Integrand size = 26, antiderivative size = 390 \[ \int x^m \left (d+c^2 d x^2\right )^{3/2} (a+b \text {arcsinh}(c x)) \, dx=-\frac {3 b c d x^{2+m} \sqrt {d+c^2 d x^2}}{(2+m)^2 (4+m) \sqrt {1+c^2 x^2}}-\frac {b c d x^{2+m} \sqrt {d+c^2 d x^2}}{\left (8+6 m+m^2\right ) \sqrt {1+c^2 x^2}}-\frac {b c^3 d x^{4+m} \sqrt {d+c^2 d x^2}}{(4+m)^2 \sqrt {1+c^2 x^2}}+\frac {3 d x^{1+m} \sqrt {d+c^2 d x^2} (a+b \text {arcsinh}(c x))}{8+6 m+m^2}+\frac {x^{1+m} \left (d+c^2 d x^2\right )^{3/2} (a+b \text {arcsinh}(c x))}{4+m}+\frac {3 d x^{1+m} \sqrt {d+c^2 d x^2} (a+b \text {arcsinh}(c x)) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},-c^2 x^2\right )}{\left (8+14 m+7 m^2+m^3\right ) \sqrt {1+c^2 x^2}}-\frac {3 b c d x^{2+m} \sqrt {d+c^2 d x^2} \, _3F_2\left (1,1+\frac {m}{2},1+\frac {m}{2};\frac {3}{2}+\frac {m}{2},2+\frac {m}{2};-c^2 x^2\right )}{(1+m) (2+m)^2 (4+m) \sqrt {1+c^2 x^2}} \]
[Out]
Time = 0.23 (sec) , antiderivative size = 390, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {5808, 5806, 5817, 30, 14} \[ \int x^m \left (d+c^2 d x^2\right )^{3/2} (a+b \text {arcsinh}(c x)) \, dx=-\frac {3 b c d x^{m+2} \sqrt {c^2 d x^2+d} \, _3F_2\left (1,\frac {m}{2}+1,\frac {m}{2}+1;\frac {m}{2}+\frac {3}{2},\frac {m}{2}+2;-c^2 x^2\right )}{(m+1) (m+2)^2 (m+4) \sqrt {c^2 x^2+1}}+\frac {3 d x^{m+1} \sqrt {c^2 d x^2+d} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+1}{2},\frac {m+3}{2},-c^2 x^2\right ) (a+b \text {arcsinh}(c x))}{\left (m^3+7 m^2+14 m+8\right ) \sqrt {c^2 x^2+1}}+\frac {3 d x^{m+1} \sqrt {c^2 d x^2+d} (a+b \text {arcsinh}(c x))}{m^2+6 m+8}+\frac {x^{m+1} \left (c^2 d x^2+d\right )^{3/2} (a+b \text {arcsinh}(c x))}{m+4}-\frac {b c d x^{m+2} \sqrt {c^2 d x^2+d}}{\left (m^2+6 m+8\right ) \sqrt {c^2 x^2+1}}-\frac {3 b c d x^{m+2} \sqrt {c^2 d x^2+d}}{(m+2)^2 (m+4) \sqrt {c^2 x^2+1}}-\frac {b c^3 d x^{m+4} \sqrt {c^2 d x^2+d}}{(m+4)^2 \sqrt {c^2 x^2+1}} \]
[In]
[Out]
Rule 14
Rule 30
Rule 5806
Rule 5808
Rule 5817
Rubi steps \begin{align*} \text {integral}& = \frac {x^{1+m} \left (d+c^2 d x^2\right )^{3/2} (a+b \text {arcsinh}(c x))}{4+m}+\frac {(3 d) \int x^m \sqrt {d+c^2 d x^2} (a+b \text {arcsinh}(c x)) \, dx}{4+m}-\frac {\left (b c d \sqrt {d+c^2 d x^2}\right ) \int x^{1+m} \left (1+c^2 x^2\right ) \, dx}{(4+m) \sqrt {1+c^2 x^2}} \\ & = \frac {3 d x^{1+m} \sqrt {d+c^2 d x^2} (a+b \text {arcsinh}(c x))}{8+6 m+m^2}+\frac {x^{1+m} \left (d+c^2 d x^2\right )^{3/2} (a+b \text {arcsinh}(c x))}{4+m}-\frac {\left (b c d \sqrt {d+c^2 d x^2}\right ) \int \left (x^{1+m}+c^2 x^{3+m}\right ) \, dx}{(4+m) \sqrt {1+c^2 x^2}}+\frac {\left (3 d \sqrt {d+c^2 d x^2}\right ) \int \frac {x^m (a+b \text {arcsinh}(c x))}{\sqrt {1+c^2 x^2}} \, dx}{(2+m) (4+m) \sqrt {1+c^2 x^2}}-\frac {\left (3 b c d \sqrt {d+c^2 d x^2}\right ) \int x^{1+m} \, dx}{(2+m) (4+m) \sqrt {1+c^2 x^2}} \\ & = -\frac {3 b c d x^{2+m} \sqrt {d+c^2 d x^2}}{(2+m)^2 (4+m) \sqrt {1+c^2 x^2}}-\frac {b c d x^{2+m} \sqrt {d+c^2 d x^2}}{\left (8+6 m+m^2\right ) \sqrt {1+c^2 x^2}}-\frac {b c^3 d x^{4+m} \sqrt {d+c^2 d x^2}}{(4+m)^2 \sqrt {1+c^2 x^2}}+\frac {3 d x^{1+m} \sqrt {d+c^2 d x^2} (a+b \text {arcsinh}(c x))}{8+6 m+m^2}+\frac {x^{1+m} \left (d+c^2 d x^2\right )^{3/2} (a+b \text {arcsinh}(c x))}{4+m}+\frac {3 d x^{1+m} \sqrt {d+c^2 d 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 )}{\left (8+14 m+7 m^2+m^3\right ) \sqrt {1+c^2 x^2}}-\frac {3 b c d x^{2+m} \sqrt {d+c^2 d 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 )}{(1+m) (2+m)^2 (4+m) \sqrt {1+c^2 x^2}} \\ \end{align*}
Time = 0.37 (sec) , antiderivative size = 233, normalized size of antiderivative = 0.60 \[ \int x^m \left (d+c^2 d x^2\right )^{3/2} (a+b \text {arcsinh}(c x)) \, dx=\frac {d x^{1+m} \sqrt {d+c^2 d x^2} \left (-\frac {b c x \left (4+m+c^2 (2+m) x^2\right )}{(2+m) (4+m) \sqrt {1+c^2 x^2}}+\left (1+c^2 x^2\right ) (a+b \text {arcsinh}(c x))-\frac {3 \left (b c (1+m) x-(1+m) (2+m) \sqrt {1+c^2 x^2} (a+b \text {arcsinh}(c x))-(2+m) (a+b \text {arcsinh}(c x)) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},-c^2 x^2\right )+b c x \, _3F_2\left (1,1+\frac {m}{2},1+\frac {m}{2};\frac {3}{2}+\frac {m}{2},2+\frac {m}{2};-c^2 x^2\right )\right )}{(1+m) (2+m)^2 \sqrt {1+c^2 x^2}}\right )}{4+m} \]
[In]
[Out]
\[\int x^{m} \left (c^{2} d \,x^{2}+d \right )^{\frac {3}{2}} \left (a +b \,\operatorname {arcsinh}\left (c x \right )\right )d x\]
[In]
[Out]
\[ \int x^m \left (d+c^2 d x^2\right )^{3/2} (a+b \text {arcsinh}(c x)) \, dx=\int { {\left (c^{2} d x^{2} + d\right )}^{\frac {3}{2}} {\left (b \operatorname {arsinh}\left (c x\right ) + a\right )} x^{m} \,d x } \]
[In]
[Out]
Timed out. \[ \int x^m \left (d+c^2 d x^2\right )^{3/2} (a+b \text {arcsinh}(c x)) \, dx=\text {Timed out} \]
[In]
[Out]
\[ \int x^m \left (d+c^2 d x^2\right )^{3/2} (a+b \text {arcsinh}(c x)) \, dx=\int { {\left (c^{2} d x^{2} + d\right )}^{\frac {3}{2}} {\left (b \operatorname {arsinh}\left (c x\right ) + a\right )} x^{m} \,d x } \]
[In]
[Out]
Exception generated. \[ \int x^m \left (d+c^2 d x^2\right )^{3/2} (a+b \text {arcsinh}(c x)) \, dx=\text {Exception raised: TypeError} \]
[In]
[Out]
Timed out. \[ \int x^m \left (d+c^2 d x^2\right )^{3/2} (a+b \text {arcsinh}(c x)) \, dx=\int x^m\,\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )\,{\left (d\,c^2\,x^2+d\right )}^{3/2} \,d x \]
[In]
[Out]