\(\int \frac {\coth ^{-1}(\tanh (a+b x))}{x^2} \, dx\) [133]

Optimal result

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

[Out]

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

[In]

Int[ArcCoth[Tanh[a + b*x]]/x^2,x]

[Out]

-(ArcCoth[Tanh[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}(\tanh (a+b x))}{x}+b \int \frac {1}{x} \, dx \\ & = -\frac {\coth ^{-1}(\tanh (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 \[ \int \frac {\coth ^{-1}(\tanh (a+b x))}{x^2} \, dx=b-\frac {\coth ^{-1}(\tanh (a+b x))}{x}+b \log (x) \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.32 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.18

method	result	size
parallelrisch	\(\frac {b \ln \left (x \right ) x -\operatorname {arccoth}\left (\tanh \left (b x +a \right )\right )}{x}\)	\(20\)
default	\(-\frac {\operatorname {arccoth}\left (\tanh \left (b x +a \right )\right )}{x}+b \ln \left (-b x \right )\)	\(21\)
parts	\(-\frac {\operatorname {arccoth}\left (\tanh \left (b x +a \right )\right )}{x}+b \ln \left (-b x \right )\)	\(21\)
risch	\(-\frac {\ln \left ({\mathrm e}^{b x +a}\right )}{x}+\frac {2 i \pi \operatorname {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right )^{3}+i \pi \operatorname {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{3}-i \pi \,\operatorname {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{2}+i \pi \operatorname {csgn}\left (i {\mathrm e}^{b x +a}\right )^{2} \operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )-2 i \pi \,\operatorname {csgn}\left (i {\mathrm e}^{b x +a}\right ) \operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )^{2}-i \pi \,\operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{2}+i \pi \,\operatorname {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right ) \operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )+i \pi \operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )^{3}-2 i \pi \operatorname {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right )^{2}+2 i \pi +4 b \ln \left (x \right ) x}{4 x}\)	\(339\)

[In]

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

[Out]

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

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.25 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.06 \[ \int \frac {\coth ^{-1}(\tanh (a+b x))}{x^2} \, dx=\frac {-i \, \pi + 2 \, b x \log \left (x\right ) - 2 \, a}{2 \, x} \]

[In]

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

[Out]

1/2*(-I*pi + 2*b*x*log(x) - 2*a)/x

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82 \[ \int \frac {\coth ^{-1}(\tanh (a+b x))}{x^2} \, dx=b \log {\left (x \right )} - \frac {\operatorname {acoth}{\left (\tanh {\left (a + b x \right )} \right )}}{x} \]

[In]

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

[Out]

b*log(x) - acoth(tanh(a + b*x))/x

Maxima [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int \frac {\coth ^{-1}(\tanh (a+b x))}{x^2} \, dx=b \log \left (x\right ) - \frac {\operatorname {arcoth}\left (\tanh \left (b x + a\right )\right )}{x} \]

[In]

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

[Out]

b*log(x) - arccoth(tanh(b*x + a))/x

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 70 vs. \(2 (17) = 34\).

Time = 0.29 (sec) , antiderivative size = 70, normalized size of antiderivative = 4.12 \[ \int \frac {\coth ^{-1}(\tanh (a+b x))}{x^2} \, dx=b \log \left ({\left | x \right |}\right ) - \frac {\log \left (-\frac {\frac {e^{\left (2 \, b x + 2 \, a\right )} + 1}{e^{\left (2 \, b x + 2 \, a\right )} - 1} + 1}{\frac {e^{\left (2 \, b x + 2 \, a\right )} + 1}{e^{\left (2 \, b x + 2 \, a\right )} - 1} - 1}\right )}{2 \, x} \]

[In]

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

[Out]

b*log(abs(x)) - 1/2*log(-((e^(2*b*x + 2*a) + 1)/(e^(2*b*x + 2*a) - 1) + 1)/((e^(2*b*x + 2*a) + 1)/(e^(2*b*x +
2*a) - 1) - 1))/x

Mupad [B] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int \frac {\coth ^{-1}(\tanh (a+b x))}{x^2} \, dx=b\,\ln \left (x\right )-\frac {\mathrm {acoth}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}{x} \]

[In]

int(acoth(tanh(a + b*x))/x^2,x)

[Out]

b*log(x) - acoth(tanh(a + b*x))/x