\(\int \frac {\sqrt {1+c^2 x^2}}{a+b \text {arcsinh}(c x)} \, dx\) [353]

Optimal result

Integrand size = 24, antiderivative size = 82 \[ \int \frac {\sqrt {1+c^2 x^2}}{a+b \text {arcsinh}(c x)} \, dx=\frac {\cosh \left (\frac {2 a}{b}\right ) \text {Chi}\left (\frac {2 (a+b \text {arcsinh}(c x))}{b}\right )}{2 b c}+\frac {\log (a+b \text {arcsinh}(c x))}{2 b c}-\frac {\sinh \left (\frac {2 a}{b}\right ) \text {Shi}\left (\frac {2 (a+b \text {arcsinh}(c x))}{b}\right )}{2 b c} \] Output:

1/2*cosh(2*a/b)*Chi(2*(a+b*arcsinh(c*x))/b)/b/c+1/2*ln(a+b*arcsinh(c*x))/b 
/c-1/2*sinh(2*a/b)*Shi(2*(a+b*arcsinh(c*x))/b)/b/c
 

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.77 \[ \int \frac {\sqrt {1+c^2 x^2}}{a+b \text {arcsinh}(c x)} \, dx=\frac {\cosh \left (\frac {2 a}{b}\right ) \text {Chi}\left (2 \left (\frac {a}{b}+\text {arcsinh}(c x)\right )\right )+\log (a+b \text {arcsinh}(c x))-\sinh \left (\frac {2 a}{b}\right ) \text {Shi}\left (2 \left (\frac {a}{b}+\text {arcsinh}(c x)\right )\right )}{2 b c} \] Input:

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

Output:

(Cosh[(2*a)/b]*CoshIntegral[2*(a/b + ArcSinh[c*x])] + Log[a + b*ArcSinh[c* 
x]] - Sinh[(2*a)/b]*SinhIntegral[2*(a/b + ArcSinh[c*x])])/(2*b*c)
 

Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.87, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {6206, 3042, 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.

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

\(\Big \downarrow \) 6206

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3793

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

((Cosh[(2*a)/b]*CoshIntegral[(2*(a + b*ArcSinh[c*x]))/b])/2 + Log[a + b*Ar 
cSinh[c*x]]/2 - (Sinh[(2*a)/b]*SinhIntegral[(2*(a + b*ArcSinh[c*x]))/b])/2 
)/(b*c)
 

Defintions of rubi rules used

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 6206

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_)^2)^(p_.), 
x_Symbol] :> Simp[(1/(b*c))*Simp[(d + e*x^2)^p/(1 + c^2*x^2)^p]   Subst[Int 
[x^n*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, 0]
 

Maple [A] (verified)

Time = 1.71 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.82

method	result	size
default	\(-\frac {{\mathrm e}^{\frac {2 a}{b}} \operatorname {expIntegral}_{1}\left (2 \,\operatorname {arcsinh}\left (x c \right )+\frac {2 a}{b}\right )+{\mathrm e}^{-\frac {2 a}{b}} \operatorname {expIntegral}_{1}\left (-2 \,\operatorname {arcsinh}\left (x c \right )-\frac {2 a}{b}\right )-2 \ln \left (a +b \,\operatorname {arcsinh}\left (x c \right )\right )}{4 c b}\)	\(67\)

Input:

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

Output:

-1/4*(exp(2*a/b)*Ei(1,2*arcsinh(x*c)+2*a/b)+exp(-2*a/b)*Ei(1,-2*arcsinh(x* 
c)-2*a/b)-2*ln(a+b*arcsinh(x*c)))/c/b
 

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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