\(\int \frac {x^3 \sqrt {1+c^2 x^2}}{(a+b \text {arcsinh}(c x))^2} \, dx\) [396]

Optimal result

Integrand size = 27, antiderivative size = 213 \[ \int \frac {x^3 \sqrt {1+c^2 x^2}}{(a+b \text {arcsinh}(c x))^2} \, dx=-\frac {x^3 \left (1+c^2 x^2\right )}{b c (a+b \text {arcsinh}(c x))}-\frac {\cosh \left (\frac {a}{b}\right ) \text {Chi}\left (\frac {a+b \text {arcsinh}(c x)}{b}\right )}{8 b^2 c^4}-\frac {3 \cosh \left (\frac {3 a}{b}\right ) \text {Chi}\left (\frac {3 (a+b \text {arcsinh}(c x))}{b}\right )}{16 b^2 c^4}+\frac {5 \cosh \left (\frac {5 a}{b}\right ) \text {Chi}\left (\frac {5 (a+b \text {arcsinh}(c x))}{b}\right )}{16 b^2 c^4}+\frac {\sinh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a+b \text {arcsinh}(c x)}{b}\right )}{8 b^2 c^4}+\frac {3 \sinh \left (\frac {3 a}{b}\right ) \text {Shi}\left (\frac {3 (a+b \text {arcsinh}(c x))}{b}\right )}{16 b^2 c^4}-\frac {5 \sinh \left (\frac {5 a}{b}\right ) \text {Shi}\left (\frac {5 (a+b \text {arcsinh}(c x))}{b}\right )}{16 b^2 c^4} \] Output:

-x^3*(c^2*x^2+1)/b/c/(a+b*arcsinh(c*x))-1/8*cosh(a/b)*Chi((a+b*arcsinh(c*x 
))/b)/b^2/c^4-3/16*cosh(3*a/b)*Chi(3*(a+b*arcsinh(c*x))/b)/b^2/c^4+5/16*co 
sh(5*a/b)*Chi(5*(a+b*arcsinh(c*x))/b)/b^2/c^4+1/8*sinh(a/b)*Shi((a+b*arcsi 
nh(c*x))/b)/b^2/c^4+3/16*sinh(3*a/b)*Shi(3*(a+b*arcsinh(c*x))/b)/b^2/c^4-5 
/16*sinh(5*a/b)*Shi(5*(a+b*arcsinh(c*x))/b)/b^2/c^4
 

Mathematica [A] (verified)

Time = 0.47 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.82 \[ \int \frac {x^3 \sqrt {1+c^2 x^2}}{(a+b \text {arcsinh}(c x))^2} \, dx=-\frac {\frac {16 b c^3 x^3}{a+b \text {arcsinh}(c x)}+\frac {16 b c^5 x^5}{a+b \text {arcsinh}(c x)}+2 \cosh \left (\frac {a}{b}\right ) \text {Chi}\left (\frac {a}{b}+\text {arcsinh}(c x)\right )+3 \cosh \left (\frac {3 a}{b}\right ) \text {Chi}\left (3 \left (\frac {a}{b}+\text {arcsinh}(c x)\right )\right )-5 \cosh \left (\frac {5 a}{b}\right ) \text {Chi}\left (5 \left (\frac {a}{b}+\text {arcsinh}(c x)\right )\right )-2 \sinh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a}{b}+\text {arcsinh}(c x)\right )-3 \sinh \left (\frac {3 a}{b}\right ) \text {Shi}\left (3 \left (\frac {a}{b}+\text {arcsinh}(c x)\right )\right )+5 \sinh \left (\frac {5 a}{b}\right ) \text {Shi}\left (5 \left (\frac {a}{b}+\text {arcsinh}(c x)\right )\right )}{16 b^2 c^4} \] Input:

Integrate[(x^3*Sqrt[1 + c^2*x^2])/(a + b*ArcSinh[c*x])^2,x]
 

Output:

-1/16*((16*b*c^3*x^3)/(a + b*ArcSinh[c*x]) + (16*b*c^5*x^5)/(a + b*ArcSinh 
[c*x]) + 2*Cosh[a/b]*CoshIntegral[a/b + ArcSinh[c*x]] + 3*Cosh[(3*a)/b]*Co 
shIntegral[3*(a/b + ArcSinh[c*x])] - 5*Cosh[(5*a)/b]*CoshIntegral[5*(a/b + 
 ArcSinh[c*x])] - 2*Sinh[a/b]*SinhIntegral[a/b + ArcSinh[c*x]] - 3*Sinh[(3 
*a)/b]*SinhIntegral[3*(a/b + ArcSinh[c*x])] + 5*Sinh[(5*a)/b]*SinhIntegral 
[5*(a/b + ArcSinh[c*x])])/(b^2*c^4)
 

Rubi [A] (verified)

Time = 1.52 (sec) , antiderivative size = 291, normalized size of antiderivative = 1.37, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {6229, 6195, 5971, 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.

\(\displaystyle \int \frac {x^3 \sqrt {c^2 x^2+1}}{(a+b \text {arcsinh}(c x))^2} \, dx\)

\(\Big \downarrow \) 6229

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

\(\Big \downarrow \) 6195

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

\(\Big \downarrow \) 5971

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {3 \left (-\frac {1}{4} \cosh \left (\frac {a}{b}\right ) \text {Chi}\left (\frac {a+b \text {arcsinh}(c x)}{b}\right )+\frac {1}{4} \cosh \left (\frac {3 a}{b}\right ) \text {Chi}\left (\frac {3 (a+b \text {arcsinh}(c x))}{b}\right )+\frac {1}{4} \sinh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a+b \text {arcsinh}(c x)}{b}\right )-\frac {1}{4} \sinh \left (\frac {3 a}{b}\right ) \text {Shi}\left (\frac {3 (a+b \text {arcsinh}(c x))}{b}\right )\right )}{b^2 c^4}+\frac {5 \left (\frac {1}{8} \cosh \left (\frac {a}{b}\right ) \text {Chi}\left (\frac {a+b \text {arcsinh}(c x)}{b}\right )-\frac {3}{16} \cosh \left (\frac {3 a}{b}\right ) \text {Chi}\left (\frac {3 (a+b \text {arcsinh}(c x))}{b}\right )+\frac {1}{16} \cosh \left (\frac {5 a}{b}\right ) \text {Chi}\left (\frac {5 (a+b \text {arcsinh}(c x))}{b}\right )-\frac {1}{8} \sinh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a+b \text {arcsinh}(c x)}{b}\right )+\frac {3}{16} \sinh \left (\frac {3 a}{b}\right ) \text {Shi}\left (\frac {3 (a+b \text {arcsinh}(c x))}{b}\right )-\frac {1}{16} \sinh \left (\frac {5 a}{b}\right ) \text {Shi}\left (\frac {5 (a+b \text {arcsinh}(c x))}{b}\right )\right )}{b^2 c^4}-\frac {x^3 \left (c^2 x^2+1\right )}{b c (a+b \text {arcsinh}(c x))}\)

Input:

Int[(x^3*Sqrt[1 + c^2*x^2])/(a + b*ArcSinh[c*x])^2,x]
 

Output:

-((x^3*(1 + c^2*x^2))/(b*c*(a + b*ArcSinh[c*x]))) + (3*(-1/4*(Cosh[a/b]*Co 
shIntegral[(a + b*ArcSinh[c*x])/b]) + (Cosh[(3*a)/b]*CoshIntegral[(3*(a + 
b*ArcSinh[c*x]))/b])/4 + (Sinh[a/b]*SinhIntegral[(a + b*ArcSinh[c*x])/b])/ 
4 - (Sinh[(3*a)/b]*SinhIntegral[(3*(a + b*ArcSinh[c*x]))/b])/4))/(b^2*c^4) 
 + (5*((Cosh[a/b]*CoshIntegral[(a + b*ArcSinh[c*x])/b])/8 - (3*Cosh[(3*a)/ 
b]*CoshIntegral[(3*(a + b*ArcSinh[c*x]))/b])/16 + (Cosh[(5*a)/b]*CoshInteg 
ral[(5*(a + b*ArcSinh[c*x]))/b])/16 - (Sinh[a/b]*SinhIntegral[(a + b*ArcSi 
nh[c*x])/b])/8 + (3*Sinh[(3*a)/b]*SinhIntegral[(3*(a + b*ArcSinh[c*x]))/b] 
)/16 - (Sinh[(5*a)/b]*SinhIntegral[(5*(a + b*ArcSinh[c*x]))/b])/16))/(b^2* 
c^4)
 

Defintions of rubi rules used

rule 2009

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

rule 5971

Int[Cosh[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sinh[(a_.) + 
(b_.)*(x_)]^(n_.), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sinh[a + 
b*x]^n*Cosh[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] & 
& IGtQ[p, 0]
 

rule 6195

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

rule 6229

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_)*((f_.)*(x_))^(m_.)*((d_) + (e_ 
.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(f*x)^m*Sqrt[1 + c^2*x^2]*(d + e*x^2)^p 
*((a + b*ArcSinh[c*x])^(n + 1)/(b*c*(n + 1))), x] + (-Simp[f*(m/(b*c*(n + 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] - Simp[c*((m + 2*p + 1)/(b*f* 
(n + 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}, x] && EqQ[e, c^2*d] && LtQ[n, -1] && IGtQ[2*p, 0] && NeQ[m + 2*p + 1 
, 0] && IGtQ[m, -3]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(632\) vs. \(2(201)=402\).

Time = 2.30 (sec) , antiderivative size = 633, normalized size of antiderivative = 2.97

method	result	size
default	\(-\frac {16 x^{5} c^{5}-16 x^{4} c^{4} \sqrt {c^{2} x^{2}+1}+20 x^{3} c^{3}-12 x^{2} c^{2} \sqrt {c^{2} x^{2}+1}+5 x c -\sqrt {c^{2} x^{2}+1}}{32 c^{4} b \left (a +b \,\operatorname {arcsinh}\left (x c \right )\right )}-\frac {5 \,{\mathrm e}^{\frac {5 a}{b}} \operatorname {expIntegral}_{1}\left (5 \,\operatorname {arcsinh}\left (x c \right )+\frac {5 a}{b}\right )}{32 c^{4} b^{2}}+\frac {-4 x^{2} c^{2} \sqrt {c^{2} x^{2}+1}+4 x^{3} c^{3}-\sqrt {c^{2} x^{2}+1}+3 x c}{32 c^{4} b \left (a +b \,\operatorname {arcsinh}\left (x c \right )\right )}+\frac {3 \,{\mathrm e}^{\frac {3 a}{b}} \operatorname {expIntegral}_{1}\left (3 \,\operatorname {arcsinh}\left (x c \right )+\frac {3 a}{b}\right )}{32 c^{4} b^{2}}+\frac {x c -\sqrt {c^{2} x^{2}+1}}{16 c^{4} b \left (a +b \,\operatorname {arcsinh}\left (x c \right )\right )}+\frac {{\mathrm e}^{\frac {a}{b}} \operatorname {expIntegral}_{1}\left (\operatorname {arcsinh}\left (x c \right )+\frac {a}{b}\right )}{16 c^{4} b^{2}}+\frac {\operatorname {arcsinh}\left (x c \right ) {\mathrm e}^{-\frac {a}{b}} \operatorname {expIntegral}_{1}\left (-\operatorname {arcsinh}\left (x c \right )-\frac {a}{b}\right ) b +{\mathrm e}^{-\frac {a}{b}} \operatorname {expIntegral}_{1}\left (-\operatorname {arcsinh}\left (x c \right )-\frac {a}{b}\right ) a +b x c +\sqrt {c^{2} x^{2}+1}\, b}{16 c^{4} b^{2} \left (a +b \,\operatorname {arcsinh}\left (x c \right )\right )}+\frac {4 b \,c^{3} x^{3}+4 \sqrt {c^{2} x^{2}+1}\, b \,c^{2} x^{2}+3 \,\operatorname {arcsinh}\left (x c \right ) {\mathrm e}^{-\frac {3 a}{b}} \operatorname {expIntegral}_{1}\left (-3 \,\operatorname {arcsinh}\left (x c \right )-\frac {3 a}{b}\right ) b +3 \,{\mathrm e}^{-\frac {3 a}{b}} \operatorname {expIntegral}_{1}\left (-3 \,\operatorname {arcsinh}\left (x c \right )-\frac {3 a}{b}\right ) a +3 b x c +\sqrt {c^{2} x^{2}+1}\, b}{32 c^{4} b^{2} \left (a +b \,\operatorname {arcsinh}\left (x c \right )\right )}-\frac {16 b \,c^{5} x^{5}+16 \sqrt {c^{2} x^{2}+1}\, b \,c^{4} x^{4}+20 b \,c^{3} x^{3}+12 \sqrt {c^{2} x^{2}+1}\, b \,c^{2} x^{2}+5 \,\operatorname {arcsinh}\left (x c \right ) \operatorname {expIntegral}_{1}\left (-5 \,\operatorname {arcsinh}\left (x c \right )-\frac {5 a}{b}\right ) {\mathrm e}^{-\frac {5 a}{b}} b +5 \,\operatorname {expIntegral}_{1}\left (-5 \,\operatorname {arcsinh}\left (x c \right )-\frac {5 a}{b}\right ) {\mathrm e}^{-\frac {5 a}{b}} a +5 b x c +\sqrt {c^{2} x^{2}+1}\, b}{32 c^{4} b^{2} \left (a +b \,\operatorname {arcsinh}\left (x c \right )\right )}\)	\(633\)

Input:

int(x^3*(c^2*x^2+1)^(1/2)/(a+b*arcsinh(x*c))^2,x,method=_RETURNVERBOSE)
 

Output:

-1/32*(16*x^5*c^5-16*x^4*c^4*(c^2*x^2+1)^(1/2)+20*x^3*c^3-12*x^2*c^2*(c^2* 
x^2+1)^(1/2)+5*x*c-(c^2*x^2+1)^(1/2))/c^4/b/(a+b*arcsinh(x*c))-5/32/c^4/b^ 
2*exp(5*a/b)*Ei(1,5*arcsinh(x*c)+5*a/b)+1/32*(-4*x^2*c^2*(c^2*x^2+1)^(1/2) 
+4*x^3*c^3-(c^2*x^2+1)^(1/2)+3*x*c)/c^4/b/(a+b*arcsinh(x*c))+3/32/c^4/b^2* 
exp(3*a/b)*Ei(1,3*arcsinh(x*c)+3*a/b)+1/16*(x*c-(c^2*x^2+1)^(1/2))/c^4/b/( 
a+b*arcsinh(x*c))+1/16/c^4/b^2*exp(a/b)*Ei(1,arcsinh(x*c)+a/b)+1/16/c^4/b^ 
2*(arcsinh(x*c)*exp(-a/b)*Ei(1,-arcsinh(x*c)-a/b)*b+exp(-a/b)*Ei(1,-arcsin 
h(x*c)-a/b)*a+b*x*c+(c^2*x^2+1)^(1/2)*b)/(a+b*arcsinh(x*c))+1/32/c^4/b^2*( 
4*b*c^3*x^3+4*(c^2*x^2+1)^(1/2)*b*c^2*x^2+3*arcsinh(x*c)*exp(-3*a/b)*Ei(1, 
-3*arcsinh(x*c)-3*a/b)*b+3*exp(-3*a/b)*Ei(1,-3*arcsinh(x*c)-3*a/b)*a+3*b*x 
*c+(c^2*x^2+1)^(1/2)*b)/(a+b*arcsinh(x*c))-1/32/c^4/b^2*(16*b*c^5*x^5+16*( 
c^2*x^2+1)^(1/2)*b*c^4*x^4+20*b*c^3*x^3+12*(c^2*x^2+1)^(1/2)*b*c^2*x^2+5*a 
rcsinh(x*c)*Ei(1,-5*arcsinh(x*c)-5*a/b)*exp(-5*a/b)*b+5*Ei(1,-5*arcsinh(x* 
c)-5*a/b)*exp(-5*a/b)*a+5*b*x*c+(c^2*x^2+1)^(1/2)*b)/(a+b*arcsinh(x*c))
 

Fricas [F]

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

integrate(x^3*(c^2*x^2+1)^(1/2)/(a+b*arcsinh(c*x))^2,x, algorithm="fricas" 
)
 

Output:

integral(sqrt(c^2*x^2 + 1)*x^3/(b^2*arcsinh(c*x)^2 + 2*a*b*arcsinh(c*x) + 
a^2), x)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

integrate(x^3*(c^2*x^2+1)^(1/2)/(a+b*arcsinh(c*x))^2,x, algorithm="maxima" 
)
 

Output:

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

Giac [F(-2)]

Exception generated. \[ \int \frac {x^3 \sqrt {1+c^2 x^2}}{(a+b \text {arcsinh}(c x))^2} \, dx=\text {Exception raised: TypeError} \] Input:

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

int((x^3*(c^2*x^2 + 1)^(1/2))/(a + b*asinh(c*x))^2,x)
 

Output:

int((x^3*(c^2*x^2 + 1)^(1/2))/(a + b*asinh(c*x))^2, x)
 

Reduce [F]

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

int(x^3*(c^2*x^2+1)^(1/2)/(a+b*asinh(c*x))^2,x)
 

Output:

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