\(\int \frac {x^3}{\sqrt {1+c^2 x^2} (a+b \text {arcsinh}(c x))} \, dx\) [381]

Optimal result

Integrand size = 27, antiderivative size = 121 \[ \int \frac {x^3}{\sqrt {1+c^2 x^2} (a+b \text {arcsinh}(c x))} \, dx=\frac {3 \text {Chi}\left (\frac {a+b \text {arcsinh}(c x)}{b}\right ) \sinh \left (\frac {a}{b}\right )}{4 b c^4}-\frac {\text {Chi}\left (\frac {3 (a+b \text {arcsinh}(c x))}{b}\right ) \sinh \left (\frac {3 a}{b}\right )}{4 b c^4}-\frac {3 \cosh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a+b \text {arcsinh}(c x)}{b}\right )}{4 b c^4}+\frac {\cosh \left (\frac {3 a}{b}\right ) \text {Shi}\left (\frac {3 (a+b \text {arcsinh}(c x))}{b}\right )}{4 b c^4} \] Output:

3/4*Chi((a+b*arcsinh(c*x))/b)*sinh(a/b)/b/c^4-1/4*Chi(3*(a+b*arcsinh(c*x)) 
/b)*sinh(3*a/b)/b/c^4-3/4*cosh(a/b)*Shi((a+b*arcsinh(c*x))/b)/b/c^4+1/4*co 
sh(3*a/b)*Shi(3*(a+b*arcsinh(c*x))/b)/b/c^4
 

Mathematica [A] (verified)

Time = 0.29 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.76 \[ \int \frac {x^3}{\sqrt {1+c^2 x^2} (a+b \text {arcsinh}(c x))} \, dx=\frac {3 \text {Chi}\left (\frac {a}{b}+\text {arcsinh}(c x)\right ) \sinh \left (\frac {a}{b}\right )-\text {Chi}\left (3 \left (\frac {a}{b}+\text {arcsinh}(c x)\right )\right ) \sinh \left (\frac {3 a}{b}\right )-3 \cosh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a}{b}+\text {arcsinh}(c x)\right )+\cosh \left (\frac {3 a}{b}\right ) \text {Shi}\left (3 \left (\frac {a}{b}+\text {arcsinh}(c x)\right )\right )}{4 b c^4} \] Input:

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

Output:

(3*CoshIntegral[a/b + ArcSinh[c*x]]*Sinh[a/b] - CoshIntegral[3*(a/b + ArcS 
inh[c*x])]*Sinh[(3*a)/b] - 3*Cosh[a/b]*SinhIntegral[a/b + ArcSinh[c*x]] + 
Cosh[(3*a)/b]*SinhIntegral[3*(a/b + ArcSinh[c*x])])/(4*b*c^4)
 

Rubi [C] (verified)

Result contains complex when optimal does not.

Time = 0.77 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.95, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {6234, 25, 3042, 26, 3793, 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.

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

\(\Big \downarrow \) 6234

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\int \frac {i \sin \left (\frac {i a}{b}-\frac {i (a+b \text {arcsinh}(c x))}{b}\right )^3}{a+b \text {arcsinh}(c x)}d(a+b \text {arcsinh}(c x))}{b c^4}\)

\(\Big \downarrow \) 26

\(\displaystyle -\frac {i \int \frac {\sin \left (\frac {i a}{b}-\frac {i (a+b \text {arcsinh}(c x))}{b}\right )^3}{a+b \text {arcsinh}(c x)}d(a+b \text {arcsinh}(c x))}{b c^4}\)

\(\Big \downarrow \) 3793

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 25

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

rule 26

Int[(Complex[0, a_])*(Fx_), x_Symbol] :> Simp[(Complex[Identity[0], a])   I 
nt[Fx, x], x] /; FreeQ[a, x] && EqQ[a^2, 1]
 

rule 2009

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

rule 3042

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

rule 3793

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> In 
t[ExpandTrigReduce[(c + d*x)^m, Sin[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f 
, m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 1]))
 

rule 6234

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

Maple [A] (verified)

Time = 1.57 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.83

method	result	size
default	\(\frac {{\mathrm e}^{\frac {3 a}{b}} \operatorname {expIntegral}_{1}\left (3 \,\operatorname {arcsinh}\left (x c \right )+\frac {3 a}{b}\right )-3 \,{\mathrm e}^{\frac {a}{b}} \operatorname {expIntegral}_{1}\left (\operatorname {arcsinh}\left (x c \right )+\frac {a}{b}\right )+3 \,{\mathrm e}^{-\frac {a}{b}} \operatorname {expIntegral}_{1}\left (-\operatorname {arcsinh}\left (x c \right )-\frac {a}{b}\right )-{\mathrm e}^{-\frac {3 a}{b}} \operatorname {expIntegral}_{1}\left (-3 \,\operatorname {arcsinh}\left (x c \right )-\frac {3 a}{b}\right )}{8 c^{4} b}\)	\(101\)

Input:

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

Output:

1/8*(exp(3*a/b)*Ei(1,3*arcsinh(x*c)+3*a/b)-3*exp(a/b)*Ei(1,arcsinh(x*c)+a/ 
b)+3*exp(-a/b)*Ei(1,-arcsinh(x*c)-a/b)-exp(-3*a/b)*Ei(1,-3*arcsinh(x*c)-3* 
a/b))/c^4/b
 

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F(-2)]

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

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

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

Output:

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

Reduce [F]

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

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

Output:

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