3.2.40 \(\int \coth ^{-1}(\tanh (a+b x))^2 \, dx\) [140]

Optimal. Leaf size=16 \[ \frac {\coth ^{-1}(\tanh (a+b x))^3}{3 b} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.00, antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {2188, 30} \begin {gather*} \frac {\coth ^{-1}(\tanh (a+b x))^3}{3 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[ArcCoth[Tanh[a + b*x]]^2,x]

[Out]

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

________________________________________________________________________________________

Mathematica [A]
time = 0.01, size = 16, normalized size = 1.00 \begin {gather*} \frac {\coth ^{-1}(\tanh (a+b x))^3}{3 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[ArcCoth[Tanh[a + b*x]]^2,x]

[Out]

ArcCoth[Tanh[a + b*x]]^3/(3*b)

________________________________________________________________________________________

Maple [A]
time = 0.50, size = 15, normalized size = 0.94

method result size
derivativedivides \(\frac {\mathrm {arccoth}\left (\tanh \left (b x +a \right )\right )^{3}}{3 b}\) \(15\)
default \(\frac {\mathrm {arccoth}\left (\tanh \left (b x +a \right )\right )^{3}}{3 b}\) \(15\)
risch \(\text {Expression too large to display}\) \(14844\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 33 vs. \(2 (14) = 28\).
time = 0.33, size = 33, normalized size = 2.06 \begin {gather*} \frac {1}{3} \, b^{2} x^{3} - b x^{2} \operatorname {arcoth}\left (\tanh \left (b x + a\right )\right ) + x \operatorname {arcoth}\left (\tanh \left (b x + a\right )\right )^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [A]
time = 0.34, size = 27, normalized size = 1.69 \begin {gather*} \frac {1}{3} \, b^{2} x^{3} + a b x^{2} - \frac {1}{4} \, {\left (\pi ^{2} - 4 \, a^{2}\right )} x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [A]
time = 0.08, size = 20, normalized size = 1.25 \begin {gather*} \begin {cases} \frac {\operatorname {acoth}^{3}{\left (\tanh {\left (a + b x \right )} \right )}}{3 b} & \text {for}\: b \neq 0 \\x \operatorname {acoth}^{2}{\left (\tanh {\left (a \right )} \right )} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((acoth(tanh(a + b*x))**3/(3*b), Ne(b, 0)), (x*acoth(tanh(a))**2, True))

________________________________________________________________________________________

Giac [C] Result contains complex when optimal does not.
time = 0.40, size = 39, normalized size = 2.44 \begin {gather*} \frac {1}{3} \, b^{2} x^{3} - \frac {1}{2} \, {\left (-i \, \pi b - 2 \, a b\right )} x^{2} - \frac {1}{4} \, {\left (\pi ^{2} - 4 i \, \pi a - 4 \, a^{2}\right )} x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Mupad [B]
time = 1.12, size = 33, normalized size = 2.06 \begin {gather*} \frac {b^2\,x^3}{3}-b\,x^2\,\mathrm {acoth}\left (\mathrm {tanh}\left (a+b\,x\right )\right )+x\,{\mathrm {acoth}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(acoth(tanh(a + b*x))^2,x)

[Out]

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

________________________________________________________________________________________