\(\int \frac {e^{n \text {arctanh}(a x)}}{\sqrt {c-\frac {c}{a^2 x^2}}} \, dx\) [783]

Optimal result

Integrand size = 24, antiderivative size = 182 \[ \int \frac {e^{n \text {arctanh}(a x)}}{\sqrt {c-\frac {c}{a^2 x^2}}} \, dx=-\frac {(1-a x)^{\frac {1-n}{2}} (1+a x)^{\frac {1+n}{2}} \sqrt {1-a^2 x^2}}{a^2 (1+n) \sqrt {c-\frac {c}{a^2 x^2}} x}-\frac {2^{\frac {3+n}{2}} n (1-a x)^{\frac {1-n}{2}} \sqrt {1-a^2 x^2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (-1-n),\frac {1-n}{2},\frac {3-n}{2},\frac {1}{2} (1-a x)\right )}{a^2 \left (1-n^2\right ) \sqrt {c-\frac {c}{a^2 x^2}} x} \] Output:

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

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.71 \[ \int \frac {e^{n \text {arctanh}(a x)}}{\sqrt {c-\frac {c}{a^2 x^2}}} \, dx=\frac {(1-a x)^{\frac {1}{2}-\frac {n}{2}} \sqrt {1-a^2 x^2} \left (-\left ((-1+n) (1+a x)^{\frac {1+n}{2}}\right )+2^{\frac {3+n}{2}} n \operatorname {Hypergeometric2F1}\left (-\frac {1}{2}-\frac {n}{2},\frac {1}{2}-\frac {n}{2},\frac {3}{2}-\frac {n}{2},\frac {1}{2}-\frac {a x}{2}\right )\right )}{a^2 (-1+n) (1+n) \sqrt {c-\frac {c}{a^2 x^2}} x} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.79 (sec) , antiderivative size = 154, normalized size of antiderivative = 0.85, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {6710, 6700, 88, 79}

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.

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

\(\Big \downarrow \) 6710

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

\(\Big \downarrow \) 6700

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

\(\Big \downarrow \) 88

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

\(\Big \downarrow \) 79

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 79

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(( 
a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c - a*d))^n))*Hypergeometric2F1[-n, m + 1 
, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m, n}, x] 
&&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] 
 ||  !(RationalQ[n] && GtQ[-d/(b*c - a*d), 0]))
 

rule 88

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p 
_.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p 
+ 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p 
+ 1)))/(f*(p + 1)*(c*f - d*e))   Int[(c + d*x)^n*(e + f*x)^Simplify[p + 1], 
 x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] &&  !RationalQ[p] && SumSimpl 
erQ[p, 1]
 

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 6710

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

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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