\(\int \frac {\text {arctanh}(\tanh (a+b x))^n}{x^3} \, dx\) [273]

Optimal result

Integrand size = 13, antiderivative size = 101 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^n}{x^3} \, dx=-\frac {b n \text {arctanh}(\tanh (a+b x))^{-1+n}}{2 x}-\frac {\text {arctanh}(\tanh (a+b x))^n}{2 x^2}+\frac {b^2 n \text {arctanh}(\tanh (a+b x))^{-1+n} \operatorname {Hypergeometric2F1}\left (1,-1+n,n,-\frac {\text {arctanh}(\tanh (a+b x))}{b x-\text {arctanh}(\tanh (a+b x))}\right )}{2 (b x-\text {arctanh}(\tanh (a+b x)))} \] Output:

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

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.66 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^n}{x^3} \, dx=\frac {\text {arctanh}(\tanh (a+b x))^n \left (\frac {\text {arctanh}(\tanh (a+b x))}{b x}\right )^{-n} \operatorname {Hypergeometric2F1}\left (2-n,-n,3-n,1-\frac {\text {arctanh}(\tanh (a+b x))}{b x}\right )}{(-2+n) x^2} \] Input:

Integrate[ArcTanh[Tanh[a + b*x]]^n/x^3,x]
 

Output:

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

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.97, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {2599, 2599, 2595}

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.

\(\displaystyle \int \frac {\text {arctanh}(\tanh (a+b x))^n}{x^3} \, dx\)

\(\Big \downarrow \) 2599

\(\displaystyle \frac {1}{2} b n \int \frac {\text {arctanh}(\tanh (a+b x))^{n-1}}{x^2}dx-\frac {\text {arctanh}(\tanh (a+b x))^n}{2 x^2}\)

\(\Big \downarrow \) 2599

\(\displaystyle \frac {1}{2} b n \left (-b (1-n) \int \frac {\text {arctanh}(\tanh (a+b x))^{n-2}}{x}dx-\frac {\text {arctanh}(\tanh (a+b x))^{n-1}}{x}\right )-\frac {\text {arctanh}(\tanh (a+b x))^n}{2 x^2}\)

\(\Big \downarrow \) 2595

\(\displaystyle \frac {1}{2} b n \left (\frac {b \text {arctanh}(\tanh (a+b x))^{n-1} \operatorname {Hypergeometric2F1}\left (1,n-1,n,-\frac {\text {arctanh}(\tanh (a+b x))}{b x-\text {arctanh}(\tanh (a+b x))}\right )}{b x-\text {arctanh}(\tanh (a+b x))}-\frac {\text {arctanh}(\tanh (a+b x))^{n-1}}{x}\right )-\frac {\text {arctanh}(\tanh (a+b x))^n}{2 x^2}\)

Input:

Int[ArcTanh[Tanh[a + b*x]]^n/x^3,x]
 

Output:

-1/2*ArcTanh[Tanh[a + b*x]]^n/x^2 + (b*n*(-(ArcTanh[Tanh[a + b*x]]^(-1 + n 
)/x) + (b*ArcTanh[Tanh[a + b*x]]^(-1 + n)*Hypergeometric2F1[1, -1 + n, n, 
-(ArcTanh[Tanh[a + b*x]]/(b*x - ArcTanh[Tanh[a + b*x]]))])/(b*x - ArcTanh[ 
Tanh[a + b*x]])))/2
 

Defintions of rubi rules used

rule 2595

Int[(v_)^(n_)/(u_), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[ 
D[v, x]]}, Simp[(v^(n + 1)/((n + 1)*(b*u - a*v)))*Hypergeometric2F1[1, n + 
1, n + 2, (-a)*(v/(b*u - a*v))], x] /; NeQ[b*u - a*v, 0]] /; PiecewiseLinea 
rQ[u, v, x] &&  !IntegerQ[n]
 

rule 2599

Int[(u_)^(m_)*(v_)^(n_.), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Sim 
plify[D[v, x]]}, Simp[u^(m + 1)*(v^n/(a*(m + 1))), x] - Simp[b*(n/(a*(m + 1 
)))   Int[u^(m + 1)*v^(n - 1), x], x] /; NeQ[b*u - a*v, 0]] /; FreeQ[{m, n} 
, x] && PiecewiseLinearQ[u, v, x] && NeQ[m, -1] && ((LtQ[m, -1] && GtQ[n, 0 
] &&  !(ILtQ[m + n, -2] && (FractionQ[m] || GeQ[2*n + m + 1, 0]))) || (IGtQ 
[n, 0] && IGtQ[m, 0] && LeQ[n, m]) || (IGtQ[n, 0] &&  !IntegerQ[m]) || (ILt 
Q[m, 0] &&  !IntegerQ[n]))
 

Maple [F]

\[\int \frac {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{n}}{x^{3}}d x\]

Input:

int(arctanh(tanh(b*x+a))^n/x^3,x)
 

Output:

int(arctanh(tanh(b*x+a))^n/x^3,x)
 

Fricas [F]

\[ \int \frac {\text {arctanh}(\tanh (a+b x))^n}{x^3} \, dx=\int { \frac {\operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{n}}{x^{3}} \,d x } \] Input:

integrate(arctanh(tanh(b*x+a))^n/x^3,x, algorithm="fricas")
 

Output:

integral(arctanh(tanh(b*x + a))^n/x^3, x)
 

Sympy [F]

\[ \int \frac {\text {arctanh}(\tanh (a+b x))^n}{x^3} \, dx=\int \frac {\operatorname {atanh}^{n}{\left (\tanh {\left (a + b x \right )} \right )}}{x^{3}}\, dx \] Input:

integrate(atanh(tanh(b*x+a))**n/x**3,x)
 

Output:

Integral(atanh(tanh(a + b*x))**n/x**3, x)
 

Maxima [F]

\[ \int \frac {\text {arctanh}(\tanh (a+b x))^n}{x^3} \, dx=\int { \frac {\operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{n}}{x^{3}} \,d x } \] Input:

integrate(arctanh(tanh(b*x+a))^n/x^3,x, algorithm="maxima")
 

Output:

integrate(arctanh(tanh(b*x + a))^n/x^3, x)
 

Giac [F]

\[ \int \frac {\text {arctanh}(\tanh (a+b x))^n}{x^3} \, dx=\int { \frac {\operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{n}}{x^{3}} \,d x } \] Input:

integrate(arctanh(tanh(b*x+a))^n/x^3,x, algorithm="giac")
 

Output:

integrate(arctanh(tanh(b*x + a))^n/x^3, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\text {arctanh}(\tanh (a+b x))^n}{x^3} \, dx=\int \frac {{\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^n}{x^3} \,d x \] Input:

int(atanh(tanh(a + b*x))^n/x^3,x)
 

Output:

int(atanh(tanh(a + b*x))^n/x^3, x)
 

Reduce [F]

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

int(atanh(tanh(b*x+a))^n/x^3,x)
 

Output:

( - atanh(tanh(a + b*x))**n + int(atanh(tanh(a + b*x))**n/(atanh(tanh(a + 
b*x))*x**2),x)*b*n*x**2)/(2*x**2)