\(\int \frac {(1+c^2 x^2)^{5/2}}{a+b \text {arcsinh}(c x)} \, dx\) [367]

Optimal result

Integrand size = 24, antiderivative size = 206 \[ \int \frac {\left (1+c^2 x^2\right )^{5/2}}{a+b \text {arcsinh}(c x)} \, dx=\frac {15 \cosh \left (\frac {2 a}{b}\right ) \text {Chi}\left (\frac {2 (a+b \text {arcsinh}(c x))}{b}\right )}{32 b c}+\frac {3 \cosh \left (\frac {4 a}{b}\right ) \text {Chi}\left (\frac {4 (a+b \text {arcsinh}(c x))}{b}\right )}{16 b c}+\frac {\cosh \left (\frac {6 a}{b}\right ) \text {Chi}\left (\frac {6 (a+b \text {arcsinh}(c x))}{b}\right )}{32 b c}+\frac {5 \log (a+b \text {arcsinh}(c x))}{16 b c}-\frac {15 \sinh \left (\frac {2 a}{b}\right ) \text {Shi}\left (\frac {2 (a+b \text {arcsinh}(c x))}{b}\right )}{32 b c}-\frac {3 \sinh \left (\frac {4 a}{b}\right ) \text {Shi}\left (\frac {4 (a+b \text {arcsinh}(c x))}{b}\right )}{16 b c}-\frac {\sinh \left (\frac {6 a}{b}\right ) \text {Shi}\left (\frac {6 (a+b \text {arcsinh}(c x))}{b}\right )}{32 b c} \] Output:

15/32*cosh(2*a/b)*Chi(2*(a+b*arcsinh(c*x))/b)/b/c+3/16*cosh(4*a/b)*Chi(4*( 
a+b*arcsinh(c*x))/b)/b/c+1/32*cosh(6*a/b)*Chi(6*(a+b*arcsinh(c*x))/b)/b/c+ 
5/16*ln(a+b*arcsinh(c*x))/b/c-15/32*sinh(2*a/b)*Shi(2*(a+b*arcsinh(c*x))/b 
)/b/c-3/16*sinh(4*a/b)*Shi(4*(a+b*arcsinh(c*x))/b)/b/c-1/32*sinh(6*a/b)*Sh 
i(6*(a+b*arcsinh(c*x))/b)/b/c
 

Mathematica [A] (verified)

Time = 0.50 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.74 \[ \int \frac {\left (1+c^2 x^2\right )^{5/2}}{a+b \text {arcsinh}(c x)} \, dx=\frac {15 \cosh \left (\frac {2 a}{b}\right ) \text {Chi}\left (2 \left (\frac {a}{b}+\text {arcsinh}(c x)\right )\right )+6 \cosh \left (\frac {4 a}{b}\right ) \text {Chi}\left (4 \left (\frac {a}{b}+\text {arcsinh}(c x)\right )\right )+\cosh \left (\frac {6 a}{b}\right ) \text {Chi}\left (6 \left (\frac {a}{b}+\text {arcsinh}(c x)\right )\right )+10 \log (a+b \text {arcsinh}(c x))-15 \sinh \left (\frac {2 a}{b}\right ) \text {Shi}\left (2 \left (\frac {a}{b}+\text {arcsinh}(c x)\right )\right )-6 \sinh \left (\frac {4 a}{b}\right ) \text {Shi}\left (4 \left (\frac {a}{b}+\text {arcsinh}(c x)\right )\right )-\sinh \left (\frac {6 a}{b}\right ) \text {Shi}\left (6 \left (\frac {a}{b}+\text {arcsinh}(c x)\right )\right )}{32 b c} \] Input:

Integrate[(1 + c^2*x^2)^(5/2)/(a + b*ArcSinh[c*x]),x]
 

Output:

(15*Cosh[(2*a)/b]*CoshIntegral[2*(a/b + ArcSinh[c*x])] + 6*Cosh[(4*a)/b]*C 
oshIntegral[4*(a/b + ArcSinh[c*x])] + Cosh[(6*a)/b]*CoshIntegral[6*(a/b + 
ArcSinh[c*x])] + 10*Log[a + b*ArcSinh[c*x]] - 15*Sinh[(2*a)/b]*SinhIntegra 
l[2*(a/b + ArcSinh[c*x])] - 6*Sinh[(4*a)/b]*SinhIntegral[4*(a/b + ArcSinh[ 
c*x])] - Sinh[(6*a)/b]*SinhIntegral[6*(a/b + ArcSinh[c*x])])/(32*b*c)
 

Rubi [A] (verified)

Time = 0.88 (sec) , antiderivative size = 171, normalized size of antiderivative = 0.83, 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.

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

\(\Big \downarrow \) 6206

\(\displaystyle \frac {\int \frac {\cosh ^6\left (\frac {a}{b}-\frac {a+b \text {arcsinh}(c x)}{b}\right )}{a+b \text {arcsinh}(c x)}d(a+b \text {arcsinh}(c x))}{b c}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {\sin \left (\frac {i a}{b}-\frac {i (a+b \text {arcsinh}(c x))}{b}+\frac {\pi }{2}\right )^6}{a+b \text {arcsinh}(c x)}d(a+b \text {arcsinh}(c x))}{b c}\)

\(\Big \downarrow \) 3793

\(\displaystyle \frac {\int \left (\frac {\cosh \left (\frac {6 a}{b}-\frac {6 (a+b \text {arcsinh}(c x))}{b}\right )}{32 (a+b \text {arcsinh}(c x))}+\frac {3 \cosh \left (\frac {4 a}{b}-\frac {4 (a+b \text {arcsinh}(c x))}{b}\right )}{16 (a+b \text {arcsinh}(c x))}+\frac {15 \cosh \left (\frac {2 a}{b}-\frac {2 (a+b \text {arcsinh}(c x))}{b}\right )}{32 (a+b \text {arcsinh}(c x))}+\frac {5}{16 (a+b \text {arcsinh}(c x))}\right )d(a+b \text {arcsinh}(c x))}{b c}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\frac {15}{32} \cosh \left (\frac {2 a}{b}\right ) \text {Chi}\left (\frac {2 (a+b \text {arcsinh}(c x))}{b}\right )+\frac {3}{16} \cosh \left (\frac {4 a}{b}\right ) \text {Chi}\left (\frac {4 (a+b \text {arcsinh}(c x))}{b}\right )+\frac {1}{32} \cosh \left (\frac {6 a}{b}\right ) \text {Chi}\left (\frac {6 (a+b \text {arcsinh}(c x))}{b}\right )-\frac {15}{32} \sinh \left (\frac {2 a}{b}\right ) \text {Shi}\left (\frac {2 (a+b \text {arcsinh}(c x))}{b}\right )-\frac {3}{16} \sinh \left (\frac {4 a}{b}\right ) \text {Shi}\left (\frac {4 (a+b \text {arcsinh}(c x))}{b}\right )-\frac {1}{32} \sinh \left (\frac {6 a}{b}\right ) \text {Shi}\left (\frac {6 (a+b \text {arcsinh}(c x))}{b}\right )+\frac {5}{16} \log (a+b \text {arcsinh}(c x))}{b c}\)

Input:

Int[(1 + c^2*x^2)^(5/2)/(a + b*ArcSinh[c*x]),x]
 

Output:

((15*Cosh[(2*a)/b]*CoshIntegral[(2*(a + b*ArcSinh[c*x]))/b])/32 + (3*Cosh[ 
(4*a)/b]*CoshIntegral[(4*(a + b*ArcSinh[c*x]))/b])/16 + (Cosh[(6*a)/b]*Cos 
hIntegral[(6*(a + b*ArcSinh[c*x]))/b])/32 + (5*Log[a + b*ArcSinh[c*x]])/16 
 - (15*Sinh[(2*a)/b]*SinhIntegral[(2*(a + b*ArcSinh[c*x]))/b])/32 - (3*Sin 
h[(4*a)/b]*SinhIntegral[(4*(a + b*ArcSinh[c*x]))/b])/16 - (Sinh[(6*a)/b]*S 
inhIntegral[(6*(a + b*ArcSinh[c*x]))/b])/32)/(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 = 3.93 (sec) , antiderivative size = 163, normalized size of antiderivative = 0.79

method	result	size
default	\(-\frac {{\mathrm e}^{\frac {6 a}{b}} \operatorname {expIntegral}_{1}\left (6 \,\operatorname {arcsinh}\left (x c \right )+\frac {6 a}{b}\right )+6 \,{\mathrm e}^{\frac {4 a}{b}} \operatorname {expIntegral}_{1}\left (4 \,\operatorname {arcsinh}\left (x c \right )+\frac {4 a}{b}\right )+15 \,{\mathrm e}^{\frac {2 a}{b}} \operatorname {expIntegral}_{1}\left (2 \,\operatorname {arcsinh}\left (x c \right )+\frac {2 a}{b}\right )+15 \,{\mathrm e}^{-\frac {2 a}{b}} \operatorname {expIntegral}_{1}\left (-2 \,\operatorname {arcsinh}\left (x c \right )-\frac {2 a}{b}\right )+6 \,{\mathrm e}^{-\frac {4 a}{b}} \operatorname {expIntegral}_{1}\left (-4 \,\operatorname {arcsinh}\left (x c \right )-\frac {4 a}{b}\right )+{\mathrm e}^{-\frac {6 a}{b}} \operatorname {expIntegral}_{1}\left (-6 \,\operatorname {arcsinh}\left (x c \right )-\frac {6 a}{b}\right )-20 \ln \left (a +b \,\operatorname {arcsinh}\left (x c \right )\right )}{64 c b}\)	\(163\)

Input:

int((c^2*x^2+1)^(5/2)/(a+b*arcsinh(x*c)),x,method=_RETURNVERBOSE)
 

Output:

-1/64*(exp(6*a/b)*Ei(1,6*arcsinh(x*c)+6*a/b)+6*exp(4*a/b)*Ei(1,4*arcsinh(x 
*c)+4*a/b)+15*exp(2*a/b)*Ei(1,2*arcsinh(x*c)+2*a/b)+15*exp(-2*a/b)*Ei(1,-2 
*arcsinh(x*c)-2*a/b)+6*exp(-4*a/b)*Ei(1,-4*arcsinh(x*c)-4*a/b)+exp(-6*a/b) 
*Ei(1,-6*arcsinh(x*c)-6*a/b)-20*ln(a+b*arcsinh(x*c)))/c/b
 

Fricas [F]

\[ \int \frac {\left (1+c^2 x^2\right )^{5/2}}{a+b \text {arcsinh}(c x)} \, dx=\int { \frac {{\left (c^{2} x^{2} + 1\right )}^{\frac {5}{2}}}{b \operatorname {arsinh}\left (c x\right ) + a} \,d x } \] Input:

integrate((c^2*x^2+1)^(5/2)/(a+b*arcsinh(c*x)),x, algorithm="fricas")
 

Output:

integral((c^4*x^4 + 2*c^2*x^2 + 1)*sqrt(c^2*x^2 + 1)/(b*arcsinh(c*x) + a), 
 x)
 

Sympy [F]

\[ \int \frac {\left (1+c^2 x^2\right )^{5/2}}{a+b \text {arcsinh}(c x)} \, dx=\int \frac {\left (c^{2} x^{2} + 1\right )^{\frac {5}{2}}}{a + b \operatorname {asinh}{\left (c x \right )}}\, dx \] Input:

integrate((c**2*x**2+1)**(5/2)/(a+b*asinh(c*x)),x)
 

Output:

Integral((c**2*x**2 + 1)**(5/2)/(a + b*asinh(c*x)), x)
 

Maxima [F]

\[ \int \frac {\left (1+c^2 x^2\right )^{5/2}}{a+b \text {arcsinh}(c x)} \, dx=\int { \frac {{\left (c^{2} x^{2} + 1\right )}^{\frac {5}{2}}}{b \operatorname {arsinh}\left (c x\right ) + a} \,d x } \] Input:

integrate((c^2*x^2+1)^(5/2)/(a+b*arcsinh(c*x)),x, algorithm="maxima")
 

Output:

integrate((c^2*x^2 + 1)^(5/2)/(b*arcsinh(c*x) + a), x)
 

Giac [F]

\[ \int \frac {\left (1+c^2 x^2\right )^{5/2}}{a+b \text {arcsinh}(c x)} \, dx=\int { \frac {{\left (c^{2} x^{2} + 1\right )}^{\frac {5}{2}}}{b \operatorname {arsinh}\left (c x\right ) + a} \,d x } \] Input:

integrate((c^2*x^2+1)^(5/2)/(a+b*arcsinh(c*x)),x, algorithm="giac")
 

Output:

integrate((c^2*x^2 + 1)^(5/2)/(b*arcsinh(c*x) + a), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (1+c^2 x^2\right )^{5/2}}{a+b \text {arcsinh}(c x)} \, dx=\int \frac {{\left (c^2\,x^2+1\right )}^{5/2}}{a+b\,\mathrm {asinh}\left (c\,x\right )} \,d x \] Input:

int((c^2*x^2 + 1)^(5/2)/(a + b*asinh(c*x)),x)
 

Output:

int((c^2*x^2 + 1)^(5/2)/(a + b*asinh(c*x)), x)
 

Reduce [F]

\[ \int \frac {\left (1+c^2 x^2\right )^{5/2}}{a+b \text {arcsinh}(c x)} \, dx=\int \frac {\sqrt {c^{2} x^{2}+1}}{\mathit {asinh} \left (c x \right ) b +a}d x +\left (\int \frac {\sqrt {c^{2} x^{2}+1}\, x^{4}}{\mathit {asinh} \left (c x \right ) b +a}d x \right ) c^{4}+2 \left (\int \frac {\sqrt {c^{2} x^{2}+1}\, x^{2}}{\mathit {asinh} \left (c x \right ) b +a}d x \right ) c^{2} \] Input:

int((c^2*x^2+1)^(5/2)/(a+b*asinh(c*x)),x)
 

Output:

int(sqrt(c**2*x**2 + 1)/(asinh(c*x)*b + a),x) + int((sqrt(c**2*x**2 + 1)*x 
**4)/(asinh(c*x)*b + a),x)*c**4 + 2*int((sqrt(c**2*x**2 + 1)*x**2)/(asinh( 
c*x)*b + a),x)*c**2