Integrand size = 26, antiderivative size = 348 \[ \int \frac {(e+f x)^3 \cosh (c+d x)}{(a+b \sinh (c+d x))^2} \, dx=\frac {3 f (e+f x)^2 \log \left (1+\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{b \sqrt {a^2+b^2} d^2}-\frac {3 f (e+f x)^2 \log \left (1+\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{b \sqrt {a^2+b^2} d^2}+\frac {6 f^2 (e+f x) \operatorname {PolyLog}\left (2,-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{b \sqrt {a^2+b^2} d^3}-\frac {6 f^2 (e+f x) \operatorname {PolyLog}\left (2,-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{b \sqrt {a^2+b^2} d^3}-\frac {6 f^3 \operatorname {PolyLog}\left (3,-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{b \sqrt {a^2+b^2} d^4}+\frac {6 f^3 \operatorname {PolyLog}\left (3,-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{b \sqrt {a^2+b^2} d^4}-\frac {(e+f x)^3}{b d (a+b \sinh (c+d x))} \]
-(f*x+e)^3/b/d/(a+b*sinh(d*x+c))+3*f*(f*x+e)^2*ln(1+b*exp(d*x+c)/(a-(a^2+b ^2)^(1/2)))/b/d^2/(a^2+b^2)^(1/2)-3*f*(f*x+e)^2*ln(1+b*exp(d*x+c)/(a+(a^2+ b^2)^(1/2)))/b/d^2/(a^2+b^2)^(1/2)+6*f^2*(f*x+e)*polylog(2,-b*exp(d*x+c)/( a-(a^2+b^2)^(1/2)))/b/d^3/(a^2+b^2)^(1/2)-6*f^2*(f*x+e)*polylog(2,-b*exp(d *x+c)/(a+(a^2+b^2)^(1/2)))/b/d^3/(a^2+b^2)^(1/2)-6*f^3*polylog(3,-b*exp(d* x+c)/(a-(a^2+b^2)^(1/2)))/b/d^4/(a^2+b^2)^(1/2)+6*f^3*polylog(3,-b*exp(d*x +c)/(a+(a^2+b^2)^(1/2)))/b/d^4/(a^2+b^2)^(1/2)
Time = 0.25 (sec) , antiderivative size = 368, normalized size of antiderivative = 1.06 \[ \int \frac {(e+f x)^3 \cosh (c+d x)}{(a+b \sinh (c+d x))^2} \, dx=\frac {3 f \left (-2 d^2 e^2 \text {arctanh}\left (\frac {a+b e^{c+d x}}{\sqrt {a^2+b^2}}\right )+2 d^2 e f x \log \left (1+\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )+d^2 f^2 x^2 \log \left (1+\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )-2 d^2 e f x \log \left (1+\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )-d^2 f^2 x^2 \log \left (1+\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )+2 d f (e+f x) \operatorname {PolyLog}\left (2,\frac {b e^{c+d x}}{-a+\sqrt {a^2+b^2}}\right )-2 d f (e+f x) \operatorname {PolyLog}\left (2,-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )-2 f^2 \operatorname {PolyLog}\left (3,\frac {b e^{c+d x}}{-a+\sqrt {a^2+b^2}}\right )+2 f^2 \operatorname {PolyLog}\left (3,-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )\right )}{b \sqrt {a^2+b^2} d^4}-\frac {(e+f x)^3}{b d (a+b \sinh (c+d x))} \]
(3*f*(-2*d^2*e^2*ArcTanh[(a + b*E^(c + d*x))/Sqrt[a^2 + b^2]] + 2*d^2*e*f* x*Log[1 + (b*E^(c + d*x))/(a - Sqrt[a^2 + b^2])] + d^2*f^2*x^2*Log[1 + (b* E^(c + d*x))/(a - Sqrt[a^2 + b^2])] - 2*d^2*e*f*x*Log[1 + (b*E^(c + d*x))/ (a + Sqrt[a^2 + b^2])] - d^2*f^2*x^2*Log[1 + (b*E^(c + d*x))/(a + Sqrt[a^2 + b^2])] + 2*d*f*(e + f*x)*PolyLog[2, (b*E^(c + d*x))/(-a + Sqrt[a^2 + b^ 2])] - 2*d*f*(e + f*x)*PolyLog[2, -((b*E^(c + d*x))/(a + Sqrt[a^2 + b^2])) ] - 2*f^2*PolyLog[3, (b*E^(c + d*x))/(-a + Sqrt[a^2 + b^2])] + 2*f^2*PolyL og[3, -((b*E^(c + d*x))/(a + Sqrt[a^2 + b^2]))]))/(b*Sqrt[a^2 + b^2]*d^4) - (e + f*x)^3/(b*d*(a + b*Sinh[c + d*x]))
Time = 1.39 (sec) , antiderivative size = 318, normalized size of antiderivative = 0.91, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {5987, 3042, 3803, 25, 2694, 27, 2620, 3011, 2720, 7143}
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.
| \(\displaystyle \int \frac {(e+f x)^3 \cosh (c+d x)}{(a+b \sinh (c+d x))^2} \, dx\) |
| \(\Big \downarrow \) 5987 |
| \(\displaystyle \frac {3 f \int \frac {(e+f x)^2}{a+b \sinh (c+d x)}dx}{b d}-\frac {(e+f x)^3}{b d (a+b \sinh (c+d x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {(e+f x)^3}{b d (a+b \sinh (c+d x))}+\frac {3 f \int \frac {(e+f x)^2}{a-i b \sin (i c+i d x)}dx}{b d}\) |
| \(\Big \downarrow \) 3803 |
| \(\displaystyle \frac {6 f \int -\frac {e^{c+d x} (e+f x)^2}{-2 e^{c+d x} a-b e^{2 (c+d x)}+b}dx}{b d}-\frac {(e+f x)^3}{b d (a+b \sinh (c+d x))}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {6 f \int \frac {e^{c+d x} (e+f x)^2}{-2 e^{c+d x} a-b e^{2 (c+d x)}+b}dx}{b d}-\frac {(e+f x)^3}{b d (a+b \sinh (c+d x))}\) |
| \(\Big \downarrow \) 2694 |
| \(\displaystyle -\frac {6 f \left (\frac {b \int -\frac {e^{c+d x} (e+f x)^2}{2 \left (a+b e^{c+d x}-\sqrt {a^2+b^2}\right )}dx}{\sqrt {a^2+b^2}}-\frac {b \int -\frac {e^{c+d x} (e+f x)^2}{2 \left (a+b e^{c+d x}+\sqrt {a^2+b^2}\right )}dx}{\sqrt {a^2+b^2}}\right )}{b d}-\frac {(e+f x)^3}{b d (a+b \sinh (c+d x))}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {6 f \left (\frac {b \int \frac {e^{c+d x} (e+f x)^2}{a+b e^{c+d x}+\sqrt {a^2+b^2}}dx}{2 \sqrt {a^2+b^2}}-\frac {b \int \frac {e^{c+d x} (e+f x)^2}{a+b e^{c+d x}-\sqrt {a^2+b^2}}dx}{2 \sqrt {a^2+b^2}}\right )}{b d}-\frac {(e+f x)^3}{b d (a+b \sinh (c+d x))}\) |
| \(\Big \downarrow \) 2620 |
| \(\displaystyle -\frac {6 f \left (\frac {b \left (\frac {(e+f x)^2 \log \left (\frac {b e^{c+d x}}{\sqrt {a^2+b^2}+a}+1\right )}{b d}-\frac {2 f \int (e+f x) \log \left (\frac {e^{c+d x} b}{a+\sqrt {a^2+b^2}}+1\right )dx}{b d}\right )}{2 \sqrt {a^2+b^2}}-\frac {b \left (\frac {(e+f x)^2 \log \left (\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}+1\right )}{b d}-\frac {2 f \int (e+f x) \log \left (\frac {e^{c+d x} b}{a-\sqrt {a^2+b^2}}+1\right )dx}{b d}\right )}{2 \sqrt {a^2+b^2}}\right )}{b d}-\frac {(e+f x)^3}{b d (a+b \sinh (c+d x))}\) |
| \(\Big \downarrow \) 3011 |
| \(\displaystyle -\frac {6 f \left (\frac {b \left (\frac {(e+f x)^2 \log \left (\frac {b e^{c+d x}}{\sqrt {a^2+b^2}+a}+1\right )}{b d}-\frac {2 f \left (\frac {f \int \operatorname {PolyLog}\left (2,-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )dx}{d}-\frac {(e+f x) \operatorname {PolyLog}\left (2,-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{d}\right )}{b d}\right )}{2 \sqrt {a^2+b^2}}-\frac {b \left (\frac {(e+f x)^2 \log \left (\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}+1\right )}{b d}-\frac {2 f \left (\frac {f \int \operatorname {PolyLog}\left (2,-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )dx}{d}-\frac {(e+f x) \operatorname {PolyLog}\left (2,-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{d}\right )}{b d}\right )}{2 \sqrt {a^2+b^2}}\right )}{b d}-\frac {(e+f x)^3}{b d (a+b \sinh (c+d x))}\) |
| \(\Big \downarrow \) 2720 |
| \(\displaystyle -\frac {6 f \left (\frac {b \left (\frac {(e+f x)^2 \log \left (\frac {b e^{c+d x}}{\sqrt {a^2+b^2}+a}+1\right )}{b d}-\frac {2 f \left (\frac {f \int e^{-c-d x} \operatorname {PolyLog}\left (2,-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )de^{c+d x}}{d^2}-\frac {(e+f x) \operatorname {PolyLog}\left (2,-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{d}\right )}{b d}\right )}{2 \sqrt {a^2+b^2}}-\frac {b \left (\frac {(e+f x)^2 \log \left (\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}+1\right )}{b d}-\frac {2 f \left (\frac {f \int e^{-c-d x} \operatorname {PolyLog}\left (2,-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )de^{c+d x}}{d^2}-\frac {(e+f x) \operatorname {PolyLog}\left (2,-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{d}\right )}{b d}\right )}{2 \sqrt {a^2+b^2}}\right )}{b d}-\frac {(e+f x)^3}{b d (a+b \sinh (c+d x))}\) |
| \(\Big \downarrow \) 7143 |
| \(\displaystyle -\frac {6 f \left (\frac {b \left (\frac {(e+f x)^2 \log \left (\frac {b e^{c+d x}}{\sqrt {a^2+b^2}+a}+1\right )}{b d}-\frac {2 f \left (\frac {f \operatorname {PolyLog}\left (3,-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{d^2}-\frac {(e+f x) \operatorname {PolyLog}\left (2,-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{d}\right )}{b d}\right )}{2 \sqrt {a^2+b^2}}-\frac {b \left (\frac {(e+f x)^2 \log \left (\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}+1\right )}{b d}-\frac {2 f \left (\frac {f \operatorname {PolyLog}\left (3,-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{d^2}-\frac {(e+f x) \operatorname {PolyLog}\left (2,-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{d}\right )}{b d}\right )}{2 \sqrt {a^2+b^2}}\right )}{b d}-\frac {(e+f x)^3}{b d (a+b \sinh (c+d x))}\) |
(-6*f*(-1/2*(b*(((e + f*x)^2*Log[1 + (b*E^(c + d*x))/(a - Sqrt[a^2 + b^2]) ])/(b*d) - (2*f*(-(((e + f*x)*PolyLog[2, -((b*E^(c + d*x))/(a - Sqrt[a^2 + b^2]))])/d) + (f*PolyLog[3, -((b*E^(c + d*x))/(a - Sqrt[a^2 + b^2]))])/d^ 2))/(b*d)))/Sqrt[a^2 + b^2] + (b*(((e + f*x)^2*Log[1 + (b*E^(c + d*x))/(a + Sqrt[a^2 + b^2])])/(b*d) - (2*f*(-(((e + f*x)*PolyLog[2, -((b*E^(c + d*x ))/(a + Sqrt[a^2 + b^2]))])/d) + (f*PolyLog[3, -((b*E^(c + d*x))/(a + Sqrt [a^2 + b^2]))])/d^2))/(b*d)))/(2*Sqrt[a^2 + b^2])))/(b*d) - (e + f*x)^3/(b *d*(a + b*Sinh[c + d*x]))
3.4.26.3.1 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[(((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/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x] - Si mp[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]
Int[((F_)^(u_)*((f_.) + (g_.)*(x_))^(m_.))/((a_.) + (b_.)*(F_)^(u_) + (c_.) *(F_)^(v_)), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[2*(c/q) Int [(f + g*x)^m*(F^u/(b - q + 2*c*F^u)), x], x] - Simp[2*(c/q) Int[(f + g*x) ^m*(F^u/(b + q + 2*c*F^u)), x], x]] /; FreeQ[{F, a, b, c, f, g}, x] && EqQ[ v, 2*u] && LinearQ[u, x] && NeQ[b^2 - 4*a*c, 0] && IGtQ[m, 0]
Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Simp[v/D[v, x] Subst[Int[FunctionOfExponentialFunction[u, x]/x, x], x, v], x]] /; Funct ionOfExponentialQ[u, x] && !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; FreeQ [{a, m, n}, x] && IntegerQ[m*n]] && !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x)) *(F_)[v_] /; FreeQ[{a, b, c}, x] && InverseFunctionQ[F[x]]]
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] + Simp[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]
Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*sin[(e_.) + (Complex[0, fz_])* (f_.)*(x_)]), x_Symbol] :> Simp[2 Int[(c + d*x)^m*(E^((-I)*e + f*fz*x)/(( -I)*b + 2*a*E^((-I)*e + f*fz*x) + I*b*E^(2*((-I)*e + f*fz*x)))), x], x] /; FreeQ[{a, b, c, d, e, f, fz}, x] && NeQ[a^2 - b^2, 0] && IGtQ[m, 0]
Int[Cosh[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)*(x_))^(m_.)*((a_) + (b_.)*Sinh[ (c_.) + (d_.)*(x_)])^(n_.), x_Symbol] :> Simp[(e + f*x)^m*((a + b*Sinh[c + d*x])^(n + 1)/(b*d*(n + 1))), x] - Simp[f*(m/(b*d*(n + 1))) Int[(e + f*x) ^(m - 1)*(a + b*Sinh[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && IGtQ[m, 0] && NeQ[n, -1]
Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_S ymbol] :> 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]
\[\int \frac {\left (f x +e \right )^{3} \cosh \left (d x +c \right )}{\left (a +b \sinh \left (d x +c \right )\right )^{2}}d x\]
Leaf count of result is larger than twice the leaf count of optimal. 2420 vs. \(2 (316) = 632\).
Time = 0.29 (sec) , antiderivative size = 2420, normalized size of antiderivative = 6.95 \[ \int \frac {(e+f x)^3 \cosh (c+d x)}{(a+b \sinh (c+d x))^2} \, dx=\text {Too large to display} \]
-(6*(b^2*d*f^3*x + b^2*d*e*f^2 - (b^2*d*f^3*x + b^2*d*e*f^2)*cosh(d*x + c) ^2 - (b^2*d*f^3*x + b^2*d*e*f^2)*sinh(d*x + c)^2 - 2*(a*b*d*f^3*x + a*b*d* e*f^2)*cosh(d*x + c) - 2*(a*b*d*f^3*x + a*b*d*e*f^2 + (b^2*d*f^3*x + b^2*d *e*f^2)*cosh(d*x + c))*sinh(d*x + c))*sqrt((a^2 + b^2)/b^2)*dilog((a*cosh( d*x + c) + a*sinh(d*x + c) + (b*cosh(d*x + c) + b*sinh(d*x + c))*sqrt((a^2 + b^2)/b^2) - b)/b + 1) - 6*(b^2*d*f^3*x + b^2*d*e*f^2 - (b^2*d*f^3*x + b ^2*d*e*f^2)*cosh(d*x + c)^2 - (b^2*d*f^3*x + b^2*d*e*f^2)*sinh(d*x + c)^2 - 2*(a*b*d*f^3*x + a*b*d*e*f^2)*cosh(d*x + c) - 2*(a*b*d*f^3*x + a*b*d*e*f ^2 + (b^2*d*f^3*x + b^2*d*e*f^2)*cosh(d*x + c))*sinh(d*x + c))*sqrt((a^2 + b^2)/b^2)*dilog((a*cosh(d*x + c) + a*sinh(d*x + c) - (b*cosh(d*x + c) + b *sinh(d*x + c))*sqrt((a^2 + b^2)/b^2) - b)/b + 1) - 3*(b^2*d^2*e^2*f - 2*b ^2*c*d*e*f^2 + b^2*c^2*f^3 - (b^2*d^2*e^2*f - 2*b^2*c*d*e*f^2 + b^2*c^2*f^ 3)*cosh(d*x + c)^2 - (b^2*d^2*e^2*f - 2*b^2*c*d*e*f^2 + b^2*c^2*f^3)*sinh( d*x + c)^2 - 2*(a*b*d^2*e^2*f - 2*a*b*c*d*e*f^2 + a*b*c^2*f^3)*cosh(d*x + c) - 2*(a*b*d^2*e^2*f - 2*a*b*c*d*e*f^2 + a*b*c^2*f^3 + (b^2*d^2*e^2*f - 2 *b^2*c*d*e*f^2 + b^2*c^2*f^3)*cosh(d*x + c))*sinh(d*x + c))*sqrt((a^2 + b^ 2)/b^2)*log(2*b*cosh(d*x + c) + 2*b*sinh(d*x + c) + 2*b*sqrt((a^2 + b^2)/b ^2) + 2*a) + 3*(b^2*d^2*e^2*f - 2*b^2*c*d*e*f^2 + b^2*c^2*f^3 - (b^2*d^2*e ^2*f - 2*b^2*c*d*e*f^2 + b^2*c^2*f^3)*cosh(d*x + c)^2 - (b^2*d^2*e^2*f - 2 *b^2*c*d*e*f^2 + b^2*c^2*f^3)*sinh(d*x + c)^2 - 2*(a*b*d^2*e^2*f - 2*a*...
Timed out. \[ \int \frac {(e+f x)^3 \cosh (c+d x)}{(a+b \sinh (c+d x))^2} \, dx=\text {Timed out} \]
\[ \int \frac {(e+f x)^3 \cosh (c+d x)}{(a+b \sinh (c+d x))^2} \, dx=\int { \frac {{\left (f x + e\right )}^{3} \cosh \left (d x + c\right )}{{\left (b \sinh \left (d x + c\right ) + a\right )}^{2}} \,d x } \]
-3*e^2*f*(2*x*e^(d*x + c)/(b^2*d*e^(2*d*x + 2*c) + 2*a*b*d*e^(d*x + c) - b ^2*d) - log((b*e^(d*x + c) + a - sqrt(a^2 + b^2))/(b*e^(d*x + c) + a + sqr t(a^2 + b^2)))/(sqrt(a^2 + b^2)*b*d^2)) - 2*e^3*e^(-d*x - c)/((2*a*b*e^(-d *x - c) - b^2*e^(-2*d*x - 2*c) + b^2)*d) - 2*(f^3*x^3*e^c + 3*e*f^2*x^2*e^ c)*e^(d*x)/(b^2*d*e^(2*d*x + 2*c) + 2*a*b*d*e^(d*x + c) - b^2*d) + integra te(6*(f^3*x^2*e^c + 2*e*f^2*x*e^c)*e^(d*x)/(b^2*d*e^(2*d*x + 2*c) + 2*a*b* d*e^(d*x + c) - b^2*d), x)
\[ \int \frac {(e+f x)^3 \cosh (c+d x)}{(a+b \sinh (c+d x))^2} \, dx=\int { \frac {{\left (f x + e\right )}^{3} \cosh \left (d x + c\right )}{{\left (b \sinh \left (d x + c\right ) + a\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {(e+f x)^3 \cosh (c+d x)}{(a+b \sinh (c+d x))^2} \, dx=\int \frac {\mathrm {cosh}\left (c+d\,x\right )\,{\left (e+f\,x\right )}^3}{{\left (a+b\,\mathrm {sinh}\left (c+d\,x\right )\right )}^2} \,d x \]