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

Optimal. Leaf size=23 \[ -\frac{\tanh ^{-1}(\tanh (a+b x))}{3 x^3}-\frac{b}{6 x^2} \]

[Out]

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

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Rubi [A]  time = 0.0081961, antiderivative size = 23, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {2168, 30} \[ -\frac{\tanh ^{-1}(\tanh (a+b x))}{3 x^3}-\frac{b}{6 x^2} \]

Antiderivative was successfully verified.

[In]

Int[ArcTanh[Tanh[a + b*x]]/x^4,x]

[Out]

-b/(6*x^2) - ArcTanh[Tanh[a + b*x]]/(3*x^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 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

\begin{align*} \int \frac{\tanh ^{-1}(\tanh (a+b x))}{x^4} \, dx &=-\frac{\tanh ^{-1}(\tanh (a+b x))}{3 x^3}+\frac{1}{3} b \int \frac{1}{x^3} \, dx\\ &=-\frac{b}{6 x^2}-\frac{\tanh ^{-1}(\tanh (a+b x))}{3 x^3}\\ \end{align*}

Mathematica [A]  time = 0.0146518, size = 20, normalized size = 0.87 \[ -\frac{2 \tanh ^{-1}(\tanh (a+b x))+b x}{6 x^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.033, size = 20, normalized size = 0.9 \begin{align*} -{\frac{b}{6\,{x}^{2}}}-{\frac{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }{3\,{x}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.15543, size = 26, normalized size = 1.13 \begin{align*} -\frac{b}{6 \, x^{2}} - \frac{\operatorname{artanh}\left (\tanh \left (b x + a\right )\right )}{3 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.87144, size = 32, normalized size = 1.39 \begin{align*} -\frac{3 \, b x + 2 \, a}{6 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(3*b*x + 2*a)/x^3

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Sympy [A]  time = 7.86199, size = 20, normalized size = 0.87 \begin{align*} - \frac{b}{6 x^{2}} - \frac{\operatorname{atanh}{\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))/x**4,x)

[Out]

-b/(6*x**2) - atanh(tanh(a + b*x))/(3*x**3)

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Giac [A]  time = 1.12839, size = 18, normalized size = 0.78 \begin{align*} -\frac{3 \, b x + 2 \, a}{6 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(3*b*x + 2*a)/x^3