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

Optimal. Leaf size=39 \[ -\frac{\tanh ^{-1}(\tanh (a+b x))^2}{x}-2 b \log (x) \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )+2 b^2 x \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0243156, antiderivative size = 39, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {2168, 2158, 29} \[ -\frac{\tanh ^{-1}(\tanh (a+b x))^2}{x}-2 b \log (x) \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )+2 b^2 x \]

Antiderivative was successfully verified.

[In]

Int[ArcTanh[Tanh[a + b*x]]^2/x^2,x]

[Out]

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

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 29

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

Rubi steps

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

Mathematica [A]  time = 0.0449575, size = 37, normalized size = 0.95 \[ -\frac{\tanh ^{-1}(\tanh (a+b x))^2}{x}+2 b (\log (x)+1) \tanh ^{-1}(\tanh (a+b x))-2 b^2 x \log (x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(ArcTanh[Tanh[a + b*x]]^2/x) - 2*b^2*x*Log[x] + 2*b*ArcTanh[Tanh[a + b*x]]*(1 + Log[x])

________________________________________________________________________________________

Maple [A]  time = 0.04, size = 42, normalized size = 1.1 \begin{align*} -{\frac{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{2}}{x}}-2\,\ln \left ( x \right ) x{b}^{2}+2\,\ln \left ( x \right ){\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) b+2\,{b}^{2}x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.19392, size = 73, normalized size = 1.87 \begin{align*} 2 \, b \operatorname{artanh}\left (\tanh \left (b x + a\right )\right ) \log \left (x\right ) - 2 \,{\left (b{\left (x + \frac{a}{b}\right )} \log \left (x\right ) - b{\left (x + \frac{a \log \left (x\right )}{b}\right )}\right )} b - \frac{\operatorname{artanh}\left (\tanh \left (b x + a\right )\right )^{2}}{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*b*arctanh(tanh(b*x + a))*log(x) - 2*(b*(x + a/b)*log(x) - b*(x + a*log(x)/b))*b - arctanh(tanh(b*x + a))^2/x

________________________________________________________________________________________

Fricas [A]  time = 1.48802, size = 49, normalized size = 1.26 \begin{align*} \frac{b^{2} x^{2} + 2 \, a b x \log \left (x\right ) - a^{2}}{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b^2*x^2 + 2*a*b*x*log(x) - a^2)/x

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [A]  time = 1.13201, size = 28, normalized size = 0.72 \begin{align*} b^{2} x + 2 \, a b \log \left ({\left | x \right |}\right ) - \frac{a^{2}}{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b^2*x + 2*a*b*log(abs(x)) - a^2/x