3.105 \(\int \frac{x^2}{\tanh ^{-1}(\tanh (a+b x))^3} \, dx\)

Optimal. Leaf size=47 \[ -\frac{x}{b^2 \tanh ^{-1}(\tanh (a+b x))}+\frac{\log \left (\tanh ^{-1}(\tanh (a+b x))\right )}{b^3}-\frac{x^2}{2 b \tanh ^{-1}(\tanh (a+b x))^2} \]

[Out]

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

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Rubi [A]  time = 0.0287172, antiderivative size = 47, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {2168, 2157, 29} \[ -\frac{x}{b^2 \tanh ^{-1}(\tanh (a+b x))}+\frac{\log \left (\tanh ^{-1}(\tanh (a+b x))\right )}{b^3}-\frac{x^2}{2 b \tanh ^{-1}(\tanh (a+b x))^2} \]

Antiderivative was successfully verified.

[In]

Int[x^2/ArcTanh[Tanh[a + b*x]]^3,x]

[Out]

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

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 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rubi steps

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

Mathematica [A]  time = 0.0316702, size = 49, normalized size = 1.04 \[ \frac{-\frac{b^2 x^2}{\tanh ^{-1}(\tanh (a+b x))^2}-\frac{2 b x}{\tanh ^{-1}(\tanh (a+b x))}+2 \log \left (\tanh ^{-1}(\tanh (a+b x))\right )+3}{2 b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2/ArcTanh[Tanh[a + b*x]]^3,x]

[Out]

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

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Maple [B]  time = 0.04, size = 136, normalized size = 2.9 \begin{align*} 2\,{\frac{a}{{b}^{3}{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }}+2\,{\frac{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a}{{b}^{3}{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }}+{\frac{\ln \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) }{{b}^{3}}}-{\frac{{a}^{2}}{2\,{b}^{3} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{2}}}-{\frac{a \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) }{{b}^{3} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{2}}}-{\frac{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) ^{2}}{2\,{b}^{3} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/arctanh(tanh(b*x+a))^3,x)

[Out]

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

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Maxima [A]  time = 3.47843, size = 65, normalized size = 1.38 \begin{align*} \frac{4 \, a b x + 3 \, a^{2}}{2 \,{\left (b^{5} x^{2} + 2 \, a b^{4} x + a^{2} b^{3}\right )}} + \frac{\log \left (b x + a\right )}{b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/arctanh(tanh(b*x+a))^3,x, algorithm="maxima")

[Out]

1/2*(4*a*b*x + 3*a^2)/(b^5*x^2 + 2*a*b^4*x + a^2*b^3) + log(b*x + a)/b^3

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Fricas [A]  time = 1.47192, size = 132, normalized size = 2.81 \begin{align*} \frac{4 \, a b x + 3 \, a^{2} + 2 \,{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \log \left (b x + a\right )}{2 \,{\left (b^{5} x^{2} + 2 \, a b^{4} x + a^{2} b^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/arctanh(tanh(b*x+a))^3,x, algorithm="fricas")

[Out]

1/2*(4*a*b*x + 3*a^2 + 2*(b^2*x^2 + 2*a*b*x + a^2)*log(b*x + a))/(b^5*x^2 + 2*a*b^4*x + a^2*b^3)

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Sympy [A]  time = 22.1723, size = 54, normalized size = 1.15 \begin{align*} \begin{cases} - \frac{x^{2}}{2 b \operatorname{atanh}^{2}{\left (\tanh{\left (a + b x \right )} \right )}} - \frac{x}{b^{2} \operatorname{atanh}{\left (\tanh{\left (a + b x \right )} \right )}} + \frac{\log{\left (\operatorname{atanh}{\left (\tanh{\left (a + b x \right )} \right )} \right )}}{b^{3}} & \text{for}\: b \neq 0 \\\frac{x^{3}}{3 \operatorname{atanh}^{3}{\left (\tanh{\left (a \right )} \right )}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2/atanh(tanh(b*x+a))**3,x)

[Out]

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

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Giac [A]  time = 1.13653, size = 50, normalized size = 1.06 \begin{align*} \frac{\log \left ({\left | b x + a \right |}\right )}{b^{3}} + \frac{4 \, a x + \frac{3 \, a^{2}}{b}}{2 \,{\left (b x + a\right )}^{2} b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/arctanh(tanh(b*x+a))^3,x, algorithm="giac")

[Out]

log(abs(b*x + a))/b^3 + 1/2*(4*a*x + 3*a^2/b)/((b*x + a)^2*b^2)