3.4.21 \(\int x^m (d+c^2 d x^2)^{5/2} (a+b \text {arcsinh}(c x))^2 \, dx\) [321]

3.4.21.1 Optimal result

Integrand size = 28, antiderivative size = 28 \[ \int x^m \left (d+c^2 d x^2\right )^{5/2} (a+b \text {arcsinh}(c x))^2 \, dx=\frac {10 b^2 c^2 d^2 x^{3+m} \sqrt {d+c^2 d x^2}}{(4+m)^3 (6+m)}+\frac {2 b^2 c^2 d^2 \left (52+15 m+m^2\right ) x^{3+m} \sqrt {d+c^2 d x^2}}{(4+m)^2 (6+m)^3}+\frac {2 b^2 c^4 d^2 x^{5+m} \sqrt {d+c^2 d x^2}}{(6+m)^3}-\frac {30 b c d^2 x^{2+m} \sqrt {d+c^2 d x^2} (a+b \text {arcsinh}(c x))}{(2+m)^2 (4+m) (6+m) \sqrt {1+c^2 x^2}}-\frac {10 b c d^2 x^{2+m} \sqrt {d+c^2 d x^2} (a+b \text {arcsinh}(c x))}{(6+m) \left (8+6 m+m^2\right ) \sqrt {1+c^2 x^2}}-\frac {2 b c d^2 x^{2+m} \sqrt {d+c^2 d x^2} (a+b \text {arcsinh}(c x))}{\left (12+8 m+m^2\right ) \sqrt {1+c^2 x^2}}-\frac {10 b c^3 d^2 x^{4+m} \sqrt {d+c^2 d x^2} (a+b \text {arcsinh}(c x))}{(4+m)^2 (6+m) \sqrt {1+c^2 x^2}}-\frac {4 b c^3 d^2 x^{4+m} \sqrt {d+c^2 d x^2} (a+b \text {arcsinh}(c x))}{(4+m) (6+m) \sqrt {1+c^2 x^2}}-\frac {2 b c^5 d^2 x^{6+m} \sqrt {d+c^2 d x^2} (a+b \text {arcsinh}(c x))}{(6+m)^2 \sqrt {1+c^2 x^2}}+\frac {15 d^2 x^{1+m} \sqrt {d+c^2 d x^2} (a+b \text {arcsinh}(c x))^2}{(6+m) \left (8+6 m+m^2\right )}+\frac {5 d x^{1+m} \left (d+c^2 d x^2\right )^{3/2} (a+b \text {arcsinh}(c x))^2}{(4+m) (6+m)}+\frac {x^{1+m} \left (d+c^2 d x^2\right )^{5/2} (a+b \text {arcsinh}(c x))^2}{6+m}+\frac {30 b^2 c^2 d^2 x^{3+m} \sqrt {d+c^2 d x^2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3+m}{2},\frac {5+m}{2},-c^2 x^2\right )}{(2+m)^2 (3+m) (4+m) (6+m) \sqrt {1+c^2 x^2}}+\frac {10 b^2 c^2 d^2 (10+3 m) x^{3+m} \sqrt {d+c^2 d x^2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3+m}{2},\frac {5+m}{2},-c^2 x^2\right )}{(2+m) (3+m) (4+m)^3 (6+m) \sqrt {1+c^2 x^2}}+\frac {2 b^2 c^2 d^2 \left (264+130 m+15 m^2\right ) x^{3+m} \sqrt {d+c^2 d x^2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3+m}{2},\frac {5+m}{2},-c^2 x^2\right )}{(2+m) (3+m) (4+m)^2 (6+m)^3 \sqrt {1+c^2 x^2}}+\frac {15 d^3 \text {Int}\left (\frac {x^m (a+b \text {arcsinh}(c x))^2}{\sqrt {d+c^2 d x^2}},x\right )}{(6+m) \left (8+6 m+m^2\right )} \]

output

5*d*x^(1+m)*(c^2*d*x^2+d)^(3/2)*(a+b*arcsinh(c*x))^2/(4+m)/(6+m)+x^(1+m)*( 
c^2*d*x^2+d)^(5/2)*(a+b*arcsinh(c*x))^2/(6+m)+10*b^2*c^2*d^2*x^(3+m)*(c^2* 
d*x^2+d)^(1/2)/(4+m)^3/(6+m)+2*b^2*c^2*d^2*(m^2+15*m+52)*x^(3+m)*(c^2*d*x^ 
2+d)^(1/2)/(4+m)^2/(6+m)^3+2*b^2*c^4*d^2*x^(5+m)*(c^2*d*x^2+d)^(1/2)/(6+m) 
^3+15*d^2*x^(1+m)*(a+b*arcsinh(c*x))^2*(c^2*d*x^2+d)^(1/2)/(6+m)/(m^2+6*m+ 
8)-30*b*c*d^2*x^(2+m)*(a+b*arcsinh(c*x))*(c^2*d*x^2+d)^(1/2)/(2+m)^2/(4+m) 
/(6+m)/(c^2*x^2+1)^(1/2)-10*b*c*d^2*x^(2+m)*(a+b*arcsinh(c*x))*(c^2*d*x^2+ 
d)^(1/2)/(6+m)/(m^2+6*m+8)/(c^2*x^2+1)^(1/2)-2*b*c*d^2*x^(2+m)*(a+b*arcsin 
h(c*x))*(c^2*d*x^2+d)^(1/2)/(m^2+8*m+12)/(c^2*x^2+1)^(1/2)-10*b*c^3*d^2*x^ 
(4+m)*(a+b*arcsinh(c*x))*(c^2*d*x^2+d)^(1/2)/(4+m)^2/(6+m)/(c^2*x^2+1)^(1/ 
2)-4*b*c^3*d^2*x^(4+m)*(a+b*arcsinh(c*x))*(c^2*d*x^2+d)^(1/2)/(4+m)/(6+m)/ 
(c^2*x^2+1)^(1/2)-2*b*c^5*d^2*x^(6+m)*(a+b*arcsinh(c*x))*(c^2*d*x^2+d)^(1/ 
2)/(6+m)^2/(c^2*x^2+1)^(1/2)+10*b^2*c^2*d^2*(10+3*m)*x^(3+m)*hypergeom([1/ 
2, 3/2+1/2*m],[5/2+1/2*m],-c^2*x^2)*(c^2*d*x^2+d)^(1/2)/(4+m)^3/(6+m)/(m^2 
+5*m+6)/(c^2*x^2+1)^(1/2)+30*b^2*c^2*d^2*x^(3+m)*hypergeom([1/2, 3/2+1/2*m 
],[5/2+1/2*m],-c^2*x^2)*(c^2*d*x^2+d)^(1/2)/(2+m)^2/(6+m)/(m^2+7*m+12)/(c^ 
2*x^2+1)^(1/2)+2*b^2*c^2*d^2*(15*m^2+130*m+264)*x^(3+m)*hypergeom([1/2, 3/ 
2+1/2*m],[5/2+1/2*m],-c^2*x^2)*(c^2*d*x^2+d)^(1/2)/(4+m)^2/(6+m)^3/(m^2+5* 
m+6)/(c^2*x^2+1)^(1/2)+15*d^3*Unintegrable(x^m*(a+b*arcsinh(c*x))^2/(c^2*d 
*x^2+d)^(1/2),x)/(6+m)/(m^2+6*m+8)
 

3.4.21.2 Mathematica [N/A]

Not integrable

Time = 3.62 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.07 \[ \int x^m \left (d+c^2 d x^2\right )^{5/2} (a+b \text {arcsinh}(c x))^2 \, dx=\int x^m \left (d+c^2 d x^2\right )^{5/2} (a+b \text {arcsinh}(c x))^2 \, dx \]

input

Integrate[x^m*(d + c^2*d*x^2)^(5/2)*(a + b*ArcSinh[c*x])^2,x]
 

output

Integrate[x^m*(d + c^2*d*x^2)^(5/2)*(a + b*ArcSinh[c*x])^2, x]
 

3.4.21.3 Rubi [N/A]

Not integrable

Time = 3.27 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {6223, 6218, 1590, 27, 363, 278, 6223, 6218, 363, 278, 6223, 6191, 278, 6239}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int x^m \left (c^2 d x^2+d\right )^{5/2} (a+b \text {arcsinh}(c x))^2 \, dx\)

\(\Big \downarrow \) 6223

\(\displaystyle -\frac {2 b c d^2 \sqrt {c^2 d x^2+d} \int x^{m+1} \left (c^2 x^2+1\right )^2 (a+b \text {arcsinh}(c x))dx}{(m+6) \sqrt {c^2 x^2+1}}+\frac {5 d \int x^m \left (c^2 d x^2+d\right )^{3/2} (a+b \text {arcsinh}(c x))^2dx}{m+6}+\frac {x^{m+1} \left (c^2 d x^2+d\right )^{5/2} (a+b \text {arcsinh}(c x))^2}{m+6}\)

\(\Big \downarrow \) 6218

\(\displaystyle \frac {5 d \int x^m \left (c^2 d x^2+d\right )^{3/2} (a+b \text {arcsinh}(c x))^2dx}{m+6}-\frac {2 b c d^2 \sqrt {c^2 d x^2+d} \left (-b c \int \frac {x^{m+2} \left (\frac {c^4 x^4}{m+6}+\frac {2 c^2 x^2}{m+4}+\frac {1}{m+2}\right )}{\sqrt {c^2 x^2+1}}dx+\frac {c^4 x^{m+6} (a+b \text {arcsinh}(c x))}{m+6}+\frac {2 c^2 x^{m+4} (a+b \text {arcsinh}(c x))}{m+4}+\frac {x^{m+2} (a+b \text {arcsinh}(c x))}{m+2}\right )}{(m+6) \sqrt {c^2 x^2+1}}+\frac {x^{m+1} \left (c^2 d x^2+d\right )^{5/2} (a+b \text {arcsinh}(c x))^2}{m+6}\)

\(\Big \downarrow \) 1590

\(\displaystyle \frac {5 d \int x^m \left (c^2 d x^2+d\right )^{3/2} (a+b \text {arcsinh}(c x))^2dx}{m+6}-\frac {2 b c d^2 \sqrt {c^2 d x^2+d} \left (-b c \left (\frac {\int \frac {c^2 x^{m+2} \left (\frac {c^2 \left (m^2+15 m+52\right ) x^2}{(m+4) (m+6)}+\frac {m+6}{m+2}\right )}{\sqrt {c^2 x^2+1}}dx}{c^2 (m+6)}+\frac {c^2 \sqrt {c^2 x^2+1} x^{m+5}}{(m+6)^2}\right )+\frac {c^4 x^{m+6} (a+b \text {arcsinh}(c x))}{m+6}+\frac {2 c^2 x^{m+4} (a+b \text {arcsinh}(c x))}{m+4}+\frac {x^{m+2} (a+b \text {arcsinh}(c x))}{m+2}\right )}{(m+6) \sqrt {c^2 x^2+1}}+\frac {x^{m+1} \left (c^2 d x^2+d\right )^{5/2} (a+b \text {arcsinh}(c x))^2}{m+6}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {5 d \int x^m \left (c^2 d x^2+d\right )^{3/2} (a+b \text {arcsinh}(c x))^2dx}{m+6}-\frac {2 b c d^2 \sqrt {c^2 d x^2+d} \left (-b c \left (\frac {\int \frac {x^{m+2} \left (\frac {c^2 \left (m^2+15 m+52\right ) x^2}{(m+4) (m+6)}+\frac {m+6}{m+2}\right )}{\sqrt {c^2 x^2+1}}dx}{m+6}+\frac {c^2 \sqrt {c^2 x^2+1} x^{m+5}}{(m+6)^2}\right )+\frac {c^4 x^{m+6} (a+b \text {arcsinh}(c x))}{m+6}+\frac {2 c^2 x^{m+4} (a+b \text {arcsinh}(c x))}{m+4}+\frac {x^{m+2} (a+b \text {arcsinh}(c x))}{m+2}\right )}{(m+6) \sqrt {c^2 x^2+1}}+\frac {x^{m+1} \left (c^2 d x^2+d\right )^{5/2} (a+b \text {arcsinh}(c x))^2}{m+6}\)

\(\Big \downarrow \) 363

\(\displaystyle \frac {5 d \int x^m \left (c^2 d x^2+d\right )^{3/2} (a+b \text {arcsinh}(c x))^2dx}{m+6}-\frac {2 b c d^2 \sqrt {c^2 d x^2+d} \left (-b c \left (\frac {\frac {\left (15 m^2+130 m+264\right ) \int \frac {x^{m+2}}{\sqrt {c^2 x^2+1}}dx}{(m+2) (m+4)^2 (m+6)}+\frac {\left (m^2+15 m+52\right ) \sqrt {c^2 x^2+1} x^{m+3}}{(m+4)^2 (m+6)}}{m+6}+\frac {c^2 \sqrt {c^2 x^2+1} x^{m+5}}{(m+6)^2}\right )+\frac {c^4 x^{m+6} (a+b \text {arcsinh}(c x))}{m+6}+\frac {2 c^2 x^{m+4} (a+b \text {arcsinh}(c x))}{m+4}+\frac {x^{m+2} (a+b \text {arcsinh}(c x))}{m+2}\right )}{(m+6) \sqrt {c^2 x^2+1}}+\frac {x^{m+1} \left (c^2 d x^2+d\right )^{5/2} (a+b \text {arcsinh}(c x))^2}{m+6}\)

\(\Big \downarrow \) 278

\(\displaystyle \frac {5 d \int x^m \left (c^2 d x^2+d\right )^{3/2} (a+b \text {arcsinh}(c x))^2dx}{m+6}+\frac {x^{m+1} \left (c^2 d x^2+d\right )^{5/2} (a+b \text {arcsinh}(c x))^2}{m+6}-\frac {2 b c d^2 \sqrt {c^2 d x^2+d} \left (\frac {c^4 x^{m+6} (a+b \text {arcsinh}(c x))}{m+6}+\frac {2 c^2 x^{m+4} (a+b \text {arcsinh}(c x))}{m+4}+\frac {x^{m+2} (a+b \text {arcsinh}(c x))}{m+2}-b c \left (\frac {\frac {\left (15 m^2+130 m+264\right ) x^{m+3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+3}{2},\frac {m+5}{2},-c^2 x^2\right )}{(m+2) (m+3) (m+4)^2 (m+6)}+\frac {\left (m^2+15 m+52\right ) \sqrt {c^2 x^2+1} x^{m+3}}{(m+4)^2 (m+6)}}{m+6}+\frac {c^2 \sqrt {c^2 x^2+1} x^{m+5}}{(m+6)^2}\right )\right )}{(m+6) \sqrt {c^2 x^2+1}}\)

\(\Big \downarrow \) 6223

\(\displaystyle \frac {5 d \left (-\frac {2 b c d \sqrt {c^2 d x^2+d} \int x^{m+1} \left (c^2 x^2+1\right ) (a+b \text {arcsinh}(c x))dx}{(m+4) \sqrt {c^2 x^2+1}}+\frac {3 d \int x^m \sqrt {c^2 d x^2+d} (a+b \text {arcsinh}(c x))^2dx}{m+4}+\frac {x^{m+1} \left (c^2 d x^2+d\right )^{3/2} (a+b \text {arcsinh}(c x))^2}{m+4}\right )}{m+6}+\frac {x^{m+1} \left (c^2 d x^2+d\right )^{5/2} (a+b \text {arcsinh}(c x))^2}{m+6}-\frac {2 b c d^2 \sqrt {c^2 d x^2+d} \left (\frac {c^4 x^{m+6} (a+b \text {arcsinh}(c x))}{m+6}+\frac {2 c^2 x^{m+4} (a+b \text {arcsinh}(c x))}{m+4}+\frac {x^{m+2} (a+b \text {arcsinh}(c x))}{m+2}-b c \left (\frac {\frac {\left (15 m^2+130 m+264\right ) x^{m+3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+3}{2},\frac {m+5}{2},-c^2 x^2\right )}{(m+2) (m+3) (m+4)^2 (m+6)}+\frac {\left (m^2+15 m+52\right ) \sqrt {c^2 x^2+1} x^{m+3}}{(m+4)^2 (m+6)}}{m+6}+\frac {c^2 \sqrt {c^2 x^2+1} x^{m+5}}{(m+6)^2}\right )\right )}{(m+6) \sqrt {c^2 x^2+1}}\)

\(\Big \downarrow \) 6218

\(\displaystyle \frac {5 d \left (-\frac {2 b c d \sqrt {c^2 d x^2+d} \left (-b c \int \frac {x^{m+2} \left (\frac {c^2 x^2}{m+4}+\frac {1}{m+2}\right )}{\sqrt {c^2 x^2+1}}dx+\frac {c^2 x^{m+4} (a+b \text {arcsinh}(c x))}{m+4}+\frac {x^{m+2} (a+b \text {arcsinh}(c x))}{m+2}\right )}{(m+4) \sqrt {c^2 x^2+1}}+\frac {3 d \int x^m \sqrt {c^2 d x^2+d} (a+b \text {arcsinh}(c x))^2dx}{m+4}+\frac {x^{m+1} \left (c^2 d x^2+d\right )^{3/2} (a+b \text {arcsinh}(c x))^2}{m+4}\right )}{m+6}+\frac {x^{m+1} \left (c^2 d x^2+d\right )^{5/2} (a+b \text {arcsinh}(c x))^2}{m+6}-\frac {2 b c d^2 \sqrt {c^2 d x^2+d} \left (\frac {c^4 x^{m+6} (a+b \text {arcsinh}(c x))}{m+6}+\frac {2 c^2 x^{m+4} (a+b \text {arcsinh}(c x))}{m+4}+\frac {x^{m+2} (a+b \text {arcsinh}(c x))}{m+2}-b c \left (\frac {\frac {\left (15 m^2+130 m+264\right ) x^{m+3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+3}{2},\frac {m+5}{2},-c^2 x^2\right )}{(m+2) (m+3) (m+4)^2 (m+6)}+\frac {\left (m^2+15 m+52\right ) \sqrt {c^2 x^2+1} x^{m+3}}{(m+4)^2 (m+6)}}{m+6}+\frac {c^2 \sqrt {c^2 x^2+1} x^{m+5}}{(m+6)^2}\right )\right )}{(m+6) \sqrt {c^2 x^2+1}}\)

\(\Big \downarrow \) 363

\(\displaystyle \frac {5 d \left (-\frac {2 b c d \sqrt {c^2 d x^2+d} \left (-b c \left (\frac {(3 m+10) \int \frac {x^{m+2}}{\sqrt {c^2 x^2+1}}dx}{(m+2) (m+4)^2}+\frac {\sqrt {c^2 x^2+1} x^{m+3}}{(m+4)^2}\right )+\frac {c^2 x^{m+4} (a+b \text {arcsinh}(c x))}{m+4}+\frac {x^{m+2} (a+b \text {arcsinh}(c x))}{m+2}\right )}{(m+4) \sqrt {c^2 x^2+1}}+\frac {3 d \int x^m \sqrt {c^2 d x^2+d} (a+b \text {arcsinh}(c x))^2dx}{m+4}+\frac {x^{m+1} \left (c^2 d x^2+d\right )^{3/2} (a+b \text {arcsinh}(c x))^2}{m+4}\right )}{m+6}+\frac {x^{m+1} \left (c^2 d x^2+d\right )^{5/2} (a+b \text {arcsinh}(c x))^2}{m+6}-\frac {2 b c d^2 \sqrt {c^2 d x^2+d} \left (\frac {c^4 x^{m+6} (a+b \text {arcsinh}(c x))}{m+6}+\frac {2 c^2 x^{m+4} (a+b \text {arcsinh}(c x))}{m+4}+\frac {x^{m+2} (a+b \text {arcsinh}(c x))}{m+2}-b c \left (\frac {\frac {\left (15 m^2+130 m+264\right ) x^{m+3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+3}{2},\frac {m+5}{2},-c^2 x^2\right )}{(m+2) (m+3) (m+4)^2 (m+6)}+\frac {\left (m^2+15 m+52\right ) \sqrt {c^2 x^2+1} x^{m+3}}{(m+4)^2 (m+6)}}{m+6}+\frac {c^2 \sqrt {c^2 x^2+1} x^{m+5}}{(m+6)^2}\right )\right )}{(m+6) \sqrt {c^2 x^2+1}}\)

\(\Big \downarrow \) 278

\(\displaystyle \frac {5 d \left (\frac {3 d \int x^m \sqrt {c^2 d x^2+d} (a+b \text {arcsinh}(c x))^2dx}{m+4}-\frac {2 b c d \sqrt {c^2 d x^2+d} \left (\frac {c^2 x^{m+4} (a+b \text {arcsinh}(c x))}{m+4}+\frac {x^{m+2} (a+b \text {arcsinh}(c x))}{m+2}-b c \left (\frac {(3 m+10) x^{m+3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+3}{2},\frac {m+5}{2},-c^2 x^2\right )}{(m+2) (m+3) (m+4)^2}+\frac {\sqrt {c^2 x^2+1} x^{m+3}}{(m+4)^2}\right )\right )}{(m+4) \sqrt {c^2 x^2+1}}+\frac {x^{m+1} \left (c^2 d x^2+d\right )^{3/2} (a+b \text {arcsinh}(c x))^2}{m+4}\right )}{m+6}+\frac {x^{m+1} \left (c^2 d x^2+d\right )^{5/2} (a+b \text {arcsinh}(c x))^2}{m+6}-\frac {2 b c d^2 \sqrt {c^2 d x^2+d} \left (\frac {c^4 x^{m+6} (a+b \text {arcsinh}(c x))}{m+6}+\frac {2 c^2 x^{m+4} (a+b \text {arcsinh}(c x))}{m+4}+\frac {x^{m+2} (a+b \text {arcsinh}(c x))}{m+2}-b c \left (\frac {\frac {\left (15 m^2+130 m+264\right ) x^{m+3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+3}{2},\frac {m+5}{2},-c^2 x^2\right )}{(m+2) (m+3) (m+4)^2 (m+6)}+\frac {\left (m^2+15 m+52\right ) \sqrt {c^2 x^2+1} x^{m+3}}{(m+4)^2 (m+6)}}{m+6}+\frac {c^2 \sqrt {c^2 x^2+1} x^{m+5}}{(m+6)^2}\right )\right )}{(m+6) \sqrt {c^2 x^2+1}}\)

\(\Big \downarrow \) 6223

\(\displaystyle \frac {5 d \left (\frac {3 d \left (-\frac {2 b c \sqrt {c^2 d x^2+d} \int x^{m+1} (a+b \text {arcsinh}(c x))dx}{(m+2) \sqrt {c^2 x^2+1}}+\frac {d \int \frac {x^m (a+b \text {arcsinh}(c x))^2}{\sqrt {c^2 d x^2+d}}dx}{m+2}+\frac {x^{m+1} \sqrt {c^2 d x^2+d} (a+b \text {arcsinh}(c x))^2}{m+2}\right )}{m+4}-\frac {2 b c d \sqrt {c^2 d x^2+d} \left (\frac {c^2 x^{m+4} (a+b \text {arcsinh}(c x))}{m+4}+\frac {x^{m+2} (a+b \text {arcsinh}(c x))}{m+2}-b c \left (\frac {(3 m+10) x^{m+3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+3}{2},\frac {m+5}{2},-c^2 x^2\right )}{(m+2) (m+3) (m+4)^2}+\frac {\sqrt {c^2 x^2+1} x^{m+3}}{(m+4)^2}\right )\right )}{(m+4) \sqrt {c^2 x^2+1}}+\frac {x^{m+1} \left (c^2 d x^2+d\right )^{3/2} (a+b \text {arcsinh}(c x))^2}{m+4}\right )}{m+6}+\frac {x^{m+1} \left (c^2 d x^2+d\right )^{5/2} (a+b \text {arcsinh}(c x))^2}{m+6}-\frac {2 b c d^2 \sqrt {c^2 d x^2+d} \left (\frac {c^4 x^{m+6} (a+b \text {arcsinh}(c x))}{m+6}+\frac {2 c^2 x^{m+4} (a+b \text {arcsinh}(c x))}{m+4}+\frac {x^{m+2} (a+b \text {arcsinh}(c x))}{m+2}-b c \left (\frac {\frac {\left (15 m^2+130 m+264\right ) x^{m+3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+3}{2},\frac {m+5}{2},-c^2 x^2\right )}{(m+2) (m+3) (m+4)^2 (m+6)}+\frac {\left (m^2+15 m+52\right ) \sqrt {c^2 x^2+1} x^{m+3}}{(m+4)^2 (m+6)}}{m+6}+\frac {c^2 \sqrt {c^2 x^2+1} x^{m+5}}{(m+6)^2}\right )\right )}{(m+6) \sqrt {c^2 x^2+1}}\)

\(\Big \downarrow \) 6191

\(\displaystyle \frac {5 d \left (\frac {3 d \left (-\frac {2 b c \sqrt {c^2 d x^2+d} \left (\frac {x^{m+2} (a+b \text {arcsinh}(c x))}{m+2}-\frac {b c \int \frac {x^{m+2}}{\sqrt {c^2 x^2+1}}dx}{m+2}\right )}{(m+2) \sqrt {c^2 x^2+1}}+\frac {d \int \frac {x^m (a+b \text {arcsinh}(c x))^2}{\sqrt {c^2 d x^2+d}}dx}{m+2}+\frac {x^{m+1} \sqrt {c^2 d x^2+d} (a+b \text {arcsinh}(c x))^2}{m+2}\right )}{m+4}-\frac {2 b c d \sqrt {c^2 d x^2+d} \left (\frac {c^2 x^{m+4} (a+b \text {arcsinh}(c x))}{m+4}+\frac {x^{m+2} (a+b \text {arcsinh}(c x))}{m+2}-b c \left (\frac {(3 m+10) x^{m+3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+3}{2},\frac {m+5}{2},-c^2 x^2\right )}{(m+2) (m+3) (m+4)^2}+\frac {\sqrt {c^2 x^2+1} x^{m+3}}{(m+4)^2}\right )\right )}{(m+4) \sqrt {c^2 x^2+1}}+\frac {x^{m+1} \left (c^2 d x^2+d\right )^{3/2} (a+b \text {arcsinh}(c x))^2}{m+4}\right )}{m+6}+\frac {x^{m+1} \left (c^2 d x^2+d\right )^{5/2} (a+b \text {arcsinh}(c x))^2}{m+6}-\frac {2 b c d^2 \sqrt {c^2 d x^2+d} \left (\frac {c^4 x^{m+6} (a+b \text {arcsinh}(c x))}{m+6}+\frac {2 c^2 x^{m+4} (a+b \text {arcsinh}(c x))}{m+4}+\frac {x^{m+2} (a+b \text {arcsinh}(c x))}{m+2}-b c \left (\frac {\frac {\left (15 m^2+130 m+264\right ) x^{m+3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+3}{2},\frac {m+5}{2},-c^2 x^2\right )}{(m+2) (m+3) (m+4)^2 (m+6)}+\frac {\left (m^2+15 m+52\right ) \sqrt {c^2 x^2+1} x^{m+3}}{(m+4)^2 (m+6)}}{m+6}+\frac {c^2 \sqrt {c^2 x^2+1} x^{m+5}}{(m+6)^2}\right )\right )}{(m+6) \sqrt {c^2 x^2+1}}\)

\(\Big \downarrow \) 278

\(\displaystyle \frac {5 d \left (\frac {3 d \left (\frac {d \int \frac {x^m (a+b \text {arcsinh}(c x))^2}{\sqrt {c^2 d x^2+d}}dx}{m+2}-\frac {2 b c \sqrt {c^2 d x^2+d} \left (\frac {x^{m+2} (a+b \text {arcsinh}(c x))}{m+2}-\frac {b c x^{m+3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+3}{2},\frac {m+5}{2},-c^2 x^2\right )}{(m+2) (m+3)}\right )}{(m+2) \sqrt {c^2 x^2+1}}+\frac {x^{m+1} \sqrt {c^2 d x^2+d} (a+b \text {arcsinh}(c x))^2}{m+2}\right )}{m+4}-\frac {2 b c d \sqrt {c^2 d x^2+d} \left (\frac {c^2 x^{m+4} (a+b \text {arcsinh}(c x))}{m+4}+\frac {x^{m+2} (a+b \text {arcsinh}(c x))}{m+2}-b c \left (\frac {(3 m+10) x^{m+3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+3}{2},\frac {m+5}{2},-c^2 x^2\right )}{(m+2) (m+3) (m+4)^2}+\frac {\sqrt {c^2 x^2+1} x^{m+3}}{(m+4)^2}\right )\right )}{(m+4) \sqrt {c^2 x^2+1}}+\frac {x^{m+1} \left (c^2 d x^2+d\right )^{3/2} (a+b \text {arcsinh}(c x))^2}{m+4}\right )}{m+6}+\frac {x^{m+1} \left (c^2 d x^2+d\right )^{5/2} (a+b \text {arcsinh}(c x))^2}{m+6}-\frac {2 b c d^2 \sqrt {c^2 d x^2+d} \left (\frac {c^4 x^{m+6} (a+b \text {arcsinh}(c x))}{m+6}+\frac {2 c^2 x^{m+4} (a+b \text {arcsinh}(c x))}{m+4}+\frac {x^{m+2} (a+b \text {arcsinh}(c x))}{m+2}-b c \left (\frac {\frac {\left (15 m^2+130 m+264\right ) x^{m+3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+3}{2},\frac {m+5}{2},-c^2 x^2\right )}{(m+2) (m+3) (m+4)^2 (m+6)}+\frac {\left (m^2+15 m+52\right ) \sqrt {c^2 x^2+1} x^{m+3}}{(m+4)^2 (m+6)}}{m+6}+\frac {c^2 \sqrt {c^2 x^2+1} x^{m+5}}{(m+6)^2}\right )\right )}{(m+6) \sqrt {c^2 x^2+1}}\)

\(\Big \downarrow \) 6239

\(\displaystyle \frac {5 d \left (\frac {3 d \left (\frac {d \int \frac {x^m (a+b \text {arcsinh}(c x))^2}{\sqrt {c^2 d x^2+d}}dx}{m+2}-\frac {2 b c \sqrt {c^2 d x^2+d} \left (\frac {x^{m+2} (a+b \text {arcsinh}(c x))}{m+2}-\frac {b c x^{m+3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+3}{2},\frac {m+5}{2},-c^2 x^2\right )}{(m+2) (m+3)}\right )}{(m+2) \sqrt {c^2 x^2+1}}+\frac {x^{m+1} \sqrt {c^2 d x^2+d} (a+b \text {arcsinh}(c x))^2}{m+2}\right )}{m+4}-\frac {2 b c d \sqrt {c^2 d x^2+d} \left (\frac {c^2 x^{m+4} (a+b \text {arcsinh}(c x))}{m+4}+\frac {x^{m+2} (a+b \text {arcsinh}(c x))}{m+2}-b c \left (\frac {(3 m+10) x^{m+3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+3}{2},\frac {m+5}{2},-c^2 x^2\right )}{(m+2) (m+3) (m+4)^2}+\frac {\sqrt {c^2 x^2+1} x^{m+3}}{(m+4)^2}\right )\right )}{(m+4) \sqrt {c^2 x^2+1}}+\frac {x^{m+1} \left (c^2 d x^2+d\right )^{3/2} (a+b \text {arcsinh}(c x))^2}{m+4}\right )}{m+6}+\frac {x^{m+1} \left (c^2 d x^2+d\right )^{5/2} (a+b \text {arcsinh}(c x))^2}{m+6}-\frac {2 b c d^2 \sqrt {c^2 d x^2+d} \left (\frac {c^4 x^{m+6} (a+b \text {arcsinh}(c x))}{m+6}+\frac {2 c^2 x^{m+4} (a+b \text {arcsinh}(c x))}{m+4}+\frac {x^{m+2} (a+b \text {arcsinh}(c x))}{m+2}-b c \left (\frac {\frac {\left (15 m^2+130 m+264\right ) x^{m+3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+3}{2},\frac {m+5}{2},-c^2 x^2\right )}{(m+2) (m+3) (m+4)^2 (m+6)}+\frac {\left (m^2+15 m+52\right ) \sqrt {c^2 x^2+1} x^{m+3}}{(m+4)^2 (m+6)}}{m+6}+\frac {c^2 \sqrt {c^2 x^2+1} x^{m+5}}{(m+6)^2}\right )\right )}{(m+6) \sqrt {c^2 x^2+1}}\)

input

Int[x^m*(d + c^2*d*x^2)^(5/2)*(a + b*ArcSinh[c*x])^2,x]
 

output

$Aborted
 

3.4.21.3.1 Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 278

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^p*(( 
c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/2, (m + 1)/2 + 1, ( 
-b)*(x^2/a)], x] /; FreeQ[{a, b, c, m, p}, x] &&  !IGtQ[p, 0] && (ILtQ[p, 0 
] || GtQ[a, 0])
 

rule 363

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x 
_Symbol] :> Simp[d*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(b*e*(m + 2*p + 3))), 
 x] - Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(b*(m + 2*p + 3))   Int[(e*x)^ 
m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b*c - a*d 
, 0] && NeQ[m + 2*p + 3, 0]
 

rule 1590

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + ( 
c_.)*(x_)^4)^(p_.), x_Symbol] :> Simp[c^p*(f*x)^(m + 4*p - 1)*((d + e*x^2)^ 
(q + 1)/(e*f^(4*p - 1)*(m + 4*p + 2*q + 1))), x] + Simp[1/(e*(m + 4*p + 2*q 
 + 1))   Int[(f*x)^m*(d + e*x^2)^q*ExpandToSum[e*(m + 4*p + 2*q + 1)*((a + 
b*x^2 + c*x^4)^p - c^p*x^(4*p)) - d*c^p*(m + 4*p - 1)*x^(4*p - 2), x], x], 
x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] && NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 
0] &&  !IntegerQ[q] && NeQ[m + 4*p + 2*q + 1, 0]
 

rule 6191

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((d_.)*(x_))^(m_.), x_Symbol] 
 :> Simp[(d*x)^(m + 1)*((a + b*ArcSinh[c*x])^n/(d*(m + 1))), x] - Simp[b*c* 
(n/(d*(m + 1)))   Int[(d*x)^(m + 1)*((a + b*ArcSinh[c*x])^(n - 1)/Sqrt[1 + 
c^2*x^2]), x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && NeQ[m, -1]
 

rule 6218

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_ 
)^2)^(p_.), x_Symbol] :> With[{u = IntHide[(f*x)^m*(d + e*x^2)^p, x]}, Simp 
[(a + b*ArcSinh[c*x])   u, x] - Simp[b*c   Int[SimplifyIntegrand[u/Sqrt[1 + 
 c^2*x^2], x], x], x]] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[e, c^2*d] 
&& IGtQ[p, 0]
 

rule 6223

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d_) + (e_ 
.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(f*x)^(m + 1)*(d + e*x^2)^p*((a + b*Arc 
Sinh[c*x])^n/(f*(m + 2*p + 1))), x] + (Simp[2*d*(p/(m + 2*p + 1))   Int[(f* 
x)^m*(d + e*x^2)^(p - 1)*(a + b*ArcSinh[c*x])^n, x], x] - Simp[b*c*(n/(f*(m 
 + 2*p + 1)))*Simp[(d + e*x^2)^p/(1 + c^2*x^2)^p]   Int[(f*x)^(m + 1)*(1 + 
c^2*x^2)^(p - 1/2)*(a + b*ArcSinh[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, 
 d, e, f, m}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && GtQ[p, 0] &&  !LtQ[m, -1]
 

rule 6239

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_.)*((d_) + (e 
_.)*(x_)^2)^(p_.), x_Symbol] :> Unintegrable[(f*x)^m*(d + e*x^2)^p*(a + b*A 
rcSinh[c*x])^n, x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x]
 

3.4.21.4 Maple [N/A] (verified)

Not integrable

Time = 1.05 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93

\[\int x^{m} \left (c^{2} d \,x^{2}+d \right )^{\frac {5}{2}} \left (a +b \,\operatorname {arcsinh}\left (c x \right )\right )^{2}d x\]

input

int(x^m*(c^2*d*x^2+d)^(5/2)*(a+b*arcsinh(c*x))^2,x)
 

output

int(x^m*(c^2*d*x^2+d)^(5/2)*(a+b*arcsinh(c*x))^2,x)
 

3.4.21.5 Fricas [N/A]

Not integrable

Time = 0.27 (sec) , antiderivative size = 133, normalized size of antiderivative = 4.75 \[ \int x^m \left (d+c^2 d x^2\right )^{5/2} (a+b \text {arcsinh}(c x))^2 \, dx=\int { {\left (c^{2} d x^{2} + d\right )}^{\frac {5}{2}} {\left (b \operatorname {arsinh}\left (c x\right ) + a\right )}^{2} x^{m} \,d x } \]

input

integrate(x^m*(c^2*d*x^2+d)^(5/2)*(a+b*arcsinh(c*x))^2,x, algorithm="frica 
s")
 

output

integral((a^2*c^4*d^2*x^4 + 2*a^2*c^2*d^2*x^2 + a^2*d^2 + (b^2*c^4*d^2*x^4 
 + 2*b^2*c^2*d^2*x^2 + b^2*d^2)*arcsinh(c*x)^2 + 2*(a*b*c^4*d^2*x^4 + 2*a* 
b*c^2*d^2*x^2 + a*b*d^2)*arcsinh(c*x))*sqrt(c^2*d*x^2 + d)*x^m, x)
 

3.4.21.6 Sympy [F(-1)]

Timed out. \[ \int x^m \left (d+c^2 d x^2\right )^{5/2} (a+b \text {arcsinh}(c x))^2 \, dx=\text {Timed out} \]

input

integrate(x**m*(c**2*d*x**2+d)**(5/2)*(a+b*asinh(c*x))**2,x)
 

output

Timed out
 

3.4.21.7 Maxima [N/A]

Not integrable

Time = 0.36 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int x^m \left (d+c^2 d x^2\right )^{5/2} (a+b \text {arcsinh}(c x))^2 \, dx=\int { {\left (c^{2} d x^{2} + d\right )}^{\frac {5}{2}} {\left (b \operatorname {arsinh}\left (c x\right ) + a\right )}^{2} x^{m} \,d x } \]

input

integrate(x^m*(c^2*d*x^2+d)^(5/2)*(a+b*arcsinh(c*x))^2,x, algorithm="maxim 
a")
 

output

integrate((c^2*d*x^2 + d)^(5/2)*(b*arcsinh(c*x) + a)^2*x^m, x)
 

3.4.21.8 Giac [F(-2)]

Exception generated. \[ \int x^m \left (d+c^2 d x^2\right )^{5/2} (a+b \text {arcsinh}(c x))^2 \, dx=\text {Exception raised: TypeError} \]

input

integrate(x^m*(c^2*d*x^2+d)^(5/2)*(a+b*arcsinh(c*x))^2,x, algorithm="giac" 
)
 

output

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const 
index_m & i,const vecteur & l) Error: Bad Argument Value
 

3.4.21.9 Mupad [N/A]

Not integrable

Time = 3.12 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int x^m \left (d+c^2 d x^2\right )^{5/2} (a+b \text {arcsinh}(c x))^2 \, dx=\int x^m\,{\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}^2\,{\left (d\,c^2\,x^2+d\right )}^{5/2} \,d x \]

input

int(x^m*(a + b*asinh(c*x))^2*(d + c^2*d*x^2)^(5/2),x)
 

output

int(x^m*(a + b*asinh(c*x))^2*(d + c^2*d*x^2)^(5/2), x)