\(\int \frac {\coth ^{-1}(\tanh (a+b x))^n}{x^3} \, dx\) [192]

Optimal result

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

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2199, 2195} \[ \int \frac {\coth ^{-1}(\tanh (a+b x))^n}{x^3} \, dx=\frac {b^2 n \coth ^{-1}(\tanh (a+b x))^{n-1} \operatorname {Hypergeometric2F1}\left (1,n-1,n,-\frac {\coth ^{-1}(\tanh (a+b x))}{b x-\coth ^{-1}(\tanh (a+b x))}\right )}{2 \left (b x-\coth ^{-1}(\tanh (a+b x))\right )}-\frac {\coth ^{-1}(\tanh (a+b x))^n}{2 x^2}-\frac {b n \coth ^{-1}(\tanh (a+b x))^{n-1}}{2 x} \]

[In]

Int[ArcCoth[Tanh[a + b*x]]^n/x^3,x]

[Out]

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

Rule 2199

Int[(u_)^(m_)*(v_)^(n_.), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, Simp[u^(m + 1)*(v^
n/(a*(m + 1))), x] - Dist[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] && (Fra
ctionQ[m] || GeQ[2*n + m + 1, 0]))) || (IGtQ[n, 0] && IGtQ[m, 0] && LeQ[n, m]) || (IGtQ[n, 0] &&  !IntegerQ[m]
) || (ILtQ[m, 0] &&  !IntegerQ[n]))

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.66 \[ \int \frac {\coth ^{-1}(\tanh (a+b x))^n}{x^3} \, dx=\frac {\coth ^{-1}(\tanh (a+b x))^n \left (\frac {\coth ^{-1}(\tanh (a+b x))}{b x}\right )^{-n} \operatorname {Hypergeometric2F1}\left (2-n,-n,3-n,1-\frac {\coth ^{-1}(\tanh (a+b x))}{b x}\right )}{(-2+n) x^2} \]

[In]

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

[Out]

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

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

\[ \int \frac {\coth ^{-1}(\tanh (a+b x))^n}{x^3} \, dx=\int { \frac {\operatorname {arcoth}\left (\tanh \left (b x + a\right )\right )^{n}}{x^{3}} \,d x } \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \frac {\coth ^{-1}(\tanh (a+b x))^n}{x^3} \, dx=\int \frac {\operatorname {acoth}^{n}{\left (\tanh {\left (a + b x \right )} \right )}}{x^{3}}\, dx \]

[In]

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

[Out]

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

Maxima [F]

\[ \int \frac {\coth ^{-1}(\tanh (a+b x))^n}{x^3} \, dx=\int { \frac {\operatorname {arcoth}\left (\tanh \left (b x + a\right )\right )^{n}}{x^{3}} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int \frac {\coth ^{-1}(\tanh (a+b x))^n}{x^3} \, dx=\int { \frac {\operatorname {arcoth}\left (\tanh \left (b x + a\right )\right )^{n}}{x^{3}} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\coth ^{-1}(\tanh (a+b x))^n}{x^3} \, dx=\int \frac {{\mathrm {acoth}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^n}{x^3} \,d x \]

[In]

int(acoth(tanh(a + b*x))^n/x^3,x)

[Out]

int(acoth(tanh(a + b*x))^n/x^3, x)