\(\int \frac {x^m}{\coth ^{-1}(\tanh (a+b x))^3} \, dx\) [175]

Optimal result

Integrand size = 13, antiderivative size = 94 \[ \int \frac {x^m}{\coth ^{-1}(\tanh (a+b x))^3} \, dx=-\frac {x^m}{2 b \coth ^{-1}(\tanh (a+b x))^2}-\frac {m x^{-1+m}}{2 b^2 \coth ^{-1}(\tanh (a+b x))}-\frac {m x^{-1+m} \operatorname {Hypergeometric2F1}\left (1,-1+m,m,\frac {b x}{b x-\coth ^{-1}(\tanh (a+b x))}\right )}{2 b^2 \left (b x-\coth ^{-1}(\tanh (a+b x))\right )} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 94, 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} \[ \int \frac {x^m}{\coth ^{-1}(\tanh (a+b x))^3} \, dx=-\frac {m x^{m-1} \operatorname {Hypergeometric2F1}\left (1,m-1,m,\frac {b x}{b x-\coth ^{-1}(\tanh (a+b x))}\right )}{2 b^2 \left (b x-\coth ^{-1}(\tanh (a+b x))\right )}-\frac {m x^{m-1}}{2 b^2 \coth ^{-1}(\tanh (a+b x))}-\frac {x^m}{2 b \coth ^{-1}(\tanh (a+b x))^2} \]

[In]

Int[x^m/ArcCoth[Tanh[a + b*x]]^3,x]

[Out]

-1/2*x^m/(b*ArcCoth[Tanh[a + b*x]]^2) - (m*x^(-1 + m))/(2*b^2*ArcCoth[Tanh[a + b*x]]) - (m*x^(-1 + m)*Hypergeo
metric2F1[1, -1 + m, m, (b*x)/(b*x - ArcCoth[Tanh[a + b*x]])])/(2*b^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 {x^m}{2 b \coth ^{-1}(\tanh (a+b x))^2}+\frac {m \int \frac {x^{-1+m}}{\coth ^{-1}(\tanh (a+b x))^2} \, dx}{2 b} \\ & = -\frac {x^m}{2 b \coth ^{-1}(\tanh (a+b x))^2}-\frac {m x^{-1+m}}{2 b^2 \coth ^{-1}(\tanh (a+b x))}-\frac {((1-m) m) \int \frac {x^{-2+m}}{\coth ^{-1}(\tanh (a+b x))} \, dx}{2 b^2} \\ & = -\frac {x^m}{2 b \coth ^{-1}(\tanh (a+b x))^2}-\frac {m x^{-1+m}}{2 b^2 \coth ^{-1}(\tanh (a+b x))}-\frac {m x^{-1+m} \operatorname {Hypergeometric2F1}\left (1,-1+m,m,\frac {b x}{b x-\coth ^{-1}(\tanh (a+b x))}\right )}{2 b^2 \left (b x-\coth ^{-1}(\tanh (a+b x))\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.88 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.54 \[ \int \frac {x^m}{\coth ^{-1}(\tanh (a+b x))^3} \, dx=\frac {x^{1+m} \operatorname {Hypergeometric2F1}\left (3,1+m,2+m,-\frac {b x}{-b x+\coth ^{-1}(\tanh (a+b x))}\right )}{(1+m) \left (-b x+\coth ^{-1}(\tanh (a+b x))\right )^3} \]

[In]

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

[Out]

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

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

int(x^m/acoth(tanh(a + b*x))^3,x)

[Out]

int(x^m/acoth(tanh(a + b*x))^3, x)