3.139 \(\int x \coth ^{-1}(\tanh (a+b x))^2 \, dx\)
Optimal. Leaf size=34 \[ \frac {x \coth ^{-1}(\tanh (a+b x))^3}{3 b}-\frac {\coth ^{-1}(\tanh (a+b x))^4}{12 b^2} \]
[Out]
1/3*x*arccoth(tanh(b*x+a))^3/b-1/12*arccoth(tanh(b*x+a))^4/b^2
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Rubi [A] time = 0.03, antiderivative size = 34, normalized size of antiderivative = 1.00,
number of steps used = 3, number of rules used = 3, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used =
{2168, 2157, 30} \[ \frac {x \coth ^{-1}(\tanh (a+b x))^3}{3 b}-\frac {\coth ^{-1}(\tanh (a+b x))^4}{12 b^2} \]
Antiderivative was successfully verified.
[In]
Int[x*ArcCoth[Tanh[a + b*x]]^2,x]
[Out]
(x*ArcCoth[Tanh[a + b*x]]^3)/(3*b) - ArcCoth[Tanh[a + b*x]]^4/(12*b^2)
Rule 30
Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]
Rule 2157
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]
Rule 2168
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*} \int x \coth ^{-1}(\tanh (a+b x))^2 \, dx &=\frac {x \coth ^{-1}(\tanh (a+b x))^3}{3 b}-\frac {\int \coth ^{-1}(\tanh (a+b x))^3 \, dx}{3 b}\\ &=\frac {x \coth ^{-1}(\tanh (a+b x))^3}{3 b}-\frac {\operatorname {Subst}\left (\int x^3 \, dx,x,\coth ^{-1}(\tanh (a+b x))\right )}{3 b^2}\\ &=\frac {x \coth ^{-1}(\tanh (a+b x))^3}{3 b}-\frac {\coth ^{-1}(\tanh (a+b x))^4}{12 b^2}\\ \end {align*}
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Mathematica [B] time = 0.08, size = 74, normalized size = 2.18 \[ \frac {(a+b x) \left (4 \left (2 a^2+a b x-b^2 x^2\right ) \coth ^{-1}(\tanh (a+b x))-\left ((3 a-b x) (a+b x)^2\right )-6 (a-b x) \coth ^{-1}(\tanh (a+b x))^2\right )}{12 b^2} \]
Antiderivative was successfully verified.
[In]
Integrate[x*ArcCoth[Tanh[a + b*x]]^2,x]
[Out]
((a + b*x)*(-((3*a - b*x)*(a + b*x)^2) + 4*(2*a^2 + a*b*x - b^2*x^2)*ArcCoth[Tanh[a + b*x]] - 6*(a - b*x)*ArcC
oth[Tanh[a + b*x]]^2))/(12*b^2)
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fricas [A] time = 0.67, size = 30, normalized size = 0.88 \[ \frac {1}{4} \, b^{2} x^{4} + \frac {2}{3} \, a b x^{3} - \frac {1}{8} \, {\left (\pi ^{2} - 4 \, a^{2}\right )} x^{2} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x*arccoth(tanh(b*x+a))^2,x, algorithm="fricas")
[Out]
1/4*b^2*x^4 + 2/3*a*b*x^3 - 1/8*(pi^2 - 4*a^2)*x^2
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int x \operatorname {arcoth}\left (\tanh \left (b x + a\right )\right )^{2}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x*arccoth(tanh(b*x+a))^2,x, algorithm="giac")
[Out]
integrate(x*arccoth(tanh(b*x + a))^2, x)
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maple [C] time = 0.40, size = 3418, normalized size = 100.53 \[ \text {output too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(x*arccoth(tanh(b*x+a))^2,x)
[Out]
1/12*I*Pi*b*x^3*csgn(I/(exp(2*b*x+2*a)+1))*csgn(I*exp(2*b*x+2*a))*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))-1/
8*Pi^2*x^2*csgn(I/(exp(2*b*x+2*a)+1))*csgn(I*exp(2*b*x+2*a))*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^4+1/16*
Pi^2*x^2*csgn(I*exp(b*x+a))^2*csgn(I*exp(2*b*x+2*a))^2*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^2+1/16*Pi^2*x
^2*csgn(I/(exp(2*b*x+2*a)+1))*csgn(I*exp(2*b*x+2*a))^2*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^3+1/16*Pi^2*x
^2*csgn(I/(exp(2*b*x+2*a)+1))^2*csgn(I*exp(2*b*x+2*a))*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^3-1/16*Pi^2*x
^2*csgn(I/(exp(2*b*x+2*a)+1))*csgn(I*exp(2*b*x+2*a))^4*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))+1/16*Pi^2*x^2
*csgn(I/(exp(2*b*x+2*a)+1))*csgn(I*exp(2*b*x+2*a))^3*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^2+1/4*Pi^2*x^2*
csgn(I*exp(b*x+a))*csgn(I*exp(2*b*x+2*a))^2*csgn(I/(exp(2*b*x+2*a)+1))^3-1/8*Pi^2*x^2*csgn(I/(exp(2*b*x+2*a)+1
))^2*csgn(I*exp(2*b*x+2*a))*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^2+1/12*I*Pi*b*x^3*csgn(I*exp(2*b*x+2*a))
^3-1/8*Pi^2*x^2*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^3*csgn(I/(exp(2*b*x+2*a)+1))^3+1/8*Pi^2*x^2*csgn(I/(
exp(2*b*x+2*a)+1))*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^2-1/8*csgn(I/(exp(2*b*x+2*a)+1))^4*Pi^2*x^2-1/16*
Pi^2*x^2*csgn(I*exp(b*x+a))^2*csgn(I*exp(2*b*x+2*a))*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^3-1/8*Pi^2*x^2*
csgn(I*exp(b*x+a))^2*csgn(I*exp(2*b*x+2*a))*csgn(I/(exp(2*b*x+2*a)+1))^3-1/8*Pi^2*x^2*csgn(I*exp(b*x+a))*csgn(
I*exp(2*b*x+2*a))^3*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^2+1/16*Pi^2*x^2*csgn(I/(exp(2*b*x+2*a)+1))*csgn(
I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^5-1/8*Pi^2*x^2*csgn(I*exp(2*b*x+2*a))^3*csgn(I/(exp(2*b*x+2*a)+1))^3-1/32
*Pi^2*x^2*csgn(I*exp(2*b*x+2*a))^2*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^4-1/8*Pi^2*x^2*csgn(I*exp(2*b*x+2
*a)/(exp(2*b*x+2*a)+1))^2*csgn(I/(exp(2*b*x+2*a)+1))^3+1/8*Pi^2*x^2*csgn(I*exp(2*b*x+2*a))^3*csgn(I/(exp(2*b*x
+2*a)+1))^2+1/12*I*Pi*b*x^3*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^3+1/8*Pi^2*x^2*csgn(I/(exp(2*b*x+2*a)+1)
)^4*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^2+1/16*Pi^2*x^2*csgn(I*exp(2*b*x+2*a))*csgn(I*exp(2*b*x+2*a)/(ex
p(2*b*x+2*a)+1))^5+1/8*Pi^2*x^2*csgn(I*exp(b*x+a))*csgn(I*exp(2*b*x+2*a))^5+1/16*Pi^2*x^2*csgn(I*exp(2*b*x+2*a
))^4*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^2-1/16*Pi^2*x^2*csgn(I*exp(2*b*x+2*a))^3*csgn(I*exp(2*b*x+2*a)/
(exp(2*b*x+2*a)+1))^3-1/32*Pi^2*x^2*csgn(I/(exp(2*b*x+2*a)+1))^2*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^4+1
/8*Pi^2*x^2*csgn(I*exp(2*b*x+2*a))*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^2*csgn(I/(exp(2*b*x+2*a)+1))^3+1/
8*Pi^2*x^2*csgn(I*exp(b*x+a))*csgn(I*exp(2*b*x+2*a))^2*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^3-1/6*I*Pi*b*
x^3*csgn(I/(exp(2*b*x+2*a)+1))^2+1/4*Pi^2*x^2*csgn(I/(exp(2*b*x+2*a)+1))^5-1/8*Pi^2*x^2+1/2*x^2*ln(exp(b*x+a))
^2+(-1/3*b*x^3-1/2*I*Pi*x^2+1/4*I*Pi*x^2*csgn(I/(exp(2*b*x+2*a)+1))*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^
2-1/4*I*Pi*x^2*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^3+1/2*I*Pi*x^2*csgn(I/(exp(2*b*x+2*a)+1))^2-1/2*I*Pi*
x^2*csgn(I/(exp(2*b*x+2*a)+1))^3+1/2*I*Pi*x^2*csgn(I*exp(b*x+a))*csgn(I*exp(2*b*x+2*a))^2-1/4*I*Pi*x^2*csgn(I/
(exp(2*b*x+2*a)+1))*csgn(I*exp(2*b*x+2*a))*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))-1/4*I*Pi*x^2*csgn(I*exp(2
*b*x+2*a))^3-1/4*I*Pi*x^2*csgn(I*exp(b*x+a))^2*csgn(I*exp(2*b*x+2*a))+1/4*I*Pi*x^2*csgn(I*exp(2*b*x+2*a))*csgn
(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^2)*ln(exp(b*x+a))-1/8*Pi^2*x^2*csgn(I/(exp(2*b*x+2*a)+1))^6-1/32*Pi^2*x^
2*csgn(I*exp(2*b*x+2*a))^6-1/8*Pi^2*x^2*csgn(I*exp(2*b*x+2*a))^3+1/4*Pi^2*x^2*csgn(I/(exp(2*b*x+2*a)+1))^2-1/4
*csgn(I/(exp(2*b*x+2*a)+1))^3*Pi^2*x^2-1/8*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^3*Pi^2*x^2-1/8*Pi^2*x^2*c
sgn(I*exp(b*x+a))^2*csgn(I*exp(2*b*x+2*a))-1/32*Pi^2*x^2*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^6-1/8*Pi^2*
x^2*csgn(I/(exp(2*b*x+2*a)+1))^4*csgn(I*exp(2*b*x+2*a))*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))+1/6*I*Pi*b*x
^3*csgn(I/(exp(2*b*x+2*a)+1))^3+1/8*Pi^2*x^2*csgn(I/(exp(2*b*x+2*a)+1))^2*csgn(I*exp(b*x+a))^2*csgn(I*exp(2*b*
x+2*a))-1/4*Pi^2*x^2*csgn(I/(exp(2*b*x+2*a)+1))^2*csgn(I*exp(b*x+a))*csgn(I*exp(2*b*x+2*a))^2-3/16*Pi^2*x^2*cs
gn(I*exp(b*x+a))^2*csgn(I*exp(2*b*x+2*a))^4-1/32*Pi^2*x^2*csgn(I/(exp(2*b*x+2*a)+1))^2*csgn(I*exp(2*b*x+2*a))^
2*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^2+1/8*Pi^2*x^2*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^3*csgn(I/
(exp(2*b*x+2*a)+1))^2-1/12*I*Pi*b*x^3*csgn(I*exp(2*b*x+2*a))*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^2+1/8*P
i^2*x^2*csgn(I/(exp(2*b*x+2*a)+1))*csgn(I*exp(b*x+a))*csgn(I*exp(2*b*x+2*a))^3*csgn(I*exp(2*b*x+2*a)/(exp(2*b*
x+2*a)+1))-1/8*Pi^2*x^2*csgn(I/(exp(2*b*x+2*a)+1))*csgn(I*exp(b*x+a))*csgn(I*exp(2*b*x+2*a))^2*csgn(I*exp(2*b*
x+2*a)/(exp(2*b*x+2*a)+1))^2-1/32*Pi^2*x^2*csgn(I*exp(b*x+a))^4*csgn(I*exp(2*b*x+2*a))^2+1/8*Pi^2*x^2*csgn(I*e
xp(b*x+a))^3*csgn(I*exp(2*b*x+2*a))^3+1/6*I*Pi*b*x^3+1/8*Pi^2*x^2*csgn(I/(exp(2*b*x+2*a)+1))^3*csgn(I*exp(2*b*
x+2*a))*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))-1/8*Pi^2*x^2*csgn(I/(exp(2*b*x+2*a)+1))*csgn(I*exp(2*b*x+2*a
))*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))+1/12*b^2*x^4+1/8*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^2*csgn
(I*exp(2*b*x+2*a))*Pi^2*x^2+1/4*csgn(I*exp(2*b*x+2*a))^2*csgn(I*exp(b*x+a))*Pi^2*x^2-1/12*I*Pi*b*x^3*csgn(I/(e
xp(2*b*x+2*a)+1))*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^2+1/12*I*Pi*b*x^3*csgn(I*exp(b*x+a))^2*csgn(I*exp(
2*b*x+2*a))-1/6*I*Pi*b*x^3*csgn(I*exp(b*x+a))*csgn(I*exp(2*b*x+2*a))^2-1/16*Pi^2*x^2*csgn(I/(exp(2*b*x+2*a)+1)
)*csgn(I*exp(b*x+a))^2*csgn(I*exp(2*b*x+2*a))^2*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))+1/16*Pi^2*x^2*csgn(I
/(exp(2*b*x+2*a)+1))*csgn(I*exp(b*x+a))^2*csgn(I*exp(2*b*x+2*a))*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^2
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maxima [A] time = 0.45, size = 36, normalized size = 1.06 \[ \frac {1}{12} \, b^{2} x^{4} - \frac {1}{3} \, b x^{3} \operatorname {arcoth}\left (\tanh \left (b x + a\right )\right ) + \frac {1}{2} \, x^{2} \operatorname {arcoth}\left (\tanh \left (b x + a\right )\right )^{2} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x*arccoth(tanh(b*x+a))^2,x, algorithm="maxima")
[Out]
1/12*b^2*x^4 - 1/3*b*x^3*arccoth(tanh(b*x + a)) + 1/2*x^2*arccoth(tanh(b*x + a))^2
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mupad [B] time = 1.15, size = 36, normalized size = 1.06 \[ \frac {b^2\,x^4}{12}-\frac {b\,x^3\,\mathrm {acoth}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}{3}+\frac {x^2\,{\mathrm {acoth}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^2}{2} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(x*acoth(tanh(a + b*x))^2,x)
[Out]
(x^2*acoth(tanh(a + b*x))^2)/2 + (b^2*x^4)/12 - (b*x^3*acoth(tanh(a + b*x)))/3
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sympy [A] time = 0.67, size = 41, normalized size = 1.21 \[ \begin {cases} \frac {x \operatorname {acoth}^{3}{\left (\tanh {\left (a + b x \right )} \right )}}{3 b} - \frac {\operatorname {acoth}^{4}{\left (\tanh {\left (a + b x \right )} \right )}}{12 b^{2}} & \text {for}\: b \neq 0 \\\frac {x^{2} \operatorname {acoth}^{2}{\left (\tanh {\relax (a )} \right )}}{2} & \text {otherwise} \end {cases} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x*acoth(tanh(b*x+a))**2,x)
[Out]
Piecewise((x*acoth(tanh(a + b*x))**3/(3*b) - acoth(tanh(a + b*x))**4/(12*b**2), Ne(b, 0)), (x**2*acoth(tanh(a)
)**2/2, True))
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