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

Optimal. Leaf size=71 \[ -\frac{2 b x^{m+2} \coth ^{-1}(\tanh (a+b x))}{m^2+3 m+2}+\frac{x^{m+1} \coth ^{-1}(\tanh (a+b x))^2}{m+1}+\frac{2 b^2 x^{m+3}}{m^3+6 m^2+11 m+6} \]

[Out]

(2*b^2*x^(3 + m))/(6 + 11*m + 6*m^2 + m^3) - (2*b*x^(2 + m)*ArcCoth[Tanh[a + b*x]])/(2 + 3*m + m^2) + (x^(1 +
m)*ArcCoth[Tanh[a + b*x]]^2)/(1 + m)

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Rubi [A]  time = 0.0352614, antiderivative size = 71, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {2168, 30} \[ -\frac{2 b x^{m+2} \coth ^{-1}(\tanh (a+b x))}{m^2+3 m+2}+\frac{x^{m+1} \coth ^{-1}(\tanh (a+b x))^2}{m+1}+\frac{2 b^2 x^{m+3}}{m^3+6 m^2+11 m+6} \]

Antiderivative was successfully verified.

[In]

Int[x^m*ArcCoth[Tanh[a + b*x]]^2,x]

[Out]

(2*b^2*x^(3 + m))/(6 + 11*m + 6*m^2 + m^3) - (2*b*x^(2 + m)*ArcCoth[Tanh[a + b*x]])/(2 + 3*m + m^2) + (x^(1 +
m)*ArcCoth[Tanh[a + b*x]]^2)/(1 + m)

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 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

\begin{align*} \int x^m \coth ^{-1}(\tanh (a+b x))^2 \, dx &=\frac{x^{1+m} \coth ^{-1}(\tanh (a+b x))^2}{1+m}-\frac{(2 b) \int x^{1+m} \coth ^{-1}(\tanh (a+b x)) \, dx}{1+m}\\ &=-\frac{2 b x^{2+m} \coth ^{-1}(\tanh (a+b x))}{2+3 m+m^2}+\frac{x^{1+m} \coth ^{-1}(\tanh (a+b x))^2}{1+m}+\frac{\left (2 b^2\right ) \int x^{2+m} \, dx}{2+3 m+m^2}\\ &=\frac{2 b^2 x^{3+m}}{6+11 m+6 m^2+m^3}-\frac{2 b x^{2+m} \coth ^{-1}(\tanh (a+b x))}{2+3 m+m^2}+\frac{x^{1+m} \coth ^{-1}(\tanh (a+b x))^2}{1+m}\\ \end{align*}

Mathematica [A]  time = 0.11794, size = 62, normalized size = 0.87 \[ \frac{x^{m+1} \left (\left (m^2+5 m+6\right ) \coth ^{-1}(\tanh (a+b x))^2-2 b (m+3) x \coth ^{-1}(\tanh (a+b x))+2 b^2 x^2\right )}{(m+1) (m+2) (m+3)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x^(1 + m)*(2*b^2*x^2 - 2*b*(3 + m)*x*ArcCoth[Tanh[a + b*x]] + (6 + 5*m + m^2)*ArcCoth[Tanh[a + b*x]]^2))/((1
+ m)*(2 + m)*(3 + m))

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Maple [C]  time = 0.773, size = 9175, normalized size = 129.2 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \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]

Exception raised: ValueError

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Fricas [A]  time = 1.74427, size = 223, normalized size = 3.14 \begin{align*} \frac{{\left (4 \,{\left (b^{2} m^{2} + 3 \, b^{2} m + 2 \, b^{2}\right )} x^{3} + 8 \,{\left (a b m^{2} + 4 \, a b m + 3 \, a b\right )} x^{2} +{\left (4 \, a^{2} m^{2} - \pi ^{2}{\left (m^{2} + 5 \, m + 6\right )} + 20 \, a^{2} m + 24 \, a^{2}\right )} x\right )} x^{m}}{4 \,{\left (m^{3} + 6 \, m^{2} + 11 \, m + 6\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]

1/4*(4*(b^2*m^2 + 3*b^2*m + 2*b^2)*x^3 + 8*(a*b*m^2 + 4*a*b*m + 3*a*b)*x^2 + (4*a^2*m^2 - pi^2*(m^2 + 5*m + 6)
+ 20*a^2*m + 24*a^2)*x)*x^m/(m^3 + 6*m^2 + 11*m + 6)

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Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int 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)