\(\int \frac {\coth ^{-1}(\tanh (a+b x))^2}{x^3} \, dx\) [143]

Optimal result

Integrand size = 13, antiderivative size = 36 \[ \int \frac {\coth ^{-1}(\tanh (a+b x))^2}{x^3} \, dx=-\frac {b \coth ^{-1}(\tanh (a+b x))}{x}-\frac {\coth ^{-1}(\tanh (a+b x))^2}{2 x^2}+b^2 \log (x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2199, 29} \[ \int \frac {\coth ^{-1}(\tanh (a+b x))^2}{x^3} \, dx=-\frac {\coth ^{-1}(\tanh (a+b x))^2}{2 x^2}-\frac {b \coth ^{-1}(\tanh (a+b x))}{x}+b^2 \log (x) \]

[In]

Int[ArcCoth[Tanh[a + b*x]]^2/x^3,x]

[Out]

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.17 \[ \int \frac {\coth ^{-1}(\tanh (a+b x))^2}{x^3} \, dx=-\frac {2 b x \coth ^{-1}(\tanh (a+b x))+\coth ^{-1}(\tanh (a+b x))^2-b^2 x^2 (3+2 \log (x))}{2 x^2} \]

[In]

Integrate[ArcCoth[Tanh[a + b*x]]^2/x^3,x]

[Out]

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

Maple [A] (verified)

Time = 0.29 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.08

method	result	size
parallelrisch	\(\frac {2 b^{2} \ln \left (x \right ) x^{2}-2 b x \,\operatorname {arccoth}\left (\tanh \left (b x +a \right )\right )-\operatorname {arccoth}\left (\tanh \left (b x +a \right )\right )^{2}}{2 x^{2}}\)	\(39\)
risch	\(\text {Expression too large to display}\)	\(3213\)

[In]

int(arccoth(tanh(b*x+a))^2/x^3,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.25 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.06 \[ \int \frac {\coth ^{-1}(\tanh (a+b x))^2}{x^3} \, dx=\frac {8 \, b^{2} x^{2} \log \left (x\right ) - 16 \, a b x + \pi ^{2} - 4 i \, \pi {\left (2 \, b x + a\right )} - 4 \, a^{2}}{8 \, x^{2}} \]

[In]

integrate(arccoth(tanh(b*x+a))^2/x^3,x, algorithm="fricas")

[Out]

1/8*(8*b^2*x^2*log(x) - 16*a*b*x + pi^2 - 4*I*pi*(2*b*x + a) - 4*a^2)/x^2

Sympy [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.89 \[ \int \frac {\coth ^{-1}(\tanh (a+b x))^2}{x^3} \, dx=b^{2} \log {\left (x \right )} - \frac {b \operatorname {acoth}{\left (\tanh {\left (a + b x \right )} \right )}}{x} - \frac {\operatorname {acoth}^{2}{\left (\tanh {\left (a + b x \right )} \right )}}{2 x^{2}} \]

[In]

integrate(acoth(tanh(b*x+a))**2/x**3,x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.94 \[ \int \frac {\coth ^{-1}(\tanh (a+b x))^2}{x^3} \, dx=b^{2} \log \left (x\right ) - \frac {b \operatorname {arcoth}\left (\tanh \left (b x + a\right )\right )}{x} - \frac {\operatorname {arcoth}\left (\tanh \left (b x + a\right )\right )^{2}}{2 \, x^{2}} \]

[In]

integrate(arccoth(tanh(b*x+a))^2/x^3,x, algorithm="maxima")

[Out]

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

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.30 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.03 \[ \int \frac {\coth ^{-1}(\tanh (a+b x))^2}{x^3} \, dx=b^{2} \log \left (x\right ) - \frac {8 i \, \pi b x + 16 \, a b x - \pi ^{2} + 4 i \, \pi a + 4 \, a^{2}}{8 \, x^{2}} \]

[In]

integrate(arccoth(tanh(b*x+a))^2/x^3,x, algorithm="giac")

[Out]

b^2*log(x) - 1/8*(8*I*pi*b*x + 16*a*b*x - pi^2 + 4*I*pi*a + 4*a^2)/x^2

Mupad [B] (verification not implemented)

Time = 4.06 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.94 \[ \int \frac {\coth ^{-1}(\tanh (a+b x))^2}{x^3} \, dx=b^2\,\ln \left (x\right )-\frac {\frac {{\mathrm {acoth}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^2}{2}+b\,x\,\mathrm {acoth}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}{x^2} \]

[In]

int(acoth(tanh(a + b*x))^2/x^3,x)

[Out]

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