\(\int \frac {e^{n \coth ^{-1}(a x)}}{(c-a c x)^{3/2}} \, dx\) [295]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [F]
Fricas [F]
Sympy [F]
Maxima [F]
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 20, antiderivative size = 96 \[ \int \frac {e^{n \coth ^{-1}(a x)}}{(c-a c x)^{3/2}} \, dx=-\frac {2 \left (\frac {a-\frac {1}{x}}{a+\frac {1}{x}}\right )^{\frac {3+n}{2}} \left (1-\frac {1}{a x}\right )^{-n/2} \left (1+\frac {1}{a x}\right )^{\frac {2+n}{2}} x \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3+n}{2},\frac {3}{2},\frac {2}{\left (a+\frac {1}{x}\right ) x}\right )}{(c-a c x)^{3/2}} \] Output:

-2*((a-1/x)/(a+1/x))^(3/2+1/2*n)*(1+1/a/x)^(1+1/2*n)*x*hypergeom([1/2, 3/2 
+1/2*n],[3/2],2/(a+1/x)/x)/((1-1/a/x)^(1/2*n))/(-a*c*x+c)^(3/2)
 

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.98 \[ \int \frac {e^{n \coth ^{-1}(a x)}}{(c-a c x)^{3/2}} \, dx=\frac {2 \left (1-\frac {1}{a x}\right )^{-n/2} \left (1+\frac {1}{a x}\right )^{n/2} \left (\frac {-1+a x}{1+a x}\right )^{\frac {1+n}{2}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3+n}{2},\frac {3}{2},\frac {2}{1+a x}\right )}{a c \sqrt {c-a c x}} \] Input:

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

Output:

(2*(1 + 1/(a*x))^(n/2)*((-1 + a*x)/(1 + a*x))^((1 + n)/2)*Hypergeometric2F 
1[1/2, (3 + n)/2, 3/2, 2/(1 + a*x)])/(a*c*(1 - 1/(a*x))^(n/2)*Sqrt[c - a*c 
*x])
 

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.08, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {6727, 142}

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.

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

\(\Big \downarrow \) 6727

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

\(\Big \downarrow \) 142

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 142
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n*((e + f*x)^(p + 1)/((b*e 
 - a*f)*(m + 1)))*Hypergeometric2F1[m + 1, -n, m + 2, (-(d*e - c*f))*((a + 
b*x)/((b*c - a*d)*(e + f*x)))])/((b*e - a*f)*((c + d*x)/((b*c - a*d)*(e + f 
*x))))^n, x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[m + n + p + 2, 
 0] &&  !IntegerQ[n]
 

rule 6727
Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*((c_) + (d_.)*(x_))^(p_), x_Symbol] :> Si 
mp[(-(1/x)^p)*((c + d*x)^p/(1 + c/(d*x))^p)   Subst[Int[((1 + c*(x/d))^p*(( 
1 + x/a)^(n/2)/x^(p + 2)))/(1 - x/a)^(n/2), x], x, 1/x], x] /; FreeQ[{a, c, 
 d, n, p}, x] && EqQ[a^2*c^2 - d^2, 0] &&  !IntegerQ[p]
 
Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

( - int(e**(acoth(a*x)*n)/(sqrt( - a*x + 1)*a*x - sqrt( - a*x + 1)),x))/(s 
qrt(c)*c)