\(\int \frac {a+b \text {arcsinh}(c x)}{x^4 (d+c^2 d x^2)^{5/2}} \, dx\) [183]

Optimal result

Integrand size = 26, antiderivative size = 295 \[ \int \frac {a+b \text {arcsinh}(c x)}{x^4 \left (d+c^2 d x^2\right )^{5/2}} \, dx=\frac {b c^3}{6 d^2 \sqrt {1+c^2 x^2} \sqrt {d+c^2 d x^2}}-\frac {b c \sqrt {1+c^2 x^2}}{6 d^2 x^2 \sqrt {d+c^2 d x^2}}+\frac {a+b \text {arcsinh}(c x)}{3 d x^3 \left (d+c^2 d x^2\right )^{3/2}}+\frac {2 (a+b \text {arcsinh}(c x))}{d^2 x^3 \sqrt {d+c^2 d x^2}}-\frac {8 \sqrt {d+c^2 d x^2} (a+b \text {arcsinh}(c x))}{3 d^3 x^3}+\frac {16 c^2 \sqrt {d+c^2 d x^2} (a+b \text {arcsinh}(c x))}{3 d^3 x}-\frac {8 b c^3 \sqrt {1+c^2 x^2} \log (x)}{3 d^2 \sqrt {d+c^2 d x^2}}-\frac {4 b c^3 \sqrt {1+c^2 x^2} \log \left (1+c^2 x^2\right )}{3 d^2 \sqrt {d+c^2 d x^2}} \] Output:

1/6*b*c^3/d^2/(c^2*x^2+1)^(1/2)/(c^2*d*x^2+d)^(1/2)-1/6*b*c*(c^2*x^2+1)^(1 
/2)/d^2/x^2/(c^2*d*x^2+d)^(1/2)+1/3*(a+b*arcsinh(c*x))/d/x^3/(c^2*d*x^2+d) 
^(3/2)+2*(a+b*arcsinh(c*x))/d^2/x^3/(c^2*d*x^2+d)^(1/2)-8/3*(c^2*d*x^2+d)^ 
(1/2)*(a+b*arcsinh(c*x))/d^3/x^3+16/3*c^2*(c^2*d*x^2+d)^(1/2)*(a+b*arcsinh 
(c*x))/d^3/x-8/3*b*c^3*(c^2*x^2+1)^(1/2)*ln(x)/d^2/(c^2*d*x^2+d)^(1/2)-4/3 
*b*c^3*(c^2*x^2+1)^(1/2)*ln(c^2*x^2+1)/d^2/(c^2*d*x^2+d)^(1/2)
 

Mathematica [A] (verified)

Time = 0.33 (sec) , antiderivative size = 267, normalized size of antiderivative = 0.91 \[ \int \frac {a+b \text {arcsinh}(c x)}{x^4 \left (d+c^2 d x^2\right )^{5/2}} \, dx=\frac {\sqrt {d+c^2 d x^2} \left (-b c x-b c^3 x^3-2 a \sqrt {1+c^2 x^2}+12 a c^2 x^2 \sqrt {1+c^2 x^2}+48 a c^4 x^4 \sqrt {1+c^2 x^2}+32 a c^6 x^6 \sqrt {1+c^2 x^2}+2 b \sqrt {1+c^2 x^2} \left (-1+6 c^2 x^2+24 c^4 x^4+16 c^6 x^6\right ) \text {arcsinh}(c x)+8 b c^3 x^3 \left (1+c^2 x^2\right )^2 \log \left (1+\frac {1}{c^2 x^2}\right )-16 b c^3 x^3 \log \left (1+c^2 x^2\right )-32 b c^5 x^5 \log \left (1+c^2 x^2\right )-16 b c^7 x^7 \log \left (1+c^2 x^2\right )\right )}{6 d^3 x^3 \left (1+c^2 x^2\right )^{5/2}} \] Input:

Integrate[(a + b*ArcSinh[c*x])/(x^4*(d + c^2*d*x^2)^(5/2)),x]
 

Output:

(Sqrt[d + c^2*d*x^2]*(-(b*c*x) - b*c^3*x^3 - 2*a*Sqrt[1 + c^2*x^2] + 12*a* 
c^2*x^2*Sqrt[1 + c^2*x^2] + 48*a*c^4*x^4*Sqrt[1 + c^2*x^2] + 32*a*c^6*x^6* 
Sqrt[1 + c^2*x^2] + 2*b*Sqrt[1 + c^2*x^2]*(-1 + 6*c^2*x^2 + 24*c^4*x^4 + 1 
6*c^6*x^6)*ArcSinh[c*x] + 8*b*c^3*x^3*(1 + c^2*x^2)^2*Log[1 + 1/(c^2*x^2)] 
 - 16*b*c^3*x^3*Log[1 + c^2*x^2] - 32*b*c^5*x^5*Log[1 + c^2*x^2] - 16*b*c^ 
7*x^7*Log[1 + c^2*x^2]))/(6*d^3*x^3*(1 + c^2*x^2)^(5/2))
 

Rubi [A] (verified)

Time = 0.65 (sec) , antiderivative size = 213, normalized size of antiderivative = 0.72, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {6219, 27, 2331, 2123, 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.

\(\displaystyle \int \frac {a+b \text {arcsinh}(c x)}{x^4 \left (c^2 d x^2+d\right )^{5/2}} \, dx\)

\(\Big \downarrow \) 6219

\(\displaystyle -\frac {b c \sqrt {c^2 d x^2+d} \int -\frac {-16 c^6 x^6-24 c^4 x^4-6 c^2 x^2+1}{3 d^3 x^3 \left (c^2 x^2+1\right )^2}dx}{\sqrt {c^2 x^2+1}}+\frac {2 c^2 (a+b \text {arcsinh}(c x))}{d x \left (c^2 d x^2+d\right )^{3/2}}-\frac {a+b \text {arcsinh}(c x)}{3 d x^3 \left (c^2 d x^2+d\right )^{3/2}}+\frac {16 c^4 x (a+b \text {arcsinh}(c x))}{3 d^2 \sqrt {c^2 d x^2+d}}+\frac {8 c^4 x (a+b \text {arcsinh}(c x))}{3 d \left (c^2 d x^2+d\right )^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {b c \sqrt {c^2 d x^2+d} \int \frac {-16 c^6 x^6-24 c^4 x^4-6 c^2 x^2+1}{x^3 \left (c^2 x^2+1\right )^2}dx}{3 d^3 \sqrt {c^2 x^2+1}}+\frac {2 c^2 (a+b \text {arcsinh}(c x))}{d x \left (c^2 d x^2+d\right )^{3/2}}-\frac {a+b \text {arcsinh}(c x)}{3 d x^3 \left (c^2 d x^2+d\right )^{3/2}}+\frac {16 c^4 x (a+b \text {arcsinh}(c x))}{3 d^2 \sqrt {c^2 d x^2+d}}+\frac {8 c^4 x (a+b \text {arcsinh}(c x))}{3 d \left (c^2 d x^2+d\right )^{3/2}}\)

\(\Big \downarrow \) 2331

\(\displaystyle \frac {b c \sqrt {c^2 d x^2+d} \int \frac {-16 c^6 x^6-24 c^4 x^4-6 c^2 x^2+1}{x^4 \left (c^2 x^2+1\right )^2}dx^2}{6 d^3 \sqrt {c^2 x^2+1}}+\frac {2 c^2 (a+b \text {arcsinh}(c x))}{d x \left (c^2 d x^2+d\right )^{3/2}}-\frac {a+b \text {arcsinh}(c x)}{3 d x^3 \left (c^2 d x^2+d\right )^{3/2}}+\frac {16 c^4 x (a+b \text {arcsinh}(c x))}{3 d^2 \sqrt {c^2 d x^2+d}}+\frac {8 c^4 x (a+b \text {arcsinh}(c x))}{3 d \left (c^2 d x^2+d\right )^{3/2}}\)

\(\Big \downarrow \) 2123

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

\(\Big \downarrow \) 2009

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

Input:

Int[(a + b*ArcSinh[c*x])/(x^4*(d + c^2*d*x^2)^(5/2)),x]
 

Output:

-1/3*(a + b*ArcSinh[c*x])/(d*x^3*(d + c^2*d*x^2)^(3/2)) + (2*c^2*(a + b*Ar 
cSinh[c*x]))/(d*x*(d + c^2*d*x^2)^(3/2)) + (8*c^4*x*(a + b*ArcSinh[c*x]))/ 
(3*d*(d + c^2*d*x^2)^(3/2)) + (16*c^4*x*(a + b*ArcSinh[c*x]))/(3*d^2*Sqrt[ 
d + c^2*d*x^2]) + (b*c*Sqrt[d + c^2*d*x^2]*(-x^(-2) + c^2/(1 + c^2*x^2) - 
8*c^2*Log[x^2] - 8*c^2*Log[1 + c^2*x^2]))/(6*d^3*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 2009

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

rule 2123

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] 
:> Int[ExpandIntegrand[Px*(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c 
, d, m, n}, x] && PolyQ[Px, x] && (IntegersQ[m, n] || IGtQ[m, -2])
 

rule 2331

Int[(Pq_)*(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[1/2   S 
ubst[Int[x^((m - 1)/2)*SubstFor[x^2, Pq, x]*(a + b*x)^p, x], x, x^2], x] /; 
 FreeQ[{a, b, p}, x] && PolyQ[Pq, x^2] && IntegerQ[(m - 1)/2]
 

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 [A] (verified)

Time = 1.02 (sec) , antiderivative size = 372, normalized size of antiderivative = 1.26

method	result	size
default	\(a \left (-\frac {1}{3 d \,x^{3} \left (c^{2} d \,x^{2}+d \right )^{\frac {3}{2}}}-2 c^{2} \left (-\frac {1}{d x \left (c^{2} d \,x^{2}+d \right )^{\frac {3}{2}}}-4 c^{2} \left (\frac {x}{3 d \left (c^{2} d \,x^{2}+d \right )^{\frac {3}{2}}}+\frac {2 x}{3 d^{2} \sqrt {c^{2} d \,x^{2}+d}}\right )\right )\right )+\frac {b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \sqrt {c^{2} x^{2}+1}\, \left (32 \,\operatorname {arcsinh}\left (x c \right ) c^{7} x^{7}-16 \ln \left (\left (x c +\sqrt {c^{2} x^{2}+1}\right )^{4}-1\right ) x^{7} c^{7}+32 \sqrt {c^{2} x^{2}+1}\, \operatorname {arcsinh}\left (x c \right ) c^{6} x^{6}+64 \,\operatorname {arcsinh}\left (x c \right ) x^{5} c^{5}-32 \ln \left (\left (x c +\sqrt {c^{2} x^{2}+1}\right )^{4}-1\right ) x^{5} c^{5}+48 \,\operatorname {arcsinh}\left (x c \right ) \sqrt {c^{2} x^{2}+1}\, x^{4} c^{4}+32 \,\operatorname {arcsinh}\left (x c \right ) x^{3} c^{3}-16 \ln \left (\left (x c +\sqrt {c^{2} x^{2}+1}\right )^{4}-1\right ) x^{3} c^{3}+12 \sqrt {c^{2} x^{2}+1}\, \operatorname {arcsinh}\left (x c \right ) x^{2} c^{2}-x^{3} c^{3}-2 \,\operatorname {arcsinh}\left (x c \right ) \sqrt {c^{2} x^{2}+1}-x c \right )}{6 \left (c^{6} x^{6}+3 c^{4} x^{4}+3 c^{2} x^{2}+1\right ) x^{3} d^{3}}\)	\(372\)
parts	\(a \left (-\frac {1}{3 d \,x^{3} \left (c^{2} d \,x^{2}+d \right )^{\frac {3}{2}}}-2 c^{2} \left (-\frac {1}{d x \left (c^{2} d \,x^{2}+d \right )^{\frac {3}{2}}}-4 c^{2} \left (\frac {x}{3 d \left (c^{2} d \,x^{2}+d \right )^{\frac {3}{2}}}+\frac {2 x}{3 d^{2} \sqrt {c^{2} d \,x^{2}+d}}\right )\right )\right )+\frac {b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \sqrt {c^{2} x^{2}+1}\, \left (32 \,\operatorname {arcsinh}\left (x c \right ) c^{7} x^{7}-16 \ln \left (\left (x c +\sqrt {c^{2} x^{2}+1}\right )^{4}-1\right ) x^{7} c^{7}+32 \sqrt {c^{2} x^{2}+1}\, \operatorname {arcsinh}\left (x c \right ) c^{6} x^{6}+64 \,\operatorname {arcsinh}\left (x c \right ) x^{5} c^{5}-32 \ln \left (\left (x c +\sqrt {c^{2} x^{2}+1}\right )^{4}-1\right ) x^{5} c^{5}+48 \,\operatorname {arcsinh}\left (x c \right ) \sqrt {c^{2} x^{2}+1}\, x^{4} c^{4}+32 \,\operatorname {arcsinh}\left (x c \right ) x^{3} c^{3}-16 \ln \left (\left (x c +\sqrt {c^{2} x^{2}+1}\right )^{4}-1\right ) x^{3} c^{3}+12 \sqrt {c^{2} x^{2}+1}\, \operatorname {arcsinh}\left (x c \right ) x^{2} c^{2}-x^{3} c^{3}-2 \,\operatorname {arcsinh}\left (x c \right ) \sqrt {c^{2} x^{2}+1}-x c \right )}{6 \left (c^{6} x^{6}+3 c^{4} x^{4}+3 c^{2} x^{2}+1\right ) x^{3} d^{3}}\)	\(372\)

Input:

int((a+b*arcsinh(x*c))/x^4/(c^2*d*x^2+d)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

a*(-1/3/d/x^3/(c^2*d*x^2+d)^(3/2)-2*c^2*(-1/d/x/(c^2*d*x^2+d)^(3/2)-4*c^2* 
(1/3*x/d/(c^2*d*x^2+d)^(3/2)+2/3/d^2*x/(c^2*d*x^2+d)^(1/2))))+1/6*b*(d*(c^ 
2*x^2+1))^(1/2)*(c^2*x^2+1)^(1/2)*(32*arcsinh(x*c)*c^7*x^7-16*ln((x*c+(c^2 
*x^2+1)^(1/2))^4-1)*x^7*c^7+32*(c^2*x^2+1)^(1/2)*arcsinh(x*c)*c^6*x^6+64*a 
rcsinh(x*c)*x^5*c^5-32*ln((x*c+(c^2*x^2+1)^(1/2))^4-1)*x^5*c^5+48*arcsinh( 
x*c)*(c^2*x^2+1)^(1/2)*x^4*c^4+32*arcsinh(x*c)*x^3*c^3-16*ln((x*c+(c^2*x^2 
+1)^(1/2))^4-1)*x^3*c^3+12*(c^2*x^2+1)^(1/2)*arcsinh(x*c)*x^2*c^2-x^3*c^3- 
2*arcsinh(x*c)*(c^2*x^2+1)^(1/2)-x*c)/(c^6*x^6+3*c^4*x^4+3*c^2*x^2+1)/x^3/ 
d^3
 

Fricas [F]

\[ \int \frac {a+b \text {arcsinh}(c x)}{x^4 \left (d+c^2 d x^2\right )^{5/2}} \, dx=\int { \frac {b \operatorname {arsinh}\left (c x\right ) + a}{{\left (c^{2} d x^{2} + d\right )}^{\frac {5}{2}} x^{4}} \,d x } \] Input:

integrate((a+b*arcsinh(c*x))/x^4/(c^2*d*x^2+d)^(5/2),x, algorithm="fricas" 
)
 

Output:

integral(sqrt(c^2*d*x^2 + d)*(b*arcsinh(c*x) + a)/(c^6*d^3*x^10 + 3*c^4*d^ 
3*x^8 + 3*c^2*d^3*x^6 + d^3*x^4), x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {a+b \text {arcsinh}(c x)}{x^4 \left (d+c^2 d x^2\right )^{5/2}} \, dx=\text {Timed out} \] Input:

integrate((a+b*asinh(c*x))/x**4/(c**2*d*x**2+d)**(5/2),x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 236, normalized size of antiderivative = 0.80 \[ \int \frac {a+b \text {arcsinh}(c x)}{x^4 \left (d+c^2 d x^2\right )^{5/2}} \, dx=-\frac {1}{6} \, b c {\left (\frac {8 \, c^{2} \log \left (c^{2} x^{2} + 1\right )}{d^{\frac {5}{2}}} + \frac {16 \, c^{2} \log \left (x\right )}{d^{\frac {5}{2}}} + \frac {1}{c^{2} d^{\frac {5}{2}} x^{4} + d^{\frac {5}{2}} x^{2}}\right )} + \frac {1}{3} \, {\left (\frac {16 \, c^{4} x}{\sqrt {c^{2} d x^{2} + d} d^{2}} + \frac {8 \, c^{4} x}{{\left (c^{2} d x^{2} + d\right )}^{\frac {3}{2}} d} + \frac {6 \, c^{2}}{{\left (c^{2} d x^{2} + d\right )}^{\frac {3}{2}} d x} - \frac {1}{{\left (c^{2} d x^{2} + d\right )}^{\frac {3}{2}} d x^{3}}\right )} b \operatorname {arsinh}\left (c x\right ) + \frac {1}{3} \, {\left (\frac {16 \, c^{4} x}{\sqrt {c^{2} d x^{2} + d} d^{2}} + \frac {8 \, c^{4} x}{{\left (c^{2} d x^{2} + d\right )}^{\frac {3}{2}} d} + \frac {6 \, c^{2}}{{\left (c^{2} d x^{2} + d\right )}^{\frac {3}{2}} d x} - \frac {1}{{\left (c^{2} d x^{2} + d\right )}^{\frac {3}{2}} d x^{3}}\right )} a \] Input:

integrate((a+b*arcsinh(c*x))/x^4/(c^2*d*x^2+d)^(5/2),x, algorithm="maxima" 
)
 

Output:

-1/6*b*c*(8*c^2*log(c^2*x^2 + 1)/d^(5/2) + 16*c^2*log(x)/d^(5/2) + 1/(c^2* 
d^(5/2)*x^4 + d^(5/2)*x^2)) + 1/3*(16*c^4*x/(sqrt(c^2*d*x^2 + d)*d^2) + 8* 
c^4*x/((c^2*d*x^2 + d)^(3/2)*d) + 6*c^2/((c^2*d*x^2 + d)^(3/2)*d*x) - 1/(( 
c^2*d*x^2 + d)^(3/2)*d*x^3))*b*arcsinh(c*x) + 1/3*(16*c^4*x/(sqrt(c^2*d*x^ 
2 + d)*d^2) + 8*c^4*x/((c^2*d*x^2 + d)^(3/2)*d) + 6*c^2/((c^2*d*x^2 + d)^( 
3/2)*d*x) - 1/((c^2*d*x^2 + d)^(3/2)*d*x^3))*a
 

Giac [F]

\[ \int \frac {a+b \text {arcsinh}(c x)}{x^4 \left (d+c^2 d x^2\right )^{5/2}} \, dx=\int { \frac {b \operatorname {arsinh}\left (c x\right ) + a}{{\left (c^{2} d x^{2} + d\right )}^{\frac {5}{2}} x^{4}} \,d x } \] Input:

integrate((a+b*arcsinh(c*x))/x^4/(c^2*d*x^2+d)^(5/2),x, algorithm="giac")
 

Output:

integrate((b*arcsinh(c*x) + a)/((c^2*d*x^2 + d)^(5/2)*x^4), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {a+b \text {arcsinh}(c x)}{x^4 \left (d+c^2 d x^2\right )^{5/2}} \, dx=\int \frac {a+b\,\mathrm {asinh}\left (c\,x\right )}{x^4\,{\left (d\,c^2\,x^2+d\right )}^{5/2}} \,d x \] Input:

int((a + b*asinh(c*x))/(x^4*(d + c^2*d*x^2)^(5/2)),x)
 

Output:

int((a + b*asinh(c*x))/(x^4*(d + c^2*d*x^2)^(5/2)), x)
 

Reduce [F]

\[ \int \frac {a+b \text {arcsinh}(c x)}{x^4 \left (d+c^2 d x^2\right )^{5/2}} \, dx=\frac {16 \sqrt {c^{2} x^{2}+1}\, a \,c^{6} x^{6}+24 \sqrt {c^{2} x^{2}+1}\, a \,c^{4} x^{4}+6 \sqrt {c^{2} x^{2}+1}\, a \,c^{2} x^{2}-\sqrt {c^{2} x^{2}+1}\, a +3 \left (\int \frac {\mathit {asinh} \left (c x \right )}{\sqrt {c^{2} x^{2}+1}\, c^{4} x^{8}+2 \sqrt {c^{2} x^{2}+1}\, c^{2} x^{6}+\sqrt {c^{2} x^{2}+1}\, x^{4}}d x \right ) b \,c^{4} x^{7}+6 \left (\int \frac {\mathit {asinh} \left (c x \right )}{\sqrt {c^{2} x^{2}+1}\, c^{4} x^{8}+2 \sqrt {c^{2} x^{2}+1}\, c^{2} x^{6}+\sqrt {c^{2} x^{2}+1}\, x^{4}}d x \right ) b \,c^{2} x^{5}+3 \left (\int \frac {\mathit {asinh} \left (c x \right )}{\sqrt {c^{2} x^{2}+1}\, c^{4} x^{8}+2 \sqrt {c^{2} x^{2}+1}\, c^{2} x^{6}+\sqrt {c^{2} x^{2}+1}\, x^{4}}d x \right ) b \,x^{3}-16 a \,c^{7} x^{7}-32 a \,c^{5} x^{5}-16 a \,c^{3} x^{3}}{3 \sqrt {d}\, d^{2} x^{3} \left (c^{4} x^{4}+2 c^{2} x^{2}+1\right )} \] Input:

int((a+b*asinh(c*x))/x^4/(c^2*d*x^2+d)^(5/2),x)
 

Output:

(16*sqrt(c**2*x**2 + 1)*a*c**6*x**6 + 24*sqrt(c**2*x**2 + 1)*a*c**4*x**4 + 
 6*sqrt(c**2*x**2 + 1)*a*c**2*x**2 - sqrt(c**2*x**2 + 1)*a + 3*int(asinh(c 
*x)/(sqrt(c**2*x**2 + 1)*c**4*x**8 + 2*sqrt(c**2*x**2 + 1)*c**2*x**6 + sqr 
t(c**2*x**2 + 1)*x**4),x)*b*c**4*x**7 + 6*int(asinh(c*x)/(sqrt(c**2*x**2 + 
 1)*c**4*x**8 + 2*sqrt(c**2*x**2 + 1)*c**2*x**6 + sqrt(c**2*x**2 + 1)*x**4 
),x)*b*c**2*x**5 + 3*int(asinh(c*x)/(sqrt(c**2*x**2 + 1)*c**4*x**8 + 2*sqr 
t(c**2*x**2 + 1)*c**2*x**6 + sqrt(c**2*x**2 + 1)*x**4),x)*b*x**3 - 16*a*c* 
*7*x**7 - 32*a*c**5*x**5 - 16*a*c**3*x**3)/(3*sqrt(d)*d**2*x**3*(c**4*x**4 
 + 2*c**2*x**2 + 1))