3.199 \(\int \frac {\coth ^{-1}(\coth (a+b x))}{x^3} \, dx\)

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

[Out]

-1/2*b/x-1/2*arccoth(coth(b*x+a))/x^2

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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} \[ -\frac {\coth ^{-1}(\coth (a+b x))}{2 x^2}-\frac {b}{2 x} \]

Antiderivative was successfully verified.

[In]

Int[ArcCoth[Coth[a + b*x]]/x^3,x]

[Out]

-b/(2*x) - ArcCoth[Coth[a + b*x]]/(2*x^2)

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

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Mathematica [A]  time = 0.02, size = 18, normalized size = 0.78 \[ -\frac {\coth ^{-1}(\coth (a+b x))+b x}{2 x^2} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcCoth[Coth[a + b*x]]/x^3,x]

[Out]

-1/2*(b*x + ArcCoth[Coth[a + b*x]])/x^2

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fricas [A]  time = 0.54, size = 11, normalized size = 0.48 \[ -\frac {2 \, b x + a}{2 \, x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccoth(coth(b*x+a))/x^3,x, algorithm="fricas")

[Out]

-1/2*(2*b*x + a)/x^2

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giac [A]  time = 0.12, size = 11, normalized size = 0.48 \[ -\frac {2 \, b x + a}{2 \, x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccoth(coth(b*x+a))/x^3,x, algorithm="giac")

[Out]

-1/2*(2*b*x + a)/x^2

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maple [A]  time = 0.38, size = 20, normalized size = 0.87 \[ -\frac {b}{2 x}-\frac {\mathrm {arccoth}\left (\coth \left (b x +a \right )\right )}{2 x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arccoth(coth(b*x+a))/x^3,x)

[Out]

-1/2*b/x-1/2*arccoth(coth(b*x+a))/x^2

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maxima [A]  time = 0.31, size = 11, normalized size = 0.48 \[ -\frac {2 \, b x + a}{2 \, x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccoth(coth(b*x+a))/x^3,x, algorithm="maxima")

[Out]

-1/2*(2*b*x + a)/x^2

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mupad [B]  time = 1.14, size = 16, normalized size = 0.70 \[ -\frac {\mathrm {acoth}\left (\mathrm {coth}\left (a+b\,x\right )\right )+b\,x}{2\,x^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(acoth(coth(a + b*x))/x^3,x)

[Out]

-(acoth(coth(a + b*x)) + b*x)/(2*x^2)

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sympy [A]  time = 10.80, size = 39, normalized size = 1.70 \[ \begin {cases} 0 & \text {for}\: a = \log {\left (- e^{- b x} \right )} \vee a = \log {\left (e^{- b x} \right )} \\- \frac {b}{2 x} - \frac {\operatorname {acoth}{\left (\frac {1}{\tanh {\left (a + b x \right )}} \right )}}{2 x^{2}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(acoth(coth(b*x+a))/x**3,x)

[Out]

Piecewise((0, Eq(a, log(exp(-b*x))) | Eq(a, log(-exp(-b*x)))), (-b/(2*x) - acoth(1/tanh(a + b*x))/(2*x**2), Tr
ue))

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