\(\int \frac {\coth ^{-1}(\tanh (a+b x))}{x^3} \, dx\) [134]

Optimal result

Integrand size = 11, antiderivative size = 23 \[ \int \frac {\coth ^{-1}(\tanh (a+b x))}{x^3} \, dx=-\frac {b}{2 x}-\frac {\coth ^{-1}(\tanh (a+b x))}{2 x^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 23, 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, 30} \[ \int \frac {\coth ^{-1}(\tanh (a+b x))}{x^3} \, dx=-\frac {\coth ^{-1}(\tanh (a+b x))}{2 x^2}-\frac {b}{2 x} \]

[In]

Int[ArcCoth[Tanh[a + b*x]]/x^3,x]

[Out]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.78 \[ \int \frac {\coth ^{-1}(\tanh (a+b x))}{x^3} \, dx=-\frac {b x+\coth ^{-1}(\tanh (a+b x))}{2 x^2} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.31 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.74

method	result	size
parallelrisch	\(-\frac {b x +\operatorname {arccoth}\left (\tanh \left (b x +a \right )\right )}{2 x^{2}}\)	\(17\)
default	\(-\frac {b}{2 x}-\frac {\operatorname {arccoth}\left (\tanh \left (b x +a \right )\right )}{2 x^{2}}\)	\(20\)
parts	\(-\frac {b}{2 x}-\frac {\operatorname {arccoth}\left (\tanh \left (b x +a \right )\right )}{2 x^{2}}\)	\(20\)
risch	\(-\frac {\ln \left ({\mathrm e}^{b x +a}\right )}{2 x^{2}}-\frac {4 b x -i \pi \operatorname {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{3}+i \pi \,\operatorname {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{2}-i \pi \operatorname {csgn}\left (i {\mathrm e}^{b x +a}\right )^{2} \operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )+2 i \pi \,\operatorname {csgn}\left (i {\mathrm e}^{b x +a}\right ) \operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )^{2}+i \pi \,\operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{2}-2 i \pi \operatorname {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right )^{3}-i \pi \,\operatorname {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right ) \operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )-i \pi \operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )^{3}-2 i \pi +2 i \pi \operatorname {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right )^{2}}{8 x^{2}}\)	\(337\)

[In]

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

[Out]

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

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.24 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.70 \[ \int \frac {\coth ^{-1}(\tanh (a+b x))}{x^3} \, dx=\frac {-i \, \pi - 4 \, b x - 2 \, a}{4 \, x^{2}} \]

[In]

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

[Out]

1/4*(-I*pi - 4*b*x - 2*a)/x^2

Sympy [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {\coth ^{-1}(\tanh (a+b x))}{x^3} \, dx=- \frac {b}{2 x} - \frac {\operatorname {acoth}{\left (\tanh {\left (a + b x \right )} \right )}}{2 x^{2}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {\coth ^{-1}(\tanh (a+b x))}{x^3} \, dx=-\frac {b}{2 \, x} - \frac {\operatorname {arcoth}\left (\tanh \left (b x + a\right )\right )}{2 \, x^{2}} \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 71 vs. \(2 (19) = 38\).

Time = 0.28 (sec) , antiderivative size = 71, normalized size of antiderivative = 3.09 \[ \int \frac {\coth ^{-1}(\tanh (a+b x))}{x^3} \, dx=-\frac {b}{2 \, x} - \frac {\log \left (-\frac {\frac {e^{\left (2 \, b x + 2 \, a\right )} + 1}{e^{\left (2 \, b x + 2 \, a\right )} - 1} + 1}{\frac {e^{\left (2 \, b x + 2 \, a\right )} + 1}{e^{\left (2 \, b x + 2 \, a\right )} - 1} - 1}\right )}{4 \, x^{2}} \]

[In]

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

[Out]

-1/2*b/x - 1/4*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^2

Mupad [B] (verification not implemented)

Time = 3.75 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.70 \[ \int \frac {\coth ^{-1}(\tanh (a+b x))}{x^3} \, dx=-\frac {\mathrm {acoth}\left (\mathrm {tanh}\left (a+b\,x\right )\right )+b\,x}{2\,x^2} \]

[In]

int(acoth(tanh(a + b*x))/x^3,x)

[Out]

-(acoth(tanh(a + b*x)) + b*x)/(2*x^2)