\(\int \coth ^{-1}(\tanh (a+b x))^3 \, dx\) [151]

Optimal result

Integrand size = 9, antiderivative size = 16 \[ \int \coth ^{-1}(\tanh (a+b x))^3 \, dx=\frac {\coth ^{-1}(\tanh (a+b x))^4}{4 b} \]

[Out]

1/4*arccoth(tanh(b*x+a))^4/b

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {2188, 30} \[ \int \coth ^{-1}(\tanh (a+b x))^3 \, dx=\frac {\coth ^{-1}(\tanh (a+b x))^4}{4 b} \]

[In]

Int[ArcCoth[Tanh[a + b*x]]^3,x]

[Out]

ArcCoth[Tanh[a + b*x]]^4/(4*b)

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2188

Int[(u_)^(m_.), x_Symbol] :> With[{c = Simplify[D[u, x]]}, Dist[1/c, Subst[Int[x^m, x], x, u], x]] /; FreeQ[m,
x] && PiecewiseLinearQ[u, x]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int x^3 \, dx,x,\coth ^{-1}(\tanh (a+b x))\right )}{b} \\ & = \frac {\coth ^{-1}(\tanh (a+b x))^4}{4 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00 \[ \int \coth ^{-1}(\tanh (a+b x))^3 \, dx=\frac {\coth ^{-1}(\tanh (a+b x))^4}{4 b} \]

[In]

Integrate[ArcCoth[Tanh[a + b*x]]^3,x]

[Out]

ArcCoth[Tanh[a + b*x]]^4/(4*b)

Maple [A] (verified)

Time = 25.86 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.94

method	result	size
derivativedivides	\(\frac {\operatorname {arccoth}\left (\tanh \left (b x +a \right )\right )^{4}}{4 b}\)	\(15\)
default	\(\frac {\operatorname {arccoth}\left (\tanh \left (b x +a \right )\right )^{4}}{4 b}\)	\(15\)
parallelrisch	\(b^{2} \operatorname {arccoth}\left (\tanh \left (b x +a \right )\right ) x^{3}-\frac {3 b \operatorname {arccoth}\left (\tanh \left (b x +a \right )\right )^{2} x^{2}}{2}+x \operatorname {arccoth}\left (\tanh \left (b x +a \right )\right )^{3}-\frac {b^{3} x^{4}}{4}\)	\(50\)
risch	\(\text {Expression too large to display}\)	\(14682\)

[In]

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

[Out]

1/4*arccoth(tanh(b*x+a))^4/b

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.24 (sec) , antiderivative size = 76, normalized size of antiderivative = 4.75 \[ \int \coth ^{-1}(\tanh (a+b x))^3 \, dx=\frac {1}{4} \, b^{3} x^{4} + a b^{2} x^{3} + \frac {3}{2} \, a^{2} b x^{2} - \frac {1}{8} i \, \pi ^{3} x + a^{3} x - \frac {3}{8} \, \pi ^{2} {\left (b x^{2} + 2 \, a x\right )} + \frac {1}{2} i \, \pi {\left (b^{2} x^{3} + 3 \, a b x^{2} + 3 \, a^{2} x\right )} \]

[In]

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

[Out]

1/4*b^3*x^4 + a*b^2*x^3 + 3/2*a^2*b*x^2 - 1/8*I*pi^3*x + a^3*x - 3/8*pi^2*(b*x^2 + 2*a*x) + 1/2*I*pi*(b^2*x^3
+ 3*a*b*x^2 + 3*a^2*x)

Sympy [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.25 \[ \int \coth ^{-1}(\tanh (a+b x))^3 \, dx=\begin {cases} \frac {\operatorname {acoth}^{4}{\left (\tanh {\left (a + b x \right )} \right )}}{4 b} & \text {for}\: b \neq 0 \\x \operatorname {acoth}^{3}{\left (\tanh {\left (a \right )} \right )} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((acoth(tanh(a + b*x))**4/(4*b), Ne(b, 0)), (x*acoth(tanh(a))**3, True))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 51 vs. \(2 (14) = 28\).

Time = 0.35 (sec) , antiderivative size = 51, normalized size of antiderivative = 3.19 \[ \int \coth ^{-1}(\tanh (a+b x))^3 \, dx=-\frac {3}{2} \, b x^{2} \operatorname {arcoth}\left (\tanh \left (b x + a\right )\right )^{2} + x \operatorname {arcoth}\left (\tanh \left (b x + a\right )\right )^{3} - \frac {1}{4} \, {\left (b^{2} x^{4} - 4 \, b x^{3} \operatorname {arcoth}\left (\tanh \left (b x + a\right )\right )\right )} b \]

[In]

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

[Out]

-3/2*b*x^2*arccoth(tanh(b*x + a))^2 + x*arccoth(tanh(b*x + a))^3 - 1/4*(b^2*x^4 - 4*b*x^3*arccoth(tanh(b*x + a
)))*b

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.28 (sec) , antiderivative size = 75, normalized size of antiderivative = 4.69 \[ \int \coth ^{-1}(\tanh (a+b x))^3 \, dx=\frac {1}{4} \, b^{3} x^{4} - \frac {1}{2} \, {\left (-i \, \pi b^{2} - 2 \, a b^{2}\right )} x^{3} - \frac {3}{8} \, {\left (\pi ^{2} b - 4 i \, \pi a b - 4 \, a^{2} b\right )} x^{2} - \frac {1}{8} \, {\left (i \, \pi ^{3} + 6 \, \pi ^{2} a - 12 i \, \pi a^{2} - 8 \, a^{3}\right )} x \]

[In]

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

[Out]

1/4*b^3*x^4 - 1/2*(-I*pi*b^2 - 2*a*b^2)*x^3 - 3/8*(pi^2*b - 4*I*pi*a*b - 4*a^2*b)*x^2 - 1/8*(I*pi^3 + 6*pi^2*a
- 12*I*pi*a^2 - 8*a^3)*x

Mupad [B] (verification not implemented)

Time = 3.83 (sec) , antiderivative size = 47, normalized size of antiderivative = 2.94 \[ \int \coth ^{-1}(\tanh (a+b x))^3 \, dx=\frac {x\,\left (2\,\mathrm {acoth}\left (\mathrm {tanh}\left (a+b\,x\right )\right )-b\,x\right )\,\left (b^2\,x^2-2\,b\,x\,\mathrm {acoth}\left (\mathrm {tanh}\left (a+b\,x\right )\right )+2\,{\mathrm {acoth}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^2\right )}{4} \]

[In]

int(acoth(tanh(a + b*x))^3,x)

[Out]

(x*(2*acoth(tanh(a + b*x)) - b*x)*(2*acoth(tanh(a + b*x))^2 + b^2*x^2 - 2*b*x*acoth(tanh(a + b*x))))/4