3.57 \(\int \sqrt{x} (a+b \text{csch}(c+d \sqrt{x}))^2 \, dx\)

Optimal. Leaf size=209 \[ -\frac{8 a b \sqrt{x} \text{PolyLog}\left (2,-e^{c+d \sqrt{x}}\right )}{d^2}+\frac{8 a b \sqrt{x} \text{PolyLog}\left (2,e^{c+d \sqrt{x}}\right )}{d^2}+\frac{8 a b \text{PolyLog}\left (3,-e^{c+d \sqrt{x}}\right )}{d^3}-\frac{8 a b \text{PolyLog}\left (3,e^{c+d \sqrt{x}}\right )}{d^3}+\frac{2 b^2 \text{PolyLog}\left (2,e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{2}{3} a^2 x^{3/2}-\frac{8 a b x \tanh ^{-1}\left (e^{c+d \sqrt{x}}\right )}{d}+\frac{4 b^2 \sqrt{x} \log \left (1-e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{2 b^2 x \coth \left (c+d \sqrt{x}\right )}{d}-\frac{2 b^2 x}{d} \]

[Out]

(-2*b^2*x)/d + (2*a^2*x^(3/2))/3 - (8*a*b*x*ArcTanh[E^(c + d*Sqrt[x])])/d - (2*b^2*x*Coth[c + d*Sqrt[x]])/d +
(4*b^2*Sqrt[x]*Log[1 - E^(2*(c + d*Sqrt[x]))])/d^2 - (8*a*b*Sqrt[x]*PolyLog[2, -E^(c + d*Sqrt[x])])/d^2 + (8*a
*b*Sqrt[x]*PolyLog[2, E^(c + d*Sqrt[x])])/d^2 + (2*b^2*PolyLog[2, E^(2*(c + d*Sqrt[x]))])/d^3 + (8*a*b*PolyLog
[3, -E^(c + d*Sqrt[x])])/d^3 - (8*a*b*PolyLog[3, E^(c + d*Sqrt[x])])/d^3

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Rubi [A]  time = 0.329456, antiderivative size = 209, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 11, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {5437, 4190, 4182, 2531, 2282, 6589, 4184, 3716, 2190, 2279, 2391} \[ -\frac{8 a b \sqrt{x} \text{PolyLog}\left (2,-e^{c+d \sqrt{x}}\right )}{d^2}+\frac{8 a b \sqrt{x} \text{PolyLog}\left (2,e^{c+d \sqrt{x}}\right )}{d^2}+\frac{8 a b \text{PolyLog}\left (3,-e^{c+d \sqrt{x}}\right )}{d^3}-\frac{8 a b \text{PolyLog}\left (3,e^{c+d \sqrt{x}}\right )}{d^3}+\frac{2 b^2 \text{PolyLog}\left (2,e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{2}{3} a^2 x^{3/2}-\frac{8 a b x \tanh ^{-1}\left (e^{c+d \sqrt{x}}\right )}{d}+\frac{4 b^2 \sqrt{x} \log \left (1-e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{2 b^2 x \coth \left (c+d \sqrt{x}\right )}{d}-\frac{2 b^2 x}{d} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[x]*(a + b*Csch[c + d*Sqrt[x]])^2,x]

[Out]

(-2*b^2*x)/d + (2*a^2*x^(3/2))/3 - (8*a*b*x*ArcTanh[E^(c + d*Sqrt[x])])/d - (2*b^2*x*Coth[c + d*Sqrt[x]])/d +
(4*b^2*Sqrt[x]*Log[1 - E^(2*(c + d*Sqrt[x]))])/d^2 - (8*a*b*Sqrt[x]*PolyLog[2, -E^(c + d*Sqrt[x])])/d^2 + (8*a
*b*Sqrt[x]*PolyLog[2, E^(c + d*Sqrt[x])])/d^2 + (2*b^2*PolyLog[2, E^(2*(c + d*Sqrt[x]))])/d^3 + (8*a*b*PolyLog
[3, -E^(c + d*Sqrt[x])])/d^3 - (8*a*b*PolyLog[3, E^(c + d*Sqrt[x])])/d^3

Rule 5437

Int[((a_.) + Csch[(c_.) + (d_.)*(x_)^(n_)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simpli
fy[(m + 1)/n] - 1)*(a + b*Csch[c + d*x])^p, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && IGtQ[Simplif
y[(m + 1)/n], 0] && IntegerQ[p]

Rule 4190

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[
(c + d*x)^m, (a + b*Csc[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[m, 0] && IGtQ[n, 0]

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 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 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 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]

Rule 4184

Int[csc[(e_.) + (f_.)*(x_)]^2*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> -Simp[((c + d*x)^m*Cot[e + f*x])/f, x]
+ Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cot[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 3716

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + Pi*(k_.) + (Complex[0, fz_])*(f_.)*(x_)], x_Symbol] :> -Simp[(I*(c
+ d*x)^(m + 1))/(d*(m + 1)), x] + Dist[2*I, Int[((c + d*x)^m*E^(2*(-(I*e) + f*fz*x)))/(E^(2*I*k*Pi)*(1 + E^(2*
(-(I*e) + f*fz*x))/E^(2*I*k*Pi))), x], x] /; FreeQ[{c, d, e, f, fz}, x] && IntegerQ[4*k] && IGtQ[m, 0]

Rule 2190

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

Rule 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rubi steps

\begin{align*} \int \sqrt{x} \left (a+b \text{csch}\left (c+d \sqrt{x}\right )\right )^2 \, dx &=2 \operatorname{Subst}\left (\int x^2 (a+b \text{csch}(c+d x))^2 \, dx,x,\sqrt{x}\right )\\ &=2 \operatorname{Subst}\left (\int \left (a^2 x^2+2 a b x^2 \text{csch}(c+d x)+b^2 x^2 \text{csch}^2(c+d x)\right ) \, dx,x,\sqrt{x}\right )\\ &=\frac{2}{3} a^2 x^{3/2}+(4 a b) \operatorname{Subst}\left (\int x^2 \text{csch}(c+d x) \, dx,x,\sqrt{x}\right )+\left (2 b^2\right ) \operatorname{Subst}\left (\int x^2 \text{csch}^2(c+d x) \, dx,x,\sqrt{x}\right )\\ &=\frac{2}{3} a^2 x^{3/2}-\frac{8 a b x \tanh ^{-1}\left (e^{c+d \sqrt{x}}\right )}{d}-\frac{2 b^2 x \coth \left (c+d \sqrt{x}\right )}{d}-\frac{(8 a b) \operatorname{Subst}\left (\int x \log \left (1-e^{c+d x}\right ) \, dx,x,\sqrt{x}\right )}{d}+\frac{(8 a b) \operatorname{Subst}\left (\int x \log \left (1+e^{c+d x}\right ) \, dx,x,\sqrt{x}\right )}{d}+\frac{\left (4 b^2\right ) \operatorname{Subst}\left (\int x \coth (c+d x) \, dx,x,\sqrt{x}\right )}{d}\\ &=-\frac{2 b^2 x}{d}+\frac{2}{3} a^2 x^{3/2}-\frac{8 a b x \tanh ^{-1}\left (e^{c+d \sqrt{x}}\right )}{d}-\frac{2 b^2 x \coth \left (c+d \sqrt{x}\right )}{d}-\frac{8 a b \sqrt{x} \text{Li}_2\left (-e^{c+d \sqrt{x}}\right )}{d^2}+\frac{8 a b \sqrt{x} \text{Li}_2\left (e^{c+d \sqrt{x}}\right )}{d^2}+\frac{(8 a b) \operatorname{Subst}\left (\int \text{Li}_2\left (-e^{c+d x}\right ) \, dx,x,\sqrt{x}\right )}{d^2}-\frac{(8 a b) \operatorname{Subst}\left (\int \text{Li}_2\left (e^{c+d x}\right ) \, dx,x,\sqrt{x}\right )}{d^2}-\frac{\left (8 b^2\right ) \operatorname{Subst}\left (\int \frac{e^{2 (c+d x)} x}{1-e^{2 (c+d x)}} \, dx,x,\sqrt{x}\right )}{d}\\ &=-\frac{2 b^2 x}{d}+\frac{2}{3} a^2 x^{3/2}-\frac{8 a b x \tanh ^{-1}\left (e^{c+d \sqrt{x}}\right )}{d}-\frac{2 b^2 x \coth \left (c+d \sqrt{x}\right )}{d}+\frac{4 b^2 \sqrt{x} \log \left (1-e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{8 a b \sqrt{x} \text{Li}_2\left (-e^{c+d \sqrt{x}}\right )}{d^2}+\frac{8 a b \sqrt{x} \text{Li}_2\left (e^{c+d \sqrt{x}}\right )}{d^2}+\frac{(8 a b) \operatorname{Subst}\left (\int \frac{\text{Li}_2(-x)}{x} \, dx,x,e^{c+d \sqrt{x}}\right )}{d^3}-\frac{(8 a b) \operatorname{Subst}\left (\int \frac{\text{Li}_2(x)}{x} \, dx,x,e^{c+d \sqrt{x}}\right )}{d^3}-\frac{\left (4 b^2\right ) \operatorname{Subst}\left (\int \log \left (1-e^{2 (c+d x)}\right ) \, dx,x,\sqrt{x}\right )}{d^2}\\ &=-\frac{2 b^2 x}{d}+\frac{2}{3} a^2 x^{3/2}-\frac{8 a b x \tanh ^{-1}\left (e^{c+d \sqrt{x}}\right )}{d}-\frac{2 b^2 x \coth \left (c+d \sqrt{x}\right )}{d}+\frac{4 b^2 \sqrt{x} \log \left (1-e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{8 a b \sqrt{x} \text{Li}_2\left (-e^{c+d \sqrt{x}}\right )}{d^2}+\frac{8 a b \sqrt{x} \text{Li}_2\left (e^{c+d \sqrt{x}}\right )}{d^2}+\frac{8 a b \text{Li}_3\left (-e^{c+d \sqrt{x}}\right )}{d^3}-\frac{8 a b \text{Li}_3\left (e^{c+d \sqrt{x}}\right )}{d^3}-\frac{\left (2 b^2\right ) \operatorname{Subst}\left (\int \frac{\log (1-x)}{x} \, dx,x,e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^3}\\ &=-\frac{2 b^2 x}{d}+\frac{2}{3} a^2 x^{3/2}-\frac{8 a b x \tanh ^{-1}\left (e^{c+d \sqrt{x}}\right )}{d}-\frac{2 b^2 x \coth \left (c+d \sqrt{x}\right )}{d}+\frac{4 b^2 \sqrt{x} \log \left (1-e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{8 a b \sqrt{x} \text{Li}_2\left (-e^{c+d \sqrt{x}}\right )}{d^2}+\frac{8 a b \sqrt{x} \text{Li}_2\left (e^{c+d \sqrt{x}}\right )}{d^2}+\frac{2 b^2 \text{Li}_2\left (e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{8 a b \text{Li}_3\left (-e^{c+d \sqrt{x}}\right )}{d^3}-\frac{8 a b \text{Li}_3\left (e^{c+d \sqrt{x}}\right )}{d^3}\\ \end{align*}

Mathematica [A]  time = 9.23274, size = 316, normalized size = 1.51 \[ \frac{4 b \left (-\left (b-2 a d \sqrt{x}\right ) \text{PolyLog}\left (2,-e^{-c-d \sqrt{x}}\right )-\left (2 a d \sqrt{x}+b\right ) \text{PolyLog}\left (2,e^{-c-d \sqrt{x}}\right )+2 a \text{PolyLog}\left (3,-e^{-c-d \sqrt{x}}\right )-2 a \text{PolyLog}\left (3,e^{-c-d \sqrt{x}}\right )+a d^2 x \log \left (1-e^{-c-d \sqrt{x}}\right )-a d^2 x \log \left (e^{-c-d \sqrt{x}}+1\right )-\frac{b d^2 x}{e^{2 c}-1}+b d \sqrt{x} \log \left (1-e^{-c-d \sqrt{x}}\right )+b d \sqrt{x} \log \left (e^{-c-d \sqrt{x}}+1\right )\right )}{d^3}+\frac{2}{3} a^2 x^{3/2}+\frac{b^2 x \text{csch}\left (\frac{c}{2}\right ) \sinh \left (\frac{d \sqrt{x}}{2}\right ) \text{csch}\left (\frac{1}{2} \left (c+d \sqrt{x}\right )\right )}{d}-\frac{b^2 x \text{sech}\left (\frac{c}{2}\right ) \sinh \left (\frac{d \sqrt{x}}{2}\right ) \text{sech}\left (\frac{1}{2} \left (c+d \sqrt{x}\right )\right )}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[x]*(a + b*Csch[c + d*Sqrt[x]])^2,x]

[Out]

(2*a^2*x^(3/2))/3 + (4*b*(-((b*d^2*x)/(-1 + E^(2*c))) + b*d*Sqrt[x]*Log[1 - E^(-c - d*Sqrt[x])] + a*d^2*x*Log[
1 - E^(-c - d*Sqrt[x])] + b*d*Sqrt[x]*Log[1 + E^(-c - d*Sqrt[x])] - a*d^2*x*Log[1 + E^(-c - d*Sqrt[x])] - (b -
2*a*d*Sqrt[x])*PolyLog[2, -E^(-c - d*Sqrt[x])] - (b + 2*a*d*Sqrt[x])*PolyLog[2, E^(-c - d*Sqrt[x])] + 2*a*Pol
yLog[3, -E^(-c - d*Sqrt[x])] - 2*a*PolyLog[3, E^(-c - d*Sqrt[x])]))/d^3 + (b^2*x*Csch[c/2]*Csch[(c + d*Sqrt[x]
)/2]*Sinh[(d*Sqrt[x])/2])/d - (b^2*x*Sech[c/2]*Sech[(c + d*Sqrt[x])/2]*Sinh[(d*Sqrt[x])/2])/d

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Maple [F]  time = 0.138, size = 0, normalized size = 0. \begin{align*} \int \left ( a+b{\rm csch} \left (c+d\sqrt{x}\right ) \right ) ^{2}\sqrt{x}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*csch(c+d*x^(1/2)))^2*x^(1/2),x)

[Out]

int((a+b*csch(c+d*x^(1/2)))^2*x^(1/2),x)

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Maxima [A]  time = 1.94531, size = 356, normalized size = 1.7 \begin{align*} \frac{2}{3} \, a^{2} x^{\frac{3}{2}} - \frac{4 \, b^{2} x}{d e^{\left (2 \, d \sqrt{x} + 2 \, c\right )} - d} - \frac{4 \,{\left (d^{2} x \log \left (e^{\left (d \sqrt{x} + c\right )} + 1\right ) + 2 \, d \sqrt{x}{\rm Li}_2\left (-e^{\left (d \sqrt{x} + c\right )}\right ) - 2 \,{\rm Li}_{3}(-e^{\left (d \sqrt{x} + c\right )})\right )} a b}{d^{3}} + \frac{4 \,{\left (d^{2} x \log \left (-e^{\left (d \sqrt{x} + c\right )} + 1\right ) + 2 \, d \sqrt{x}{\rm Li}_2\left (e^{\left (d \sqrt{x} + c\right )}\right ) - 2 \,{\rm Li}_{3}(e^{\left (d \sqrt{x} + c\right )})\right )} a b}{d^{3}} + \frac{4 \,{\left (d \sqrt{x} \log \left (e^{\left (d \sqrt{x} + c\right )} + 1\right ) +{\rm Li}_2\left (-e^{\left (d \sqrt{x} + c\right )}\right )\right )} b^{2}}{d^{3}} + \frac{4 \,{\left (d \sqrt{x} \log \left (-e^{\left (d \sqrt{x} + c\right )} + 1\right ) +{\rm Li}_2\left (e^{\left (d \sqrt{x} + c\right )}\right )\right )} b^{2}}{d^{3}} - \frac{2 \,{\left (2 \, a b d^{3} x^{\frac{3}{2}} + 3 \, b^{2} d^{2} x\right )}}{3 \, d^{3}} + \frac{2 \,{\left (2 \, a b d^{3} x^{\frac{3}{2}} - 3 \, b^{2} d^{2} x\right )}}{3 \, d^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*csch(c+d*x^(1/2)))^2*x^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (b^{2} \sqrt{x} \operatorname{csch}\left (d \sqrt{x} + c\right )^{2} + 2 \, a b \sqrt{x} \operatorname{csch}\left (d \sqrt{x} + c\right ) + a^{2} \sqrt{x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*csch(c+d*x^(1/2)))^2*x^(1/2),x, algorithm="fricas")

[Out]

integral(b^2*sqrt(x)*csch(d*sqrt(x) + c)^2 + 2*a*b*sqrt(x)*csch(d*sqrt(x) + c) + a^2*sqrt(x), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{x} \left (a + b \operatorname{csch}{\left (c + d \sqrt{x} \right )}\right )^{2}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*csch(c+d*x**(1/2)))**2*x**(1/2),x)

[Out]

Integral(sqrt(x)*(a + b*csch(c + d*sqrt(x)))**2, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \operatorname{csch}\left (d \sqrt{x} + c\right ) + a\right )}^{2} \sqrt{x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*csch(c+d*x^(1/2)))^2*x^(1/2),x, algorithm="giac")

[Out]

integrate((b*csch(d*sqrt(x) + c) + a)^2*sqrt(x), x)