\(\int \frac {\coth ^{-1}(\coth (a+b x))}{x^2} \, dx\) [198]

Optimal result

Integrand size = 11, antiderivative size = 17 \[ \int \frac {\coth ^{-1}(\coth (a+b x))}{x^2} \, dx=-\frac {\coth ^{-1}(\coth (a+b x))}{x}+b \log (x) \]

[Out]

-arccoth(coth(b*x+a))/x+b*ln(x)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {2199, 29} \[ \int \frac {\coth ^{-1}(\coth (a+b x))}{x^2} \, dx=b \log (x)-\frac {\coth ^{-1}(\coth (a+b x))}{x} \]

[In]

Int[ArcCoth[Coth[a + b*x]]/x^2,x]

[Out]

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

Rule 29

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

Rule 2199

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.06 \[ \int \frac {\coth ^{-1}(\coth (a+b x))}{x^2} \, dx=b-\frac {\coth ^{-1}(\coth (a+b x))}{x}+b \log (x) \]

[In]

Integrate[ArcCoth[Coth[a + b*x]]/x^2,x]

[Out]

b - ArcCoth[Coth[a + b*x]]/x + b*Log[x]

Maple [A] (verified)

Time = 0.26 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.24

method	result	size
default	\(-\frac {\operatorname {arccoth}\left (\coth \left (b x +a \right )\right )}{x}+b \ln \left (-b x \right )\)	\(21\)
parts	\(-\frac {\operatorname {arccoth}\left (\coth \left (b x +a \right )\right )}{x}+b \ln \left (-b x \right )\)	\(21\)
parallelrisch	\(\frac {b \ln \left (x \right ) x -\operatorname {arccoth}\left (\frac {1}{\tanh \left (b x +a \right )}\right )}{x}\)	\(22\)
risch	\(-\frac {\ln \left ({\mathrm e}^{b x +a}\right )}{x}+\frac {i \pi \operatorname {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}-1}\right )^{3}+i \pi \operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )^{3}-i \pi \,\operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}-1}\right )^{2}+i \pi \operatorname {csgn}\left (i {\mathrm e}^{b x +a}\right )^{2} \operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )-2 i \pi \,\operatorname {csgn}\left (i {\mathrm e}^{b x +a}\right ) \operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )^{2}+i \pi \,\operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right ) \operatorname {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}-1}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}-1}\right )-i \pi \,\operatorname {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}-1}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}-1}\right )^{2}+4 b \ln \left (x \right ) x}{4 x}\)	\(289\)

[In]

int(arccoth(coth(b*x+a))/x^2,x,method=_RETURNVERBOSE)

[Out]

-arccoth(coth(b*x+a))/x+b*ln(-b*x)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.76 \[ \int \frac {\coth ^{-1}(\coth (a+b x))}{x^2} \, dx=\frac {b x \log \left (x\right ) - a}{x} \]

[In]

integrate(arccoth(coth(b*x+a))/x^2,x, algorithm="fricas")

[Out]

(b*x*log(x) - a)/x

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 68 vs. \(2 (14) = 28\).

Time = 2.61 (sec) , antiderivative size = 68, normalized size of antiderivative = 4.00 \[ \int \frac {\coth ^{-1}(\coth (a+b x))}{x^2} \, dx=\begin {cases} - \frac {\operatorname {acoth}{\left (\coth {\left (b x + \log {\left (- e^{- b x} \right )} \right )} \right )}}{x} & \text {for}\: a = \log {\left (- e^{- b x} \right )} \\- \frac {\operatorname {acoth}{\left (\coth {\left (b x + \log {\left (e^{- b x} \right )} \right )} \right )}}{x} & \text {for}\: a = \log {\left (e^{- b x} \right )} \\b \log {\left (x \right )} - \frac {\operatorname {acoth}{\left (\frac {1}{\tanh {\left (a + b x \right )}} \right )}}{x} & \text {otherwise} \end {cases} \]

[In]

integrate(acoth(coth(b*x+a))/x**2,x)

[Out]

Piecewise((-acoth(coth(b*x + log(-exp(-b*x))))/x, Eq(a, log(-exp(-b*x)))), (-acoth(coth(b*x + log(exp(-b*x))))
/x, Eq(a, log(exp(-b*x)))), (b*log(x) - acoth(1/tanh(a + b*x))/x, True))

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.65 \[ \int \frac {\coth ^{-1}(\coth (a+b x))}{x^2} \, dx=b \log \left (x\right ) - \frac {a}{x} \]

[In]

integrate(arccoth(coth(b*x+a))/x^2,x, algorithm="maxima")

[Out]

b*log(x) - a/x

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.71 \[ \int \frac {\coth ^{-1}(\coth (a+b x))}{x^2} \, dx=b \log \left ({\left | x \right |}\right ) - \frac {a}{x} \]

[In]

integrate(arccoth(coth(b*x+a))/x^2,x, algorithm="giac")

[Out]

b*log(abs(x)) - a/x

Mupad [B] (verification not implemented)

Time = 3.75 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int \frac {\coth ^{-1}(\coth (a+b x))}{x^2} \, dx=b\,\ln \left (x\right )-\frac {\mathrm {acoth}\left (\mathrm {coth}\left (a+b\,x\right )\right )}{x} \]

[In]

int(acoth(coth(a + b*x))/x^2,x)

[Out]

b*log(x) - acoth(coth(a + b*x))/x