\(\int \frac {e^{n \coth ^{-1}(a x)}}{\sqrt {c-a c x}} \, dx\) [294]

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

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

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.00 \[ \int \frac {e^{n \coth ^{-1}(a x)}}{\sqrt {c-a c x}} \, dx=\frac {2 \left (1-\frac {1}{a x}\right )^{-n/2} \left (1+\frac {1}{a x}\right )^{n/2} \left (\frac {-1+a x}{1+a x}\right )^{\frac {1+n}{2}} (1+a x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1+n}{2},\frac {1}{2},\frac {2}{1+a x}\right )}{a \sqrt {c-a c x}} \] Input:

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

Output:

(2*(1 + 1/(a*x))^(n/2)*((-1 + a*x)/(1 + a*x))^((1 + n)/2)*(1 + a*x)*Hyperg 
eometric2F1[-1/2, (1 + n)/2, 1/2, 2/(1 + a*x)])/(a*(1 - 1/(a*x))^(n/2)*Sqr 
t[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.

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

\(\Big \downarrow \) 6727

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

\(\Big \downarrow \) 142

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

Input:

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

Output:

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

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]

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

Input:

int(exp(n*arccoth(a*x))/(-a*c*x+c)^(1/2),x)
 

Output:

int(exp(n*arccoth(a*x))/(-a*c*x+c)^(1/2),x)
 

Fricas [F]

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

integrate(exp(n*arccoth(a*x))/(-a*c*x+c)^(1/2),x, algorithm="fricas")
 

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

integrate(exp(n*arccoth(a*x))/(-a*c*x+c)^(1/2),x, algorithm="maxima")
 

Output:

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

Giac [F]

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

integrate(exp(n*arccoth(a*x))/(-a*c*x+c)^(1/2),x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

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

int(exp(n*acoth(a*x))/(c - a*c*x)^(1/2),x)
 

Output:

int(exp(n*acoth(a*x))/(c - a*c*x)^(1/2), x)
 

Reduce [F]

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

int(exp(n*acoth(a*x))/(-a*c*x+c)^(1/2),x)
 

Output:

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