\left (b \operatorname {arsinh}\left (d x + c\right ) + a\right )}^{2} {\left (d e x + c e\right )}^{m} \,d x } \]
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Timed out. \[ \int (c e+d e x)^m (a+b \text {arcsinh}(c+d x))^2 \, dx=\int {\left (c\,e+d\,e\,x\right )}^m\,{\left (a+b\,\mathrm {asinh}\left (c+d\,x\right )\right )}^2 \,d x \]
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