\(\int \frac {e^{n \text {arctanh}(a x)}}{x (c-a^2 c x^2)} \, dx\) [1354]

Optimal result

Integrand size = 25, antiderivative size = 90 \[ \int \frac {e^{n \text {arctanh}(a x)}}{x \left (c-a^2 c x^2\right )} \, dx=\frac {(1-a x)^{-n/2} (1+a x)^{n/2}}{c n}-\frac {2 (1-a x)^{-n/2} (1+a x)^{n/2} \operatorname {Hypergeometric2F1}\left (1,\frac {n}{2},\frac {2+n}{2},\frac {1+a x}{1-a x}\right )}{c n} \] Output:

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

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.94 \[ \int \frac {e^{n \text {arctanh}(a x)}}{x \left (c-a^2 c x^2\right )} \, dx=\frac {(1-a x)^{-n/2} (1+a x)^{-1+\frac {n}{2}} \left ((-2+n) (1+a x)-2 n (-1+a x) \operatorname {Hypergeometric2F1}\left (1,1-\frac {n}{2},2-\frac {n}{2},\frac {1-a x}{1+a x}\right )\right )}{c (-2+n) n} \] Input:

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

Output:

((1 + a*x)^(-1 + n/2)*((-2 + n)*(1 + a*x) - 2*n*(-1 + a*x)*Hypergeometric2 
F1[1, 1 - n/2, 2 - n/2, (1 - a*x)/(1 + a*x)]))/(c*(-2 + n)*n*(1 - a*x)^(n/ 
2))
 

Rubi [A] (verified)

Time = 0.52 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.98, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {6700, 107, 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.

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

\(\Big \downarrow \) 6700

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

\(\Big \downarrow \) 107

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

\(\Big \downarrow \) 141

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 107

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Simp[b*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1 
)/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + Simp[(a*d*f*(m + 1) + b*c*f*(n + 
 1) + b*d*e*(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f))   Int[(a + b*x)^(m + 
 1)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x 
] && EqQ[Simplify[m + n + p + 3], 0] && (LtQ[m, -1] || SumSimplerQ[m, 1])
 

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])
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

\[ \int \frac {e^{n \text {arctanh}(a x)}}{x \left (c-a^2 c x^2\right )} \, dx=- \frac {\int \frac {e^{n \operatorname {atanh}{\left (a x \right )}}}{a^{2} x^{3} - x}\, dx}{c} \] Input:

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {e^{n \text {arctanh}(a x)}}{x \left (c-a^2 c x^2\right )} \, dx=-\frac {\int \frac {e^{\mathit {atanh} \left (a x \right ) n}}{a^{2} x^{3}-x}d x}{c} \] Input:

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

Output:

( - int(e**(atanh(a*x)*n)/(a**2*x**3 - x),x))/c