3.146 \(\int \frac {\coth ^3(c+d x)}{(a+b \text {sech}(c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=316 \[ \frac {2 \tanh ^{-1}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a}}\right )}{a^{3/2} d}-\frac {2 b^4}{a d \left (a^2-b^2\right )^2 \sqrt {a+b \text {sech}(c+d x)}}-\frac {\sqrt {a+b \text {sech}(c+d x)}}{4 d (a+b)^2 (1-\text {sech}(c+d x))}-\frac {\sqrt {a+b \text {sech}(c+d x)}}{4 d (a-b)^2 (\text {sech}(c+d x)+1)}+\frac {b \tanh ^{-1}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a-b}}\right )}{4 d (a-b)^{5/2}}-\frac {b \tanh ^{-1}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a+b}}\right )}{4 d (a+b)^{5/2}}-\frac {(2 a-3 b) \tanh ^{-1}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a-b}}\right )}{2 d (a-b)^{5/2}}-\frac {(2 a+3 b) \tanh ^{-1}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a+b}}\right )}{2 d (a+b)^{5/2}} \]

[Out]

2*arctanh((a+b*sech(d*x+c))^(1/2)/a^(1/2))/a^(3/2)/d-1/2*(2*a-3*b)*arctanh((a+b*sech(d*x+c))^(1/2)/(a-b)^(1/2)
)/(a-b)^(5/2)/d+1/4*b*arctanh((a+b*sech(d*x+c))^(1/2)/(a-b)^(1/2))/(a-b)^(5/2)/d-1/4*b*arctanh((a+b*sech(d*x+c
))^(1/2)/(a+b)^(1/2))/(a+b)^(5/2)/d-1/2*(2*a+3*b)*arctanh((a+b*sech(d*x+c))^(1/2)/(a+b)^(1/2))/(a+b)^(5/2)/d-2
*b^4/a/(a^2-b^2)^2/d/(a+b*sech(d*x+c))^(1/2)-1/4*(a+b*sech(d*x+c))^(1/2)/(a+b)^2/d/(1-sech(d*x+c))-1/4*(a+b*se
ch(d*x+c))^(1/2)/(a-b)^2/d/(1+sech(d*x+c))

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Rubi [A]  time = 0.43, antiderivative size = 316, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3885, 898, 1335, 206, 199} \[ -\frac {2 b^4}{a d \left (a^2-b^2\right )^2 \sqrt {a+b \text {sech}(c+d x)}}+\frac {2 \tanh ^{-1}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a}}\right )}{a^{3/2} d}-\frac {\sqrt {a+b \text {sech}(c+d x)}}{4 d (a+b)^2 (1-\text {sech}(c+d x))}-\frac {\sqrt {a+b \text {sech}(c+d x)}}{4 d (a-b)^2 (\text {sech}(c+d x)+1)}+\frac {b \tanh ^{-1}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a-b}}\right )}{4 d (a-b)^{5/2}}-\frac {b \tanh ^{-1}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a+b}}\right )}{4 d (a+b)^{5/2}}-\frac {(2 a-3 b) \tanh ^{-1}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a-b}}\right )}{2 d (a-b)^{5/2}}-\frac {(2 a+3 b) \tanh ^{-1}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a+b}}\right )}{2 d (a+b)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[Coth[c + d*x]^3/(a + b*Sech[c + d*x])^(3/2),x]

[Out]

(2*ArcTanh[Sqrt[a + b*Sech[c + d*x]]/Sqrt[a]])/(a^(3/2)*d) - ((2*a - 3*b)*ArcTanh[Sqrt[a + b*Sech[c + d*x]]/Sq
rt[a - b]])/(2*(a - b)^(5/2)*d) + (b*ArcTanh[Sqrt[a + b*Sech[c + d*x]]/Sqrt[a - b]])/(4*(a - b)^(5/2)*d) - (b*
ArcTanh[Sqrt[a + b*Sech[c + d*x]]/Sqrt[a + b]])/(4*(a + b)^(5/2)*d) - ((2*a + 3*b)*ArcTanh[Sqrt[a + b*Sech[c +
d*x]]/Sqrt[a + b]])/(2*(a + b)^(5/2)*d) - (2*b^4)/(a*(a^2 - b^2)^2*d*Sqrt[a + b*Sech[c + d*x]]) - Sqrt[a + b*
Sech[c + d*x]]/(4*(a + b)^2*d*(1 - Sech[c + d*x])) - Sqrt[a + b*Sech[c + d*x]]/(4*(a - b)^2*d*(1 + Sech[c + d*
x]))

Rule 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 898

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> With[{q = De
nominator[m]}, Dist[q/e, Subst[Int[x^(q*(m + 1) - 1)*((e*f - d*g)/e + (g*x^q)/e)^n*((c*d^2 + a*e^2)/e^2 - (2*c
*d*x^q)/e^2 + (c*x^(2*q))/e^2)^p, x], x, (d + e*x)^(1/q)], x]] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*
g, 0] && NeQ[c*d^2 + a*e^2, 0] && IntegersQ[n, p] && FractionQ[m]

Rule 1335

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> In
t[ExpandIntegrand[(f*x)^m*(d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, p, q}, x]
&& NeQ[b^2 - 4*a*c, 0] && (IGtQ[p, 0] || IGtQ[q, 0] || IntegersQ[m, q])

Rule 3885

Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> -Dist[(-1)^((m - 1
)/2)/(d*b^(m - 1)), Subst[Int[((b^2 - x^2)^((m - 1)/2)*(a + x)^n)/x, x], x, b*Csc[c + d*x]], x] /; FreeQ[{a, b
, c, d, n}, x] && IntegerQ[(m - 1)/2] && NeQ[a^2 - b^2, 0]

Rubi steps

\begin {align*} \int \frac {\coth ^3(c+d x)}{(a+b \text {sech}(c+d x))^{3/2}} \, dx &=-\frac {b^4 \operatorname {Subst}\left (\int \frac {1}{x (a+x)^{3/2} \left (b^2-x^2\right )^2} \, dx,x,b \text {sech}(c+d x)\right )}{d}\\ &=-\frac {\left (2 b^4\right ) \operatorname {Subst}\left (\int \frac {1}{x^2 \left (-a+x^2\right ) \left (-a^2+b^2+2 a x^2-x^4\right )^2} \, dx,x,\sqrt {a+b \text {sech}(c+d x)}\right )}{d}\\ &=-\frac {\left (2 b^4\right ) \operatorname {Subst}\left (\int \left (-\frac {1}{a (a-b)^2 (a+b)^2 x^2}-\frac {1}{a b^4 \left (a-x^2\right )}-\frac {1}{4 (a-b) b^3 \left (a-b-x^2\right )^2}+\frac {2 a-3 b}{4 (a-b)^2 b^4 \left (a-b-x^2\right )}+\frac {1}{4 b^3 (a+b) \left (a+b-x^2\right )^2}+\frac {2 a+3 b}{4 b^4 (a+b)^2 \left (a+b-x^2\right )}\right ) \, dx,x,\sqrt {a+b \text {sech}(c+d x)}\right )}{d}\\ &=-\frac {2 b^4}{a \left (a^2-b^2\right )^2 d \sqrt {a+b \text {sech}(c+d x)}}+\frac {2 \operatorname {Subst}\left (\int \frac {1}{a-x^2} \, dx,x,\sqrt {a+b \text {sech}(c+d x)}\right )}{a d}-\frac {(2 a-3 b) \operatorname {Subst}\left (\int \frac {1}{a-b-x^2} \, dx,x,\sqrt {a+b \text {sech}(c+d x)}\right )}{2 (a-b)^2 d}+\frac {b \operatorname {Subst}\left (\int \frac {1}{\left (a-b-x^2\right )^2} \, dx,x,\sqrt {a+b \text {sech}(c+d x)}\right )}{2 (a-b) d}-\frac {b \operatorname {Subst}\left (\int \frac {1}{\left (a+b-x^2\right )^2} \, dx,x,\sqrt {a+b \text {sech}(c+d x)}\right )}{2 (a+b) d}-\frac {(2 a+3 b) \operatorname {Subst}\left (\int \frac {1}{a+b-x^2} \, dx,x,\sqrt {a+b \text {sech}(c+d x)}\right )}{2 (a+b)^2 d}\\ &=\frac {2 \tanh ^{-1}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a}}\right )}{a^{3/2} d}-\frac {(2 a-3 b) \tanh ^{-1}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a-b}}\right )}{2 (a-b)^{5/2} d}-\frac {(2 a+3 b) \tanh ^{-1}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a+b}}\right )}{2 (a+b)^{5/2} d}-\frac {2 b^4}{a \left (a^2-b^2\right )^2 d \sqrt {a+b \text {sech}(c+d x)}}-\frac {\sqrt {a+b \text {sech}(c+d x)}}{4 (a+b)^2 d (1-\text {sech}(c+d x))}-\frac {\sqrt {a+b \text {sech}(c+d x)}}{4 (a-b)^2 d (1+\text {sech}(c+d x))}+\frac {b \operatorname {Subst}\left (\int \frac {1}{a-b-x^2} \, dx,x,\sqrt {a+b \text {sech}(c+d x)}\right )}{4 (a-b)^2 d}-\frac {b \operatorname {Subst}\left (\int \frac {1}{a+b-x^2} \, dx,x,\sqrt {a+b \text {sech}(c+d x)}\right )}{4 (a+b)^2 d}\\ &=\frac {2 \tanh ^{-1}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a}}\right )}{a^{3/2} d}-\frac {(2 a-3 b) \tanh ^{-1}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a-b}}\right )}{2 (a-b)^{5/2} d}+\frac {b \tanh ^{-1}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a-b}}\right )}{4 (a-b)^{5/2} d}-\frac {b \tanh ^{-1}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a+b}}\right )}{4 (a+b)^{5/2} d}-\frac {(2 a+3 b) \tanh ^{-1}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a+b}}\right )}{2 (a+b)^{5/2} d}-\frac {2 b^4}{a \left (a^2-b^2\right )^2 d \sqrt {a+b \text {sech}(c+d x)}}-\frac {\sqrt {a+b \text {sech}(c+d x)}}{4 (a+b)^2 d (1-\text {sech}(c+d x))}-\frac {\sqrt {a+b \text {sech}(c+d x)}}{4 (a-b)^2 d (1+\text {sech}(c+d x))}\\ \end {align*}

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Mathematica [B]  time = 7.61, size = 996, normalized size = 3.15 \[ \frac {(b+a \cosh (c+d x))^2 \left (\frac {2 b^5}{a^2 \left (a^2-b^2\right )^2 (b+a \cosh (c+d x))}+\frac {\left (-a^2+2 b \cosh (c+d x) a-b^2\right ) \text {csch}^2(c+d x)}{2 \left (b^2-a^2\right )^2}-\frac {a^4+b^2 a^2+4 b^4}{2 a^2 \left (b^2-a^2\right )^2}\right ) \text {sech}^2(c+d x)}{d (a+b \text {sech}(c+d x))^{3/2}}+\frac {(b+a \cosh (c+d x))^{3/2} \left (\frac {\left (2 a^4-4 b^2 a^2+2 b^4\right ) \left (\sqrt {a} \left (\sqrt {a+b} \tan ^{-1}\left (\frac {\sqrt {a} \sqrt {b+a \cosh (c+d x)}}{\sqrt {a-b} \sqrt {-a \cosh (c+d x)}}\right )+\sqrt {a-b} \tan ^{-1}\left (\frac {\sqrt {a} \sqrt {b+a \cosh (c+d x)}}{\sqrt {a+b} \sqrt {-a \cosh (c+d x)}}\right )\right )-4 \sqrt {a-b} \sqrt {a+b} \tan ^{-1}\left (\frac {\sqrt {b+a \cosh (c+d x)}}{\sqrt {-a \cosh (c+d x)}}\right )\right ) \sqrt {-a \cosh (c+d x)} \sqrt {\frac {a \cosh (c+d x)-a}{\cosh (c+d x) a+a}} \cosh (2 (c+d x)) \sqrt {\text {sech}(c+d x)} (\cosh (c+d x) a+a)}{\sqrt {a-b} \sqrt {a+b} \sqrt {\cosh (c+d x)-1} \sqrt {\cosh (c+d x)+1} \left (a^2-2 b^2-2 (b+a \cosh (c+d x))^2+4 b (b+a \cosh (c+d x))\right )}-\frac {\left (2 a^4-6 b^2 a^2-2 b^4\right ) \left (\sqrt {a+b} \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {b+a \cosh (c+d x)}}{\sqrt {a-b} \sqrt {a \cosh (c+d x)}}\right )+\sqrt {a-b} \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {b+a \cosh (c+d x)}}{\sqrt {a+b} \sqrt {a \cosh (c+d x)}}\right )\right ) \sqrt {a \cosh (c+d x)} \sqrt {\frac {a \cosh (c+d x)-a}{\cosh (c+d x) a+a}} \sqrt {\text {sech}(c+d x)} (\cosh (c+d x) a+a)}{a^{3/2} \sqrt {a-b} \sqrt {a+b} \sqrt {\cosh (c+d x)-1} \sqrt {\cosh (c+d x)+1}}+\frac {\left (7 a b^3-a^3 b\right ) \left (\sqrt {a-b} \tan ^{-1}\left (\frac {\sqrt {a} \sqrt {b+a \cosh (c+d x)}}{\sqrt {-a-b} \sqrt {a \cosh (c+d x)}}\right )+\sqrt {-a-b} \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {b+a \cosh (c+d x)}}{\sqrt {a-b} \sqrt {a \cosh (c+d x)}}\right )\right ) \sqrt {\frac {a \cosh (c+d x)-a}{\cosh (c+d x) a+a}} (\cosh (c+d x) a+a)}{\sqrt {a} \sqrt {-a-b} \sqrt {a-b} \sqrt {\cosh (c+d x)-1} \sqrt {a \cosh (c+d x)} \sqrt {\cosh (c+d x)+1} \sqrt {\text {sech}(c+d x)}}\right ) \text {sech}^{\frac {3}{2}}(c+d x)}{4 a (a-b)^2 (a+b)^2 d (a+b \text {sech}(c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Coth[c + d*x]^3/(a + b*Sech[c + d*x])^(3/2),x]

[Out]

((b + a*Cosh[c + d*x])^(3/2)*(((-(a^3*b) + 7*a*b^3)*(Sqrt[a - b]*ArcTan[(Sqrt[a]*Sqrt[b + a*Cosh[c + d*x]])/(S
qrt[-a - b]*Sqrt[a*Cosh[c + d*x]])] + Sqrt[-a - b]*ArcTanh[(Sqrt[a]*Sqrt[b + a*Cosh[c + d*x]])/(Sqrt[a - b]*Sq
rt[a*Cosh[c + d*x]])])*Sqrt[(-a + a*Cosh[c + d*x])/(a + a*Cosh[c + d*x])]*(a + a*Cosh[c + d*x]))/(Sqrt[a]*Sqrt
[-a - b]*Sqrt[a - b]*Sqrt[-1 + Cosh[c + d*x]]*Sqrt[a*Cosh[c + d*x]]*Sqrt[1 + Cosh[c + d*x]]*Sqrt[Sech[c + d*x]
]) - ((2*a^4 - 6*a^2*b^2 - 2*b^4)*(Sqrt[a + b]*ArcTanh[(Sqrt[a]*Sqrt[b + a*Cosh[c + d*x]])/(Sqrt[a - b]*Sqrt[a
*Cosh[c + d*x]])] + Sqrt[a - b]*ArcTanh[(Sqrt[a]*Sqrt[b + a*Cosh[c + d*x]])/(Sqrt[a + b]*Sqrt[a*Cosh[c + d*x]]
)])*Sqrt[a*Cosh[c + d*x]]*Sqrt[(-a + a*Cosh[c + d*x])/(a + a*Cosh[c + d*x])]*(a + a*Cosh[c + d*x])*Sqrt[Sech[c
+ d*x]])/(a^(3/2)*Sqrt[a - b]*Sqrt[a + b]*Sqrt[-1 + Cosh[c + d*x]]*Sqrt[1 + Cosh[c + d*x]]) + ((2*a^4 - 4*a^2
*b^2 + 2*b^4)*(-4*Sqrt[a - b]*Sqrt[a + b]*ArcTan[Sqrt[b + a*Cosh[c + d*x]]/Sqrt[-(a*Cosh[c + d*x])]] + Sqrt[a]
*(Sqrt[a + b]*ArcTan[(Sqrt[a]*Sqrt[b + a*Cosh[c + d*x]])/(Sqrt[a - b]*Sqrt[-(a*Cosh[c + d*x])])] + Sqrt[a - b]
*ArcTan[(Sqrt[a]*Sqrt[b + a*Cosh[c + d*x]])/(Sqrt[a + b]*Sqrt[-(a*Cosh[c + d*x])])]))*Sqrt[-(a*Cosh[c + d*x])]
*Sqrt[(-a + a*Cosh[c + d*x])/(a + a*Cosh[c + d*x])]*(a + a*Cosh[c + d*x])*Cosh[2*(c + d*x)]*Sqrt[Sech[c + d*x]
])/(Sqrt[a - b]*Sqrt[a + b]*Sqrt[-1 + Cosh[c + d*x]]*Sqrt[1 + Cosh[c + d*x]]*(a^2 - 2*b^2 + 4*b*(b + a*Cosh[c
+ d*x]) - 2*(b + a*Cosh[c + d*x])^2)))*Sech[c + d*x]^(3/2))/(4*a*(a - b)^2*(a + b)^2*d*(a + b*Sech[c + d*x])^(
3/2)) + ((b + a*Cosh[c + d*x])^2*(-1/2*(a^4 + a^2*b^2 + 4*b^4)/(a^2*(-a^2 + b^2)^2) + (2*b^5)/(a^2*(a^2 - b^2)
^2*(b + a*Cosh[c + d*x])) + ((-a^2 - b^2 + 2*a*b*Cosh[c + d*x])*Csch[c + d*x]^2)/(2*(-a^2 + b^2)^2))*Sech[c +
d*x]^2)/(d*(a + b*Sech[c + d*x])^(3/2))

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fricas [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(coth(d*x+c)^3/(a+b*sech(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(d*x+c)^3/(a+b*sech(d*x+c))^(3/2),x, algorithm="giac")

[Out]

integrate(coth(d*x + c)^3/(b*sech(d*x + c) + a)^(3/2), x)

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maple [F]  time = 0.70, size = 0, normalized size = 0.00 \[ \int \frac {\coth ^{3}\left (d x +c \right )}{\left (a +b \,\mathrm {sech}\left (d x +c \right )\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(coth(d*x+c)^3/(a+b*sech(d*x+c))^(3/2),x)

[Out]

int(coth(d*x+c)^3/(a+b*sech(d*x+c))^(3/2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\coth \left (d x + c\right )^{3}}{{\left (b \operatorname {sech}\left (d x + c\right ) + a\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(d*x+c)^3/(a+b*sech(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

integrate(coth(d*x + c)^3/(b*sech(d*x + c) + a)^(3/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(coth(c + d*x)^3/(a + b/cosh(c + d*x))^(3/2),x)

[Out]

int(coth(c + d*x)^3/(a + b/cosh(c + d*x))^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\coth ^{3}{\left (c + d x \right )}}{\left (a + b \operatorname {sech}{\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(d*x+c)**3/(a+b*sech(d*x+c))**(3/2),x)

[Out]

Integral(coth(c + d*x)**3/(a + b*sech(c + d*x))**(3/2), x)

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