\(\int x (d+c^2 d x^2)^{3/2} (a+b \text {arcsinh}(c x))^2 \, dx\) [274]

Optimal result

Integrand size = 26, antiderivative size = 248 \[ \int x \left (d+c^2 d x^2\right )^{3/2} (a+b \text {arcsinh}(c x))^2 \, dx=\frac {16 b^2 d \sqrt {d+c^2 d x^2}}{75 c^2}+\frac {8 b^2 \left (d+c^2 d x^2\right )^{3/2}}{225 c^2}+\frac {2 b^2 \left (d+c^2 d x^2\right )^{5/2}}{125 c^2 d}-\frac {2 b d x \sqrt {d+c^2 d x^2} (a+b \text {arcsinh}(c x))}{5 c \sqrt {1+c^2 x^2}}-\frac {4 b c d x^3 \sqrt {d+c^2 d x^2} (a+b \text {arcsinh}(c x))}{15 \sqrt {1+c^2 x^2}}-\frac {2 b c^3 d x^5 \sqrt {d+c^2 d x^2} (a+b \text {arcsinh}(c x))}{25 \sqrt {1+c^2 x^2}}+\frac {\left (d+c^2 d x^2\right )^{5/2} (a+b \text {arcsinh}(c x))^2}{5 c^2 d} \] Output:

16/75*b^2*d*(c^2*d*x^2+d)^(1/2)/c^2+8/225*b^2*(c^2*d*x^2+d)^(3/2)/c^2+2/12 
5*b^2*(c^2*d*x^2+d)^(5/2)/c^2/d-2/5*b*d*x*(c^2*d*x^2+d)^(1/2)*(a+b*arcsinh 
(c*x))/c/(c^2*x^2+1)^(1/2)-4/15*b*c*d*x^3*(c^2*d*x^2+d)^(1/2)*(a+b*arcsinh 
(c*x))/(c^2*x^2+1)^(1/2)-2/25*b*c^3*d*x^5*(c^2*d*x^2+d)^(1/2)*(a+b*arcsinh 
(c*x))/(c^2*x^2+1)^(1/2)+1/5*(c^2*d*x^2+d)^(5/2)*(a+b*arcsinh(c*x))^2/c^2/ 
d
 

Mathematica [A] (verified)

Time = 1.12 (sec) , antiderivative size = 198, normalized size of antiderivative = 0.80 \[ \int x \left (d+c^2 d x^2\right )^{3/2} (a+b \text {arcsinh}(c x))^2 \, dx=\frac {d \sqrt {d+c^2 d x^2} \left (225 a^2 \left (1+c^2 x^2\right )^3-30 a b c x \sqrt {1+c^2 x^2} \left (15+10 c^2 x^2+3 c^4 x^4\right )+2 b^2 \left (149+187 c^2 x^2+47 c^4 x^4+9 c^6 x^6\right )+30 b \left (15 a \left (1+c^2 x^2\right )^3-b c x \sqrt {1+c^2 x^2} \left (15+10 c^2 x^2+3 c^4 x^4\right )\right ) \text {arcsinh}(c x)+225 b^2 \left (1+c^2 x^2\right )^3 \text {arcsinh}(c x)^2\right )}{1125 c^2 \left (1+c^2 x^2\right )} \] Input:

Integrate[x*(d + c^2*d*x^2)^(3/2)*(a + b*ArcSinh[c*x])^2,x]
 

Output:

(d*Sqrt[d + c^2*d*x^2]*(225*a^2*(1 + c^2*x^2)^3 - 30*a*b*c*x*Sqrt[1 + c^2* 
x^2]*(15 + 10*c^2*x^2 + 3*c^4*x^4) + 2*b^2*(149 + 187*c^2*x^2 + 47*c^4*x^4 
 + 9*c^6*x^6) + 30*b*(15*a*(1 + c^2*x^2)^3 - b*c*x*Sqrt[1 + c^2*x^2]*(15 + 
 10*c^2*x^2 + 3*c^4*x^4))*ArcSinh[c*x] + 225*b^2*(1 + c^2*x^2)^3*ArcSinh[c 
*x]^2))/(1125*c^2*(1 + c^2*x^2))
 

Rubi [A] (verified)

Time = 0.58 (sec) , antiderivative size = 183, normalized size of antiderivative = 0.74, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {6213, 6199, 27, 1576, 1140, 2009}

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.

\(\displaystyle \int x \left (c^2 d x^2+d\right )^{3/2} (a+b \text {arcsinh}(c x))^2 \, dx\)

\(\Big \downarrow \) 6213

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

\(\Big \downarrow \) 6199

\(\displaystyle \frac {\left (c^2 d x^2+d\right )^{5/2} (a+b \text {arcsinh}(c x))^2}{5 c^2 d}-\frac {2 b d \sqrt {c^2 d x^2+d} \left (-b c \int \frac {x \left (3 c^4 x^4+10 c^2 x^2+15\right )}{15 \sqrt {c^2 x^2+1}}dx+\frac {1}{5} c^4 x^5 (a+b \text {arcsinh}(c x))+\frac {2}{3} c^2 x^3 (a+b \text {arcsinh}(c x))+x (a+b \text {arcsinh}(c x))\right )}{5 c \sqrt {c^2 x^2+1}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\left (c^2 d x^2+d\right )^{5/2} (a+b \text {arcsinh}(c x))^2}{5 c^2 d}-\frac {2 b d \sqrt {c^2 d x^2+d} \left (-\frac {1}{15} b c \int \frac {x \left (3 c^4 x^4+10 c^2 x^2+15\right )}{\sqrt {c^2 x^2+1}}dx+\frac {1}{5} c^4 x^5 (a+b \text {arcsinh}(c x))+\frac {2}{3} c^2 x^3 (a+b \text {arcsinh}(c x))+x (a+b \text {arcsinh}(c x))\right )}{5 c \sqrt {c^2 x^2+1}}\)

\(\Big \downarrow \) 1576

\(\displaystyle \frac {\left (c^2 d x^2+d\right )^{5/2} (a+b \text {arcsinh}(c x))^2}{5 c^2 d}-\frac {2 b d \sqrt {c^2 d x^2+d} \left (-\frac {1}{30} b c \int \frac {3 c^4 x^4+10 c^2 x^2+15}{\sqrt {c^2 x^2+1}}dx^2+\frac {1}{5} c^4 x^5 (a+b \text {arcsinh}(c x))+\frac {2}{3} c^2 x^3 (a+b \text {arcsinh}(c x))+x (a+b \text {arcsinh}(c x))\right )}{5 c \sqrt {c^2 x^2+1}}\)

\(\Big \downarrow \) 1140

\(\displaystyle \frac {\left (c^2 d x^2+d\right )^{5/2} (a+b \text {arcsinh}(c x))^2}{5 c^2 d}-\frac {2 b d \sqrt {c^2 d x^2+d} \left (-\frac {1}{30} b c \int \left (3 \left (c^2 x^2+1\right )^{3/2}+4 \sqrt {c^2 x^2+1}+\frac {8}{\sqrt {c^2 x^2+1}}\right )dx^2+\frac {1}{5} c^4 x^5 (a+b \text {arcsinh}(c x))+\frac {2}{3} c^2 x^3 (a+b \text {arcsinh}(c x))+x (a+b \text {arcsinh}(c x))\right )}{5 c \sqrt {c^2 x^2+1}}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\left (c^2 d x^2+d\right )^{5/2} (a+b \text {arcsinh}(c x))^2}{5 c^2 d}-\frac {2 b d \sqrt {c^2 d x^2+d} \left (\frac {1}{5} c^4 x^5 (a+b \text {arcsinh}(c x))+\frac {2}{3} c^2 x^3 (a+b \text {arcsinh}(c x))+x (a+b \text {arcsinh}(c x))-\frac {1}{30} b c \left (\frac {6 \left (c^2 x^2+1\right )^{5/2}}{5 c^2}+\frac {8 \left (c^2 x^2+1\right )^{3/2}}{3 c^2}+\frac {16 \sqrt {c^2 x^2+1}}{c^2}\right )\right )}{5 c \sqrt {c^2 x^2+1}}\)

Input:

Int[x*(d + c^2*d*x^2)^(3/2)*(a + b*ArcSinh[c*x])^2,x]
 

Output:

((d + c^2*d*x^2)^(5/2)*(a + b*ArcSinh[c*x])^2)/(5*c^2*d) - (2*b*d*Sqrt[d + 
 c^2*d*x^2]*(-1/30*(b*c*((16*Sqrt[1 + c^2*x^2])/c^2 + (8*(1 + c^2*x^2)^(3/ 
2))/(3*c^2) + (6*(1 + c^2*x^2)^(5/2))/(5*c^2))) + x*(a + b*ArcSinh[c*x]) + 
 (2*c^2*x^3*(a + b*ArcSinh[c*x]))/3 + (c^4*x^5*(a + b*ArcSinh[c*x]))/5))/( 
5*c*Sqrt[1 + c^2*x^2])
 

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 1140

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

rule 1576

Int[(x_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^( 
p_.), x_Symbol] :> Simp[1/2   Subst[Int[(d + e*x)^q*(a + b*x + c*x^2)^p, x] 
, x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 6199

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))*((d_) + (e_.)*(x_)^2)^(p_.), x_Symb 
ol] :> With[{u = IntHide[(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}, x] && EqQ[e, c^2*d] && IGtQ[p, 0]
 

rule 6213

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*(x_)*((d_) + (e_.)*(x_)^2)^(p 
_.), x_Symbol] :> Simp[(d + e*x^2)^(p + 1)*((a + b*ArcSinh[c*x])^n/(2*e*(p 
+ 1))), x] - Simp[b*(n/(2*c*(p + 1)))*Simp[(d + e*x^2)^p/(1 + c^2*x^2)^p] 
 Int[(1 + c^2*x^2)^(p + 1/2)*(a + b*ArcSinh[c*x])^(n - 1), x], x] /; FreeQ[ 
{a, b, c, d, e, p}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && NeQ[p, -1]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(449\) vs. \(2(214)=428\).

Time = 1.98 (sec) , antiderivative size = 450, normalized size of antiderivative = 1.81

method	result	size
orering	\(\frac {\left (549 c^{8} x^{8}+1982 c^{6} x^{6}+4355 c^{4} x^{4}+1420 c^{2} x^{2}+298\right ) \left (c^{2} d \,x^{2}+d \right )^{\frac {3}{2}} \left (a +b \,\operatorname {arcsinh}\left (x c \right )\right )^{2}}{1125 c^{4} x^{2} \left (c^{2} x^{2}+1\right )^{2}}-\frac {2 \left (54 c^{6} x^{6}+217 c^{4} x^{4}+672 c^{2} x^{2}+149\right ) \left (\left (c^{2} d \,x^{2}+d \right )^{\frac {3}{2}} \left (a +b \,\operatorname {arcsinh}\left (x c \right )\right )^{2}+3 x^{2} \sqrt {c^{2} d \,x^{2}+d}\, \left (a +b \,\operatorname {arcsinh}\left (x c \right )\right )^{2} c^{2} d +\frac {2 x \left (c^{2} d \,x^{2}+d \right )^{\frac {3}{2}} \left (a +b \,\operatorname {arcsinh}\left (x c \right )\right ) b c}{\sqrt {c^{2} x^{2}+1}}\right )}{1125 c^{4} x^{2} \left (c^{2} x^{2}+1\right )}+\frac {\left (9 c^{4} x^{4}+38 c^{2} x^{2}+149\right ) \left (9 c^{2} d x \sqrt {c^{2} d \,x^{2}+d}\, \left (a +b \,\operatorname {arcsinh}\left (x c \right )\right )^{2}+\frac {4 \left (c^{2} d \,x^{2}+d \right )^{\frac {3}{2}} \left (a +b \,\operatorname {arcsinh}\left (x c \right )\right ) b c}{\sqrt {c^{2} x^{2}+1}}+\frac {3 x^{3} \left (a +b \,\operatorname {arcsinh}\left (x c \right )\right )^{2} c^{4} d^{2}}{\sqrt {c^{2} d \,x^{2}+d}}+\frac {12 b \,c^{3} d \,x^{2} \sqrt {c^{2} d \,x^{2}+d}\, \left (a +b \,\operatorname {arcsinh}\left (x c \right )\right )}{\sqrt {c^{2} x^{2}+1}}+\frac {2 x \left (c^{2} d \,x^{2}+d \right )^{\frac {3}{2}} b^{2} c^{2}}{c^{2} x^{2}+1}-\frac {2 x^{2} \left (c^{2} d \,x^{2}+d \right )^{\frac {3}{2}} \left (a +b \,\operatorname {arcsinh}\left (x c \right )\right ) b \,c^{3}}{\left (c^{2} x^{2}+1\right )^{\frac {3}{2}}}\right )}{1125 c^{4} x}\)	\(450\)
default	\(\text {Expression too large to display}\)	\(1149\)
parts	\(\text {Expression too large to display}\)	\(1149\)

Input:

int(x*(c^2*d*x^2+d)^(3/2)*(a+b*arcsinh(x*c))^2,x,method=_RETURNVERBOSE)
 

Output:

1/1125*(549*c^8*x^8+1982*c^6*x^6+4355*c^4*x^4+1420*c^2*x^2+298)/c^4/x^2/(c 
^2*x^2+1)^2*(c^2*d*x^2+d)^(3/2)*(a+b*arcsinh(x*c))^2-2/1125*(54*c^6*x^6+21 
7*c^4*x^4+672*c^2*x^2+149)/c^4/x^2/(c^2*x^2+1)*((c^2*d*x^2+d)^(3/2)*(a+b*a 
rcsinh(x*c))^2+3*x^2*(c^2*d*x^2+d)^(1/2)*(a+b*arcsinh(x*c))^2*c^2*d+2*x*(c 
^2*d*x^2+d)^(3/2)*(a+b*arcsinh(x*c))*b*c/(c^2*x^2+1)^(1/2))+1/1125*(9*c^4* 
x^4+38*c^2*x^2+149)/c^4/x*(9*c^2*d*x*(c^2*d*x^2+d)^(1/2)*(a+b*arcsinh(x*c) 
)^2+4*(c^2*d*x^2+d)^(3/2)*(a+b*arcsinh(x*c))*b*c/(c^2*x^2+1)^(1/2)+3*x^3/( 
c^2*d*x^2+d)^(1/2)*(a+b*arcsinh(x*c))^2*c^4*d^2+12*b*c^3*d*x^2*(c^2*d*x^2+ 
d)^(1/2)*(a+b*arcsinh(x*c))/(c^2*x^2+1)^(1/2)+2*x*(c^2*d*x^2+d)^(3/2)*b^2* 
c^2/(c^2*x^2+1)-2*x^2*(c^2*d*x^2+d)^(3/2)*(a+b*arcsinh(x*c))*b*c^3/(c^2*x^ 
2+1)^(3/2))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 332, normalized size of antiderivative = 1.34 \[ \int x \left (d+c^2 d x^2\right )^{3/2} (a+b \text {arcsinh}(c x))^2 \, dx=\frac {225 \, {\left (b^{2} c^{6} d x^{6} + 3 \, b^{2} c^{4} d x^{4} + 3 \, b^{2} c^{2} d x^{2} + b^{2} d\right )} \sqrt {c^{2} d x^{2} + d} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right )^{2} + 30 \, {\left (15 \, a b c^{6} d x^{6} + 45 \, a b c^{4} d x^{4} + 45 \, a b c^{2} d x^{2} + 15 \, a b d - {\left (3 \, b^{2} c^{5} d x^{5} + 10 \, b^{2} c^{3} d x^{3} + 15 \, b^{2} c d x\right )} \sqrt {c^{2} x^{2} + 1}\right )} \sqrt {c^{2} d x^{2} + d} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right ) + {\left (9 \, {\left (25 \, a^{2} + 2 \, b^{2}\right )} c^{6} d x^{6} + {\left (675 \, a^{2} + 94 \, b^{2}\right )} c^{4} d x^{4} + {\left (675 \, a^{2} + 374 \, b^{2}\right )} c^{2} d x^{2} + {\left (225 \, a^{2} + 298 \, b^{2}\right )} d - 30 \, {\left (3 \, a b c^{5} d x^{5} + 10 \, a b c^{3} d x^{3} + 15 \, a b c d x\right )} \sqrt {c^{2} x^{2} + 1}\right )} \sqrt {c^{2} d x^{2} + d}}{1125 \, {\left (c^{4} x^{2} + c^{2}\right )}} \] Input:

integrate(x*(c^2*d*x^2+d)^(3/2)*(a+b*arcsinh(c*x))^2,x, algorithm="fricas" 
)
 

Output:

1/1125*(225*(b^2*c^6*d*x^6 + 3*b^2*c^4*d*x^4 + 3*b^2*c^2*d*x^2 + b^2*d)*sq 
rt(c^2*d*x^2 + d)*log(c*x + sqrt(c^2*x^2 + 1))^2 + 30*(15*a*b*c^6*d*x^6 + 
45*a*b*c^4*d*x^4 + 45*a*b*c^2*d*x^2 + 15*a*b*d - (3*b^2*c^5*d*x^5 + 10*b^2 
*c^3*d*x^3 + 15*b^2*c*d*x)*sqrt(c^2*x^2 + 1))*sqrt(c^2*d*x^2 + d)*log(c*x 
+ sqrt(c^2*x^2 + 1)) + (9*(25*a^2 + 2*b^2)*c^6*d*x^6 + (675*a^2 + 94*b^2)* 
c^4*d*x^4 + (675*a^2 + 374*b^2)*c^2*d*x^2 + (225*a^2 + 298*b^2)*d - 30*(3* 
a*b*c^5*d*x^5 + 10*a*b*c^3*d*x^3 + 15*a*b*c*d*x)*sqrt(c^2*x^2 + 1))*sqrt(c 
^2*d*x^2 + d))/(c^4*x^2 + c^2)
 

Sympy [F]

\[ \int x \left (d+c^2 d x^2\right )^{3/2} (a+b \text {arcsinh}(c x))^2 \, dx=\int x \left (d \left (c^{2} x^{2} + 1\right )\right )^{\frac {3}{2}} \left (a + b \operatorname {asinh}{\left (c x \right )}\right )^{2}\, dx \] Input:

integrate(x*(c**2*d*x**2+d)**(3/2)*(a+b*asinh(c*x))**2,x)
 

Output:

Integral(x*(d*(c**2*x**2 + 1))**(3/2)*(a + b*asinh(c*x))**2, x)
 

Maxima [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 230, normalized size of antiderivative = 0.93 \[ \int x \left (d+c^2 d x^2\right )^{3/2} (a+b \text {arcsinh}(c x))^2 \, dx=\frac {{\left (c^{2} d x^{2} + d\right )}^{\frac {5}{2}} b^{2} \operatorname {arsinh}\left (c x\right )^{2}}{5 \, c^{2} d} + \frac {2}{1125} \, b^{2} {\left (\frac {9 \, \sqrt {c^{2} x^{2} + 1} c^{2} d^{\frac {5}{2}} x^{4} + 38 \, \sqrt {c^{2} x^{2} + 1} d^{\frac {5}{2}} x^{2} + \frac {149 \, \sqrt {c^{2} x^{2} + 1} d^{\frac {5}{2}}}{c^{2}}}{d} - \frac {15 \, {\left (3 \, c^{4} d^{\frac {5}{2}} x^{5} + 10 \, c^{2} d^{\frac {5}{2}} x^{3} + 15 \, d^{\frac {5}{2}} x\right )} \operatorname {arsinh}\left (c x\right )}{c d}\right )} + \frac {2 \, {\left (c^{2} d x^{2} + d\right )}^{\frac {5}{2}} a b \operatorname {arsinh}\left (c x\right )}{5 \, c^{2} d} + \frac {{\left (c^{2} d x^{2} + d\right )}^{\frac {5}{2}} a^{2}}{5 \, c^{2} d} - \frac {2 \, {\left (3 \, c^{4} d^{\frac {5}{2}} x^{5} + 10 \, c^{2} d^{\frac {5}{2}} x^{3} + 15 \, d^{\frac {5}{2}} x\right )} a b}{75 \, c d} \] Input:

integrate(x*(c^2*d*x^2+d)^(3/2)*(a+b*arcsinh(c*x))^2,x, algorithm="maxima" 
)
 

Output:

1/5*(c^2*d*x^2 + d)^(5/2)*b^2*arcsinh(c*x)^2/(c^2*d) + 2/1125*b^2*((9*sqrt 
(c^2*x^2 + 1)*c^2*d^(5/2)*x^4 + 38*sqrt(c^2*x^2 + 1)*d^(5/2)*x^2 + 149*sqr 
t(c^2*x^2 + 1)*d^(5/2)/c^2)/d - 15*(3*c^4*d^(5/2)*x^5 + 10*c^2*d^(5/2)*x^3 
 + 15*d^(5/2)*x)*arcsinh(c*x)/(c*d)) + 2/5*(c^2*d*x^2 + d)^(5/2)*a*b*arcsi 
nh(c*x)/(c^2*d) + 1/5*(c^2*d*x^2 + d)^(5/2)*a^2/(c^2*d) - 2/75*(3*c^4*d^(5 
/2)*x^5 + 10*c^2*d^(5/2)*x^3 + 15*d^(5/2)*x)*a*b/(c*d)
 

Giac [F(-2)]

Exception generated. \[ \int x \left (d+c^2 d x^2\right )^{3/2} (a+b \text {arcsinh}(c x))^2 \, dx=\text {Exception raised: TypeError} \] Input:

integrate(x*(c^2*d*x^2+d)^(3/2)*(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
 

Mupad [F(-1)]

Timed out. \[ \int x \left (d+c^2 d x^2\right )^{3/2} (a+b \text {arcsinh}(c x))^2 \, dx=\int x\,{\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}^2\,{\left (d\,c^2\,x^2+d\right )}^{3/2} \,d x \] Input:

int(x*(a + b*asinh(c*x))^2*(d + c^2*d*x^2)^(3/2),x)
 

Output:

int(x*(a + b*asinh(c*x))^2*(d + c^2*d*x^2)^(3/2), x)
                                                                                    
                                                                                    
 

Reduce [F]

\[ \int x \left (d+c^2 d x^2\right )^{3/2} (a+b \text {arcsinh}(c x))^2 \, dx=\frac {\sqrt {d}\, d \left (\sqrt {c^{2} x^{2}+1}\, a^{2} c^{4} x^{4}+2 \sqrt {c^{2} x^{2}+1}\, a^{2} c^{2} x^{2}+\sqrt {c^{2} x^{2}+1}\, a^{2}+10 \left (\int \sqrt {c^{2} x^{2}+1}\, \mathit {asinh} \left (c x \right ) x^{3}d x \right ) a b \,c^{4}+10 \left (\int \sqrt {c^{2} x^{2}+1}\, \mathit {asinh} \left (c x \right ) x d x \right ) a b \,c^{2}+5 \left (\int \sqrt {c^{2} x^{2}+1}\, \mathit {asinh} \left (c x \right )^{2} x^{3}d x \right ) b^{2} c^{4}+5 \left (\int \sqrt {c^{2} x^{2}+1}\, \mathit {asinh} \left (c x \right )^{2} x d x \right ) b^{2} c^{2}\right )}{5 c^{2}} \] Input:

int(x*(c^2*d*x^2+d)^(3/2)*(a+b*asinh(c*x))^2,x)
 

Output:

(sqrt(d)*d*(sqrt(c**2*x**2 + 1)*a**2*c**4*x**4 + 2*sqrt(c**2*x**2 + 1)*a** 
2*c**2*x**2 + sqrt(c**2*x**2 + 1)*a**2 + 10*int(sqrt(c**2*x**2 + 1)*asinh( 
c*x)*x**3,x)*a*b*c**4 + 10*int(sqrt(c**2*x**2 + 1)*asinh(c*x)*x,x)*a*b*c** 
2 + 5*int(sqrt(c**2*x**2 + 1)*asinh(c*x)**2*x**3,x)*b**2*c**4 + 5*int(sqrt 
(c**2*x**2 + 1)*asinh(c*x)**2*x,x)*b**2*c**2))/(5*c**2)