\(\int \frac {x (1+c^2 x^2)^{3/2}}{a+b \text {arcsinh}(c x)} \, dx\) [359]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [A] (verified)
Fricas [F]
Sympy [F]
Maxima [F]
Giac [F(-2)]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 25, antiderivative size = 183 \[ \int \frac {x \left (1+c^2 x^2\right )^{3/2}}{a+b \text {arcsinh}(c x)} \, dx=-\frac {\text {Chi}\left (\frac {a+b \text {arcsinh}(c x)}{b}\right ) \sinh \left (\frac {a}{b}\right )}{8 b c^2}-\frac {3 \text {Chi}\left (\frac {3 (a+b \text {arcsinh}(c x))}{b}\right ) \sinh \left (\frac {3 a}{b}\right )}{16 b c^2}-\frac {\text {Chi}\left (\frac {5 (a+b \text {arcsinh}(c x))}{b}\right ) \sinh \left (\frac {5 a}{b}\right )}{16 b c^2}+\frac {\cosh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a+b \text {arcsinh}(c x)}{b}\right )}{8 b c^2}+\frac {3 \cosh \left (\frac {3 a}{b}\right ) \text {Shi}\left (\frac {3 (a+b \text {arcsinh}(c x))}{b}\right )}{16 b c^2}+\frac {\cosh \left (\frac {5 a}{b}\right ) \text {Shi}\left (\frac {5 (a+b \text {arcsinh}(c x))}{b}\right )}{16 b c^2} \] Output:

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

Mathematica [A] (verified)

Time = 0.36 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.74 \[ \int \frac {x \left (1+c^2 x^2\right )^{3/2}}{a+b \text {arcsinh}(c x)} \, dx=\frac {-2 \text {Chi}\left (\frac {a}{b}+\text {arcsinh}(c x)\right ) \sinh \left (\frac {a}{b}\right )-3 \text {Chi}\left (3 \left (\frac {a}{b}+\text {arcsinh}(c x)\right )\right ) \sinh \left (\frac {3 a}{b}\right )-\text {Chi}\left (5 \left (\frac {a}{b}+\text {arcsinh}(c x)\right )\right ) \sinh \left (\frac {5 a}{b}\right )+2 \cosh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a}{b}+\text {arcsinh}(c x)\right )+3 \cosh \left (\frac {3 a}{b}\right ) \text {Shi}\left (3 \left (\frac {a}{b}+\text {arcsinh}(c x)\right )\right )+\cosh \left (\frac {5 a}{b}\right ) \text {Shi}\left (5 \left (\frac {a}{b}+\text {arcsinh}(c x)\right )\right )}{16 b c^2} \] Input:

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

Output:

(-2*CoshIntegral[a/b + ArcSinh[c*x]]*Sinh[a/b] - 3*CoshIntegral[3*(a/b + A 
rcSinh[c*x])]*Sinh[(3*a)/b] - CoshIntegral[5*(a/b + ArcSinh[c*x])]*Sinh[(5 
*a)/b] + 2*Cosh[a/b]*SinhIntegral[a/b + ArcSinh[c*x]] + 3*Cosh[(3*a)/b]*Si 
nhIntegral[3*(a/b + ArcSinh[c*x])] + Cosh[(5*a)/b]*SinhIntegral[5*(a/b + A 
rcSinh[c*x])])/(16*b*c^2)
 

Rubi [A] (verified)

Time = 0.60 (sec) , antiderivative size = 154, normalized size of antiderivative = 0.84, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {6234, 25, 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.

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

\(\Big \downarrow \) 6234

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 5971

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

(-1/8*(CoshIntegral[(a + b*ArcSinh[c*x])/b]*Sinh[a/b]) - (3*CoshIntegral[( 
3*(a + b*ArcSinh[c*x]))/b]*Sinh[(3*a)/b])/16 - (CoshIntegral[(5*(a + b*Arc 
Sinh[c*x]))/b]*Sinh[(5*a)/b])/16 + (Cosh[a/b]*SinhIntegral[(a + b*ArcSinh[ 
c*x])/b])/8 + (3*Cosh[(3*a)/b]*SinhIntegral[(3*(a + b*ArcSinh[c*x]))/b])/1 
6 + (Cosh[(5*a)/b]*SinhIntegral[(5*(a + b*ArcSinh[c*x]))/b])/16)/(b*c^2)
 

Defintions of rubi rules used

rule 25
Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 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 = 1.76 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.81

method result size
default \(\frac {{\mathrm e}^{\frac {5 a}{b}} \operatorname {expIntegral}_{1}\left (5 \,\operatorname {arcsinh}\left (x c \right )+\frac {5 a}{b}\right )+3 \,{\mathrm e}^{\frac {3 a}{b}} \operatorname {expIntegral}_{1}\left (3 \,\operatorname {arcsinh}\left (x c \right )+\frac {3 a}{b}\right )+2 \,{\mathrm e}^{\frac {a}{b}} \operatorname {expIntegral}_{1}\left (\operatorname {arcsinh}\left (x c \right )+\frac {a}{b}\right )-2 \,{\mathrm e}^{-\frac {a}{b}} \operatorname {expIntegral}_{1}\left (-\operatorname {arcsinh}\left (x c \right )-\frac {a}{b}\right )-3 \,{\mathrm e}^{-\frac {3 a}{b}} \operatorname {expIntegral}_{1}\left (-3 \,\operatorname {arcsinh}\left (x c \right )-\frac {3 a}{b}\right )-{\mathrm e}^{-\frac {5 a}{b}} \operatorname {expIntegral}_{1}\left (-5 \,\operatorname {arcsinh}\left (x c \right )-\frac {5 a}{b}\right )}{32 c^{2} b}\) \(149\)

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F(-2)]

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

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

Output:

Exception raised: RuntimeError >> an error occurred running a Giac command 
:INPUT:sage2OUTPUT:sym2poly/r2sym(const gen & e,const index_m & i,const ve 
cteur & l) Error: Bad Argument Value
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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