3.454 \(\int x^2 \coth ^2(a+b x) \text {csch}(a+b x) \, dx\)

Optimal. Leaf size=123 \[ \frac {\text {Li}_3\left (-e^{a+b x}\right )}{b^3}-\frac {\text {Li}_3\left (e^{a+b x}\right )}{b^3}-\frac {\tanh ^{-1}(\cosh (a+b x))}{b^3}-\frac {x \text {Li}_2\left (-e^{a+b x}\right )}{b^2}+\frac {x \text {Li}_2\left (e^{a+b x}\right )}{b^2}-\frac {x \text {csch}(a+b x)}{b^2}-\frac {x^2 \tanh ^{-1}\left (e^{a+b x}\right )}{b}-\frac {x^2 \coth (a+b x) \text {csch}(a+b x)}{2 b} \]

[Out]

-x^2*arctanh(exp(b*x+a))/b-arctanh(cosh(b*x+a))/b^3-x*csch(b*x+a)/b^2-1/2*x^2*coth(b*x+a)*csch(b*x+a)/b-x*poly
log(2,-exp(b*x+a))/b^2+x*polylog(2,exp(b*x+a))/b^2+polylog(3,-exp(b*x+a))/b^3-polylog(3,exp(b*x+a))/b^3

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Rubi [A]  time = 0.24, antiderivative size = 123, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 7, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.389, Rules used = {5457, 4182, 2531, 2282, 6589, 4186, 3770} \[ -\frac {x \text {PolyLog}\left (2,-e^{a+b x}\right )}{b^2}+\frac {x \text {PolyLog}\left (2,e^{a+b x}\right )}{b^2}+\frac {\text {PolyLog}\left (3,-e^{a+b x}\right )}{b^3}-\frac {\text {PolyLog}\left (3,e^{a+b x}\right )}{b^3}-\frac {x \text {csch}(a+b x)}{b^2}-\frac {\tanh ^{-1}(\cosh (a+b x))}{b^3}-\frac {x^2 \tanh ^{-1}\left (e^{a+b x}\right )}{b}-\frac {x^2 \coth (a+b x) \text {csch}(a+b x)}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[x^2*Coth[a + b*x]^2*Csch[a + b*x],x]

[Out]

-((x^2*ArcTanh[E^(a + b*x)])/b) - ArcTanh[Cosh[a + b*x]]/b^3 - (x*Csch[a + b*x])/b^2 - (x^2*Coth[a + b*x]*Csch
[a + b*x])/(2*b) - (x*PolyLog[2, -E^(a + b*x)])/b^2 + (x*PolyLog[2, E^(a + b*x)])/b^2 + PolyLog[3, -E^(a + b*x
)]/b^3 - PolyLog[3, E^(a + b*x)]/b^3

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 2531

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> -Simp[((
f + g*x)^m*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)])/(b*c*n*Log[F]), x] + Dist[(g*m)/(b*c*n*Log[F]), Int[(f + g*x)
^(m - 1)*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rule 4182

Int[csc[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c + d*x)^m*Ar
cTanh[E^(-(I*e) + f*fz*x)])/(f*fz*I), x] + (-Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*Log[1 - E^(-(I*e) + f*
fz*x)], x], x] + Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*Log[1 + E^(-(I*e) + f*fz*x)], x], x]) /; FreeQ[{c,
d, e, f, fz}, x] && IGtQ[m, 0]

Rule 4186

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> -Simp[(b^2*(c + d*x)^m*Cot[e
+ f*x]*(b*Csc[e + f*x])^(n - 2))/(f*(n - 1)), x] + (Dist[(b^2*d^2*m*(m - 1))/(f^2*(n - 1)*(n - 2)), Int[(c + d
*x)^(m - 2)*(b*Csc[e + f*x])^(n - 2), x], x] + Dist[(b^2*(n - 2))/(n - 1), Int[(c + d*x)^m*(b*Csc[e + f*x])^(n
- 2), x], x] - Simp[(b^2*d*m*(c + d*x)^(m - 1)*(b*Csc[e + f*x])^(n - 2))/(f^2*(n - 1)*(n - 2)), x]) /; FreeQ[
{b, c, d, e, f}, x] && GtQ[n, 1] && NeQ[n, 2] && GtQ[m, 1]

Rule 5457

Int[Coth[(a_.) + (b_.)*(x_)]^(p_)*Csch[(a_.) + (b_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Int[(c + d
*x)^m*Csch[a + b*x]*Coth[a + b*x]^(p - 2), x] + Int[(c + d*x)^m*Csch[a + b*x]^3*Coth[a + b*x]^(p - 2), x] /; F
reeQ[{a, b, c, d, m}, x] && IGtQ[p/2, 0]

Rule 6589

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

Rubi steps

\begin {align*} \int x^2 \coth ^2(a+b x) \text {csch}(a+b x) \, dx &=\int x^2 \text {csch}(a+b x) \, dx+\int x^2 \text {csch}^3(a+b x) \, dx\\ &=-\frac {2 x^2 \tanh ^{-1}\left (e^{a+b x}\right )}{b}-\frac {x \text {csch}(a+b x)}{b^2}-\frac {x^2 \coth (a+b x) \text {csch}(a+b x)}{2 b}-\frac {1}{2} \int x^2 \text {csch}(a+b x) \, dx+\frac {\int \text {csch}(a+b x) \, dx}{b^2}-\frac {2 \int x \log \left (1-e^{a+b x}\right ) \, dx}{b}+\frac {2 \int x \log \left (1+e^{a+b x}\right ) \, dx}{b}\\ &=-\frac {x^2 \tanh ^{-1}\left (e^{a+b x}\right )}{b}-\frac {\tanh ^{-1}(\cosh (a+b x))}{b^3}-\frac {x \text {csch}(a+b x)}{b^2}-\frac {x^2 \coth (a+b x) \text {csch}(a+b x)}{2 b}-\frac {2 x \text {Li}_2\left (-e^{a+b x}\right )}{b^2}+\frac {2 x \text {Li}_2\left (e^{a+b x}\right )}{b^2}+\frac {2 \int \text {Li}_2\left (-e^{a+b x}\right ) \, dx}{b^2}-\frac {2 \int \text {Li}_2\left (e^{a+b x}\right ) \, dx}{b^2}+\frac {\int x \log \left (1-e^{a+b x}\right ) \, dx}{b}-\frac {\int x \log \left (1+e^{a+b x}\right ) \, dx}{b}\\ &=-\frac {x^2 \tanh ^{-1}\left (e^{a+b x}\right )}{b}-\frac {\tanh ^{-1}(\cosh (a+b x))}{b^3}-\frac {x \text {csch}(a+b x)}{b^2}-\frac {x^2 \coth (a+b x) \text {csch}(a+b x)}{2 b}-\frac {x \text {Li}_2\left (-e^{a+b x}\right )}{b^2}+\frac {x \text {Li}_2\left (e^{a+b x}\right )}{b^2}+\frac {2 \operatorname {Subst}\left (\int \frac {\text {Li}_2(-x)}{x} \, dx,x,e^{a+b x}\right )}{b^3}-\frac {2 \operatorname {Subst}\left (\int \frac {\text {Li}_2(x)}{x} \, dx,x,e^{a+b x}\right )}{b^3}-\frac {\int \text {Li}_2\left (-e^{a+b x}\right ) \, dx}{b^2}+\frac {\int \text {Li}_2\left (e^{a+b x}\right ) \, dx}{b^2}\\ &=-\frac {x^2 \tanh ^{-1}\left (e^{a+b x}\right )}{b}-\frac {\tanh ^{-1}(\cosh (a+b x))}{b^3}-\frac {x \text {csch}(a+b x)}{b^2}-\frac {x^2 \coth (a+b x) \text {csch}(a+b x)}{2 b}-\frac {x \text {Li}_2\left (-e^{a+b x}\right )}{b^2}+\frac {x \text {Li}_2\left (e^{a+b x}\right )}{b^2}+\frac {2 \text {Li}_3\left (-e^{a+b x}\right )}{b^3}-\frac {2 \text {Li}_3\left (e^{a+b x}\right )}{b^3}-\frac {\operatorname {Subst}\left (\int \frac {\text {Li}_2(-x)}{x} \, dx,x,e^{a+b x}\right )}{b^3}+\frac {\operatorname {Subst}\left (\int \frac {\text {Li}_2(x)}{x} \, dx,x,e^{a+b x}\right )}{b^3}\\ &=-\frac {x^2 \tanh ^{-1}\left (e^{a+b x}\right )}{b}-\frac {\tanh ^{-1}(\cosh (a+b x))}{b^3}-\frac {x \text {csch}(a+b x)}{b^2}-\frac {x^2 \coth (a+b x) \text {csch}(a+b x)}{2 b}-\frac {x \text {Li}_2\left (-e^{a+b x}\right )}{b^2}+\frac {x \text {Li}_2\left (e^{a+b x}\right )}{b^2}+\frac {\text {Li}_3\left (-e^{a+b x}\right )}{b^3}-\frac {\text {Li}_3\left (e^{a+b x}\right )}{b^3}\\ \end {align*}

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Mathematica [A]  time = 4.15, size = 222, normalized size = 1.80 \[ -\frac {-4 b^2 x^2 \log \left (1-e^{a+b x}\right )+4 b^2 x^2 \log \left (e^{a+b x}+1\right )+b^2 x^2 \text {csch}^2\left (\frac {1}{2} (a+b x)\right )+b^2 x^2 \text {sech}^2\left (\frac {1}{2} (a+b x)\right )+8 b x \text {Li}_2\left (-e^{a+b x}\right )-8 b x \text {Li}_2\left (e^{a+b x}\right )-8 \text {Li}_3\left (-e^{a+b x}\right )+8 \text {Li}_3\left (e^{a+b x}\right )-8 \log \left (1-e^{a+b x}\right )+8 \log \left (e^{a+b x}+1\right )+8 b x \text {csch}(a)-4 b x \text {csch}\left (\frac {a}{2}\right ) \sinh \left (\frac {b x}{2}\right ) \text {csch}\left (\frac {1}{2} (a+b x)\right )-4 b x \text {sech}\left (\frac {a}{2}\right ) \sinh \left (\frac {b x}{2}\right ) \text {sech}\left (\frac {1}{2} (a+b x)\right )}{8 b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*Coth[a + b*x]^2*Csch[a + b*x],x]

[Out]

-1/8*(8*b*x*Csch[a] + b^2*x^2*Csch[(a + b*x)/2]^2 - 8*Log[1 - E^(a + b*x)] - 4*b^2*x^2*Log[1 - E^(a + b*x)] +
8*Log[1 + E^(a + b*x)] + 4*b^2*x^2*Log[1 + E^(a + b*x)] + 8*b*x*PolyLog[2, -E^(a + b*x)] - 8*b*x*PolyLog[2, E^
(a + b*x)] - 8*PolyLog[3, -E^(a + b*x)] + 8*PolyLog[3, E^(a + b*x)] + b^2*x^2*Sech[(a + b*x)/2]^2 - 4*b*x*Csch
[a/2]*Csch[(a + b*x)/2]*Sinh[(b*x)/2] - 4*b*x*Sech[a/2]*Sech[(a + b*x)/2]*Sinh[(b*x)/2])/b^3

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fricas [C]  time = 0.45, size = 1311, normalized size = 10.66 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*cosh(b*x+a)^2*csch(b*x+a)^3,x, algorithm="fricas")

[Out]

-1/2*(2*(b^2*x^2 + 2*b*x)*cosh(b*x + a)^3 + 6*(b^2*x^2 + 2*b*x)*cosh(b*x + a)*sinh(b*x + a)^2 + 2*(b^2*x^2 + 2
*b*x)*sinh(b*x + a)^3 + 2*(b^2*x^2 - 2*b*x)*cosh(b*x + a) - 2*(b*x*cosh(b*x + a)^4 + 4*b*x*cosh(b*x + a)*sinh(
b*x + a)^3 + b*x*sinh(b*x + a)^4 - 2*b*x*cosh(b*x + a)^2 + 2*(3*b*x*cosh(b*x + a)^2 - b*x)*sinh(b*x + a)^2 + b
*x + 4*(b*x*cosh(b*x + a)^3 - b*x*cosh(b*x + a))*sinh(b*x + a))*dilog(cosh(b*x + a) + sinh(b*x + a)) + 2*(b*x*
cosh(b*x + a)^4 + 4*b*x*cosh(b*x + a)*sinh(b*x + a)^3 + b*x*sinh(b*x + a)^4 - 2*b*x*cosh(b*x + a)^2 + 2*(3*b*x
*cosh(b*x + a)^2 - b*x)*sinh(b*x + a)^2 + b*x + 4*(b*x*cosh(b*x + a)^3 - b*x*cosh(b*x + a))*sinh(b*x + a))*dil
og(-cosh(b*x + a) - sinh(b*x + a)) + ((b^2*x^2 + 2)*cosh(b*x + a)^4 + 4*(b^2*x^2 + 2)*cosh(b*x + a)*sinh(b*x +
a)^3 + (b^2*x^2 + 2)*sinh(b*x + a)^4 + b^2*x^2 - 2*(b^2*x^2 + 2)*cosh(b*x + a)^2 - 2*(b^2*x^2 - 3*(b^2*x^2 +
2)*cosh(b*x + a)^2 + 2)*sinh(b*x + a)^2 + 4*((b^2*x^2 + 2)*cosh(b*x + a)^3 - (b^2*x^2 + 2)*cosh(b*x + a))*sinh
(b*x + a) + 2)*log(cosh(b*x + a) + sinh(b*x + a) + 1) - ((a^2 + 2)*cosh(b*x + a)^4 + 4*(a^2 + 2)*cosh(b*x + a)
*sinh(b*x + a)^3 + (a^2 + 2)*sinh(b*x + a)^4 - 2*(a^2 + 2)*cosh(b*x + a)^2 + 2*(3*(a^2 + 2)*cosh(b*x + a)^2 -
a^2 - 2)*sinh(b*x + a)^2 + a^2 + 4*((a^2 + 2)*cosh(b*x + a)^3 - (a^2 + 2)*cosh(b*x + a))*sinh(b*x + a) + 2)*lo
g(cosh(b*x + a) + sinh(b*x + a) - 1) - ((b^2*x^2 - a^2)*cosh(b*x + a)^4 + 4*(b^2*x^2 - a^2)*cosh(b*x + a)*sinh
(b*x + a)^3 + (b^2*x^2 - a^2)*sinh(b*x + a)^4 + b^2*x^2 - 2*(b^2*x^2 - a^2)*cosh(b*x + a)^2 - 2*(b^2*x^2 - 3*(
b^2*x^2 - a^2)*cosh(b*x + a)^2 - a^2)*sinh(b*x + a)^2 - a^2 + 4*((b^2*x^2 - a^2)*cosh(b*x + a)^3 - (b^2*x^2 -
a^2)*cosh(b*x + a))*sinh(b*x + a))*log(-cosh(b*x + a) - sinh(b*x + a) + 1) + 2*(cosh(b*x + a)^4 + 4*cosh(b*x +
a)*sinh(b*x + a)^3 + sinh(b*x + a)^4 + 2*(3*cosh(b*x + a)^2 - 1)*sinh(b*x + a)^2 - 2*cosh(b*x + a)^2 + 4*(cos
h(b*x + a)^3 - cosh(b*x + a))*sinh(b*x + a) + 1)*polylog(3, cosh(b*x + a) + sinh(b*x + a)) - 2*(cosh(b*x + a)^
4 + 4*cosh(b*x + a)*sinh(b*x + a)^3 + sinh(b*x + a)^4 + 2*(3*cosh(b*x + a)^2 - 1)*sinh(b*x + a)^2 - 2*cosh(b*x
+ a)^2 + 4*(cosh(b*x + a)^3 - cosh(b*x + a))*sinh(b*x + a) + 1)*polylog(3, -cosh(b*x + a) - sinh(b*x + a)) +
2*(b^2*x^2 + 3*(b^2*x^2 + 2*b*x)*cosh(b*x + a)^2 - 2*b*x)*sinh(b*x + a))/(b^3*cosh(b*x + a)^4 + 4*b^3*cosh(b*x
+ a)*sinh(b*x + a)^3 + b^3*sinh(b*x + a)^4 - 2*b^3*cosh(b*x + a)^2 + b^3 + 2*(3*b^3*cosh(b*x + a)^2 - b^3)*si
nh(b*x + a)^2 + 4*(b^3*cosh(b*x + a)^3 - b^3*cosh(b*x + a))*sinh(b*x + a))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{2} \cosh \left (b x + a\right )^{2} \operatorname {csch}\left (b x + a\right )^{3}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*cosh(b*x+a)^2*csch(b*x+a)^3,x, algorithm="giac")

[Out]

integrate(x^2*cosh(b*x + a)^2*csch(b*x + a)^3, x)

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maple [A]  time = 0.72, size = 210, normalized size = 1.71 \[ -\frac {x \,{\mathrm e}^{b x +a} \left (b x \,{\mathrm e}^{2 b x +2 a}+b x +2 \,{\mathrm e}^{2 b x +2 a}-2\right )}{b^{2} \left ({\mathrm e}^{2 b x +2 a}-1\right )^{2}}-\frac {a^{2} \arctanh \left ({\mathrm e}^{b x +a}\right )}{b^{3}}-\frac {\ln \left (1+{\mathrm e}^{b x +a}\right ) x^{2}}{2 b}+\frac {\ln \left (1+{\mathrm e}^{b x +a}\right ) a^{2}}{2 b^{3}}-\frac {\polylog \left (2, -{\mathrm e}^{b x +a}\right ) x}{b^{2}}+\frac {\polylog \left (3, -{\mathrm e}^{b x +a}\right )}{b^{3}}+\frac {\ln \left (1-{\mathrm e}^{b x +a}\right ) x^{2}}{2 b}-\frac {\ln \left (1-{\mathrm e}^{b x +a}\right ) a^{2}}{2 b^{3}}+\frac {\polylog \left (2, {\mathrm e}^{b x +a}\right ) x}{b^{2}}-\frac {\polylog \left (3, {\mathrm e}^{b x +a}\right )}{b^{3}}-\frac {2 \arctanh \left ({\mathrm e}^{b x +a}\right )}{b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*cosh(b*x+a)^2*csch(b*x+a)^3,x)

[Out]

-x*exp(b*x+a)*(b*x*exp(2*b*x+2*a)+b*x+2*exp(2*b*x+2*a)-2)/b^2/(exp(2*b*x+2*a)-1)^2-1/b^3*a^2*arctanh(exp(b*x+a
))-1/2/b*ln(1+exp(b*x+a))*x^2+1/2/b^3*ln(1+exp(b*x+a))*a^2-1/b^2*polylog(2,-exp(b*x+a))*x+1/b^3*polylog(3,-exp
(b*x+a))+1/2/b*ln(1-exp(b*x+a))*x^2-1/2/b^3*ln(1-exp(b*x+a))*a^2+1/b^2*polylog(2,exp(b*x+a))*x-1/b^3*polylog(3
,exp(b*x+a))-2/b^3*arctanh(exp(b*x+a))

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maxima [A]  time = 0.49, size = 197, normalized size = 1.60 \[ -\frac {{\left (b x^{2} e^{\left (3 \, a\right )} + 2 \, x e^{\left (3 \, a\right )}\right )} e^{\left (3 \, b x\right )} + {\left (b x^{2} e^{a} - 2 \, x e^{a}\right )} e^{\left (b x\right )}}{b^{2} e^{\left (4 \, b x + 4 \, a\right )} - 2 \, b^{2} e^{\left (2 \, b x + 2 \, a\right )} + b^{2}} - \frac {b^{2} x^{2} \log \left (e^{\left (b x + a\right )} + 1\right ) + 2 \, b x {\rm Li}_2\left (-e^{\left (b x + a\right )}\right ) - 2 \, {\rm Li}_{3}(-e^{\left (b x + a\right )})}{2 \, b^{3}} + \frac {b^{2} x^{2} \log \left (-e^{\left (b x + a\right )} + 1\right ) + 2 \, b x {\rm Li}_2\left (e^{\left (b x + a\right )}\right ) - 2 \, {\rm Li}_{3}(e^{\left (b x + a\right )})}{2 \, b^{3}} - \frac {\log \left (e^{\left (b x + a\right )} + 1\right )}{b^{3}} + \frac {\log \left (e^{\left (b x + a\right )} - 1\right )}{b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*cosh(b*x+a)^2*csch(b*x+a)^3,x, algorithm="maxima")

[Out]

-((b*x^2*e^(3*a) + 2*x*e^(3*a))*e^(3*b*x) + (b*x^2*e^a - 2*x*e^a)*e^(b*x))/(b^2*e^(4*b*x + 4*a) - 2*b^2*e^(2*b
*x + 2*a) + b^2) - 1/2*(b^2*x^2*log(e^(b*x + a) + 1) + 2*b*x*dilog(-e^(b*x + a)) - 2*polylog(3, -e^(b*x + a)))
/b^3 + 1/2*(b^2*x^2*log(-e^(b*x + a) + 1) + 2*b*x*dilog(e^(b*x + a)) - 2*polylog(3, e^(b*x + a)))/b^3 - log(e^
(b*x + a) + 1)/b^3 + log(e^(b*x + a) - 1)/b^3

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^2\,{\mathrm {cosh}\left (a+b\,x\right )}^2}{{\mathrm {sinh}\left (a+b\,x\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2*cosh(a + b*x)^2)/sinh(a + b*x)^3,x)

[Out]

int((x^2*cosh(a + b*x)^2)/sinh(a + b*x)^3, x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*cosh(b*x+a)**2*csch(b*x+a)**3,x)

[Out]

Timed out

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