\(\int \frac {a+b \text {arcsinh}(c x)}{(d+c^2 d x^2)^3} \, dx\) [50]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [A] (verified)
Fricas [F]
Sympy [F]
Maxima [F]
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 21, antiderivative size = 178 \[ \int \frac {a+b \text {arcsinh}(c x)}{\left (d+c^2 d x^2\right )^3} \, dx=\frac {b}{12 c d^3 \left (1+c^2 x^2\right )^{3/2}}+\frac {3 b}{8 c d^3 \sqrt {1+c^2 x^2}}+\frac {x (a+b \text {arcsinh}(c x))}{4 d^3 \left (1+c^2 x^2\right )^2}+\frac {3 x (a+b \text {arcsinh}(c x))}{8 d^3 \left (1+c^2 x^2\right )}+\frac {3 (a+b \text {arcsinh}(c x)) \arctan \left (e^{\text {arcsinh}(c x)}\right )}{4 c d^3}-\frac {3 i b \operatorname {PolyLog}\left (2,-i e^{\text {arcsinh}(c x)}\right )}{8 c d^3}+\frac {3 i b \operatorname {PolyLog}\left (2,i e^{\text {arcsinh}(c x)}\right )}{8 c d^3} \] Output:

1/12*b/c/d^3/(c^2*x^2+1)^(3/2)+3/8*b/c/d^3/(c^2*x^2+1)^(1/2)+1/4*x*(a+b*ar 
csinh(c*x))/d^3/(c^2*x^2+1)^2+3/8*x*(a+b*arcsinh(c*x))/d^3/(c^2*x^2+1)+3/4 
*(a+b*arcsinh(c*x))*arctan(c*x+(c^2*x^2+1)^(1/2))/c/d^3-3/8*I*b*polylog(2, 
-I*(c*x+(c^2*x^2+1)^(1/2)))/c/d^3+3/8*I*b*polylog(2,I*(c*x+(c^2*x^2+1)^(1/ 
2)))/c/d^3
 

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 341, normalized size of antiderivative = 1.92 \[ \int \frac {a+b \text {arcsinh}(c x)}{\left (d+c^2 d x^2\right )^3} \, dx=\frac {15 a c x+9 a c^3 x^3+11 b \sqrt {1+c^2 x^2}+9 b c^2 x^2 \sqrt {1+c^2 x^2}+15 b c x \text {arcsinh}(c x)+9 b c^3 x^3 \text {arcsinh}(c x)+9 a \arctan (c x)+18 a c^2 x^2 \arctan (c x)+9 a c^4 x^4 \arctan (c x)+9 i b \text {arcsinh}(c x) \log \left (1-i e^{\text {arcsinh}(c x)}\right )+18 i b c^2 x^2 \text {arcsinh}(c x) \log \left (1-i e^{\text {arcsinh}(c x)}\right )+9 i b c^4 x^4 \text {arcsinh}(c x) \log \left (1-i e^{\text {arcsinh}(c x)}\right )-9 i b \text {arcsinh}(c x) \log \left (1+i e^{\text {arcsinh}(c x)}\right )-18 i b c^2 x^2 \text {arcsinh}(c x) \log \left (1+i e^{\text {arcsinh}(c x)}\right )-9 i b c^4 x^4 \text {arcsinh}(c x) \log \left (1+i e^{\text {arcsinh}(c x)}\right )-9 i b \left (1+c^2 x^2\right )^2 \operatorname {PolyLog}\left (2,-i e^{\text {arcsinh}(c x)}\right )+9 i b \left (1+c^2 x^2\right )^2 \operatorname {PolyLog}\left (2,i e^{\text {arcsinh}(c x)}\right )}{24 c d^3 \left (1+c^2 x^2\right )^2} \] Input:

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

Output:

(15*a*c*x + 9*a*c^3*x^3 + 11*b*Sqrt[1 + c^2*x^2] + 9*b*c^2*x^2*Sqrt[1 + c^ 
2*x^2] + 15*b*c*x*ArcSinh[c*x] + 9*b*c^3*x^3*ArcSinh[c*x] + 9*a*ArcTan[c*x 
] + 18*a*c^2*x^2*ArcTan[c*x] + 9*a*c^4*x^4*ArcTan[c*x] + (9*I)*b*ArcSinh[c 
*x]*Log[1 - I*E^ArcSinh[c*x]] + (18*I)*b*c^2*x^2*ArcSinh[c*x]*Log[1 - I*E^ 
ArcSinh[c*x]] + (9*I)*b*c^4*x^4*ArcSinh[c*x]*Log[1 - I*E^ArcSinh[c*x]] - ( 
9*I)*b*ArcSinh[c*x]*Log[1 + I*E^ArcSinh[c*x]] - (18*I)*b*c^2*x^2*ArcSinh[c 
*x]*Log[1 + I*E^ArcSinh[c*x]] - (9*I)*b*c^4*x^4*ArcSinh[c*x]*Log[1 + I*E^A 
rcSinh[c*x]] - (9*I)*b*(1 + c^2*x^2)^2*PolyLog[2, (-I)*E^ArcSinh[c*x]] + ( 
9*I)*b*(1 + c^2*x^2)^2*PolyLog[2, I*E^ArcSinh[c*x]])/(24*c*d^3*(1 + c^2*x^ 
2)^2)
 

Rubi [A] (verified)

Time = 0.72 (sec) , antiderivative size = 164, normalized size of antiderivative = 0.92, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.476, Rules used = {6203, 27, 241, 6203, 241, 6204, 3042, 4668, 2715, 2838}

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.

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

\(\Big \downarrow \) 6203

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 241

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

\(\Big \downarrow \) 6203

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

\(\Big \downarrow \) 241

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

\(\Big \downarrow \) 6204

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

\(\Big \downarrow \) 3042

\(\displaystyle \frac {3 \left (\frac {\int (a+b \text {arcsinh}(c x)) \csc \left (i \text {arcsinh}(c x)+\frac {\pi }{2}\right )d\text {arcsinh}(c x)}{2 c}+\frac {x (a+b \text {arcsinh}(c x))}{2 \left (c^2 x^2+1\right )}+\frac {b}{2 c \sqrt {c^2 x^2+1}}\right )}{4 d^3}+\frac {x (a+b \text {arcsinh}(c x))}{4 d^3 \left (c^2 x^2+1\right )^2}+\frac {b}{12 c d^3 \left (c^2 x^2+1\right )^{3/2}}\)

\(\Big \downarrow \) 4668

\(\displaystyle \frac {3 \left (\frac {-i b \int \log \left (1-i e^{\text {arcsinh}(c x)}\right )d\text {arcsinh}(c x)+i b \int \log \left (1+i e^{\text {arcsinh}(c x)}\right )d\text {arcsinh}(c x)+2 \arctan \left (e^{\text {arcsinh}(c x)}\right ) (a+b \text {arcsinh}(c x))}{2 c}+\frac {x (a+b \text {arcsinh}(c x))}{2 \left (c^2 x^2+1\right )}+\frac {b}{2 c \sqrt {c^2 x^2+1}}\right )}{4 d^3}+\frac {x (a+b \text {arcsinh}(c x))}{4 d^3 \left (c^2 x^2+1\right )^2}+\frac {b}{12 c d^3 \left (c^2 x^2+1\right )^{3/2}}\)

\(\Big \downarrow \) 2715

\(\displaystyle \frac {3 \left (\frac {-i b \int e^{-\text {arcsinh}(c x)} \log \left (1-i e^{\text {arcsinh}(c x)}\right )de^{\text {arcsinh}(c x)}+i b \int e^{-\text {arcsinh}(c x)} \log \left (1+i e^{\text {arcsinh}(c x)}\right )de^{\text {arcsinh}(c x)}+2 \arctan \left (e^{\text {arcsinh}(c x)}\right ) (a+b \text {arcsinh}(c x))}{2 c}+\frac {x (a+b \text {arcsinh}(c x))}{2 \left (c^2 x^2+1\right )}+\frac {b}{2 c \sqrt {c^2 x^2+1}}\right )}{4 d^3}+\frac {x (a+b \text {arcsinh}(c x))}{4 d^3 \left (c^2 x^2+1\right )^2}+\frac {b}{12 c d^3 \left (c^2 x^2+1\right )^{3/2}}\)

\(\Big \downarrow \) 2838

\(\displaystyle \frac {3 \left (\frac {2 \arctan \left (e^{\text {arcsinh}(c x)}\right ) (a+b \text {arcsinh}(c x))-i b \operatorname {PolyLog}\left (2,-i e^{\text {arcsinh}(c x)}\right )+i b \operatorname {PolyLog}\left (2,i e^{\text {arcsinh}(c x)}\right )}{2 c}+\frac {x (a+b \text {arcsinh}(c x))}{2 \left (c^2 x^2+1\right )}+\frac {b}{2 c \sqrt {c^2 x^2+1}}\right )}{4 d^3}+\frac {x (a+b \text {arcsinh}(c x))}{4 d^3 \left (c^2 x^2+1\right )^2}+\frac {b}{12 c d^3 \left (c^2 x^2+1\right )^{3/2}}\)

Input:

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

Output:

b/(12*c*d^3*(1 + c^2*x^2)^(3/2)) + (x*(a + b*ArcSinh[c*x]))/(4*d^3*(1 + c^ 
2*x^2)^2) + (3*(b/(2*c*Sqrt[1 + c^2*x^2]) + (x*(a + b*ArcSinh[c*x]))/(2*(1 
 + c^2*x^2)) + (2*(a + b*ArcSinh[c*x])*ArcTan[E^ArcSinh[c*x]] - I*b*PolyLo 
g[2, (-I)*E^ArcSinh[c*x]] + I*b*PolyLog[2, I*E^ArcSinh[c*x]])/(2*c)))/(4*d 
^3)
 

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 241
Int[(x_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(a + b*x^2)^(p + 1)/ 
(2*b*(p + 1)), x] /; FreeQ[{a, b, p}, x] && NeQ[p, -1]
 

rule 2715
Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] 
:> Simp[1/(d*e*n*Log[F])   Subst[Int[Log[a + b*x]/x, x], x, (F^(e*(c + d*x) 
))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]
 

rule 2838
Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2 
, (-c)*e*x^n]/n, x] /; FreeQ[{c, d, e, n}, x] && EqQ[c*d, 1]
 

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4668
Int[csc[(e_.) + Pi*(k_.) + (Complex[0, fz_])*(f_.)*(x_)]*((c_.) + (d_.)*(x_ 
))^(m_.), x_Symbol] :> Simp[-2*(c + d*x)^m*(ArcTanh[E^((-I)*e + f*fz*x)/E^( 
I*k*Pi)]/(f*fz*I)), x] + (-Simp[d*(m/(f*fz*I))   Int[(c + d*x)^(m - 1)*Log[ 
1 - E^((-I)*e + f*fz*x)/E^(I*k*Pi)], x], x] + Simp[d*(m/(f*fz*I))   Int[(c 
+ d*x)^(m - 1)*Log[1 + E^((-I)*e + f*fz*x)/E^(I*k*Pi)], x], x]) /; FreeQ[{c 
, d, e, f, fz}, x] && IntegerQ[2*k] && IGtQ[m, 0]
 

rule 6203
Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_)^2)^(p_), x 
_Symbol] :> Simp[(-x)*(d + e*x^2)^(p + 1)*((a + b*ArcSinh[c*x])^n/(2*d*(p + 
 1))), x] + (Simp[(2*p + 3)/(2*d*(p + 1))   Int[(d + e*x^2)^(p + 1)*(a + b* 
ArcSinh[c*x])^n, x], x] + Simp[b*c*(n/(2*(p + 1)))*Simp[(d + e*x^2)^p/(1 + 
c^2*x^2)^p]   Int[x*(1 + c^2*x^2)^(p + 1/2)*(a + b*ArcSinh[c*x])^(n - 1), x 
], x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && LtQ[p, 
 -1] && NeQ[p, -3/2]
 

rule 6204
Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)/((d_) + (e_.)*(x_)^2), x_Symb 
ol] :> Simp[1/(c*d)   Subst[Int[(a + b*x)^n*Sech[x], x], x, ArcSinh[c*x]], 
x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && IGtQ[n, 0]
 
Maple [A] (verified)

Time = 0.57 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.39

method result size
derivativedivides \(\frac {\frac {a \left (\frac {x c}{4 \left (c^{2} x^{2}+1\right )^{2}}+\frac {3 x c}{8 \left (c^{2} x^{2}+1\right )}+\frac {3 \arctan \left (x c \right )}{8}\right )}{d^{3}}+\frac {b \left (\frac {\operatorname {arcsinh}\left (x c \right ) x c}{4 \left (c^{2} x^{2}+1\right )^{2}}+\frac {3 \,\operatorname {arcsinh}\left (x c \right ) x c}{8 \left (c^{2} x^{2}+1\right )}+\frac {3 \,\operatorname {arcsinh}\left (x c \right ) \arctan \left (x c \right )}{8}+\frac {11}{24 \left (c^{2} x^{2}+1\right )^{\frac {3}{2}}}+\frac {3 c^{2} x^{2}}{8 \left (c^{2} x^{2}+1\right )^{\frac {3}{2}}}+\frac {3 \arctan \left (x c \right ) \ln \left (1+\frac {i \left (i x c +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )}{8}-\frac {3 \arctan \left (x c \right ) \ln \left (1-\frac {i \left (i x c +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )}{8}-\frac {3 i \operatorname {dilog}\left (1+\frac {i \left (i x c +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )}{8}+\frac {3 i \operatorname {dilog}\left (1-\frac {i \left (i x c +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )}{8}\right )}{d^{3}}}{c}\) \(248\)
default \(\frac {\frac {a \left (\frac {x c}{4 \left (c^{2} x^{2}+1\right )^{2}}+\frac {3 x c}{8 \left (c^{2} x^{2}+1\right )}+\frac {3 \arctan \left (x c \right )}{8}\right )}{d^{3}}+\frac {b \left (\frac {\operatorname {arcsinh}\left (x c \right ) x c}{4 \left (c^{2} x^{2}+1\right )^{2}}+\frac {3 \,\operatorname {arcsinh}\left (x c \right ) x c}{8 \left (c^{2} x^{2}+1\right )}+\frac {3 \,\operatorname {arcsinh}\left (x c \right ) \arctan \left (x c \right )}{8}+\frac {11}{24 \left (c^{2} x^{2}+1\right )^{\frac {3}{2}}}+\frac {3 c^{2} x^{2}}{8 \left (c^{2} x^{2}+1\right )^{\frac {3}{2}}}+\frac {3 \arctan \left (x c \right ) \ln \left (1+\frac {i \left (i x c +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )}{8}-\frac {3 \arctan \left (x c \right ) \ln \left (1-\frac {i \left (i x c +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )}{8}-\frac {3 i \operatorname {dilog}\left (1+\frac {i \left (i x c +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )}{8}+\frac {3 i \operatorname {dilog}\left (1-\frac {i \left (i x c +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )}{8}\right )}{d^{3}}}{c}\) \(248\)
parts \(\frac {a \left (\frac {x}{4 \left (c^{2} x^{2}+1\right )^{2}}+\frac {3 x}{8 \left (c^{2} x^{2}+1\right )}+\frac {3 \arctan \left (x c \right )}{8 c}\right )}{d^{3}}+\frac {b \left (\frac {\operatorname {arcsinh}\left (x c \right ) x c}{4 \left (c^{2} x^{2}+1\right )^{2}}+\frac {3 \,\operatorname {arcsinh}\left (x c \right ) x c}{8 \left (c^{2} x^{2}+1\right )}+\frac {3 \,\operatorname {arcsinh}\left (x c \right ) \arctan \left (x c \right )}{8}+\frac {11}{24 \left (c^{2} x^{2}+1\right )^{\frac {3}{2}}}+\frac {3 c^{2} x^{2}}{8 \left (c^{2} x^{2}+1\right )^{\frac {3}{2}}}+\frac {3 \arctan \left (x c \right ) \ln \left (1+\frac {i \left (i x c +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )}{8}-\frac {3 \arctan \left (x c \right ) \ln \left (1-\frac {i \left (i x c +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )}{8}-\frac {3 i \operatorname {dilog}\left (1+\frac {i \left (i x c +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )}{8}+\frac {3 i \operatorname {dilog}\left (1-\frac {i \left (i x c +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )}{8}\right )}{d^{3} c}\) \(248\)

Input:

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

Output:

1/c*(a/d^3*(1/4/(c^2*x^2+1)^2*x*c+3/8*x*c/(c^2*x^2+1)+3/8*arctan(x*c))+b/d 
^3*(1/4*arcsinh(x*c)/(c^2*x^2+1)^2*x*c+3/8*arcsinh(x*c)*x*c/(c^2*x^2+1)+3/ 
8*arcsinh(x*c)*arctan(x*c)+11/24/(c^2*x^2+1)^(3/2)+3/8*c^2*x^2/(c^2*x^2+1) 
^(3/2)+3/8*arctan(x*c)*ln(1+I*(1+I*x*c)/(c^2*x^2+1)^(1/2))-3/8*arctan(x*c) 
*ln(1-I*(1+I*x*c)/(c^2*x^2+1)^(1/2))-3/8*I*dilog(1+I*(1+I*x*c)/(c^2*x^2+1) 
^(1/2))+3/8*I*dilog(1-I*(1+I*x*c)/(c^2*x^2+1)^(1/2))))
 

Fricas [F]

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

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

Output:

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

Sympy [F]

\[ \int \frac {a+b \text {arcsinh}(c x)}{\left (d+c^2 d x^2\right )^3} \, dx=\frac {\int \frac {a}{c^{6} x^{6} + 3 c^{4} x^{4} + 3 c^{2} x^{2} + 1}\, dx + \int \frac {b \operatorname {asinh}{\left (c x \right )}}{c^{6} x^{6} + 3 c^{4} x^{4} + 3 c^{2} x^{2} + 1}\, dx}{d^{3}} \] Input:

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

Output:

(Integral(a/(c**6*x**6 + 3*c**4*x**4 + 3*c**2*x**2 + 1), x) + Integral(b*a 
sinh(c*x)/(c**6*x**6 + 3*c**4*x**4 + 3*c**2*x**2 + 1), x))/d**3
 

Maxima [F]

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

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

Output:

1/8*a*((3*c^2*x^3 + 5*x)/(c^4*d^3*x^4 + 2*c^2*d^3*x^2 + d^3) + 3*arctan(c* 
x)/(c*d^3)) + b*integrate(log(c*x + sqrt(c^2*x^2 + 1))/(c^6*d^3*x^6 + 3*c^ 
4*d^3*x^4 + 3*c^2*d^3*x^2 + d^3), x)
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(3*atan(c*x)*a*c**4*x**4 + 6*atan(c*x)*a*c**2*x**2 + 3*atan(c*x)*a + 8*int 
(asinh(c*x)/(c**6*x**6 + 3*c**4*x**4 + 3*c**2*x**2 + 1),x)*b*c**5*x**4 + 1 
6*int(asinh(c*x)/(c**6*x**6 + 3*c**4*x**4 + 3*c**2*x**2 + 1),x)*b*c**3*x** 
2 + 8*int(asinh(c*x)/(c**6*x**6 + 3*c**4*x**4 + 3*c**2*x**2 + 1),x)*b*c + 
3*a*c**3*x**3 + 5*a*c*x)/(8*c*d**3*(c**4*x**4 + 2*c**2*x**2 + 1))