\(\int \frac {x^2 \sqrt {1-c^2 x^2}}{a+b \text {arccosh}(c x)} \, dx\) [233]

Optimal result

Integrand size = 28, antiderivative size = 139 \[ \int \frac {x^2 \sqrt {1-c^2 x^2}}{a+b \text {arccosh}(c x)} \, dx=\frac {\sqrt {1-c x} \cosh \left (\frac {4 a}{b}\right ) \text {Chi}\left (\frac {4 (a+b \text {arccosh}(c x))}{b}\right )}{8 b c^3 \sqrt {-1+c x}}-\frac {\sqrt {1-c x} \log (a+b \text {arccosh}(c x))}{8 b c^3 \sqrt {-1+c x}}-\frac {\sqrt {1-c x} \sinh \left (\frac {4 a}{b}\right ) \text {Shi}\left (\frac {4 (a+b \text {arccosh}(c x))}{b}\right )}{8 b c^3 \sqrt {-1+c x}} \] Output:

1/8*(-c*x+1)^(1/2)*cosh(4*a/b)*Chi(4*(a+b*arccosh(c*x))/b)/b/c^3/(c*x-1)^( 
1/2)-1/8*(-c*x+1)^(1/2)*ln(a+b*arccosh(c*x))/b/c^3/(c*x-1)^(1/2)-1/8*(-c*x 
+1)^(1/2)*sinh(4*a/b)*Shi(4*(a+b*arccosh(c*x))/b)/b/c^3/(c*x-1)^(1/2)
 

Mathematica [A] (verified)

Time = 0.29 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.74 \[ \int \frac {x^2 \sqrt {1-c^2 x^2}}{a+b \text {arccosh}(c x)} \, dx=-\frac {\sqrt {-((-1+c x) (1+c x))} \left (-\cosh \left (\frac {4 a}{b}\right ) \text {Chi}\left (4 \left (\frac {a}{b}+\text {arccosh}(c x)\right )\right )+\log (a+b \text {arccosh}(c x))+\sinh \left (\frac {4 a}{b}\right ) \text {Shi}\left (4 \left (\frac {a}{b}+\text {arccosh}(c x)\right )\right )\right )}{8 b c^3 \sqrt {\frac {-1+c x}{1+c x}} (1+c x)} \] Input:

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

Output:

-1/8*(Sqrt[-((-1 + c*x)*(1 + c*x))]*(-(Cosh[(4*a)/b]*CoshIntegral[4*(a/b + 
 ArcCosh[c*x])]) + Log[a + b*ArcCosh[c*x]] + Sinh[(4*a)/b]*SinhIntegral[4* 
(a/b + ArcCosh[c*x])]))/(b*c^3*Sqrt[(-1 + c*x)/(1 + c*x)]*(1 + c*x))
 

Rubi [A] (verified)

Time = 0.83 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.65, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {6367, 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.

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

\(\Big \downarrow \) 6367

\(\displaystyle \frac {\sqrt {1-c x} \int \frac {\cosh ^2\left (\frac {a}{b}-\frac {a+b \text {arccosh}(c x)}{b}\right ) \sinh ^2\left (\frac {a}{b}-\frac {a+b \text {arccosh}(c x)}{b}\right )}{a+b \text {arccosh}(c x)}d(a+b \text {arccosh}(c x))}{b c^3 \sqrt {c x-1}}\)

\(\Big \downarrow \) 5971

\(\displaystyle \frac {\sqrt {1-c x} \int \left (\frac {\cosh \left (\frac {4 a}{b}-\frac {4 (a+b \text {arccosh}(c x))}{b}\right )}{8 (a+b \text {arccosh}(c x))}-\frac {1}{8 (a+b \text {arccosh}(c x))}\right )d(a+b \text {arccosh}(c x))}{b c^3 \sqrt {c x-1}}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\sqrt {1-c x} \left (\frac {1}{8} \cosh \left (\frac {4 a}{b}\right ) \text {Chi}\left (\frac {4 (a+b \text {arccosh}(c x))}{b}\right )-\frac {1}{8} \sinh \left (\frac {4 a}{b}\right ) \text {Shi}\left (\frac {4 (a+b \text {arccosh}(c x))}{b}\right )-\frac {1}{8} \log (a+b \text {arccosh}(c x))\right )}{b c^3 \sqrt {c x-1}}\)

Input:

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

Output:

(Sqrt[1 - c*x]*((Cosh[(4*a)/b]*CoshIntegral[(4*(a + b*ArcCosh[c*x]))/b])/8 
 - Log[a + b*ArcCosh[c*x]]/8 - (Sinh[(4*a)/b]*SinhIntegral[(4*(a + b*ArcCo 
sh[c*x]))/b])/8))/(b*c^3*Sqrt[-1 + c*x])
 

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 6367

Int[((a_.) + ArcCosh[(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*x 
)^p*(-1 + c*x)^p)]   Subst[Int[x^n*Cosh[-a/b + x/b]^m*Sinh[-a/b + x/b]^(2*p 
 + 1), x], x, a + b*ArcCosh[c*x]], x] /; FreeQ[{a, b, c, d, e, n}, x] && Eq 
Q[c^2*d + e, 0] && IGtQ[2*p + 2, 0] && IGtQ[m, 0]
 

Maple [A] (verified)

Time = 0.20 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.19

method	result	size
default	\(\frac {\sqrt {-c^{2} x^{2}+1}\, \left (-\sqrt {c x -1}\, \sqrt {c x +1}\, c x +c^{2} x^{2}-1\right ) \left (2 \sqrt {c x -1}\, \sqrt {c x +1}\, \ln \left (a +b \,\operatorname {arccosh}\left (c x \right )\right )+2 \ln \left (a +b \,\operatorname {arccosh}\left (c x \right )\right ) c x +\operatorname {expIntegral}_{1}\left (4 \,\operatorname {arccosh}\left (c x \right )+\frac {4 a}{b}\right ) {\mathrm e}^{\frac {b \,\operatorname {arccosh}\left (c x \right )+4 a}{b}}+\operatorname {expIntegral}_{1}\left (-4 \,\operatorname {arccosh}\left (c x \right )-\frac {4 a}{b}\right ) {\mathrm e}^{-\frac {-b \,\operatorname {arccosh}\left (c x \right )+4 a}{b}}\right )}{16 \left (c x +1\right ) c^{3} \left (c x -1\right ) b}\)	\(165\)

Input:

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

Output:

1/16*(-c^2*x^2+1)^(1/2)*(-(c*x-1)^(1/2)*(c*x+1)^(1/2)*c*x+c^2*x^2-1)*(2*(c 
*x-1)^(1/2)*(c*x+1)^(1/2)*ln(a+b*arccosh(c*x))+2*ln(a+b*arccosh(c*x))*c*x+ 
Ei(1,4*arccosh(c*x)+4*a/b)*exp((b*arccosh(c*x)+4*a)/b)+Ei(1,-4*arccosh(c*x 
)-4*a/b)*exp(-(-b*arccosh(c*x)+4*a)/b))/(c*x+1)/c^3/(c*x-1)/b
 

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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