\(\int \frac {x^2 (1+c^2 x^2)^{5/2}}{a+b \text {arcsinh}(c x)} \, dx\) [365]

Optimal result

Integrand size = 27, antiderivative size = 268 \[ \int \frac {x^2 \left (1+c^2 x^2\right )^{5/2}}{a+b \text {arcsinh}(c x)} \, dx=-\frac {\cosh \left (\frac {2 a}{b}\right ) \text {Chi}\left (\frac {2 (a+b \text {arcsinh}(c x))}{b}\right )}{32 b c^3}+\frac {\cosh \left (\frac {4 a}{b}\right ) \text {Chi}\left (\frac {4 (a+b \text {arcsinh}(c x))}{b}\right )}{32 b c^3}+\frac {\cosh \left (\frac {6 a}{b}\right ) \text {Chi}\left (\frac {6 (a+b \text {arcsinh}(c x))}{b}\right )}{32 b c^3}+\frac {\cosh \left (\frac {8 a}{b}\right ) \text {Chi}\left (\frac {8 (a+b \text {arcsinh}(c x))}{b}\right )}{128 b c^3}-\frac {5 \log (a+b \text {arcsinh}(c x))}{128 b c^3}+\frac {\sinh \left (\frac {2 a}{b}\right ) \text {Shi}\left (\frac {2 (a+b \text {arcsinh}(c x))}{b}\right )}{32 b c^3}-\frac {\sinh \left (\frac {4 a}{b}\right ) \text {Shi}\left (\frac {4 (a+b \text {arcsinh}(c x))}{b}\right )}{32 b c^3}-\frac {\sinh \left (\frac {6 a}{b}\right ) \text {Shi}\left (\frac {6 (a+b \text {arcsinh}(c x))}{b}\right )}{32 b c^3}-\frac {\sinh \left (\frac {8 a}{b}\right ) \text {Shi}\left (\frac {8 (a+b \text {arcsinh}(c x))}{b}\right )}{128 b c^3} \] Output:

-1/32*cosh(2*a/b)*Chi(2*(a+b*arcsinh(c*x))/b)/b/c^3+1/32*cosh(4*a/b)*Chi(4 
*(a+b*arcsinh(c*x))/b)/b/c^3+1/32*cosh(6*a/b)*Chi(6*(a+b*arcsinh(c*x))/b)/ 
b/c^3+1/128*cosh(8*a/b)*Chi(8*(a+b*arcsinh(c*x))/b)/b/c^3-5/128*ln(a+b*arc 
sinh(c*x))/b/c^3+1/32*sinh(2*a/b)*Shi(2*(a+b*arcsinh(c*x))/b)/b/c^3-1/32*s 
inh(4*a/b)*Shi(4*(a+b*arcsinh(c*x))/b)/b/c^3-1/32*sinh(6*a/b)*Shi(6*(a+b*a 
rcsinh(c*x))/b)/b/c^3-1/128*sinh(8*a/b)*Shi(8*(a+b*arcsinh(c*x))/b)/b/c^3
 

Mathematica [A] (verified)

Time = 0.66 (sec) , antiderivative size = 197, normalized size of antiderivative = 0.74 \[ \int \frac {x^2 \left (1+c^2 x^2\right )^{5/2}}{a+b \text {arcsinh}(c x)} \, dx=\frac {-4 \cosh \left (\frac {2 a}{b}\right ) \text {Chi}\left (2 \left (\frac {a}{b}+\text {arcsinh}(c x)\right )\right )+4 \cosh \left (\frac {4 a}{b}\right ) \text {Chi}\left (4 \left (\frac {a}{b}+\text {arcsinh}(c x)\right )\right )+4 \cosh \left (\frac {6 a}{b}\right ) \text {Chi}\left (6 \left (\frac {a}{b}+\text {arcsinh}(c x)\right )\right )+\cosh \left (\frac {8 a}{b}\right ) \text {Chi}\left (8 \left (\frac {a}{b}+\text {arcsinh}(c x)\right )\right )-5 \log (a+b \text {arcsinh}(c x))+4 \sinh \left (\frac {2 a}{b}\right ) \text {Shi}\left (2 \left (\frac {a}{b}+\text {arcsinh}(c x)\right )\right )-4 \sinh \left (\frac {4 a}{b}\right ) \text {Shi}\left (4 \left (\frac {a}{b}+\text {arcsinh}(c x)\right )\right )-4 \sinh \left (\frac {6 a}{b}\right ) \text {Shi}\left (6 \left (\frac {a}{b}+\text {arcsinh}(c x)\right )\right )-\sinh \left (\frac {8 a}{b}\right ) \text {Shi}\left (8 \left (\frac {a}{b}+\text {arcsinh}(c x)\right )\right )}{128 b c^3} \] Input:

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

Output:

(-4*Cosh[(2*a)/b]*CoshIntegral[2*(a/b + ArcSinh[c*x])] + 4*Cosh[(4*a)/b]*C 
oshIntegral[4*(a/b + ArcSinh[c*x])] + 4*Cosh[(6*a)/b]*CoshIntegral[6*(a/b 
+ ArcSinh[c*x])] + Cosh[(8*a)/b]*CoshIntegral[8*(a/b + ArcSinh[c*x])] - 5* 
Log[a + b*ArcSinh[c*x]] + 4*Sinh[(2*a)/b]*SinhIntegral[2*(a/b + ArcSinh[c* 
x])] - 4*Sinh[(4*a)/b]*SinhIntegral[4*(a/b + ArcSinh[c*x])] - 4*Sinh[(6*a) 
/b]*SinhIntegral[6*(a/b + ArcSinh[c*x])] - Sinh[(8*a)/b]*SinhIntegral[8*(a 
/b + ArcSinh[c*x])])/(128*b*c^3)
 

Rubi [A] (verified)

Time = 1.02 (sec) , antiderivative size = 221, normalized size of antiderivative = 0.82, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {6234, 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.

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

\(\Big \downarrow \) 6234

\(\displaystyle \frac {\int \frac {\cosh ^6\left (\frac {a}{b}-\frac {a+b \text {arcsinh}(c x)}{b}\right ) \sinh ^2\left (\frac {a}{b}-\frac {a+b \text {arcsinh}(c x)}{b}\right )}{a+b \text {arcsinh}(c x)}d(a+b \text {arcsinh}(c x))}{b c^3}\)

\(\Big \downarrow \) 5971

\(\displaystyle \frac {\int \left (\frac {\cosh \left (\frac {8 a}{b}-\frac {8 (a+b \text {arcsinh}(c x))}{b}\right )}{128 (a+b \text {arcsinh}(c x))}+\frac {\cosh \left (\frac {6 a}{b}-\frac {6 (a+b \text {arcsinh}(c x))}{b}\right )}{32 (a+b \text {arcsinh}(c x))}+\frac {\cosh \left (\frac {4 a}{b}-\frac {4 (a+b \text {arcsinh}(c x))}{b}\right )}{32 (a+b \text {arcsinh}(c x))}-\frac {\cosh \left (\frac {2 a}{b}-\frac {2 (a+b \text {arcsinh}(c x))}{b}\right )}{32 (a+b \text {arcsinh}(c x))}-\frac {5}{128 (a+b \text {arcsinh}(c x))}\right )d(a+b \text {arcsinh}(c x))}{b c^3}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {-\frac {1}{32} \cosh \left (\frac {2 a}{b}\right ) \text {Chi}\left (\frac {2 (a+b \text {arcsinh}(c x))}{b}\right )+\frac {1}{32} \cosh \left (\frac {4 a}{b}\right ) \text {Chi}\left (\frac {4 (a+b \text {arcsinh}(c x))}{b}\right )+\frac {1}{32} \cosh \left (\frac {6 a}{b}\right ) \text {Chi}\left (\frac {6 (a+b \text {arcsinh}(c x))}{b}\right )+\frac {1}{128} \cosh \left (\frac {8 a}{b}\right ) \text {Chi}\left (\frac {8 (a+b \text {arcsinh}(c x))}{b}\right )+\frac {1}{32} \sinh \left (\frac {2 a}{b}\right ) \text {Shi}\left (\frac {2 (a+b \text {arcsinh}(c x))}{b}\right )-\frac {1}{32} \sinh \left (\frac {4 a}{b}\right ) \text {Shi}\left (\frac {4 (a+b \text {arcsinh}(c x))}{b}\right )-\frac {1}{32} \sinh \left (\frac {6 a}{b}\right ) \text {Shi}\left (\frac {6 (a+b \text {arcsinh}(c x))}{b}\right )-\frac {1}{128} \sinh \left (\frac {8 a}{b}\right ) \text {Shi}\left (\frac {8 (a+b \text {arcsinh}(c x))}{b}\right )-\frac {5}{128} \log (a+b \text {arcsinh}(c x))}{b c^3}\)

Input:

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

Output:

(-1/32*(Cosh[(2*a)/b]*CoshIntegral[(2*(a + b*ArcSinh[c*x]))/b]) + (Cosh[(4 
*a)/b]*CoshIntegral[(4*(a + b*ArcSinh[c*x]))/b])/32 + (Cosh[(6*a)/b]*CoshI 
ntegral[(6*(a + b*ArcSinh[c*x]))/b])/32 + (Cosh[(8*a)/b]*CoshIntegral[(8*( 
a + b*ArcSinh[c*x]))/b])/128 - (5*Log[a + b*ArcSinh[c*x]])/128 + (Sinh[(2* 
a)/b]*SinhIntegral[(2*(a + b*ArcSinh[c*x]))/b])/32 - (Sinh[(4*a)/b]*SinhIn 
tegral[(4*(a + b*ArcSinh[c*x]))/b])/32 - (Sinh[(6*a)/b]*SinhIntegral[(6*(a 
 + b*ArcSinh[c*x]))/b])/32 - (Sinh[(8*a)/b]*SinhIntegral[(8*(a + b*ArcSinh 
[c*x]))/b])/128)/(b*c^3)
 

Defintions of rubi rules used

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 = 8.41 (sec) , antiderivative size = 211, normalized size of antiderivative = 0.79

method	result	size
default	\(-\frac {{\mathrm e}^{\frac {8 a}{b}} \operatorname {expIntegral}_{1}\left (8 \,\operatorname {arcsinh}\left (x c \right )+\frac {8 a}{b}\right )+4 \,{\mathrm e}^{\frac {6 a}{b}} \operatorname {expIntegral}_{1}\left (6 \,\operatorname {arcsinh}\left (x c \right )+\frac {6 a}{b}\right )+4 \,{\mathrm e}^{\frac {4 a}{b}} \operatorname {expIntegral}_{1}\left (4 \,\operatorname {arcsinh}\left (x c \right )+\frac {4 a}{b}\right )-4 \,{\mathrm e}^{\frac {2 a}{b}} \operatorname {expIntegral}_{1}\left (2 \,\operatorname {arcsinh}\left (x c \right )+\frac {2 a}{b}\right )-4 \,{\mathrm e}^{-\frac {2 a}{b}} \operatorname {expIntegral}_{1}\left (-2 \,\operatorname {arcsinh}\left (x c \right )-\frac {2 a}{b}\right )+4 \,{\mathrm e}^{-\frac {4 a}{b}} \operatorname {expIntegral}_{1}\left (-4 \,\operatorname {arcsinh}\left (x c \right )-\frac {4 a}{b}\right )+4 \,{\mathrm e}^{-\frac {6 a}{b}} \operatorname {expIntegral}_{1}\left (-6 \,\operatorname {arcsinh}\left (x c \right )-\frac {6 a}{b}\right )+{\mathrm e}^{-\frac {8 a}{b}} \operatorname {expIntegral}_{1}\left (-8 \,\operatorname {arcsinh}\left (x c \right )-\frac {8 a}{b}\right )+10 \ln \left (a +b \,\operatorname {arcsinh}\left (x c \right )\right )}{256 c^{3} b}\)	\(211\)

Input:

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

Output:

-1/256*(exp(8*a/b)*Ei(1,8*arcsinh(x*c)+8*a/b)+4*exp(6*a/b)*Ei(1,6*arcsinh( 
x*c)+6*a/b)+4*exp(4*a/b)*Ei(1,4*arcsinh(x*c)+4*a/b)-4*exp(2*a/b)*Ei(1,2*ar 
csinh(x*c)+2*a/b)-4*exp(-2*a/b)*Ei(1,-2*arcsinh(x*c)-2*a/b)+4*exp(-4*a/b)* 
Ei(1,-4*arcsinh(x*c)-4*a/b)+4*exp(-6*a/b)*Ei(1,-6*arcsinh(x*c)-6*a/b)+exp( 
-8*a/b)*Ei(1,-8*arcsinh(x*c)-8*a/b)+10*ln(a+b*arcsinh(x*c)))/c^3/b
 

Fricas [F]

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

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

Output:

integral((c^4*x^6 + 2*c^2*x^4 + x^2)*sqrt(c^2*x^2 + 1)/(b*arcsinh(c*x) + a 
), x)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {x^2 \left (1+c^2 x^2\right )^{5/2}}{a+b \text {arcsinh}(c x)} \, dx=\left (\int \frac {\sqrt {c^{2} x^{2}+1}\, x^{6}}{\mathit {asinh} \left (c x \right ) b +a}d x \right ) c^{4}+2 \left (\int \frac {\sqrt {c^{2} x^{2}+1}\, x^{4}}{\mathit {asinh} \left (c x \right ) b +a}d x \right ) c^{2}+\int \frac {\sqrt {c^{2} x^{2}+1}\, x^{2}}{\mathit {asinh} \left (c x \right ) b +a}d x \] Input:

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

Output:

int((sqrt(c**2*x**2 + 1)*x**6)/(asinh(c*x)*b + a),x)*c**4 + 2*int((sqrt(c* 
*2*x**2 + 1)*x**4)/(asinh(c*x)*b + a),x)*c**2 + int((sqrt(c**2*x**2 + 1)*x 
**2)/(asinh(c*x)*b + a),x)