\(\int e^{n \coth ^{-1}(a x)} (c-a^2 c x^2)^{3/2} \, dx\) [718]

Optimal result

Integrand size = 24, antiderivative size = 116 \[ \int e^{n \coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^{3/2} \, dx=\frac {32 \left (1-\frac {1}{a x}\right )^{\frac {5-n}{2}} \left (1+\frac {1}{a x}\right )^{\frac {1}{2} (-5+n)} \left (c-a^2 c x^2\right )^{3/2} \operatorname {Hypergeometric2F1}\left (5,\frac {5-n}{2},\frac {7-n}{2},\frac {a-\frac {1}{x}}{a+\frac {1}{x}}\right )}{a^4 (5-n) \left (1-\frac {1}{a^2 x^2}\right )^{3/2} x^3} \] Output:

32*(1-1/a/x)^(5/2-1/2*n)*(1+1/a/x)^(-5/2+1/2*n)*(-a^2*c*x^2+c)^(3/2)*hyper 
geom([5, 5/2-1/2*n],[7/2-1/2*n],(a-1/x)/(a+1/x))/a^4/(5-n)/(1-1/a^2/x^2)^( 
3/2)/x^3
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(280\) vs. \(2(116)=232\).

Time = 3.06 (sec) , antiderivative size = 280, normalized size of antiderivative = 2.41 \[ \int e^{n \coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^{3/2} \, dx=\frac {c^2 \left (96 a^3 c \left (1-\frac {1}{a^2 x^2}\right )^{3/2} x^3 \left (a e^{n \coth ^{-1}(a x)} \sqrt {1-\frac {1}{a^2 x^2}} x (n+a x)+2 e^{(1+n) \coth ^{-1}(a x)} (-1+n) \operatorname {Hypergeometric2F1}\left (1,\frac {1+n}{2},\frac {3+n}{2},e^{2 \coth ^{-1}(a x)}\right )\right )-c \left (-1+a^2 x^2\right ) \left (2 e^{n \coth ^{-1}(a x)} \left (-1+a^2 x^2\right )^2 \left (-a \left (-21+n^2\right ) x+2 n \left (1-n^2+\left (3+n^2\right ) \cosh \left (2 \coth ^{-1}(a x)\right )\right )+a \left (3+n^2\right ) \sqrt {1-\frac {1}{a^2 x^2}} x \cosh \left (3 \coth ^{-1}(a x)\right )\right )+16 a e^{(1+n) \coth ^{-1}(a x)} \left (-3+3 n-n^2+n^3\right ) \sqrt {1-\frac {1}{a^2 x^2}} x \operatorname {Hypergeometric2F1}\left (1,\frac {1+n}{2},\frac {3+n}{2},e^{2 \coth ^{-1}(a x)}\right )\right )\right )}{192 a \left (c-a^2 c x^2\right )^{3/2}} \] Input:

Integrate[E^(n*ArcCoth[a*x])*(c - a^2*c*x^2)^(3/2),x]
 

Output:

(c^2*(96*a^3*c*(1 - 1/(a^2*x^2))^(3/2)*x^3*(a*E^(n*ArcCoth[a*x])*Sqrt[1 - 
1/(a^2*x^2)]*x*(n + a*x) + 2*E^((1 + n)*ArcCoth[a*x])*(-1 + n)*Hypergeomet 
ric2F1[1, (1 + n)/2, (3 + n)/2, E^(2*ArcCoth[a*x])]) - c*(-1 + a^2*x^2)*(2 
*E^(n*ArcCoth[a*x])*(-1 + a^2*x^2)^2*(-(a*(-21 + n^2)*x) + 2*n*(1 - n^2 + 
(3 + n^2)*Cosh[2*ArcCoth[a*x]]) + a*(3 + n^2)*Sqrt[1 - 1/(a^2*x^2)]*x*Cosh 
[3*ArcCoth[a*x]]) + 16*a*E^((1 + n)*ArcCoth[a*x])*(-3 + 3*n - n^2 + n^3)*S 
qrt[1 - 1/(a^2*x^2)]*x*Hypergeometric2F1[1, (1 + n)/2, (3 + n)/2, E^(2*Arc 
Coth[a*x])])))/(192*a*(c - a^2*c*x^2)^(3/2))
 

Rubi [A] (verified)

Time = 0.80 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {6746, 6749, 141}

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.

\(\displaystyle \int \left (c-a^2 c x^2\right )^{3/2} e^{n \coth ^{-1}(a x)} \, dx\)

\(\Big \downarrow \) 6746

\(\displaystyle \frac {\left (c-a^2 c x^2\right )^{3/2} \int e^{n \coth ^{-1}(a x)} \left (1-\frac {1}{a^2 x^2}\right )^{3/2} x^3dx}{x^3 \left (1-\frac {1}{a^2 x^2}\right )^{3/2}}\)

\(\Big \downarrow \) 6749

\(\displaystyle -\frac {\left (c-a^2 c x^2\right )^{3/2} \int \left (1-\frac {1}{a x}\right )^{\frac {3-n}{2}} \left (1+\frac {1}{a x}\right )^{\frac {n+3}{2}} x^5d\frac {1}{x}}{x^3 \left (1-\frac {1}{a^2 x^2}\right )^{3/2}}\)

\(\Big \downarrow \) 141

\(\displaystyle \frac {32 \left (c-a^2 c x^2\right )^{3/2} \left (1-\frac {1}{a x}\right )^{\frac {5-n}{2}} \left (\frac {1}{a x}+1\right )^{\frac {n-5}{2}} \operatorname {Hypergeometric2F1}\left (5,\frac {5-n}{2},\frac {7-n}{2},\frac {a-\frac {1}{x}}{a+\frac {1}{x}}\right )}{a^4 (5-n) x^3 \left (1-\frac {1}{a^2 x^2}\right )^{3/2}}\)

Input:

Int[E^(n*ArcCoth[a*x])*(c - a^2*c*x^2)^(3/2),x]
 

Output:

(32*(1 - 1/(a*x))^((5 - n)/2)*(1 + 1/(a*x))^((-5 + n)/2)*(c - a^2*c*x^2)^( 
3/2)*Hypergeometric2F1[5, (5 - n)/2, (7 - n)/2, (a - x^(-1))/(a + x^(-1))] 
)/(a^4*(5 - n)*(1 - 1/(a^2*x^2))^(3/2)*x^3)
 

Defintions of rubi rules used

rule 141

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Simp[(b*c - a*d)^n*((a + b*x)^(m + 1)/((m + 1)*(b*e - a*f)^( 
n + 1)*(e + f*x)^(m + 1)))*Hypergeometric2F1[m + 1, -n, m + 2, (-(d*e - c*f 
))*((a + b*x)/((b*c - a*d)*(e + f*x)))], x] /; FreeQ[{a, b, c, d, e, f, m, 
p}, x] && EqQ[m + n + p + 2, 0] && ILtQ[n, 0] && (SumSimplerQ[m, 1] ||  !Su 
mSimplerQ[p, 1]) &&  !ILtQ[m, 0]
 

rule 6746

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)*(x_)^2)^(p_), x_Symbo 
l] :> Simp[(c + d*x^2)^p/(x^(2*p)*(1 - 1/(a^2*x^2))^p)   Int[u*x^(2*p)*(1 - 
 1/(a^2*x^2))^p*E^(n*ArcCoth[a*x]), x], x] /; FreeQ[{a, c, d, n, p}, x] && 
EqQ[a^2*c + d, 0] &&  !IntegerQ[n/2] &&  !IntegerQ[p]
 

rule 6749

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*((c_) + (d_.)/(x_)^2)^(p_.)*(x_)^(m_.), x 
_Symbol] :> Simp[-c^p   Subst[Int[(1 - x/a)^(p - n/2)*((1 + x/a)^(p + n/2)/ 
x^(m + 2)), x], x, 1/x], x] /; FreeQ[{a, c, d, n, p}, x] && EqQ[c + a^2*d, 
0] &&  !IntegerQ[n/2] && (IntegerQ[p] || GtQ[c, 0]) &&  !IntegersQ[2*p, p + 
 n/2] && IntegerQ[m]
 

Maple [F]

\[\int {\mathrm e}^{n \,\operatorname {arccoth}\left (a x \right )} \left (-a^{2} c \,x^{2}+c \right )^{\frac {3}{2}}d x\]

Input:

int(exp(n*arccoth(a*x))*(-a^2*c*x^2+c)^(3/2),x)
 

Output:

int(exp(n*arccoth(a*x))*(-a^2*c*x^2+c)^(3/2),x)
 

Fricas [F]

\[ \int e^{n \coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^{3/2} \, dx=\int { {\left (-a^{2} c x^{2} + c\right )}^{\frac {3}{2}} \left (\frac {a x + 1}{a x - 1}\right )^{\frac {1}{2} \, n} \,d x } \] Input:

integrate(exp(n*arccoth(a*x))*(-a^2*c*x^2+c)^(3/2),x, algorithm="fricas")
 

Output:

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

Sympy [F]

\[ \int e^{n \coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^{3/2} \, dx=\int \left (- c \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {3}{2}} e^{n \operatorname {acoth}{\left (a x \right )}}\, dx \] Input:

integrate(exp(n*acoth(a*x))*(-a**2*c*x**2+c)**(3/2),x)
 

Output:

Integral((-c*(a*x - 1)*(a*x + 1))**(3/2)*exp(n*acoth(a*x)), x)
 

Maxima [F]

\[ \int e^{n \coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^{3/2} \, dx=\int { {\left (-a^{2} c x^{2} + c\right )}^{\frac {3}{2}} \left (\frac {a x + 1}{a x - 1}\right )^{\frac {1}{2} \, n} \,d x } \] Input:

integrate(exp(n*arccoth(a*x))*(-a^2*c*x^2+c)^(3/2),x, algorithm="maxima")
 

Output:

integrate((-a^2*c*x^2 + c)^(3/2)*((a*x + 1)/(a*x - 1))^(1/2*n), x)
 

Giac [F(-2)]

Exception generated. \[ \int e^{n \coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^{3/2} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(exp(n*arccoth(a*x))*(-a^2*c*x^2+c)^(3/2),x, algorithm="giac")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const 
index_m & i,const vecteur & l) Error: Bad Argument Value
 

Mupad [F(-1)]

Timed out. \[ \int e^{n \coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^{3/2} \, dx=\int {\mathrm {e}}^{n\,\mathrm {acoth}\left (a\,x\right )}\,{\left (c-a^2\,c\,x^2\right )}^{3/2} \,d x \] Input:

int(exp(n*acoth(a*x))*(c - a^2*c*x^2)^(3/2),x)
 

Output:

int(exp(n*acoth(a*x))*(c - a^2*c*x^2)^(3/2), x)
 

Reduce [F]

\[ \int e^{n \coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^{3/2} \, dx=\sqrt {c}\, c \left (-\left (\int e^{\mathit {acoth} \left (a x \right ) n} \sqrt {-a^{2} x^{2}+1}\, x^{2}d x \right ) a^{2}+\int e^{\mathit {acoth} \left (a x \right ) n} \sqrt {-a^{2} x^{2}+1}d x \right ) \] Input:

int(exp(n*acoth(a*x))*(-a^2*c*x^2+c)^(3/2),x)
 

Output:

sqrt(c)*c*( - int(e**(acoth(a*x)*n)*sqrt( - a**2*x**2 + 1)*x**2,x)*a**2 + 
int(e**(acoth(a*x)*n)*sqrt( - a**2*x**2 + 1),x))