3.62 \(\int \frac{\tanh ^{-1}(\tanh (a+b x))^3}{x^4} \, dx\)

Optimal. Leaf size=55 \[ -\frac{b^2 \tanh ^{-1}(\tanh (a+b x))}{x}-\frac{b \tanh ^{-1}(\tanh (a+b x))^2}{2 x^2}-\frac{\tanh ^{-1}(\tanh (a+b x))^3}{3 x^3}+b^3 \log (x) \]

[Out]

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

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Rubi [A]  time = 0.0368214, antiderivative size = 55, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {2168, 29} \[ -\frac{b^2 \tanh ^{-1}(\tanh (a+b x))}{x}-\frac{b \tanh ^{-1}(\tanh (a+b x))^2}{2 x^2}-\frac{\tanh ^{-1}(\tanh (a+b x))^3}{3 x^3}+b^3 \log (x) \]

Antiderivative was successfully verified.

[In]

Int[ArcTanh[Tanh[a + b*x]]^3/x^4,x]

[Out]

-((b^2*ArcTanh[Tanh[a + b*x]])/x) - (b*ArcTanh[Tanh[a + b*x]]^2)/(2*x^2) - ArcTanh[Tanh[a + b*x]]^3/(3*x^3) +
b^3*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 29

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

Rubi steps

\begin{align*} \int \frac{\tanh ^{-1}(\tanh (a+b x))^3}{x^4} \, dx &=-\frac{\tanh ^{-1}(\tanh (a+b x))^3}{3 x^3}+b \int \frac{\tanh ^{-1}(\tanh (a+b x))^2}{x^3} \, dx\\ &=-\frac{b \tanh ^{-1}(\tanh (a+b x))^2}{2 x^2}-\frac{\tanh ^{-1}(\tanh (a+b x))^3}{3 x^3}+b^2 \int \frac{\tanh ^{-1}(\tanh (a+b x))}{x^2} \, dx\\ &=-\frac{b^2 \tanh ^{-1}(\tanh (a+b x))}{x}-\frac{b \tanh ^{-1}(\tanh (a+b x))^2}{2 x^2}-\frac{\tanh ^{-1}(\tanh (a+b x))^3}{3 x^3}+b^3 \int \frac{1}{x} \, dx\\ &=-\frac{b^2 \tanh ^{-1}(\tanh (a+b x))}{x}-\frac{b \tanh ^{-1}(\tanh (a+b x))^2}{2 x^2}-\frac{\tanh ^{-1}(\tanh (a+b x))^3}{3 x^3}+b^3 \log (x)\\ \end{align*}

Mathematica [A]  time = 0.0241028, size = 60, normalized size = 1.09 \[ \frac{-6 b^2 x^2 \tanh ^{-1}(\tanh (a+b x))-3 b x \tanh ^{-1}(\tanh (a+b x))^2-2 \tanh ^{-1}(\tanh (a+b x))^3+b^3 x^3 (6 \log (x)+11)}{6 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcTanh[Tanh[a + b*x]]^3/x^4,x]

[Out]

(-6*b^2*x^2*ArcTanh[Tanh[a + b*x]] - 3*b*x*ArcTanh[Tanh[a + b*x]]^2 - 2*ArcTanh[Tanh[a + b*x]]^3 + b^3*x^3*(11
+ 6*Log[x]))/(6*x^3)

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Maple [A]  time = 0.037, size = 52, normalized size = 1. \begin{align*} -{\frac{{b}^{2}{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }{x}}-{\frac{b \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{2}}{2\,{x}^{2}}}-{\frac{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{3}}{3\,{x}^{3}}}+{b}^{3}\ln \left ( x \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctanh(tanh(b*x+a))^3/x^4,x)

[Out]

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

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Maxima [A]  time = 1.59454, size = 70, normalized size = 1.27 \begin{align*}{\left (b^{2} \log \left (x\right ) - \frac{b \operatorname{artanh}\left (\tanh \left (b x + a\right )\right )}{x}\right )} b - \frac{b \operatorname{artanh}\left (\tanh \left (b x + a\right )\right )^{2}}{2 \, x^{2}} - \frac{\operatorname{artanh}\left (\tanh \left (b x + a\right )\right )^{3}}{3 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^3/x^4,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.44155, size = 85, normalized size = 1.55 \begin{align*} \frac{6 \, b^{3} x^{3} \log \left (x\right ) - 18 \, a b^{2} x^{2} - 9 \, a^{2} b x - 2 \, a^{3}}{6 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^3/x^4,x, algorithm="fricas")

[Out]

1/6*(6*b^3*x^3*log(x) - 18*a*b^2*x^2 - 9*a^2*b*x - 2*a^3)/x^3

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Sympy [A]  time = 6.21766, size = 51, normalized size = 0.93 \begin{align*} b^{3} \log{\left (x \right )} - \frac{b^{2} \operatorname{atanh}{\left (\tanh{\left (a + b x \right )} \right )}}{x} - \frac{b \operatorname{atanh}^{2}{\left (\tanh{\left (a + b x \right )} \right )}}{2 x^{2}} - \frac{\operatorname{atanh}^{3}{\left (\tanh{\left (a + b x \right )} \right )}}{3 x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atanh(tanh(b*x+a))**3/x**4,x)

[Out]

b**3*log(x) - b**2*atanh(tanh(a + b*x))/x - b*atanh(tanh(a + b*x))**2/(2*x**2) - atanh(tanh(a + b*x))**3/(3*x*
*3)

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Giac [A]  time = 1.14549, size = 47, normalized size = 0.85 \begin{align*} b^{3} \log \left ({\left | x \right |}\right ) - \frac{18 \, a b^{2} x^{2} + 9 \, a^{2} b x + 2 \, a^{3}}{6 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^3/x^4,x, algorithm="giac")

[Out]

b^3*log(abs(x)) - 1/6*(18*a*b^2*x^2 + 9*a^2*b*x + 2*a^3)/x^3