\(\int \frac {x \sqrt {1+c^2 x^2}}{a+b \text {arcsinh}(c x)} \, dx\) [352]

Optimal result

Integrand size = 25, antiderivative size = 121 \[ \int \frac {x \sqrt {1+c^2 x^2}}{a+b \text {arcsinh}(c x)} \, dx=-\frac {\text {Chi}\left (\frac {a+b \text {arcsinh}(c x)}{b}\right ) \sinh \left (\frac {a}{b}\right )}{4 b c^2}-\frac {\text {Chi}\left (\frac {3 (a+b \text {arcsinh}(c x))}{b}\right ) \sinh \left (\frac {3 a}{b}\right )}{4 b c^2}+\frac {\cosh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a+b \text {arcsinh}(c x)}{b}\right )}{4 b c^2}+\frac {\cosh \left (\frac {3 a}{b}\right ) \text {Shi}\left (\frac {3 (a+b \text {arcsinh}(c x))}{b}\right )}{4 b c^2} \] Output:

-1/4*Chi((a+b*arcsinh(c*x))/b)*sinh(a/b)/b/c^2-1/4*Chi(3*(a+b*arcsinh(c*x) 
)/b)*sinh(3*a/b)/b/c^2+1/4*cosh(a/b)*Shi((a+b*arcsinh(c*x))/b)/b/c^2+1/4*c 
osh(3*a/b)*Shi(3*(a+b*arcsinh(c*x))/b)/b/c^2
 

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.75 \[ \int \frac {x \sqrt {1+c^2 x^2}}{a+b \text {arcsinh}(c x)} \, dx=\frac {-\text {Chi}\left (\frac {a}{b}+\text {arcsinh}(c x)\right ) \sinh \left (\frac {a}{b}\right )-\text {Chi}\left (3 \left (\frac {a}{b}+\text {arcsinh}(c x)\right )\right ) \sinh \left (\frac {3 a}{b}\right )+\cosh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a}{b}+\text {arcsinh}(c x)\right )+\cosh \left (\frac {3 a}{b}\right ) \text {Shi}\left (3 \left (\frac {a}{b}+\text {arcsinh}(c x)\right )\right )}{4 b c^2} \] Input:

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

Output:

(-(CoshIntegral[a/b + ArcSinh[c*x]]*Sinh[a/b]) - CoshIntegral[3*(a/b + Arc 
Sinh[c*x])]*Sinh[(3*a)/b] + Cosh[a/b]*SinhIntegral[a/b + ArcSinh[c*x]] + C 
osh[(3*a)/b]*SinhIntegral[3*(a/b + ArcSinh[c*x])])/(4*b*c^2)
 

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.86, 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.

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

\(\Big \downarrow \) 6234

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 5971

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

(-1/4*(CoshIntegral[(a + b*ArcSinh[c*x])/b]*Sinh[a/b]) - (CoshIntegral[(3* 
(a + b*ArcSinh[c*x]))/b]*Sinh[(3*a)/b])/4 + (Cosh[a/b]*SinhIntegral[(a + b 
*ArcSinh[c*x])/b])/4 + (Cosh[(3*a)/b]*SinhIntegral[(3*(a + b*ArcSinh[c*x]) 
)/b])/4)/(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 = 1.40 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.83

method	result	size
default	\(\frac {{\mathrm e}^{\frac {3 a}{b}} \operatorname {expIntegral}_{1}\left (3 \,\operatorname {arcsinh}\left (x c \right )+\frac {3 a}{b}\right )+{\mathrm e}^{\frac {a}{b}} \operatorname {expIntegral}_{1}\left (\operatorname {arcsinh}\left (x c \right )+\frac {a}{b}\right )-{\mathrm e}^{-\frac {a}{b}} \operatorname {expIntegral}_{1}\left (-\operatorname {arcsinh}\left (x c \right )-\frac {a}{b}\right )-{\mathrm e}^{-\frac {3 a}{b}} \operatorname {expIntegral}_{1}\left (-3 \,\operatorname {arcsinh}\left (x c \right )-\frac {3 a}{b}\right )}{8 c^{2} b}\)	\(100\)

Input:

int(x*(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)+exp(a/b)*Ei(1,arcsinh(x*c)+a/b) 
-exp(-a/b)*Ei(1,-arcsinh(x*c)-a/b)-exp(-3*a/b)*Ei(1,-3*arcsinh(x*c)-3*a/b) 
)/c^2/b
 

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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