3.191 \(\int \frac{\coth ^{-1}(\tanh (a+b x))^n}{x^2} \, dx\)

Optimal. Leaf size=71 \[ \frac{b \coth ^{-1}(\tanh (a+b x))^n \text{Hypergeometric2F1}\left (1,n,n+1,-\frac{\coth ^{-1}(\tanh (a+b x))}{b x-\coth ^{-1}(\tanh (a+b x))}\right )}{b x-\coth ^{-1}(\tanh (a+b x))}-\frac{\coth ^{-1}(\tanh (a+b x))^n}{x} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0412581, antiderivative size = 71, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {2168, 2164} \[ \frac{b \coth ^{-1}(\tanh (a+b x))^n \, _2F_1\left (1,n;n+1;-\frac{\coth ^{-1}(\tanh (a+b x))}{b x-\coth ^{-1}(\tanh (a+b x))}\right )}{b x-\coth ^{-1}(\tanh (a+b x))}-\frac{\coth ^{-1}(\tanh (a+b x))^n}{x} \]

Antiderivative was successfully verified.

[In]

Int[ArcCoth[Tanh[a + b*x]]^n/x^2,x]

[Out]

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

Rule 2168

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]))

Rule 2164

Int[(v_)^(n_)/(u_), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, Simp[(v^(n + 1)*Hypergeo
metric2F1[1, n + 1, n + 2, -((a*v)/(b*u - a*v))])/((n + 1)*(b*u - a*v)), x] /; NeQ[b*u - a*v, 0]] /; Piecewise
LinearQ[u, v, x] &&  !IntegerQ[n]

Rubi steps

\begin{align*} \int \frac{\coth ^{-1}(\tanh (a+b x))^n}{x^2} \, dx &=-\frac{\coth ^{-1}(\tanh (a+b x))^n}{x}+(b n) \int \frac{\coth ^{-1}(\tanh (a+b x))^{-1+n}}{x} \, dx\\ &=-\frac{\coth ^{-1}(\tanh (a+b x))^n}{x}+\frac{b \coth ^{-1}(\tanh (a+b x))^n \, _2F_1\left (1,n;1+n;-\frac{\coth ^{-1}(\tanh (a+b x))}{b x-\coth ^{-1}(\tanh (a+b x))}\right )}{b x-\coth ^{-1}(\tanh (a+b x))}\\ \end{align*}

Mathematica [A]  time = 0.0404399, size = 67, normalized size = 0.94 \[ \frac{\coth ^{-1}(\tanh (a+b x))^n \left (\frac{\coth ^{-1}(\tanh (a+b x))}{b x}\right )^{-n} \text{Hypergeometric2F1}\left (1-n,-n,2-n,1-\frac{\coth ^{-1}(\tanh (a+b x))}{b x}\right )}{(n-1) x} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcCoth[Tanh[a + b*x]]^n/x^2,x]

[Out]

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

________________________________________________________________________________________

Maple [F]  time = 1.964, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ({\rm arccoth} \left (\tanh \left ( bx+a \right ) \right ) \right ) ^{n}}{{x}^{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arccoth(tanh(b*x+a))^n/x^2,x)

[Out]

int(arccoth(tanh(b*x+a))^n/x^2,x)

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{arcoth}\left (\tanh \left (b x + a\right )\right )^{n}}{x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccoth(tanh(b*x+a))^n/x^2,x, algorithm="maxima")

[Out]

integrate(arccoth(tanh(b*x + a))^n/x^2, x)

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\operatorname{arcoth}\left (\tanh \left (b x + a\right )\right )^{n}}{x^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccoth(tanh(b*x+a))^n/x^2,x, algorithm="fricas")

[Out]

integral(arccoth(tanh(b*x + a))^n/x^2, x)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{acoth}^{n}{\left (\tanh{\left (a + b x \right )} \right )}}{x^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(acoth(tanh(b*x+a))**n/x**2,x)

[Out]

Integral(acoth(tanh(a + b*x))**n/x**2, x)

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{arcoth}\left (\tanh \left (b x + a\right )\right )^{n}}{x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccoth(tanh(b*x+a))^n/x^2,x, algorithm="giac")

[Out]

integrate(arccoth(tanh(b*x + a))^n/x^2, x)