\(\int \frac {e^{n \text {arctanh}(a x)}}{x \sqrt {c-a^2 c x^2}} \, dx\) [1361]

Optimal result

Integrand size = 27, antiderivative size = 101 \[ \int \frac {e^{n \text {arctanh}(a x)}}{x \sqrt {c-a^2 c x^2}} \, dx=-\frac {2 (1-a x)^{\frac {1-n}{2}} (1+a x)^{\frac {1}{2} (-1+n)} \sqrt {1-a^2 x^2} \operatorname {Hypergeometric2F1}\left (1,\frac {1-n}{2},\frac {3-n}{2},\frac {1-a x}{1+a x}\right )}{(1-n) \sqrt {c-a^2 c x^2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.98 \[ \int \frac {e^{n \text {arctanh}(a x)}}{x \sqrt {c-a^2 c x^2}} \, dx=\frac {2 (1-a x)^{\frac {1}{2}-\frac {n}{2}} (1+a x)^{\frac {1}{2} (-1+n)} \sqrt {1-a^2 x^2} \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2}-\frac {n}{2},\frac {3}{2}-\frac {n}{2},\frac {1-a x}{1+a x}\right )}{(-1+n) \sqrt {c-a^2 c x^2}} \] Input:

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

Output:

(2*(1 - a*x)^(1/2 - n/2)*(1 + a*x)^((-1 + n)/2)*Sqrt[1 - a^2*x^2]*Hypergeo 
metric2F1[1, 1/2 - n/2, 3/2 - n/2, (1 - a*x)/(1 + a*x)])/((-1 + n)*Sqrt[c 
- a^2*c*x^2])
 

Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {6703, 6700, 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.

\(\displaystyle \int \frac {e^{n \text {arctanh}(a x)}}{x \sqrt {c-a^2 c x^2}} \, dx\)

\(\Big \downarrow \) 6703

\(\displaystyle \frac {\sqrt {1-a^2 x^2} \int \frac {e^{n \text {arctanh}(a x)}}{x \sqrt {1-a^2 x^2}}dx}{\sqrt {c-a^2 c x^2}}\)

\(\Big \downarrow \) 6700

\(\displaystyle \frac {\sqrt {1-a^2 x^2} \int \frac {(1-a x)^{\frac {1}{2} (-n-1)} (a x+1)^{\frac {n-1}{2}}}{x}dx}{\sqrt {c-a^2 c x^2}}\)

\(\Big \downarrow \) 141

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

Input:

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

Output:

(-2*(1 - a*x)^((1 - n)/2)*(1 + a*x)^((-1 + n)/2)*Sqrt[1 - a^2*x^2]*Hyperge 
ometric2F1[1, (1 - n)/2, (3 - n)/2, (1 - a*x)/(1 + a*x)])/((1 - n)*Sqrt[c 
- a^2*c*x^2])
 

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 6700

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

rule 6703

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_), x_ 
Symbol] :> Simp[c^IntPart[p]*((c + d*x^2)^FracPart[p]/(1 - a^2*x^2)^FracPar 
t[p])   Int[x^m*(1 - a^2*x^2)^p*E^(n*ArcTanh[a*x]), x], x] /; FreeQ[{a, c, 
d, m, n, p}, x] && EqQ[a^2*c + d, 0] &&  !(IntegerQ[p] || GtQ[c, 0]) &&  !I 
ntegerQ[n/2]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {e^{n \text {arctanh}(a x)}}{x \sqrt {c-a^2 c x^2}} \, dx=\int \frac {{\mathrm {e}}^{n\,\mathrm {atanh}\left (a\,x\right )}}{x\,\sqrt {c-a^2\,c\,x^2}} \,d x \] Input:

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

Output:

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

Reduce [F]

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

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

Output:

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