Integrand size = 18, antiderivative size = 240 \[ \int x^3 \text {csch}^3(a+b x) \text {sech}(a+b x) \, dx=-\frac {3 x^2}{2 b^2}+\frac {x^3}{2 b}+\frac {2 x^3 \text {arctanh}\left (e^{2 a+2 b x}\right )}{b}-\frac {3 x^2 \coth (a+b x)}{2 b^2}-\frac {x^3 \coth ^2(a+b x)}{2 b}+\frac {3 x \log \left (1-e^{2 (a+b x)}\right )}{b^3}+\frac {3 \operatorname {PolyLog}\left (2,e^{2 (a+b x)}\right )}{2 b^4}+\frac {3 x^2 \operatorname {PolyLog}\left (2,-e^{2 a+2 b x}\right )}{2 b^2}-\frac {3 x^2 \operatorname {PolyLog}\left (2,e^{2 a+2 b x}\right )}{2 b^2}-\frac {3 x \operatorname {PolyLog}\left (3,-e^{2 a+2 b x}\right )}{2 b^3}+\frac {3 x \operatorname {PolyLog}\left (3,e^{2 a+2 b x}\right )}{2 b^3}+\frac {3 \operatorname {PolyLog}\left (4,-e^{2 a+2 b x}\right )}{4 b^4}-\frac {3 \operatorname {PolyLog}\left (4,e^{2 a+2 b x}\right )}{4 b^4} \]
-3/2*x^2/b^2+1/2*x^3/b+2*x^3*arctanh(exp(2*b*x+2*a))/b-3/2*x^2*coth(b*x+a) /b^2-1/2*x^3*coth(b*x+a)^2/b+3*x*ln(1-exp(2*b*x+2*a))/b^3+3/2*polylog(2,ex p(2*b*x+2*a))/b^4+3/2*x^2*polylog(2,-exp(2*b*x+2*a))/b^2-3/2*x^2*polylog(2 ,exp(2*b*x+2*a))/b^2-3/2*x*polylog(3,-exp(2*b*x+2*a))/b^3+3/2*x*polylog(3, exp(2*b*x+2*a))/b^3+3/4*polylog(4,-exp(2*b*x+2*a))/b^4-3/4*polylog(4,exp(2 *b*x+2*a))/b^4
Leaf count is larger than twice the leaf count of optimal. \(565\) vs. \(2(240)=480\).
Time = 6.51 (sec) , antiderivative size = 565, normalized size of antiderivative = 2.35 \[ \int x^3 \text {csch}^3(a+b x) \text {sech}(a+b x) \, dx=-\frac {x^3 \text {csch}^2(a+b x)}{2 b}+\frac {e^{2 a} \left (-6 b^2 e^{-2 a} x^2+b^4 e^{-2 a} x^4+6 b \left (1-e^{-2 a}\right ) x \log \left (1-e^{-a-b x}\right )-2 b^3 e^{-2 a} \left (-1+e^{2 a}\right ) x^3 \log \left (1-e^{-a-b x}\right )+6 b \left (1-e^{-2 a}\right ) x \log \left (1+e^{-a-b x}\right )-2 b^3 e^{-2 a} \left (-1+e^{2 a}\right ) x^3 \log \left (1+e^{-a-b x}\right )-6 \left (1-e^{-2 a}\right ) \operatorname {PolyLog}\left (2,-e^{-a-b x}\right )+6 b^2 \left (1-e^{-2 a}\right ) x^2 \operatorname {PolyLog}\left (2,-e^{-a-b x}\right )-6 \left (1-e^{-2 a}\right ) \operatorname {PolyLog}\left (2,e^{-a-b x}\right )+6 b^2 \left (1-e^{-2 a}\right ) x^2 \operatorname {PolyLog}\left (2,e^{-a-b x}\right )+12 b \left (1-e^{-2 a}\right ) x \operatorname {PolyLog}\left (3,-e^{-a-b x}\right )+12 b \left (1-e^{-2 a}\right ) x \operatorname {PolyLog}\left (3,e^{-a-b x}\right )+12 \left (1-e^{-2 a}\right ) \operatorname {PolyLog}\left (4,-e^{-a-b x}\right )+12 \left (1-e^{-2 a}\right ) \operatorname {PolyLog}\left (4,e^{-a-b x}\right )\right )}{2 b^4 \left (-1+e^{2 a}\right )}+\frac {e^{2 a} \left (2 b^4 e^{-2 a} x^4+4 b^3 \left (1+e^{-2 a}\right ) x^3 \log \left (1+e^{-2 (a+b x)}\right )-6 b^2 \left (1+e^{-2 a}\right ) x^2 \operatorname {PolyLog}\left (2,-e^{-2 (a+b x)}\right )-6 b \left (1+e^{-2 a}\right ) x \operatorname {PolyLog}\left (3,-e^{-2 (a+b x)}\right )-3 \left (1+e^{-2 a}\right ) \operatorname {PolyLog}\left (4,-e^{-2 (a+b x)}\right )\right )}{4 b^4 \left (1+e^{2 a}\right )}-\frac {1}{4} x^4 \text {csch}(a) \text {sech}(a)+\frac {3 x^2 \text {csch}(a) \text {csch}(a+b x) \sinh (b x)}{2 b^2} \]
-1/2*(x^3*Csch[a + b*x]^2)/b + (E^(2*a)*((-6*b^2*x^2)/E^(2*a) + (b^4*x^4)/ E^(2*a) + 6*b*(1 - E^(-2*a))*x*Log[1 - E^(-a - b*x)] - (2*b^3*(-1 + E^(2*a ))*x^3*Log[1 - E^(-a - b*x)])/E^(2*a) + 6*b*(1 - E^(-2*a))*x*Log[1 + E^(-a - b*x)] - (2*b^3*(-1 + E^(2*a))*x^3*Log[1 + E^(-a - b*x)])/E^(2*a) - 6*(1 - E^(-2*a))*PolyLog[2, -E^(-a - b*x)] + 6*b^2*(1 - E^(-2*a))*x^2*PolyLog[ 2, -E^(-a - b*x)] - 6*(1 - E^(-2*a))*PolyLog[2, E^(-a - b*x)] + 6*b^2*(1 - E^(-2*a))*x^2*PolyLog[2, E^(-a - b*x)] + 12*b*(1 - E^(-2*a))*x*PolyLog[3, -E^(-a - b*x)] + 12*b*(1 - E^(-2*a))*x*PolyLog[3, E^(-a - b*x)] + 12*(1 - E^(-2*a))*PolyLog[4, -E^(-a - b*x)] + 12*(1 - E^(-2*a))*PolyLog[4, E^(-a - b*x)]))/(2*b^4*(-1 + E^(2*a))) + (E^(2*a)*((2*b^4*x^4)/E^(2*a) + 4*b^3*( 1 + E^(-2*a))*x^3*Log[1 + E^(-2*(a + b*x))] - 6*b^2*(1 + E^(-2*a))*x^2*Pol yLog[2, -E^(-2*(a + b*x))] - 6*b*(1 + E^(-2*a))*x*PolyLog[3, -E^(-2*(a + b *x))] - 3*(1 + E^(-2*a))*PolyLog[4, -E^(-2*(a + b*x))]))/(4*b^4*(1 + E^(2* a))) - (x^4*Csch[a]*Sech[a])/4 + (3*x^2*Csch[a]*Csch[a + b*x]*Sinh[b*x])/( 2*b^2)
Time = 0.70 (sec) , antiderivative size = 262, normalized size of antiderivative = 1.09, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {5985, 27, 2010, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int x^3 \text {csch}^3(a+b x) \text {sech}(a+b x) \, dx\) |
| \(\Big \downarrow \) 5985 |
| \(\displaystyle -3 \int -\frac {1}{2} x^2 \left (\frac {\coth ^2(a+b x)}{b}+\frac {2 \log (\tanh (a+b x))}{b}\right )dx-\frac {x^3 \coth ^2(a+b x)}{2 b}-\frac {x^3 \log (\tanh (a+b x))}{b}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {3}{2} \int x^2 \left (\frac {\coth ^2(a+b x)}{b}+\frac {2 \log (\tanh (a+b x))}{b}\right )dx-\frac {x^3 \coth ^2(a+b x)}{2 b}-\frac {x^3 \log (\tanh (a+b x))}{b}\) |
| \(\Big \downarrow \) 2010 |
| \(\displaystyle \frac {3}{2} \int \left (\frac {\coth ^2(a+b x) x^2}{b}+\frac {2 \log (\tanh (a+b x)) x^2}{b}\right )dx-\frac {x^3 \coth ^2(a+b x)}{2 b}-\frac {x^3 \log (\tanh (a+b x))}{b}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {3}{2} \left (\frac {4 x^3 \text {arctanh}\left (e^{2 a+2 b x}\right )}{3 b}+\frac {\operatorname {PolyLog}\left (2,e^{2 (a+b x)}\right )}{b^4}+\frac {\operatorname {PolyLog}\left (4,-e^{2 a+2 b x}\right )}{2 b^4}-\frac {\operatorname {PolyLog}\left (4,e^{2 a+2 b x}\right )}{2 b^4}-\frac {x \operatorname {PolyLog}\left (3,-e^{2 a+2 b x}\right )}{b^3}+\frac {x \operatorname {PolyLog}\left (3,e^{2 a+2 b x}\right )}{b^3}+\frac {2 x \log \left (1-e^{2 (a+b x)}\right )}{b^3}+\frac {x^2 \operatorname {PolyLog}\left (2,-e^{2 a+2 b x}\right )}{b^2}-\frac {x^2 \operatorname {PolyLog}\left (2,e^{2 a+2 b x}\right )}{b^2}-\frac {x^2 \coth (a+b x)}{b^2}+\frac {2 x^3 \log (\tanh (a+b x))}{3 b}-\frac {x^2}{b^2}+\frac {x^3}{3 b}\right )-\frac {x^3 \coth ^2(a+b x)}{2 b}-\frac {x^3 \log (\tanh (a+b x))}{b}\) |
-1/2*(x^3*Coth[a + b*x]^2)/b - (x^3*Log[Tanh[a + b*x]])/b + (3*(-(x^2/b^2) + x^3/(3*b) + (4*x^3*ArcTanh[E^(2*a + 2*b*x)])/(3*b) - (x^2*Coth[a + b*x] )/b^2 + (2*x*Log[1 - E^(2*(a + b*x))])/b^3 + (2*x^3*Log[Tanh[a + b*x]])/(3 *b) + PolyLog[2, E^(2*(a + b*x))]/b^4 + (x^2*PolyLog[2, -E^(2*a + 2*b*x)]) /b^2 - (x^2*PolyLog[2, E^(2*a + 2*b*x)])/b^2 - (x*PolyLog[3, -E^(2*a + 2*b *x)])/b^3 + (x*PolyLog[3, E^(2*a + 2*b*x)])/b^3 + PolyLog[4, -E^(2*a + 2*b *x)]/(2*b^4) - PolyLog[4, E^(2*a + 2*b*x)]/(2*b^4)))/2
3.6.8.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x] , x] /; FreeQ[{c, m}, x] && SumQ[u] && !LinearQ[u, x] && !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]
Int[Csch[(a_.) + (b_.)*(x_)]^(n_.)*((c_.) + (d_.)*(x_))^(m_.)*Sech[(a_.) + (b_.)*(x_)]^(p_.), x_Symbol] :> With[{u = IntHide[Csch[a + b*x]^n*Sech[a + b*x]^p, x]}, Simp[(c + d*x)^m u, x] - Simp[d*m Int[(c + d*x)^(m - 1)*u, x], x]] /; FreeQ[{a, b, c, d}, x] && IntegersQ[n, p] && GtQ[m, 0] && NeQ[n , p]
Time = 3.83 (sec) , antiderivative size = 417, normalized size of antiderivative = 1.74
| method | result | size |
| risch | \(-\frac {3 a^{2}}{b^{4}}-\frac {\ln \left ({\mathrm e}^{b x +a}+1\right ) x^{3}}{b}-\frac {3 x^{2}}{b^{2}}-\frac {3 a \ln \left ({\mathrm e}^{b x +a}-1\right )}{b^{4}}+\frac {a^{3} \ln \left ({\mathrm e}^{b x +a}-1\right )}{b^{4}}+\frac {3 \ln \left ({\mathrm e}^{b x +a}+1\right ) x}{b^{3}}+\frac {3 \ln \left (1-{\mathrm e}^{b x +a}\right ) x}{b^{3}}+\frac {3 \ln \left (1-{\mathrm e}^{b x +a}\right ) a}{b^{4}}-\frac {x^{2} \left (2 \,{\mathrm e}^{2 b x +2 a} b x +3 \,{\mathrm e}^{2 b x +2 a}-3\right )}{b^{2} \left ({\mathrm e}^{2 b x +2 a}-1\right )^{2}}+\frac {6 a \ln \left ({\mathrm e}^{b x +a}\right )}{b^{4}}-\frac {6 a x}{b^{3}}-\frac {3 x^{2} \operatorname {polylog}\left (2, -{\mathrm e}^{b x +a}\right )}{b^{2}}-\frac {3 x^{2} \operatorname {polylog}\left (2, {\mathrm e}^{b x +a}\right )}{b^{2}}+\frac {6 x \operatorname {polylog}\left (3, -{\mathrm e}^{b x +a}\right )}{b^{3}}+\frac {6 x \operatorname {polylog}\left (3, {\mathrm e}^{b x +a}\right )}{b^{3}}+\frac {3 x^{2} \operatorname {polylog}\left (2, -{\mathrm e}^{2 b x +2 a}\right )}{2 b^{2}}-\frac {3 x \operatorname {polylog}\left (3, -{\mathrm e}^{2 b x +2 a}\right )}{2 b^{3}}-\frac {6 \operatorname {polylog}\left (4, -{\mathrm e}^{b x +a}\right )}{b^{4}}-\frac {6 \operatorname {polylog}\left (4, {\mathrm e}^{b x +a}\right )}{b^{4}}+\frac {x^{3} \ln \left (1+{\mathrm e}^{2 b x +2 a}\right )}{b}-\frac {\ln \left (1-{\mathrm e}^{b x +a}\right ) x^{3}}{b}-\frac {\ln \left (1-{\mathrm e}^{b x +a}\right ) a^{3}}{b^{4}}+\frac {3 \operatorname {polylog}\left (4, -{\mathrm e}^{2 b x +2 a}\right )}{4 b^{4}}+\frac {3 \operatorname {polylog}\left (2, -{\mathrm e}^{b x +a}\right )}{b^{4}}+\frac {3 \operatorname {polylog}\left (2, {\mathrm e}^{b x +a}\right )}{b^{4}}\) | \(417\) |
-3/b^4*a^2-1/b*ln(exp(b*x+a)+1)*x^3-3/b^2*x^2-3/b^4*a*ln(exp(b*x+a)-1)+1/b ^4*a^3*ln(exp(b*x+a)-1)+3/b^3*ln(exp(b*x+a)+1)*x+3/b^3*ln(1-exp(b*x+a))*x+ 3/b^4*ln(1-exp(b*x+a))*a-x^2*(2*exp(2*b*x+2*a)*b*x+3*exp(2*b*x+2*a)-3)/b^2 /(exp(2*b*x+2*a)-1)^2+6/b^4*a*ln(exp(b*x+a))-6/b^3*a*x-3*x^2*polylog(2,-ex p(b*x+a))/b^2-3*x^2*polylog(2,exp(b*x+a))/b^2+6*x*polylog(3,-exp(b*x+a))/b ^3+6*x*polylog(3,exp(b*x+a))/b^3+3/2*x^2*polylog(2,-exp(2*b*x+2*a))/b^2-3/ 2*x*polylog(3,-exp(2*b*x+2*a))/b^3-6*polylog(4,-exp(b*x+a))/b^4-6*polylog( 4,exp(b*x+a))/b^4+x^3*ln(1+exp(2*b*x+2*a))/b-1/b*ln(1-exp(b*x+a))*x^3-1/b^ 4*ln(1-exp(b*x+a))*a^3+3/4*polylog(4,-exp(2*b*x+2*a))/b^4+3*polylog(2,-exp (b*x+a))/b^4+3*polylog(2,exp(b*x+a))/b^4
Result contains complex when optimal does not.
Time = 0.30 (sec) , antiderivative size = 3394, normalized size of antiderivative = 14.14 \[ \int x^3 \text {csch}^3(a+b x) \text {sech}(a+b x) \, dx=\text {Too large to display} \]
-(3*(b^2*x^2 - a^2)*cosh(b*x + a)^4 + 12*(b^2*x^2 - a^2)*cosh(b*x + a)*sin h(b*x + a)^3 + 3*(b^2*x^2 - a^2)*sinh(b*x + a)^4 + (2*b^3*x^3 - 3*b^2*x^2 + 6*a^2)*cosh(b*x + a)^2 + (2*b^3*x^3 - 3*b^2*x^2 + 18*(b^2*x^2 - a^2)*cos h(b*x + a)^2 + 6*a^2)*sinh(b*x + a)^2 - 3*a^2 + 3*((b^2*x^2 - 1)*cosh(b*x + a)^4 + 4*(b^2*x^2 - 1)*cosh(b*x + a)*sinh(b*x + a)^3 + (b^2*x^2 - 1)*sin h(b*x + a)^4 + b^2*x^2 - 2*(b^2*x^2 - 1)*cosh(b*x + a)^2 - 2*(b^2*x^2 - 3* (b^2*x^2 - 1)*cosh(b*x + a)^2 - 1)*sinh(b*x + a)^2 + 4*((b^2*x^2 - 1)*cosh (b*x + a)^3 - (b^2*x^2 - 1)*cosh(b*x + a))*sinh(b*x + a) - 1)*dilog(cosh(b *x + a) + sinh(b*x + a)) - 3*(b^2*x^2*cosh(b*x + a)^4 + 4*b^2*x^2*cosh(b*x + a)*sinh(b*x + a)^3 + b^2*x^2*sinh(b*x + a)^4 - 2*b^2*x^2*cosh(b*x + a)^ 2 + b^2*x^2 + 2*(3*b^2*x^2*cosh(b*x + a)^2 - b^2*x^2)*sinh(b*x + a)^2 + 4* (b^2*x^2*cosh(b*x + a)^3 - b^2*x^2*cosh(b*x + a))*sinh(b*x + a))*dilog(I*c osh(b*x + a) + I*sinh(b*x + a)) - 3*(b^2*x^2*cosh(b*x + a)^4 + 4*b^2*x^2*c osh(b*x + a)*sinh(b*x + a)^3 + b^2*x^2*sinh(b*x + a)^4 - 2*b^2*x^2*cosh(b* x + a)^2 + b^2*x^2 + 2*(3*b^2*x^2*cosh(b*x + a)^2 - b^2*x^2)*sinh(b*x + a) ^2 + 4*(b^2*x^2*cosh(b*x + a)^3 - b^2*x^2*cosh(b*x + a))*sinh(b*x + a))*di log(-I*cosh(b*x + a) - I*sinh(b*x + a)) + 3*((b^2*x^2 - 1)*cosh(b*x + a)^4 + 4*(b^2*x^2 - 1)*cosh(b*x + a)*sinh(b*x + a)^3 + (b^2*x^2 - 1)*sinh(b*x + a)^4 + b^2*x^2 - 2*(b^2*x^2 - 1)*cosh(b*x + a)^2 - 2*(b^2*x^2 - 3*(b^2*x ^2 - 1)*cosh(b*x + a)^2 - 1)*sinh(b*x + a)^2 + 4*((b^2*x^2 - 1)*cosh(b*...
\[ \int x^3 \text {csch}^3(a+b x) \text {sech}(a+b x) \, dx=\int x^{3} \operatorname {csch}^{3}{\left (a + b x \right )} \operatorname {sech}{\left (a + b x \right )}\, dx \]
Time = 0.22 (sec) , antiderivative size = 352, normalized size of antiderivative = 1.47 \[ \int x^3 \text {csch}^3(a+b x) \text {sech}(a+b x) \, dx=-\frac {1}{2} \, x^{4} + \frac {3 \, x^{2} - {\left (2 \, b x^{3} e^{\left (2 \, a\right )} + 3 \, x^{2} e^{\left (2 \, a\right )}\right )} e^{\left (2 \, b x\right )}}{b^{2} e^{\left (4 \, b x + 4 \, a\right )} - 2 \, b^{2} e^{\left (2 \, b x + 2 \, a\right )} + b^{2}} + \frac {b^{4} x^{4} - 6 \, b^{2} x^{2}}{2 \, b^{4}} + \frac {4 \, b^{3} x^{3} \log \left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right ) + 6 \, b^{2} x^{2} {\rm Li}_2\left (-e^{\left (2 \, b x + 2 \, a\right )}\right ) - 6 \, b x {\rm Li}_{3}(-e^{\left (2 \, b x + 2 \, a\right )}) + 3 \, {\rm Li}_{4}(-e^{\left (2 \, b x + 2 \, a\right )})}{3 \, b^{4}} - \frac {b^{3} x^{3} \log \left (e^{\left (b x + a\right )} + 1\right ) + 3 \, b^{2} x^{2} {\rm Li}_2\left (-e^{\left (b x + a\right )}\right ) - 6 \, b x {\rm Li}_{3}(-e^{\left (b x + a\right )}) + 6 \, {\rm Li}_{4}(-e^{\left (b x + a\right )})}{b^{4}} - \frac {b^{3} x^{3} \log \left (-e^{\left (b x + a\right )} + 1\right ) + 3 \, b^{2} x^{2} {\rm Li}_2\left (e^{\left (b x + a\right )}\right ) - 6 \, b x {\rm Li}_{3}(e^{\left (b x + a\right )}) + 6 \, {\rm Li}_{4}(e^{\left (b x + a\right )})}{b^{4}} + \frac {3 \, {\left (b x \log \left (e^{\left (b x + a\right )} + 1\right ) + {\rm Li}_2\left (-e^{\left (b x + a\right )}\right )\right )}}{b^{4}} + \frac {3 \, {\left (b x \log \left (-e^{\left (b x + a\right )} + 1\right ) + {\rm Li}_2\left (e^{\left (b x + a\right )}\right )\right )}}{b^{4}} \]
-1/2*x^4 + (3*x^2 - (2*b*x^3*e^(2*a) + 3*x^2*e^(2*a))*e^(2*b*x))/(b^2*e^(4 *b*x + 4*a) - 2*b^2*e^(2*b*x + 2*a) + b^2) + 1/2*(b^4*x^4 - 6*b^2*x^2)/b^4 + 1/3*(4*b^3*x^3*log(e^(2*b*x + 2*a) + 1) + 6*b^2*x^2*dilog(-e^(2*b*x + 2 *a)) - 6*b*x*polylog(3, -e^(2*b*x + 2*a)) + 3*polylog(4, -e^(2*b*x + 2*a)) )/b^4 - (b^3*x^3*log(e^(b*x + a) + 1) + 3*b^2*x^2*dilog(-e^(b*x + a)) - 6* b*x*polylog(3, -e^(b*x + a)) + 6*polylog(4, -e^(b*x + a)))/b^4 - (b^3*x^3* log(-e^(b*x + a) + 1) + 3*b^2*x^2*dilog(e^(b*x + a)) - 6*b*x*polylog(3, e^ (b*x + a)) + 6*polylog(4, e^(b*x + a)))/b^4 + 3*(b*x*log(e^(b*x + a) + 1) + dilog(-e^(b*x + a)))/b^4 + 3*(b*x*log(-e^(b*x + a) + 1) + dilog(e^(b*x + a)))/b^4
\[ \int x^3 \text {csch}^3(a+b x) \text {sech}(a+b x) \, dx=\int { x^{3} \operatorname {csch}\left (b x + a\right )^{3} \operatorname {sech}\left (b x + a\right ) \,d x } \]
Timed out. \[ \int x^3 \text {csch}^3(a+b x) \text {sech}(a+b x) \, dx=\int \frac {x^3}{\mathrm {cosh}\left (a+b\,x\right )\,{\mathrm {sinh}\left (a+b\,x\right )}^3} \,d x \]