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

Optimal. Leaf size=71 \[ \frac {3 \left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \log \left (\coth ^{-1}(\tanh (a+b x))\right )}{b^4}-\frac {3 x^2}{2 b^2 \coth ^{-1}(\tanh (a+b x))}-\frac {x^3}{2 b \coth ^{-1}(\tanh (a+b x))^2}+\frac {3 x}{b^3} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {2168, 2158, 2157, 29} \[ -\frac {3 x^2}{2 b^2 \coth ^{-1}(\tanh (a+b x))}+\frac {3 \left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \log \left (\coth ^{-1}(\tanh (a+b x))\right )}{b^4}-\frac {x^3}{2 b \coth ^{-1}(\tanh (a+b x))^2}+\frac {3 x}{b^3} \]

Antiderivative was successfully verified.

[In]

Int[x^3/ArcCoth[Tanh[a + b*x]]^3,x]

[Out]

(3*x)/b^3 - x^3/(2*b*ArcCoth[Tanh[a + b*x]]^2) - (3*x^2)/(2*b^2*ArcCoth[Tanh[a + b*x]]) + (3*(b*x - ArcCoth[Ta
nh[a + b*x]])*Log[ArcCoth[Tanh[a + b*x]]])/b^4

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

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 2158

Int[(v_)/(u_), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, Simp[(b*x)/a, x] - Dist[(b*u
- a*v)/a, Int[1/u, x], x] /; NeQ[b*u - a*v, 0]] /; PiecewiseLinearQ[u, v, 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 \frac {x^3}{\coth ^{-1}(\tanh (a+b x))^3} \, dx &=-\frac {x^3}{2 b \coth ^{-1}(\tanh (a+b x))^2}+\frac {3 \int \frac {x^2}{\coth ^{-1}(\tanh (a+b x))^2} \, dx}{2 b}\\ &=-\frac {x^3}{2 b \coth ^{-1}(\tanh (a+b x))^2}-\frac {3 x^2}{2 b^2 \coth ^{-1}(\tanh (a+b x))}+\frac {3 \int \frac {x}{\coth ^{-1}(\tanh (a+b x))} \, dx}{b^2}\\ &=\frac {3 x}{b^3}-\frac {x^3}{2 b \coth ^{-1}(\tanh (a+b x))^2}-\frac {3 x^2}{2 b^2 \coth ^{-1}(\tanh (a+b x))}-\frac {\left (3 \left (-b x+\coth ^{-1}(\tanh (a+b x))\right )\right ) \int \frac {1}{\coth ^{-1}(\tanh (a+b x))} \, dx}{b^3}\\ &=\frac {3 x}{b^3}-\frac {x^3}{2 b \coth ^{-1}(\tanh (a+b x))^2}-\frac {3 x^2}{2 b^2 \coth ^{-1}(\tanh (a+b x))}-\frac {\left (3 \left (-b x+\coth ^{-1}(\tanh (a+b x))\right )\right ) \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,\coth ^{-1}(\tanh (a+b x))\right )}{b^4}\\ &=\frac {3 x}{b^3}-\frac {x^3}{2 b \coth ^{-1}(\tanh (a+b x))^2}-\frac {3 x^2}{2 b^2 \coth ^{-1}(\tanh (a+b x))}+\frac {3 \left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \log \left (\coth ^{-1}(\tanh (a+b x))\right )}{b^4}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 86, normalized size = 1.21 \[ -\frac {3 b^2 x^2 \coth ^{-1}(\tanh (a+b x))-b x \coth ^{-1}(\tanh (a+b x))^2 \left (6 \log \left (\coth ^{-1}(\tanh (a+b x))\right )+11\right )+\coth ^{-1}(\tanh (a+b x))^3 \left (6 \log \left (\coth ^{-1}(\tanh (a+b x))\right )+5\right )+b^3 x^3}{2 b^4 \coth ^{-1}(\tanh (a+b x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(b^3*x^3 + 3*b^2*x^2*ArcCoth[Tanh[a + b*x]] + ArcCoth[Tanh[a + b*x]]^3*(5 + 6*Log[ArcCoth[Tanh[a + b*x]]]
) - b*x*ArcCoth[Tanh[a + b*x]]^2*(11 + 6*Log[ArcCoth[Tanh[a + b*x]]]))/(b^4*ArcCoth[Tanh[a + b*x]]^2)

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fricas [B]  time = 0.65, size = 418, normalized size = 5.89 \[ \frac {32 \, b^{5} x^{5} + 128 \, a b^{4} x^{4} - 5 \, \pi ^{4} a - 40 \, \pi ^{2} a^{3} - 80 \, a^{5} + 8 \, {\left (5 \, \pi ^{2} b^{3} + 12 \, a^{2} b^{3}\right )} x^{3} + 4 \, {\left (11 \, \pi ^{2} a b^{2} - 36 \, a^{3} b^{2}\right )} x^{2} + 2 \, {\left (3 \, \pi ^{4} b - 16 \, \pi ^{2} a^{2} b - 112 \, a^{4} b\right )} x + 6 \, {\left (16 \, \pi b^{4} x^{4} + 64 \, \pi a b^{3} x^{3} + \pi ^{5} + 8 \, \pi ^{3} a^{2} + 16 \, \pi a^{4} + 8 \, {\left (\pi ^{3} b^{2} + 12 \, \pi a^{2} b^{2}\right )} x^{2} + 16 \, {\left (\pi ^{3} a b + 4 \, \pi a^{3} b\right )} x\right )} \arctan \left (-\frac {2 \, b x + 2 \, a - \sqrt {4 \, b^{2} x^{2} + 8 \, a b x + \pi ^{2} + 4 \, a^{2}}}{\pi }\right ) - 3 \, {\left (16 \, a b^{4} x^{4} + 64 \, a^{2} b^{3} x^{3} + \pi ^{4} a + 8 \, \pi ^{2} a^{3} + 16 \, a^{5} + 8 \, {\left (\pi ^{2} a b^{2} + 12 \, a^{3} b^{2}\right )} x^{2} + 16 \, {\left (\pi ^{2} a^{2} b + 4 \, a^{4} b\right )} x\right )} \log \left (4 \, b^{2} x^{2} + 8 \, a b x + \pi ^{2} + 4 \, a^{2}\right )}{2 \, {\left (16 \, b^{8} x^{4} + 64 \, a b^{7} x^{3} + \pi ^{4} b^{4} + 8 \, \pi ^{2} a^{2} b^{4} + 16 \, a^{4} b^{4} + 8 \, {\left (\pi ^{2} b^{6} + 12 \, a^{2} b^{6}\right )} x^{2} + 16 \, {\left (\pi ^{2} a b^{5} + 4 \, a^{3} b^{5}\right )} x\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(32*b^5*x^5 + 128*a*b^4*x^4 - 5*pi^4*a - 40*pi^2*a^3 - 80*a^5 + 8*(5*pi^2*b^3 + 12*a^2*b^3)*x^3 + 4*(11*pi
^2*a*b^2 - 36*a^3*b^2)*x^2 + 2*(3*pi^4*b - 16*pi^2*a^2*b - 112*a^4*b)*x + 6*(16*pi*b^4*x^4 + 64*pi*a*b^3*x^3 +
 pi^5 + 8*pi^3*a^2 + 16*pi*a^4 + 8*(pi^3*b^2 + 12*pi*a^2*b^2)*x^2 + 16*(pi^3*a*b + 4*pi*a^3*b)*x)*arctan(-(2*b
*x + 2*a - sqrt(4*b^2*x^2 + 8*a*b*x + pi^2 + 4*a^2))/pi) - 3*(16*a*b^4*x^4 + 64*a^2*b^3*x^3 + pi^4*a + 8*pi^2*
a^3 + 16*a^5 + 8*(pi^2*a*b^2 + 12*a^3*b^2)*x^2 + 16*(pi^2*a^2*b + 4*a^4*b)*x)*log(4*b^2*x^2 + 8*a*b*x + pi^2 +
 4*a^2))/(16*b^8*x^4 + 64*a*b^7*x^3 + pi^4*b^4 + 8*pi^2*a^2*b^4 + 16*a^4*b^4 + 8*(pi^2*b^6 + 12*a^2*b^6)*x^2 +
 16*(pi^2*a*b^5 + 4*a^3*b^5)*x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{3}}{\operatorname {arcoth}\left (\tanh \left (b x + a\right )\right )^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.78, size = 4977, normalized size = 70.10 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*I*(-6*Pi*x^2*csgn(I/(exp(2*b*x+2*a)+1))^2+3*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))-3*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+6*Pi*x^2*csgn(I/(exp(2*b*x+2*a)+1))^3+3*Pi*x^2*csgn(I*exp(b*x+a))^2*csgn(I*exp(2*b*x+2*a))-6*Pi*x^2*csgn
(I*exp(b*x+a))*csgn(I*exp(2*b*x+2*a))^2+3*Pi*x^2*csgn(I*exp(2*b*x+2*a))^3-3*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+3*Pi*x^2*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^3+4*I*x^3*b+6*Pi*x^
2+12*I*x^2*ln(exp(b*x+a)))/b^2/(-2*Pi*csgn(I/(exp(2*b*x+2*a)+1))^2+Pi*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))-Pi*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+2*Pi*csgn(I/(exp(2*b*x+2*a)+1))^3+Pi*csgn(I*exp(b*x+a))^2*csgn(I*exp(2*b*x+2*a))-2*Pi*csgn(I*e
xp(b*x+a))*csgn(I*exp(2*b*x+2*a))^2+Pi*csgn(I*exp(2*b*x+2*a))^3-Pi*csgn(I*exp(2*b*x+2*a))*csgn(I*exp(2*b*x+2*a
)/(exp(2*b*x+2*a)+1))^2+Pi*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^3+2*Pi+4*I*ln(exp(b*x+a)))^2+3*x/b^3-3/2*
I/b^4*ln(-2*Pi*csgn(I/(exp(2*b*x+2*a)+1))^2+Pi*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))-Pi*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+2*Pi*cs
gn(I/(exp(2*b*x+2*a)+1))^3+Pi*csgn(I*exp(b*x+a))^2*csgn(I*exp(2*b*x+2*a))-2*Pi*csgn(I*exp(b*x+a))*csgn(I*exp(2
*b*x+2*a))^2+Pi*csgn(I*exp(2*b*x+2*a))^3-Pi*csgn(I*exp(2*b*x+2*a))*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^2
+Pi*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^3+4*I*(ln(exp(b*x+a))-b*x-a)+4*I*b*x+4*I*a+2*Pi)*Pi*csgn(I/(exp(
2*b*x+2*a)+1))^2+3/4*I/b^4*ln(-2*Pi*csgn(I/(exp(2*b*x+2*a)+1))^2+Pi*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))-Pi*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+2*Pi*csgn(I/(exp(2*b*x+2*a)+1))^3+Pi*csgn(I*exp(b*x+a))^2*csgn(I*exp(2*b*x+2*a))-2*Pi*csgn(I*exp
(b*x+a))*csgn(I*exp(2*b*x+2*a))^2+Pi*csgn(I*exp(2*b*x+2*a))^3-Pi*csgn(I*exp(2*b*x+2*a))*csgn(I*exp(2*b*x+2*a)/
(exp(2*b*x+2*a)+1))^2+Pi*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^3+4*I*(ln(exp(b*x+a))-b*x-a)+4*I*b*x+4*I*a+
2*Pi)*Pi*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))-3/4*I/b^4
*ln(-2*Pi*csgn(I/(exp(2*b*x+2*a)+1))^2+Pi*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))-Pi*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+2*Pi*csgn(I/
(exp(2*b*x+2*a)+1))^3+Pi*csgn(I*exp(b*x+a))^2*csgn(I*exp(2*b*x+2*a))-2*Pi*csgn(I*exp(b*x+a))*csgn(I*exp(2*b*x+
2*a))^2+Pi*csgn(I*exp(2*b*x+2*a))^3-Pi*csgn(I*exp(2*b*x+2*a))*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^2+Pi*c
sgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^3+4*I*(ln(exp(b*x+a))-b*x-a)+4*I*b*x+4*I*a+2*Pi)*Pi*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+3/2*I/b^4*ln(-2*Pi*csgn(I/(exp(2*b*x+2*a)+1))^2+Pi*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))-Pi*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+2*Pi*csgn(I/(exp(2*b*x+2*a)+1))^3+Pi*csgn(I*exp(b*x+a))^2*c
sgn(I*exp(2*b*x+2*a))-2*Pi*csgn(I*exp(b*x+a))*csgn(I*exp(2*b*x+2*a))^2+Pi*csgn(I*exp(2*b*x+2*a))^3-Pi*csgn(I*e
xp(2*b*x+2*a))*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^2+Pi*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^3+4*I*
(ln(exp(b*x+a))-b*x-a)+4*I*b*x+4*I*a+2*Pi)*Pi*csgn(I/(exp(2*b*x+2*a)+1))^3+3/4*I/b^4*ln(-2*Pi*csgn(I/(exp(2*b*
x+2*a)+1))^2+Pi*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))-Pi
*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+2*Pi*csgn(I/(exp(2*b*x+2*a)+1))^3+Pi*c
sgn(I*exp(b*x+a))^2*csgn(I*exp(2*b*x+2*a))-2*Pi*csgn(I*exp(b*x+a))*csgn(I*exp(2*b*x+2*a))^2+Pi*csgn(I*exp(2*b*
x+2*a))^3-Pi*csgn(I*exp(2*b*x+2*a))*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^2+Pi*csgn(I*exp(2*b*x+2*a)/(exp(
2*b*x+2*a)+1))^3+4*I*(ln(exp(b*x+a))-b*x-a)+4*I*b*x+4*I*a+2*Pi)*Pi*csgn(I*exp(b*x+a))^2*csgn(I*exp(2*b*x+2*a))
-3/2*I/b^4*ln(-2*Pi*csgn(I/(exp(2*b*x+2*a)+1))^2+Pi*csgn(I/(exp(2*b*x+2*a)+1))*csgn(I*exp(2*b*x+2*a))*csgn(I*e
xp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))-Pi*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+2*
Pi*csgn(I/(exp(2*b*x+2*a)+1))^3+Pi*csgn(I*exp(b*x+a))^2*csgn(I*exp(2*b*x+2*a))-2*Pi*csgn(I*exp(b*x+a))*csgn(I*
exp(2*b*x+2*a))^2+Pi*csgn(I*exp(2*b*x+2*a))^3-Pi*csgn(I*exp(2*b*x+2*a))*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+
1))^2+Pi*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^3+4*I*(ln(exp(b*x+a))-b*x-a)+4*I*b*x+4*I*a+2*Pi)*Pi*csgn(I*
exp(b*x+a))*csgn(I*exp(2*b*x+2*a))^2+3/4*I/b^4*ln(-2*Pi*csgn(I/(exp(2*b*x+2*a)+1))^2+Pi*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))-Pi*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+2*Pi*csgn(I/(exp(2*b*x+2*a)+1))^3+Pi*csgn(I*exp(b*x+a))^2*csgn(I*exp(2*b*x+2
*a))-2*Pi*csgn(I*exp(b*x+a))*csgn(I*exp(2*b*x+2*a))^2+Pi*csgn(I*exp(2*b*x+2*a))^3-Pi*csgn(I*exp(2*b*x+2*a))*cs
gn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^2+Pi*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^3+4*I*(ln(exp(b*x+a))-b
*x-a)+4*I*b*x+4*I*a+2*Pi)*Pi*csgn(I*exp(2*b*x+2*a))^3-3/4*I/b^4*ln(-2*Pi*csgn(I/(exp(2*b*x+2*a)+1))^2+Pi*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))-Pi*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+2*Pi*csgn(I/(exp(2*b*x+2*a)+1))^3+Pi*csgn(I*exp(b*x+a))^2*c
sgn(I*exp(2*b*x+2*a))-2*Pi*csgn(I*exp(b*x+a))*csgn(I*exp(2*b*x+2*a))^2+Pi*csgn(I*exp(2*b*x+2*a))^3-Pi*csgn(I*e
xp(2*b*x+2*a))*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^2+Pi*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^3+4*I*
(ln(exp(b*x+a))-b*x-a)+4*I*b*x+4*I*a+2*Pi)*Pi*csgn(I*exp(2*b*x+2*a))*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))
^2+3/4*I/b^4*ln(-2*Pi*csgn(I/(exp(2*b*x+2*a)+1))^2+Pi*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))-Pi*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+
2*Pi*csgn(I/(exp(2*b*x+2*a)+1))^3+Pi*csgn(I*exp(b*x+a))^2*csgn(I*exp(2*b*x+2*a))-2*Pi*csgn(I*exp(b*x+a))*csgn(
I*exp(2*b*x+2*a))^2+Pi*csgn(I*exp(2*b*x+2*a))^3-Pi*csgn(I*exp(2*b*x+2*a))*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a
)+1))^2+Pi*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^3+4*I*(ln(exp(b*x+a))-b*x-a)+4*I*b*x+4*I*a+2*Pi)*Pi*csgn(
I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^3+3/2*I/b^4*Pi*ln(-2*Pi*csgn(I/(exp(2*b*x+2*a)+1))^2+Pi*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))-Pi*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+2*Pi*csgn(I/(exp(2*b*x+2*a)+1))^3+Pi*csgn(I*exp(b*x+a))^2*csgn(I*exp(2*
b*x+2*a))-2*Pi*csgn(I*exp(b*x+a))*csgn(I*exp(2*b*x+2*a))^2+Pi*csgn(I*exp(2*b*x+2*a))^3-Pi*csgn(I*exp(2*b*x+2*a
))*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^2+Pi*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^3+4*I*(ln(exp(b*x+
a))-b*x-a)+4*I*b*x+4*I*a+2*Pi)+3/b^3*ln(-2*Pi*csgn(I/(exp(2*b*x+2*a)+1))^2+Pi*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))-Pi*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+2*Pi*csgn(I/(exp(2*b*x+2*a)+1))^3+Pi*csgn(I*exp(b*x+a))^2*csgn(I*exp(2*b*x+2*a))-2*Pi*
csgn(I*exp(b*x+a))*csgn(I*exp(2*b*x+2*a))^2+Pi*csgn(I*exp(2*b*x+2*a))^3-Pi*csgn(I*exp(2*b*x+2*a))*csgn(I*exp(2
*b*x+2*a)/(exp(2*b*x+2*a)+1))^2+Pi*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^3+4*I*(ln(exp(b*x+a))-b*x-a)+4*I*
b*x+4*I*a+2*Pi)*x-3/b^4*ln(-2*Pi*csgn(I/(exp(2*b*x+2*a)+1))^2+Pi*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))-Pi*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+2*Pi*csgn(I/(exp(2*b*x+2*a)+1))^3+Pi*csgn(I*exp(b*x+a))^2*csgn(I*exp(2*b*x+2*a))-2*Pi*csgn(I*exp(b*
x+a))*csgn(I*exp(2*b*x+2*a))^2+Pi*csgn(I*exp(2*b*x+2*a))^3-Pi*csgn(I*exp(2*b*x+2*a))*csgn(I*exp(2*b*x+2*a)/(ex
p(2*b*x+2*a)+1))^2+Pi*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^3+4*I*(ln(exp(b*x+a))-b*x-a)+4*I*b*x+4*I*a+2*P
i)*ln(exp(b*x+a))

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maxima [C]  time = 1.15, size = 144, normalized size = 2.03 \[ \frac {8 \, {\left (16 \, b^{3} x^{3} - 5 i \, \pi ^{3} + 30 \, \pi ^{2} a + 60 i \, \pi a^{2} - 40 \, a^{3} + {\left (-16 i \, \pi b^{2} + 32 \, a b^{2}\right )} x^{2} + {\left (8 \, \pi ^{2} b + 32 i \, \pi a b - 32 \, a^{2} b\right )} x\right )}}{128 \, b^{6} x^{2} - 32 \, \pi ^{2} b^{4} - 128 i \, \pi a b^{4} + 128 \, a^{2} b^{4} + {\left (-128 i \, \pi b^{5} + 256 \, a b^{5}\right )} x} - \frac {3 \, {\left (-i \, \pi + 2 \, a\right )} \log \left (-i \, \pi + 2 \, b x + 2 \, a\right )}{2 \, b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

8*(16*b^3*x^3 - 5*I*pi^3 + 30*pi^2*a + 60*I*pi*a^2 - 40*a^3 + (-16*I*pi*b^2 + 32*a*b^2)*x^2 + (8*pi^2*b + 32*I
*pi*a*b - 32*a^2*b)*x)/(128*b^6*x^2 - 32*pi^2*b^4 - 128*I*pi*a*b^4 + 128*a^2*b^4 + (-128*I*pi*b^5 + 256*a*b^5)
*x) - 3/2*(-I*pi + 2*a)*log(-I*pi + 2*b*x + 2*a)/b^4

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mupad [B]  time = 1.43, size = 620, normalized size = 8.73 \[ \frac {x}{b^3}-\frac {x\,\left (3\,{\left (2\,a-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+2\,b\,x\right )}^2-12\,a\,\left (2\,a-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+2\,b\,x\right )+12\,a^2\right )-\frac {5\,\left ({\left (2\,a-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+2\,b\,x\right )}^3-8\,a^3-6\,a\,{\left (2\,a-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+2\,b\,x\right )}^2+12\,a^2\,\left (2\,a-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+2\,b\,x\right )\right )}{4\,b}}{b^3\,{\left (2\,a-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+2\,b\,x\right )}^2+x\,\left (8\,a\,b^4-4\,b^4\,\left (2\,a-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+2\,b\,x\right )\right )+4\,a^2\,b^3+4\,b^5\,x^2-4\,a\,b^3\,\left (2\,a-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+2\,b\,x\right )}+\frac {\ln \left (\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )-\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )\right )\,\left (3\,\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )-3\,\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+6\,b\,x\right )}{2\,b^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/acoth(tanh(a + b*x))^3,x)

[Out]

x/b^3 - (x*(3*(2*a - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) - 1)) + log(-2/(exp(2*a)*exp(2*b*x) - 1)
) + 2*b*x)^2 - 12*a*(2*a - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) - 1)) + log(-2/(exp(2*a)*exp(2*b*x
) - 1)) + 2*b*x) + 12*a^2) - (5*((2*a - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) - 1)) + log(-2/(exp(2
*a)*exp(2*b*x) - 1)) + 2*b*x)^3 - 8*a^3 - 6*a*(2*a - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) - 1)) +
log(-2/(exp(2*a)*exp(2*b*x) - 1)) + 2*b*x)^2 + 12*a^2*(2*a - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x)
- 1)) + log(-2/(exp(2*a)*exp(2*b*x) - 1)) + 2*b*x)))/(4*b))/(b^3*(2*a - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*
exp(2*b*x) - 1)) + log(-2/(exp(2*a)*exp(2*b*x) - 1)) + 2*b*x)^2 + x*(8*a*b^4 - 4*b^4*(2*a - log((2*exp(2*a)*ex
p(2*b*x))/(exp(2*a)*exp(2*b*x) - 1)) + log(-2/(exp(2*a)*exp(2*b*x) - 1)) + 2*b*x)) + 4*a^2*b^3 + 4*b^5*x^2 - 4
*a*b^3*(2*a - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) - 1)) + log(-2/(exp(2*a)*exp(2*b*x) - 1)) + 2*b
*x)) + (log(log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) - 1)) - log(-2/(exp(2*a)*exp(2*b*x) - 1)))*(3*log
(-2/(exp(2*a)*exp(2*b*x) - 1)) - 3*log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) - 1)) + 6*b*x))/(2*b^4)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{3}}{\operatorname {acoth}^{3}{\left (\tanh {\left (a + b x \right )} \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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