\(\int \frac {x^5 (a+b \text {arcsinh}(c x))}{(d+c^2 d x^2)^{5/2}} \, dx\) [174]

Optimal result

Integrand size = 26, antiderivative size = 210 \[ \int \frac {x^5 (a+b \text {arcsinh}(c x))}{\left (d+c^2 d x^2\right )^{5/2}} \, dx=\frac {b x}{6 c^5 d^2 \sqrt {1+c^2 x^2} \sqrt {d+c^2 d x^2}}-\frac {b x \sqrt {1+c^2 x^2}}{c^5 d^2 \sqrt {d+c^2 d x^2}}-\frac {a+b \text {arcsinh}(c x)}{3 c^6 d \left (d+c^2 d x^2\right )^{3/2}}+\frac {2 (a+b \text {arcsinh}(c x))}{c^6 d^2 \sqrt {d+c^2 d x^2}}+\frac {\sqrt {d+c^2 d x^2} (a+b \text {arcsinh}(c x))}{c^6 d^3}-\frac {11 b \sqrt {1+c^2 x^2} \arctan (c x)}{6 c^6 d^2 \sqrt {d+c^2 d x^2}} \] Output:

1/6*b*x/c^5/d^2/(c^2*x^2+1)^(1/2)/(c^2*d*x^2+d)^(1/2)-b*x*(c^2*x^2+1)^(1/2 
)/c^5/d^2/(c^2*d*x^2+d)^(1/2)-1/3*(a+b*arcsinh(c*x))/c^6/d/(c^2*d*x^2+d)^( 
3/2)+2*(a+b*arcsinh(c*x))/c^6/d^2/(c^2*d*x^2+d)^(1/2)+(c^2*d*x^2+d)^(1/2)* 
(a+b*arcsinh(c*x))/c^6/d^3-11/6*b*(c^2*x^2+1)^(1/2)*arctan(c*x)/c^6/d^2/(c 
^2*d*x^2+d)^(1/2)
 

Mathematica [A] (verified)

Time = 0.38 (sec) , antiderivative size = 177, normalized size of antiderivative = 0.84 \[ \int \frac {x^5 (a+b \text {arcsinh}(c x))}{\left (d+c^2 d x^2\right )^{5/2}} \, dx=\frac {\sqrt {d+c^2 d x^2} \left (-5 b c x-11 b c^3 x^3-6 b c^5 x^5+16 a \sqrt {1+c^2 x^2}+24 a c^2 x^2 \sqrt {1+c^2 x^2}+6 a c^4 x^4 \sqrt {1+c^2 x^2}+2 b \sqrt {1+c^2 x^2} \left (8+12 c^2 x^2+3 c^4 x^4\right ) \text {arcsinh}(c x)-11 b \left (1+c^2 x^2\right )^2 \arctan (c x)\right )}{6 c^6 d^3 \left (1+c^2 x^2\right )^{5/2}} \] Input:

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

Output:

(Sqrt[d + c^2*d*x^2]*(-5*b*c*x - 11*b*c^3*x^3 - 6*b*c^5*x^5 + 16*a*Sqrt[1 
+ c^2*x^2] + 24*a*c^2*x^2*Sqrt[1 + c^2*x^2] + 6*a*c^4*x^4*Sqrt[1 + c^2*x^2 
] + 2*b*Sqrt[1 + c^2*x^2]*(8 + 12*c^2*x^2 + 3*c^4*x^4)*ArcSinh[c*x] - 11*b 
*(1 + c^2*x^2)^2*ArcTan[c*x]))/(6*c^6*d^3*(1 + c^2*x^2)^(5/2))
 

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 164, normalized size of antiderivative = 0.78, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {6219, 27, 1471, 25, 299, 216}

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.

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

\(\Big \downarrow \) 6219

\(\displaystyle -\frac {b c \sqrt {c^2 d x^2+d} \int \frac {3 c^4 x^4+12 c^2 x^2+8}{3 c^6 d^3 \left (c^2 x^2+1\right )^2}dx}{\sqrt {c^2 x^2+1}}+\frac {\sqrt {c^2 d x^2+d} (a+b \text {arcsinh}(c x))}{c^6 d^3}+\frac {2 (a+b \text {arcsinh}(c x))}{c^6 d^2 \sqrt {c^2 d x^2+d}}-\frac {a+b \text {arcsinh}(c x)}{3 c^6 d \left (c^2 d x^2+d\right )^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {b \sqrt {c^2 d x^2+d} \int \frac {3 c^4 x^4+12 c^2 x^2+8}{\left (c^2 x^2+1\right )^2}dx}{3 c^5 d^3 \sqrt {c^2 x^2+1}}+\frac {\sqrt {c^2 d x^2+d} (a+b \text {arcsinh}(c x))}{c^6 d^3}+\frac {2 (a+b \text {arcsinh}(c x))}{c^6 d^2 \sqrt {c^2 d x^2+d}}-\frac {a+b \text {arcsinh}(c x)}{3 c^6 d \left (c^2 d x^2+d\right )^{3/2}}\)

\(\Big \downarrow \) 1471

\(\displaystyle -\frac {b \sqrt {c^2 d x^2+d} \left (-\frac {1}{2} \int -\frac {6 c^2 x^2+17}{c^2 x^2+1}dx-\frac {x}{2 \left (c^2 x^2+1\right )}\right )}{3 c^5 d^3 \sqrt {c^2 x^2+1}}+\frac {\sqrt {c^2 d x^2+d} (a+b \text {arcsinh}(c x))}{c^6 d^3}+\frac {2 (a+b \text {arcsinh}(c x))}{c^6 d^2 \sqrt {c^2 d x^2+d}}-\frac {a+b \text {arcsinh}(c x)}{3 c^6 d \left (c^2 d x^2+d\right )^{3/2}}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {b \sqrt {c^2 d x^2+d} \left (\frac {1}{2} \int \frac {6 c^2 x^2+17}{c^2 x^2+1}dx-\frac {x}{2 \left (c^2 x^2+1\right )}\right )}{3 c^5 d^3 \sqrt {c^2 x^2+1}}+\frac {\sqrt {c^2 d x^2+d} (a+b \text {arcsinh}(c x))}{c^6 d^3}+\frac {2 (a+b \text {arcsinh}(c x))}{c^6 d^2 \sqrt {c^2 d x^2+d}}-\frac {a+b \text {arcsinh}(c x)}{3 c^6 d \left (c^2 d x^2+d\right )^{3/2}}\)

\(\Big \downarrow \) 299

\(\displaystyle -\frac {b \sqrt {c^2 d x^2+d} \left (\frac {1}{2} \left (11 \int \frac {1}{c^2 x^2+1}dx+6 x\right )-\frac {x}{2 \left (c^2 x^2+1\right )}\right )}{3 c^5 d^3 \sqrt {c^2 x^2+1}}+\frac {\sqrt {c^2 d x^2+d} (a+b \text {arcsinh}(c x))}{c^6 d^3}+\frac {2 (a+b \text {arcsinh}(c x))}{c^6 d^2 \sqrt {c^2 d x^2+d}}-\frac {a+b \text {arcsinh}(c x)}{3 c^6 d \left (c^2 d x^2+d\right )^{3/2}}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {\sqrt {c^2 d x^2+d} (a+b \text {arcsinh}(c x))}{c^6 d^3}+\frac {2 (a+b \text {arcsinh}(c x))}{c^6 d^2 \sqrt {c^2 d x^2+d}}-\frac {a+b \text {arcsinh}(c x)}{3 c^6 d \left (c^2 d x^2+d\right )^{3/2}}-\frac {b \left (\frac {1}{2} \left (\frac {11 \arctan (c x)}{c}+6 x\right )-\frac {x}{2 \left (c^2 x^2+1\right )}\right ) \sqrt {c^2 d x^2+d}}{3 c^5 d^3 \sqrt {c^2 x^2+1}}\)

Input:

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

Output:

-1/3*(a + b*ArcSinh[c*x])/(c^6*d*(d + c^2*d*x^2)^(3/2)) + (2*(a + b*ArcSin 
h[c*x]))/(c^6*d^2*Sqrt[d + c^2*d*x^2]) + (Sqrt[d + c^2*d*x^2]*(a + b*ArcSi 
nh[c*x]))/(c^6*d^3) - (b*Sqrt[d + c^2*d*x^2]*(-1/2*x/(1 + c^2*x^2) + (6*x 
+ (11*ArcTan[c*x])/c)/2))/(3*c^5*d^3*Sqrt[1 + c^2*x^2])
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 216

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 0])
 

rule 299

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

rule 1471

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), 
x_Symbol] :> With[{Qx = PolynomialQuotient[(a + b*x^2 + c*x^4)^p, d + e*x^2 
, x], R = Coeff[PolynomialRemainder[(a + b*x^2 + c*x^4)^p, d + e*x^2, x], x 
, 0]}, Simp[(-R)*x*((d + e*x^2)^(q + 1)/(2*d*(q + 1))), x] + Simp[1/(2*d*(q 
 + 1))   Int[(d + e*x^2)^(q + 1)*ExpandToSum[2*d*(q + 1)*Qx + R*(2*q + 3), 
x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^ 
2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && LtQ[q, -1]
 

rule 6219

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))*(x_)^(m_)*((d_) + (e_.)*(x_)^2)^(p_ 
), x_Symbol] :> With[{u = IntHide[x^m*(d + e*x^2)^p, x]}, Simp[(a + b*ArcSi 
nh[c*x])   u, x] - Simp[b*c*Simp[Sqrt[d + e*x^2]/Sqrt[1 + c^2*x^2]]   Int[S 
implifyIntegrand[u/Sqrt[d + e*x^2], x], x], x]] /; FreeQ[{a, b, c, d, e}, x 
] && EqQ[e, c^2*d] && IntegerQ[p - 1/2] && NeQ[p, -2^(-1)] && (IGtQ[(m + 1) 
/2, 0] || ILtQ[(m + 2*p + 3)/2, 0])
 

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.99 (sec) , antiderivative size = 400, normalized size of antiderivative = 1.90

method	result	size
default	\(a \left (\frac {x^{4}}{c^{2} d \left (c^{2} d \,x^{2}+d \right )^{\frac {3}{2}}}-\frac {4 \left (-\frac {x^{2}}{c^{2} d \left (c^{2} d \,x^{2}+d \right )^{\frac {3}{2}}}-\frac {2}{3 d \,c^{4} \left (c^{2} d \,x^{2}+d \right )^{\frac {3}{2}}}\right )}{c^{2}}\right )+\frac {b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \operatorname {arcsinh}\left (x c \right ) x^{2}}{d^{3} c^{4} \left (c^{2} x^{2}+1\right )}-\frac {b \sqrt {d \left (c^{2} x^{2}+1\right )}\, x}{d^{3} c^{5} \sqrt {c^{2} x^{2}+1}}+\frac {b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \operatorname {arcsinh}\left (x c \right )}{d^{3} c^{6} \left (c^{2} x^{2}+1\right )}+\frac {2 b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \operatorname {arcsinh}\left (x c \right ) x^{2}}{\left (c^{2} x^{2}+1\right )^{2} d^{3} c^{4}}+\frac {b \sqrt {d \left (c^{2} x^{2}+1\right )}\, x}{6 \left (c^{2} x^{2}+1\right )^{\frac {3}{2}} d^{3} c^{5}}+\frac {5 b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \operatorname {arcsinh}\left (x c \right )}{3 \left (c^{2} x^{2}+1\right )^{2} d^{3} c^{6}}+\frac {11 i b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \ln \left (x c +\sqrt {c^{2} x^{2}+1}-i\right )}{6 \sqrt {c^{2} x^{2}+1}\, d^{3} c^{6}}-\frac {11 i b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \ln \left (x c +\sqrt {c^{2} x^{2}+1}+i\right )}{6 \sqrt {c^{2} x^{2}+1}\, d^{3} c^{6}}\)	\(400\)
parts	\(a \left (\frac {x^{4}}{c^{2} d \left (c^{2} d \,x^{2}+d \right )^{\frac {3}{2}}}-\frac {4 \left (-\frac {x^{2}}{c^{2} d \left (c^{2} d \,x^{2}+d \right )^{\frac {3}{2}}}-\frac {2}{3 d \,c^{4} \left (c^{2} d \,x^{2}+d \right )^{\frac {3}{2}}}\right )}{c^{2}}\right )+\frac {b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \operatorname {arcsinh}\left (x c \right ) x^{2}}{d^{3} c^{4} \left (c^{2} x^{2}+1\right )}-\frac {b \sqrt {d \left (c^{2} x^{2}+1\right )}\, x}{d^{3} c^{5} \sqrt {c^{2} x^{2}+1}}+\frac {b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \operatorname {arcsinh}\left (x c \right )}{d^{3} c^{6} \left (c^{2} x^{2}+1\right )}+\frac {2 b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \operatorname {arcsinh}\left (x c \right ) x^{2}}{\left (c^{2} x^{2}+1\right )^{2} d^{3} c^{4}}+\frac {b \sqrt {d \left (c^{2} x^{2}+1\right )}\, x}{6 \left (c^{2} x^{2}+1\right )^{\frac {3}{2}} d^{3} c^{5}}+\frac {5 b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \operatorname {arcsinh}\left (x c \right )}{3 \left (c^{2} x^{2}+1\right )^{2} d^{3} c^{6}}+\frac {11 i b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \ln \left (x c +\sqrt {c^{2} x^{2}+1}-i\right )}{6 \sqrt {c^{2} x^{2}+1}\, d^{3} c^{6}}-\frac {11 i b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \ln \left (x c +\sqrt {c^{2} x^{2}+1}+i\right )}{6 \sqrt {c^{2} x^{2}+1}\, d^{3} c^{6}}\)	\(400\)

Input:

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

Output:

a*(x^4/c^2/d/(c^2*d*x^2+d)^(3/2)-4/c^2*(-x^2/c^2/d/(c^2*d*x^2+d)^(3/2)-2/3 
/d/c^4/(c^2*d*x^2+d)^(3/2)))+b*(d*(c^2*x^2+1))^(1/2)/d^3/c^4/(c^2*x^2+1)*a 
rcsinh(x*c)*x^2-b*(d*(c^2*x^2+1))^(1/2)/d^3/c^5/(c^2*x^2+1)^(1/2)*x+b*(d*( 
c^2*x^2+1))^(1/2)/d^3/c^6/(c^2*x^2+1)*arcsinh(x*c)+2*b*(d*(c^2*x^2+1))^(1/ 
2)/(c^2*x^2+1)^2/d^3/c^4*arcsinh(x*c)*x^2+1/6*b*(d*(c^2*x^2+1))^(1/2)/(c^2 
*x^2+1)^(3/2)/d^3/c^5*x+5/3*b*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^2/d^3/c^6* 
arcsinh(x*c)+11/6*I*b*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)/d^3/c^6*ln(x 
*c+(c^2*x^2+1)^(1/2)-I)-11/6*I*b*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)/d 
^3/c^6*ln(x*c+(c^2*x^2+1)^(1/2)+I)
 

Fricas [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.04 \[ \int \frac {x^5 (a+b \text {arcsinh}(c x))}{\left (d+c^2 d x^2\right )^{5/2}} \, dx=\frac {11 \, {\left (b c^{4} x^{4} + 2 \, b c^{2} x^{2} + b\right )} \sqrt {d} \arctan \left (\frac {2 \, \sqrt {c^{2} d x^{2} + d} \sqrt {c^{2} x^{2} + 1} c \sqrt {d} x}{c^{4} d x^{4} - d}\right ) + 4 \, {\left (3 \, b c^{4} x^{4} + 12 \, b c^{2} x^{2} + 8 \, b\right )} \sqrt {c^{2} d x^{2} + d} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right ) + 2 \, {\left (6 \, a c^{4} x^{4} + 24 \, a c^{2} x^{2} - {\left (6 \, b c^{3} x^{3} + 5 \, b c x\right )} \sqrt {c^{2} x^{2} + 1} + 16 \, a\right )} \sqrt {c^{2} d x^{2} + d}}{12 \, {\left (c^{10} d^{3} x^{4} + 2 \, c^{8} d^{3} x^{2} + c^{6} d^{3}\right )}} \] Input:

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

Output:

1/12*(11*(b*c^4*x^4 + 2*b*c^2*x^2 + b)*sqrt(d)*arctan(2*sqrt(c^2*d*x^2 + d 
)*sqrt(c^2*x^2 + 1)*c*sqrt(d)*x/(c^4*d*x^4 - d)) + 4*(3*b*c^4*x^4 + 12*b*c 
^2*x^2 + 8*b)*sqrt(c^2*d*x^2 + d)*log(c*x + sqrt(c^2*x^2 + 1)) + 2*(6*a*c^ 
4*x^4 + 24*a*c^2*x^2 - (6*b*c^3*x^3 + 5*b*c*x)*sqrt(c^2*x^2 + 1) + 16*a)*s 
qrt(c^2*d*x^2 + d))/(c^10*d^3*x^4 + 2*c^8*d^3*x^2 + c^6*d^3)
 

Sympy [F]

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

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

Output:

Integral(x**5*(a + b*asinh(c*x))/(d*(c**2*x**2 + 1))**(5/2), x)
 

Maxima [F]

\[ \int \frac {x^5 (a+b \text {arcsinh}(c x))}{\left (d+c^2 d x^2\right )^{5/2}} \, dx=\int { \frac {{\left (b \operatorname {arsinh}\left (c x\right ) + a\right )} x^{5}}{{\left (c^{2} d x^{2} + d\right )}^{\frac {5}{2}}} \,d x } \] Input:

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

Output:

1/3*b*((3*c^4*sqrt(d)*x^4 + 12*c^2*sqrt(d)*x^2 + 8*sqrt(d))*log(c*x + sqrt 
(c^2*x^2 + 1))/((c^8*d^3*x^2 + c^6*d^3)*sqrt(c^2*x^2 + 1)) + 3*integrate(1 
/3*(3*c^4*sqrt(d)*x^4 + 12*c^2*sqrt(d)*x^2 + 8*sqrt(d))/(c^11*d^3*x^6 + 2* 
c^9*d^3*x^4 + c^7*d^3*x^2 + (c^10*d^3*x^5 + 2*c^8*d^3*x^3 + c^6*d^3*x)*sqr 
t(c^2*x^2 + 1)), x) - 3*integrate(1/3*(3*c^4*sqrt(d)*x^4 + 12*c^2*sqrt(d)* 
x^2 + 8*sqrt(d))/((c^8*d^3*x^3 + c^6*d^3*x)*sqrt(c^2*x^2 + 1)), x)) + 1/3* 
a*(3*x^4/((c^2*d*x^2 + d)^(3/2)*c^2*d) + 12*x^2/((c^2*d*x^2 + d)^(3/2)*c^4 
*d) + 8/((c^2*d*x^2 + d)^(3/2)*c^6*d))
 

Giac [F(-2)]

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

integrate(x^5*(a+b*arcsinh(c*x))/(c^2*d*x^2+d)^(5/2),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. \[ \int \frac {x^5 (a+b \text {arcsinh}(c x))}{\left (d+c^2 d x^2\right )^{5/2}} \, dx=\int \frac {x^5\,\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}{{\left (d\,c^2\,x^2+d\right )}^{5/2}} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {x^5 (a+b \text {arcsinh}(c x))}{\left (d+c^2 d x^2\right )^{5/2}} \, dx=\frac {3 \sqrt {c^{2} x^{2}+1}\, a \,c^{4} x^{4}+12 \sqrt {c^{2} x^{2}+1}\, a \,c^{2} x^{2}+8 \sqrt {c^{2} x^{2}+1}\, a +3 \left (\int \frac {\mathit {asinh} \left (c x \right ) x^{5}}{\sqrt {c^{2} x^{2}+1}\, c^{4} x^{4}+2 \sqrt {c^{2} x^{2}+1}\, c^{2} x^{2}+\sqrt {c^{2} x^{2}+1}}d x \right ) b \,c^{10} x^{4}+6 \left (\int \frac {\mathit {asinh} \left (c x \right ) x^{5}}{\sqrt {c^{2} x^{2}+1}\, c^{4} x^{4}+2 \sqrt {c^{2} x^{2}+1}\, c^{2} x^{2}+\sqrt {c^{2} x^{2}+1}}d x \right ) b \,c^{8} x^{2}+3 \left (\int \frac {\mathit {asinh} \left (c x \right ) x^{5}}{\sqrt {c^{2} x^{2}+1}\, c^{4} x^{4}+2 \sqrt {c^{2} x^{2}+1}\, c^{2} x^{2}+\sqrt {c^{2} x^{2}+1}}d x \right ) b \,c^{6}}{3 \sqrt {d}\, c^{6} d^{2} \left (c^{4} x^{4}+2 c^{2} x^{2}+1\right )} \] Input:

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

Output:

(3*sqrt(c**2*x**2 + 1)*a*c**4*x**4 + 12*sqrt(c**2*x**2 + 1)*a*c**2*x**2 + 
8*sqrt(c**2*x**2 + 1)*a + 3*int((asinh(c*x)*x**5)/(sqrt(c**2*x**2 + 1)*c** 
4*x**4 + 2*sqrt(c**2*x**2 + 1)*c**2*x**2 + sqrt(c**2*x**2 + 1)),x)*b*c**10 
*x**4 + 6*int((asinh(c*x)*x**5)/(sqrt(c**2*x**2 + 1)*c**4*x**4 + 2*sqrt(c* 
*2*x**2 + 1)*c**2*x**2 + sqrt(c**2*x**2 + 1)),x)*b*c**8*x**2 + 3*int((asin 
h(c*x)*x**5)/(sqrt(c**2*x**2 + 1)*c**4*x**4 + 2*sqrt(c**2*x**2 + 1)*c**2*x 
**2 + sqrt(c**2*x**2 + 1)),x)*b*c**6)/(3*sqrt(d)*c**6*d**2*(c**4*x**4 + 2* 
c**2*x**2 + 1))