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

Optimal. Leaf size=65 \[ -\frac{x^m \text{Hypergeometric2F1}\left (1,m,m+1,\frac{b x}{b x-\coth ^{-1}(\tanh (a+b x))}\right )}{b \left (b x-\coth ^{-1}(\tanh (a+b x))\right )}-\frac{x^m}{b \coth ^{-1}(\tanh (a+b x))} \]

[Out]

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

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Rubi [A]  time = 0.0390504, antiderivative size = 65, 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} \[ -\frac{x^m \, _2F_1\left (1,m;m+1;\frac{b x}{b x-\coth ^{-1}(\tanh (a+b x))}\right )}{b \left (b x-\coth ^{-1}(\tanh (a+b x))\right )}-\frac{x^m}{b \coth ^{-1}(\tanh (a+b x))} \]

Antiderivative was successfully verified.

[In]

Int[x^m/ArcCoth[Tanh[a + b*x]]^2,x]

[Out]

-(x^m/(b*ArcCoth[Tanh[a + b*x]])) - (x^m*Hypergeometric2F1[1, m, 1 + m, (b*x)/(b*x - ArcCoth[Tanh[a + b*x]])])
/(b*(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{x^m}{\coth ^{-1}(\tanh (a+b x))^2} \, dx &=-\frac{x^m}{b \coth ^{-1}(\tanh (a+b x))}+\frac{m \int \frac{x^{-1+m}}{\coth ^{-1}(\tanh (a+b x))} \, dx}{b}\\ &=-\frac{x^m}{b \coth ^{-1}(\tanh (a+b x))}-\frac{x^m \, _2F_1\left (1,m;1+m;\frac{b x}{b x-\coth ^{-1}(\tanh (a+b x))}\right )}{b \left (b x-\coth ^{-1}(\tanh (a+b x))\right )}\\ \end{align*}

Mathematica [A]  time = 0.521421, size = 51, normalized size = 0.78 \[ \frac{x^{m+1} \text{Hypergeometric2F1}\left (2,m+1,m+2,-\frac{b x}{\coth ^{-1}(\tanh (a+b x))-b x}\right )}{(m+1) \left (\coth ^{-1}(\tanh (a+b x))-b x\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 1.83, size = 0, normalized size = 0. \begin{align*} \int{\frac{{x}^{m}}{ \left ({\rm arccoth} \left (\tanh \left ( bx+a \right ) \right ) \right ) ^{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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