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

Optimal. Leaf size=50 \[ \frac{2 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \log \left (\tanh ^{-1}(\tanh (a+b x))\right )}{b^3}-\frac{x^2}{b \tanh ^{-1}(\tanh (a+b x))}+\frac{2 x}{b^2} \]

[Out]

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

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Rubi [A]  time = 0.0340511, antiderivative size = 50, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.308, Rules used = {2168, 2158, 2157, 29} \[ \frac{2 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \log \left (\tanh ^{-1}(\tanh (a+b x))\right )}{b^3}-\frac{x^2}{b \tanh ^{-1}(\tanh (a+b x))}+\frac{2 x}{b^2} \]

Antiderivative was successfully verified.

[In]

Int[x^2/ArcTanh[Tanh[a + b*x]]^2,x]

[Out]

(2*x)/b^2 - x^2/(b*ArcTanh[Tanh[a + b*x]]) + (2*(b*x - 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 2158

Int[(v_)/(u_), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, Simp[(b*x)/a, x] - Dist[(b*u
- a*v)/a, Int[1/u, x], x] /; NeQ[b*u - a*v, 0]] /; PiecewiseLinearQ[u, v, x]

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

Mathematica [A]  time = 0.0634883, size = 56, normalized size = 1.12 \[ \frac{-\frac{\left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^2}{\tanh ^{-1}(\tanh (a+b x))}+2 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \log \left (\tanh ^{-1}(\tanh (a+b x))\right )+b x}{b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.04, size = 127, normalized size = 2.5 \begin{align*}{\frac{x}{{b}^{2}}}-2\,{\frac{\ln \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) a}{{b}^{3}}}-2\,{\frac{\ln \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) }{{b}^{3}}}-{\frac{{a}^{2}}{{b}^{3}{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }}-2\,{\frac{a \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) }{{b}^{3}{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }}-{\frac{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) ^{2}}{{b}^{3}{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x/b^2-2/b^3*ln(arctanh(tanh(b*x+a)))*a-2/b^3*ln(arctanh(tanh(b*x+a)))*(arctanh(tanh(b*x+a))-b*x-a)-1/b^3/arcta
nh(tanh(b*x+a))*a^2-2/b^3/arctanh(tanh(b*x+a))*a*(arctanh(tanh(b*x+a))-b*x-a)-1/b^3/arctanh(tanh(b*x+a))*(arct
anh(tanh(b*x+a))-b*x-a)^2

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Maxima [A]  time = 2.40512, size = 59, normalized size = 1.18 \begin{align*} \frac{b^{2} x^{2} + a b x - a^{2}}{b^{4} x + a b^{3}} - \frac{2 \, a \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))^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.52317, size = 97, normalized size = 1.94 \begin{align*} \frac{b^{2} x^{2} + a b x - a^{2} - 2 \,{\left (a b x + a^{2}\right )} \log \left (b x + a\right )}{b^{4} x + a b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{2}}{\operatorname{atanh}^{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**2/atanh(tanh(b*x+a))**2,x)

[Out]

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

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Giac [A]  time = 1.15878, size = 46, normalized size = 0.92 \begin{align*} \frac{x}{b^{2}} - \frac{2 \, a \log \left ({\left | b x + a \right |}\right )}{b^{3}} - \frac{a^{2}}{{\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))^2,x, algorithm="giac")

[Out]

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