\(\int \frac {\coth ^{-1}(\tanh (a+b x))^3}{x^4} \, dx\) [155]

Optimal result

Integrand size = 13, antiderivative size = 55 \[ \int \frac {\coth ^{-1}(\tanh (a+b x))^3}{x^4} \, dx=-\frac {b^2 \coth ^{-1}(\tanh (a+b x))}{x}-\frac {b \coth ^{-1}(\tanh (a+b x))^2}{2 x^2}-\frac {\coth ^{-1}(\tanh (a+b x))^3}{3 x^3}+b^3 \log (x) \]

[Out]

-b^2*arccoth(tanh(b*x+a))/x-1/2*b*arccoth(tanh(b*x+a))^2/x^2-1/3*arccoth(tanh(b*x+a))^3/x^3+b^3*ln(x)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2199, 29} \[ \int \frac {\coth ^{-1}(\tanh (a+b x))^3}{x^4} \, dx=-\frac {b^2 \coth ^{-1}(\tanh (a+b x))}{x}-\frac {\coth ^{-1}(\tanh (a+b x))^3}{3 x^3}-\frac {b \coth ^{-1}(\tanh (a+b x))^2}{2 x^2}+b^3 \log (x) \]

[In]

Int[ArcCoth[Tanh[a + b*x]]^3/x^4,x]

[Out]

-((b^2*ArcCoth[Tanh[a + b*x]])/x) - (b*ArcCoth[Tanh[a + b*x]]^2)/(2*x^2) - ArcCoth[Tanh[a + b*x]]^3/(3*x^3) +
b^3*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))^3}{3 x^3}+b \int \frac {\coth ^{-1}(\tanh (a+b x))^2}{x^3} \, dx \\ & = -\frac {b \coth ^{-1}(\tanh (a+b x))^2}{2 x^2}-\frac {\coth ^{-1}(\tanh (a+b x))^3}{3 x^3}+b^2 \int \frac {\coth ^{-1}(\tanh (a+b x))}{x^2} \, dx \\ & = -\frac {b^2 \coth ^{-1}(\tanh (a+b x))}{x}-\frac {b \coth ^{-1}(\tanh (a+b x))^2}{2 x^2}-\frac {\coth ^{-1}(\tanh (a+b x))^3}{3 x^3}+b^3 \int \frac {1}{x} \, dx \\ & = -\frac {b^2 \coth ^{-1}(\tanh (a+b x))}{x}-\frac {b \coth ^{-1}(\tanh (a+b x))^2}{2 x^2}-\frac {\coth ^{-1}(\tanh (a+b x))^3}{3 x^3}+b^3 \log (x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.09 \[ \int \frac {\coth ^{-1}(\tanh (a+b x))^3}{x^4} \, dx=\frac {-6 b^2 x^2 \coth ^{-1}(\tanh (a+b x))-3 b x \coth ^{-1}(\tanh (a+b x))^2-2 \coth ^{-1}(\tanh (a+b x))^3+b^3 x^3 (11+6 \log (x))}{6 x^3} \]

[In]

Integrate[ArcCoth[Tanh[a + b*x]]^3/x^4,x]

[Out]

(-6*b^2*x^2*ArcCoth[Tanh[a + b*x]] - 3*b*x*ArcCoth[Tanh[a + b*x]]^2 - 2*ArcCoth[Tanh[a + b*x]]^3 + b^3*x^3*(11
+ 6*Log[x]))/(6*x^3)

Maple [A] (verified)

Time = 0.99 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.02

method	result	size
parallelrisch	\(\frac {6 b^{3} \ln \left (x \right ) x^{3}-6 b^{2} x^{2} \operatorname {arccoth}\left (\tanh \left (b x +a \right )\right )-3 b x \operatorname {arccoth}\left (\tanh \left (b x +a \right )\right )^{2}-2 \operatorname {arccoth}\left (\tanh \left (b x +a \right )\right )^{3}}{6 x^{3}}\)	\(56\)
risch	\(\text {Expression too large to display}\)	\(17237\)

[In]

int(arccoth(tanh(b*x+a))^3/x^4,x,method=_RETURNVERBOSE)

[Out]

1/6*(6*b^3*ln(x)*x^3-6*b^2*x^2*arccoth(tanh(b*x+a))-3*b*x*arccoth(tanh(b*x+a))^2-2*arccoth(tanh(b*x+a))^3)/x^3

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.24 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.36 \[ \int \frac {\coth ^{-1}(\tanh (a+b x))^3}{x^4} \, dx=\frac {24 \, b^{3} x^{3} \log \left (x\right ) - 72 \, a b^{2} x^{2} - 36 \, a^{2} b x + i \, \pi ^{3} + 3 \, \pi ^{2} {\left (3 \, b x + 2 \, a\right )} - 8 \, a^{3} - 12 i \, \pi {\left (3 \, b^{2} x^{2} + 3 \, a b x + a^{2}\right )}}{24 \, x^{3}} \]

[In]

integrate(arccoth(tanh(b*x+a))^3/x^4,x, algorithm="fricas")

[Out]

1/24*(24*b^3*x^3*log(x) - 72*a*b^2*x^2 - 36*a^2*b*x + I*pi^3 + 3*pi^2*(3*b*x + 2*a) - 8*a^3 - 12*I*pi*(3*b^2*x
^2 + 3*a*b*x + a^2))/x^3

Sympy [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.93 \[ \int \frac {\coth ^{-1}(\tanh (a+b x))^3}{x^4} \, dx=b^{3} \log {\left (x \right )} - \frac {b^{2} \operatorname {acoth}{\left (\tanh {\left (a + b x \right )} \right )}}{x} - \frac {b \operatorname {acoth}^{2}{\left (\tanh {\left (a + b x \right )} \right )}}{2 x^{2}} - \frac {\operatorname {acoth}^{3}{\left (\tanh {\left (a + b x \right )} \right )}}{3 x^{3}} \]

[In]

integrate(acoth(tanh(b*x+a))**3/x**4,x)

[Out]

b**3*log(x) - b**2*acoth(tanh(a + b*x))/x - b*acoth(tanh(a + b*x))**2/(2*x**2) - acoth(tanh(a + b*x))**3/(3*x*
*3)

Maxima [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.95 \[ \int \frac {\coth ^{-1}(\tanh (a+b x))^3}{x^4} \, dx={\left (b^{2} \log \left (x\right ) - \frac {b \operatorname {arcoth}\left (\tanh \left (b x + a\right )\right )}{x}\right )} b - \frac {b \operatorname {arcoth}\left (\tanh \left (b x + a\right )\right )^{2}}{2 \, x^{2}} - \frac {\operatorname {arcoth}\left (\tanh \left (b x + a\right )\right )^{3}}{3 \, x^{3}} \]

[In]

integrate(arccoth(tanh(b*x+a))^3/x^4,x, algorithm="maxima")

[Out]

(b^2*log(x) - b*arccoth(tanh(b*x + a))/x)*b - 1/2*b*arccoth(tanh(b*x + a))^2/x^2 - 1/3*arccoth(tanh(b*x + a))^
3/x^3

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.29 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.33 \[ \int \frac {\coth ^{-1}(\tanh (a+b x))^3}{x^4} \, dx=b^{3} \log \left (x\right ) - \frac {36 i \, \pi b^{2} x^{2} + 72 \, a b^{2} x^{2} - 9 \, \pi ^{2} b x + 36 i \, \pi a b x + 36 \, a^{2} b x - i \, \pi ^{3} - 6 \, \pi ^{2} a + 12 i \, \pi a^{2} + 8 \, a^{3}}{24 \, x^{3}} \]

[In]

integrate(arccoth(tanh(b*x+a))^3/x^4,x, algorithm="giac")

[Out]

b^3*log(x) - 1/24*(36*I*pi*b^2*x^2 + 72*a*b^2*x^2 - 9*pi^2*b*x + 36*I*pi*a*b*x + 36*a^2*b*x - I*pi^3 - 6*pi^2*
a + 12*I*pi*a^2 + 8*a^3)/x^3

Mupad [B] (verification not implemented)

Time = 3.86 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.93 \[ \int \frac {\coth ^{-1}(\tanh (a+b x))^3}{x^4} \, dx=b^3\,\ln \left (x\right )-\frac {b^2\,x^2\,\mathrm {acoth}\left (\mathrm {tanh}\left (a+b\,x\right )\right )+\frac {b\,x\,{\mathrm {acoth}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^2}{2}+\frac {{\mathrm {acoth}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^3}{3}}{x^3} \]

[In]

int(acoth(tanh(a + b*x))^3/x^4,x)

[Out]

b^3*log(x) - (acoth(tanh(a + b*x))^3/3 + (b*x*acoth(tanh(a + b*x))^2)/2 + b^2*x^2*acoth(tanh(a + b*x)))/x^3