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

Optimal. Leaf size=23 \[ \frac{1}{2} x^2 \tanh ^{-1}(\tanh (a+b x))-\frac{b x^3}{6} \]

[Out]

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

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Rubi [A]  time = 0.0068111, antiderivative size = 23, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 9, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {6239, 30} \[ \frac{1}{2} x^2 \tanh ^{-1}(\tanh (a+b x))-\frac{b x^3}{6} \]

Antiderivative was successfully verified.

[In]

Int[x*ArcTanh[Tanh[a + b*x]],x]

[Out]

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

Rule 6239

Int[ArcTanh[(c_.) + (d_.)*Tanh[(a_.) + (b_.)*(x_)]]*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[((e + f*x)^(
m + 1)*ArcTanh[c + d*Tanh[a + b*x]])/(f*(m + 1)), x] + Dist[b/(f*(m + 1)), Int[(e + f*x)^(m + 1)/(c - d + c*E^
(2*a + 2*b*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[m, 0] && EqQ[(c - d)^2, 1]

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

Mathematica [A]  time = 0.0150803, size = 20, normalized size = 0.87 \[ -\frac{1}{6} x^2 \left (b x-3 \tanh ^{-1}(\tanh (a+b x))\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x*ArcTanh[Tanh[a + b*x]],x]

[Out]

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

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Maple [A]  time = 0.03, size = 20, normalized size = 0.9 \begin{align*} -{\frac{b{x}^{3}}{6}}+{\frac{{x}^{2}{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*b*x^3+1/2*x^2*arctanh(tanh(b*x+a))

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Maxima [A]  time = 1.14261, size = 26, normalized size = 1.13 \begin{align*} -\frac{1}{6} \, b x^{3} + \frac{1}{2} \, x^{2} \operatorname{artanh}\left (\tanh \left (b x + a\right )\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*b*x^3 + 1/2*x^2*arctanh(tanh(b*x + a))

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Fricas [A]  time = 1.61921, size = 31, normalized size = 1.35 \begin{align*} \frac{1}{3} \, b x^{3} + \frac{1}{2} \, a x^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*b*x^3 + 1/2*a*x^2

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Sympy [A]  time = 3.21176, size = 39, normalized size = 1.7 \begin{align*} \begin{cases} \frac{x \operatorname{atanh}^{2}{\left (\tanh{\left (a + b x \right )} \right )}}{2 b} - \frac{\operatorname{atanh}^{3}{\left (\tanh{\left (a + b x \right )} \right )}}{6 b^{2}} & \text{for}\: b \neq 0 \\\frac{x^{2} \operatorname{atanh}{\left (\tanh{\left (a \right )} \right )}}{2} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((x*atanh(tanh(a + b*x))**2/(2*b) - atanh(tanh(a + b*x))**3/(6*b**2), Ne(b, 0)), (x**2*atanh(tanh(a))
/2, True))

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Giac [A]  time = 1.12741, size = 18, normalized size = 0.78 \begin{align*} \frac{1}{3} \, b x^{3} + \frac{1}{2} \, a x^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*b*x^3 + 1/2*a*x^2