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

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

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 23, normalized size of antiderivative = 1.00, 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} \[ -\frac {\coth ^{-1}(\tanh (a+b x))}{3 x^3}-\frac {b}{6 x^2} \]

Antiderivative was successfully verified.

[In]

Int[ArcCoth[Tanh[a + b*x]]/x^4,x]

[Out]

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

Rule 30

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

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]))

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 20, normalized size = 0.87 \[ -\frac {2 \coth ^{-1}(\tanh (a+b x))+b x}{6 x^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.57, size = 13, normalized size = 0.57 \[ -\frac {3 \, b x + 2 \, a}{6 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(3*b*x + 2*a)/x^3

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {arcoth}\left (\tanh \left (b x + a\right )\right )}{x^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arccoth(tanh(b*x + a))/x^4, x)

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maple [A]  time = 0.40, size = 20, normalized size = 0.87 \[ -\frac {b}{6 x^{2}}-\frac {\mathrm {arccoth}\left (\tanh \left (b x +a \right )\right )}{3 x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.38, size = 19, normalized size = 0.83 \[ -\frac {b}{6 \, x^{2}} - \frac {\operatorname {arcoth}\left (\tanh \left (b x + a\right )\right )}{3 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.12, size = 19, normalized size = 0.83 \[ -\frac {\mathrm {acoth}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}{3\,x^3}-\frac {b}{6\,x^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(acoth(tanh(a + b*x))/x^4,x)

[Out]

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

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sympy [A]  time = 0.79, size = 20, normalized size = 0.87 \[ - \frac {b}{6 x^{2}} - \frac {\operatorname {acoth}{\left (\tanh {\left (a + b x \right )} \right )}}{3 x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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