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

Optimal. Leaf size=68 \[ -3 b^2 x \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )-\frac{\tanh ^{-1}(\tanh (a+b x))^3}{x}+\frac{3}{2} b \tanh ^{-1}(\tanh (a+b x))^2+3 b \log (x) \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0412704, antiderivative size = 68, 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, 2159, 2158, 29} \[ -3 b^2 x \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )-\frac{\tanh ^{-1}(\tanh (a+b x))^3}{x}+\frac{3}{2} b \tanh ^{-1}(\tanh (a+b x))^2+3 b \log (x) \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \]

Antiderivative was successfully verified.

[In]

Int[ArcTanh[Tanh[a + b*x]]^3/x^2,x]

[Out]

-3*b^2*x*(b*x - ArcTanh[Tanh[a + b*x]]) + (3*b*ArcTanh[Tanh[a + b*x]]^2)/2 - ArcTanh[Tanh[a + b*x]]^3/x + 3*b*
(b*x - ArcTanh[Tanh[a + b*x]])^2*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 2159

Int[(v_)^(n_)/(u_), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, Simp[v^n/(a*n), x] - Dis
t[(b*u - a*v)/a, Int[v^(n - 1)/u, x], x] /; NeQ[b*u - a*v, 0]] /; PiecewiseLinearQ[u, v, x] && GtQ[n, 0] && Ne
Q[n, 1]

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

Mathematica [A]  time = 0.0381505, size = 62, normalized size = 0.91 \[ -6 b^2 x \log (x) \tanh ^{-1}(\tanh (a+b x))-\frac{\tanh ^{-1}(\tanh (a+b x))^3}{x}+3 b (\log (x)+1) \tanh ^{-1}(\tanh (a+b x))^2+\frac{3}{2} b^3 x^2 (2 \log (x)-1) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(ArcTanh[Tanh[a + b*x]]^3/x) - 6*b^2*x*ArcTanh[Tanh[a + b*x]]*Log[x] + 3*b*ArcTanh[Tanh[a + b*x]]^2*(1 + Log[
x]) + (3*b^3*x^2*(-1 + 2*Log[x]))/2

________________________________________________________________________________________

Maple [A]  time = 0.04, size = 76, normalized size = 1.1 \begin{align*} -{\frac{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{3}}{x}}+3\,\ln \left ( x \right ) \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{2}b+3\,{b}^{3}{x}^{2}\ln \left ( x \right ) -{\frac{9\,{b}^{3}{x}^{2}}{2}}-6\,{b}^{2}{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \ln \left ( x \right ) x+6\,{b}^{2}{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-arctanh(tanh(b*x+a))^3/x+3*ln(x)*arctanh(tanh(b*x+a))^2*b+3*b^3*x^2*ln(x)-9/2*b^3*x^2-6*b^2*arctanh(tanh(b*x+
a))*ln(x)*x+6*b^2*arctanh(tanh(b*x+a))*x

________________________________________________________________________________________

Maxima [A]  time = 2.58652, size = 88, normalized size = 1.29 \begin{align*} 3 \, b \operatorname{artanh}\left (\tanh \left (b x + a\right )\right )^{2} \log \left (x\right ) + \frac{3}{2} \,{\left (b^{2} x^{2} + 4 \, a b x + 2 \, a^{2} \log \left (x\right ) - 2 \, \operatorname{artanh}\left (\tanh \left (b x + a\right )\right )^{2} \log \left (x\right )\right )} b - \frac{\operatorname{artanh}\left (\tanh \left (b x + a\right )\right )^{3}}{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 1.50572, size = 78, normalized size = 1.15 \begin{align*} \frac{b^{3} x^{3} + 6 \, a b^{2} x^{2} + 6 \, a^{2} b x \log \left (x\right ) - 2 \, a^{3}}{2 \, x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{atanh}^{3}{\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))**3/x**2,x)

[Out]

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

________________________________________________________________________________________

Giac [A]  time = 1.10957, size = 45, normalized size = 0.66 \begin{align*} \frac{1}{2} \, b^{3} x^{2} + 3 \, a b^{2} x + 3 \, a^{2} b \log \left ({\left | x \right |}\right ) - \frac{a^{3}}{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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