\(\int \frac {\coth ^{-1}(\tanh (a+b x))^n}{x} \, dx\) [190]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [F]
   Fricas [F]
   Sympy [F]
   Maxima [F]
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 13, antiderivative size = 64 \[ \int \frac {\coth ^{-1}(\tanh (a+b x))^n}{x} \, dx=\frac {\coth ^{-1}(\tanh (a+b x))^{1+n} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,-\frac {\coth ^{-1}(\tanh (a+b x))}{b x-\coth ^{-1}(\tanh (a+b x))}\right )}{(1+n) \left (b x-\coth ^{-1}(\tanh (a+b x))\right )} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {2195} \[ \int \frac {\coth ^{-1}(\tanh (a+b x))^n}{x} \, dx=\frac {\coth ^{-1}(\tanh (a+b x))^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,-\frac {\coth ^{-1}(\tanh (a+b x))}{b x-\coth ^{-1}(\tanh (a+b x))}\right )}{(n+1) \left (b x-\coth ^{-1}(\tanh (a+b x))\right )} \]

[In]

Int[ArcCoth[Tanh[a + b*x]]^n/x,x]

[Out]

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

Rule 2195

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]] /; Piecewise
LinearQ[u, v, x] &&  !IntegerQ[n]

Rubi steps \begin{align*} \text {integral}& = \frac {\coth ^{-1}(\tanh (a+b x))^{1+n} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,-\frac {\coth ^{-1}(\tanh (a+b x))}{b x-\coth ^{-1}(\tanh (a+b x))}\right )}{(1+n) \left (b x-\coth ^{-1}(\tanh (a+b x))\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.94 \[ \int \frac {\coth ^{-1}(\tanh (a+b x))^n}{x} \, dx=\frac {\coth ^{-1}(\tanh (a+b x))^n \left (\frac {\coth ^{-1}(\tanh (a+b x))}{b x}\right )^{-n} \operatorname {Hypergeometric2F1}\left (-n,-n,1-n,1-\frac {\coth ^{-1}(\tanh (a+b x))}{b x}\right )}{n} \]

[In]

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

[Out]

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

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

\[ \int \frac {\coth ^{-1}(\tanh (a+b x))^n}{x} \, dx=\int { \frac {\operatorname {arcoth}\left (\tanh \left (b x + a\right )\right )^{n}}{x} \,d x } \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \frac {\coth ^{-1}(\tanh (a+b x))^n}{x} \, dx=\int \frac {\operatorname {acoth}^{n}{\left (\tanh {\left (a + b x \right )} \right )}}{x}\, dx \]

[In]

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

[Out]

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

Maxima [F]

\[ \int \frac {\coth ^{-1}(\tanh (a+b x))^n}{x} \, dx=\int { \frac {\operatorname {arcoth}\left (\tanh \left (b x + a\right )\right )^{n}}{x} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int \frac {\coth ^{-1}(\tanh (a+b x))^n}{x} \, dx=\int { \frac {\operatorname {arcoth}\left (\tanh \left (b x + a\right )\right )^{n}}{x} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\coth ^{-1}(\tanh (a+b x))^n}{x} \, dx=\int \frac {{\mathrm {acoth}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^n}{x} \,d x \]

[In]

int(acoth(tanh(a + b*x))^n/x,x)

[Out]

int(acoth(tanh(a + b*x))^n/x, x)