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

Optimal. Leaf size=34 \[ \frac{x \tanh ^{-1}(\tanh (a+b x))^3}{3 b}-\frac{\tanh ^{-1}(\tanh (a+b x))^4}{12 b^2} \]

[Out]

(x*ArcTanh[Tanh[a + b*x]]^3)/(3*b) - ArcTanh[Tanh[a + b*x]]^4/(12*b^2)

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Rubi [A]  time = 0.020971, antiderivative size = 34, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {2168, 2157, 30} \[ \frac{x \tanh ^{-1}(\tanh (a+b x))^3}{3 b}-\frac{\tanh ^{-1}(\tanh (a+b x))^4}{12 b^2} \]

Antiderivative was successfully verified.

[In]

Int[x*ArcTanh[Tanh[a + b*x]]^2,x]

[Out]

(x*ArcTanh[Tanh[a + b*x]]^3)/(3*b) - ArcTanh[Tanh[a + b*x]]^4/(12*b^2)

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 2157

Int[(u_)^(m_.), x_Symbol] :> With[{c = Simplify[D[u, x]]}, Dist[1/c, Subst[Int[x^m, x], x, u], x]] /; FreeQ[m,
x] && PiecewiseLinearQ[u, x]

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 \tanh ^{-1}(\tanh (a+b x))^2 \, dx &=\frac{x \tanh ^{-1}(\tanh (a+b x))^3}{3 b}-\frac{\int \tanh ^{-1}(\tanh (a+b x))^3 \, dx}{3 b}\\ &=\frac{x \tanh ^{-1}(\tanh (a+b x))^3}{3 b}-\frac{\operatorname{Subst}\left (\int x^3 \, dx,x,\tanh ^{-1}(\tanh (a+b x))\right )}{3 b^2}\\ &=\frac{x \tanh ^{-1}(\tanh (a+b x))^3}{3 b}-\frac{\tanh ^{-1}(\tanh (a+b x))^4}{12 b^2}\\ \end{align*}

Mathematica [B]  time = 0.0751046, size = 74, normalized size = 2.18 \[ \frac{(a+b x) \left (4 \left (2 a^2+a b x-b^2 x^2\right ) \tanh ^{-1}(\tanh (a+b x))-(3 a-b x) (a+b x)^2-6 (a-b x) \tanh ^{-1}(\tanh (a+b x))^2\right )}{12 b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b*x)*(-((3*a - b*x)*(a + b*x)^2) + 4*(2*a^2 + a*b*x - b^2*x^2)*ArcTanh[Tanh[a + b*x]] - 6*(a - b*x)*ArcT
anh[Tanh[a + b*x]]^2))/(12*b^2)

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Maple [A]  time = 0.035, size = 38, normalized size = 1.1 \begin{align*}{\frac{{x}^{2} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{2}}{2}}-b \left ( -{\frac{b{x}^{4}}{12}}+{\frac{{x}^{3}{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }{3}} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x^2*arctanh(tanh(b*x+a))^2-b*(-1/12*b*x^4+1/3*x^3*arctanh(tanh(b*x+a)))

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Maxima [A]  time = 1.35548, size = 49, normalized size = 1.44 \begin{align*} \frac{1}{12} \, b^{2} x^{4} - \frac{1}{3} \, b x^{3} \operatorname{artanh}\left (\tanh \left (b x + a\right )\right ) + \frac{1}{2} \, x^{2} \operatorname{artanh}\left (\tanh \left (b x + a\right )\right )^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*b^2*x^4 - 1/3*b*x^3*arctanh(tanh(b*x + a)) + 1/2*x^2*arctanh(tanh(b*x + a))^2

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Fricas [A]  time = 1.43912, size = 55, normalized size = 1.62 \begin{align*} \frac{1}{4} \, b^{2} x^{4} + \frac{2}{3} \, a b x^{3} + \frac{1}{2} \, a^{2} x^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*b^2*x^4 + 2/3*a*b*x^3 + 1/2*a^2*x^2

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Sympy [A]  time = 0.825643, size = 37, normalized size = 1.09 \begin{align*} \frac{b^{2} x^{4}}{12} - \frac{b x^{3} \operatorname{atanh}{\left (\tanh{\left (a + b x \right )} \right )}}{3} + \frac{x^{2} \operatorname{atanh}^{2}{\left (\tanh{\left (a + b x \right )} \right )}}{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*atanh(tanh(b*x+a))**2,x)

[Out]

b**2*x**4/12 - b*x**3*atanh(tanh(a + b*x))/3 + x**2*atanh(tanh(a + b*x))**2/2

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Giac [A]  time = 1.14194, size = 32, normalized size = 0.94 \begin{align*} \frac{1}{4} \, b^{2} x^{4} + \frac{2}{3} \, a b x^{3} + \frac{1}{2} \, a^{2} x^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*b^2*x^4 + 2/3*a*b*x^3 + 1/2*a^2*x^2