Integrand size = 18, antiderivative size = 143 \[ \int \frac {1}{(a+b \cosh (c+d x) \sinh (c+d x))^3} \, dx=-\frac {4 \left (8 a^2-b^2\right ) \text {arctanh}\left (\frac {b-2 a \tanh (c+d x)}{\sqrt {4 a^2+b^2}}\right )}{\left (4 a^2+b^2\right )^{5/2} d}-\frac {2 b \cosh (2 c+2 d x)}{\left (4 a^2+b^2\right ) d (2 a+b \sinh (2 c+2 d x))^2}-\frac {12 a b \cosh (2 c+2 d x)}{\left (4 a^2+b^2\right )^2 d (2 a+b \sinh (2 c+2 d x))} \]
-4*(8*a^2-b^2)*arctanh((b-2*a*tanh(d*x+c))/(4*a^2+b^2)^(1/2))/(4*a^2+b^2)^ (5/2)/d-2*b*cosh(2*d*x+2*c)/(4*a^2+b^2)/d/(2*a+b*sinh(2*d*x+2*c))^2-12*a*b *cosh(2*d*x+2*c)/(4*a^2+b^2)^2/d/(2*a+b*sinh(2*d*x+2*c))
Time = 0.71 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.85 \[ \int \frac {1}{(a+b \cosh (c+d x) \sinh (c+d x))^3} \, dx=\frac {2 \left (\frac {2 \left (8 a^2-b^2\right ) \arctan \left (\frac {b-2 a \tanh (c+d x)}{\sqrt {-4 a^2-b^2}}\right )}{\sqrt {-4 a^2-b^2}}-\frac {b \cosh (2 (c+d x)) \left (16 a^2+b^2+6 a b \sinh (2 (c+d x))\right )}{(2 a+b \sinh (2 (c+d x)))^2}\right )}{\left (4 a^2+b^2\right )^2 d} \]
(2*((2*(8*a^2 - b^2)*ArcTan[(b - 2*a*Tanh[c + d*x])/Sqrt[-4*a^2 - b^2]])/S qrt[-4*a^2 - b^2] - (b*Cosh[2*(c + d*x)]*(16*a^2 + b^2 + 6*a*b*Sinh[2*(c + d*x)]))/(2*a + b*Sinh[2*(c + d*x)])^2))/((4*a^2 + b^2)^2*d)
Time = 0.62 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.08, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.722, Rules used = {3042, 3145, 3042, 3143, 27, 3042, 3233, 25, 27, 3042, 3139, 1083, 217}
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.
| \(\displaystyle \int \frac {1}{(a+b \sinh (c+d x) \cosh (c+d x))^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{(a-i b \sin (i c+i d x) \cos (i c+i d x))^3}dx\) |
| \(\Big \downarrow \) 3145 |
| \(\displaystyle \int \frac {1}{\left (a+\frac {1}{2} b \sinh (2 c+2 d x)\right )^3}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\left (a-\frac {1}{2} i b \sin (2 i c+2 i d x)\right )^3}dx\) |
| \(\Big \downarrow \) 3143 |
| \(\displaystyle -\frac {2 \int -\frac {2 (4 a-b \sinh (2 c+2 d x))}{(2 a+b \sinh (2 c+2 d x))^2}dx}{4 a^2+b^2}-\frac {2 b \cosh (2 c+2 d x)}{d \left (4 a^2+b^2\right ) (2 a+b \sinh (2 c+2 d x))^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {4 \int \frac {4 a-b \sinh (2 c+2 d x)}{(2 a+b \sinh (2 c+2 d x))^2}dx}{4 a^2+b^2}-\frac {2 b \cosh (2 c+2 d x)}{d \left (4 a^2+b^2\right ) (2 a+b \sinh (2 c+2 d x))^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {2 b \cosh (2 c+2 d x)}{d \left (4 a^2+b^2\right ) (2 a+b \sinh (2 c+2 d x))^2}+\frac {4 \int \frac {4 a+i b \sin (2 i c+2 i d x)}{(2 a-i b \sin (2 i c+2 i d x))^2}dx}{4 a^2+b^2}\) |
| \(\Big \downarrow \) 3233 |
| \(\displaystyle \frac {4 \left (-\frac {\int -\frac {8 a^2-b^2}{2 a+b \sinh (2 c+2 d x)}dx}{4 a^2+b^2}-\frac {3 a b \cosh (2 c+2 d x)}{d \left (4 a^2+b^2\right ) (2 a+b \sinh (2 c+2 d x))}\right )}{4 a^2+b^2}-\frac {2 b \cosh (2 c+2 d x)}{d \left (4 a^2+b^2\right ) (2 a+b \sinh (2 c+2 d x))^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {4 \left (\frac {\int \frac {8 a^2-b^2}{2 a+b \sinh (2 c+2 d x)}dx}{4 a^2+b^2}-\frac {3 a b \cosh (2 c+2 d x)}{d \left (4 a^2+b^2\right ) (2 a+b \sinh (2 c+2 d x))}\right )}{4 a^2+b^2}-\frac {2 b \cosh (2 c+2 d x)}{d \left (4 a^2+b^2\right ) (2 a+b \sinh (2 c+2 d x))^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {4 \left (\frac {\left (8 a^2-b^2\right ) \int \frac {1}{2 a+b \sinh (2 c+2 d x)}dx}{4 a^2+b^2}-\frac {3 a b \cosh (2 c+2 d x)}{d \left (4 a^2+b^2\right ) (2 a+b \sinh (2 c+2 d x))}\right )}{4 a^2+b^2}-\frac {2 b \cosh (2 c+2 d x)}{d \left (4 a^2+b^2\right ) (2 a+b \sinh (2 c+2 d x))^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {2 b \cosh (2 c+2 d x)}{d \left (4 a^2+b^2\right ) (2 a+b \sinh (2 c+2 d x))^2}+\frac {4 \left (-\frac {3 a b \cosh (2 c+2 d x)}{d \left (4 a^2+b^2\right ) (2 a+b \sinh (2 c+2 d x))}+\frac {\left (8 a^2-b^2\right ) \int \frac {1}{2 a-i b \sin (2 i c+2 i d x)}dx}{4 a^2+b^2}\right )}{4 a^2+b^2}\) |
| \(\Big \downarrow \) 3139 |
| \(\displaystyle -\frac {2 b \cosh (2 c+2 d x)}{d \left (4 a^2+b^2\right ) (2 a+b \sinh (2 c+2 d x))^2}+\frac {4 \left (-\frac {3 a b \cosh (2 c+2 d x)}{d \left (4 a^2+b^2\right ) (2 a+b \sinh (2 c+2 d x))}-\frac {i \left (8 a^2-b^2\right ) \int \frac {1}{-2 a \tanh ^2(c+d x)+2 b \tanh (c+d x)+2 a}d(i \tanh (c+d x))}{d \left (4 a^2+b^2\right )}\right )}{4 a^2+b^2}\) |
| \(\Big \downarrow \) 1083 |
| \(\displaystyle -\frac {2 b \cosh (2 c+2 d x)}{d \left (4 a^2+b^2\right ) (2 a+b \sinh (2 c+2 d x))^2}+\frac {4 \left (-\frac {3 a b \cosh (2 c+2 d x)}{d \left (4 a^2+b^2\right ) (2 a+b \sinh (2 c+2 d x))}+\frac {2 i \left (8 a^2-b^2\right ) \int \frac {1}{\tanh ^2(c+d x)-4 \left (4 a^2+b^2\right )}d(4 i a \tanh (c+d x)-2 i b)}{d \left (4 a^2+b^2\right )}\right )}{4 a^2+b^2}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {4 \left (\frac {\left (8 a^2-b^2\right ) \text {arctanh}\left (\frac {\tanh (c+d x)}{2 \sqrt {4 a^2+b^2}}\right )}{d \left (4 a^2+b^2\right )^{3/2}}-\frac {3 a b \cosh (2 c+2 d x)}{d \left (4 a^2+b^2\right ) (2 a+b \sinh (2 c+2 d x))}\right )}{4 a^2+b^2}-\frac {2 b \cosh (2 c+2 d x)}{d \left (4 a^2+b^2\right ) (2 a+b \sinh (2 c+2 d x))^2}\) |
(-2*b*Cosh[2*c + 2*d*x])/((4*a^2 + b^2)*d*(2*a + b*Sinh[2*c + 2*d*x])^2) + (4*(((8*a^2 - b^2)*ArcTanh[Tanh[c + d*x]/(2*Sqrt[4*a^2 + b^2])])/((4*a^2 + b^2)^(3/2)*d) - (3*a*b*Cosh[2*c + 2*d*x])/((4*a^2 + b^2)*d*(2*a + b*Sinh [2*c + 2*d*x]))))/(4*a^2 + b^2)
3.9.60.3.1 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[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( -1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & & (LtQ[a, 0] || LtQ[b, 0])
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2 Subst[I nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x]
Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = Fre eFactors[Tan[(c + d*x)/2], x]}, Simp[2*(e/d) Subst[Int[1/(a + 2*b*e*x + a *e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] && NeQ [a^2 - b^2, 0]
Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos [c + d*x]*((a + b*Sin[c + d*x])^(n + 1)/(d*(n + 1)*(a^2 - b^2))), x] + Simp [1/((n + 1)*(a^2 - b^2)) Int[(a + b*Sin[c + d*x])^(n + 1)*Simp[a*(n + 1) - b*(n + 2)*Sin[c + d*x], x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && LtQ[n, -1] && IntegerQ[2*n]
Int[((a_) + cos[(c_.) + (d_.)*(x_)]*(b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_ Symbol] :> Int[(a + b*(Sin[2*c + 2*d*x]/2))^n, x] /; FreeQ[{a, b, c, d, n}, x]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-(b*c - a*d))*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(f*(m + 1)*(a^2 - b^2))), x] + Simp[1/((m + 1)*(a^2 - b^2)) Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[(a*c - b*d)*(m + 1) - (b*c - a*d)*( m + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && IntegerQ[2*m]
Leaf count of result is larger than twice the leaf count of optimal. \(480\) vs. \(2(139)=278\).
Time = 211.64 (sec) , antiderivative size = 481, normalized size of antiderivative = 3.36
| method | result | size |
| risch | \(\frac {32 a^{2} b \,{\mathrm e}^{6 d x +6 c}-4 b^{3} {\mathrm e}^{6 d x +6 c}+192 a^{3} {\mathrm e}^{4 d x +4 c}-24 a \,b^{2} {\mathrm e}^{4 d x +4 c}-160 a^{2} b \,{\mathrm e}^{2 d x +2 c}-4 b^{3} {\mathrm e}^{2 d x +2 c}+24 a \,b^{2}}{d \left (4 a^{2}+b^{2}\right )^{2} \left (4 a \,{\mathrm e}^{2 d x +2 c}+{\mathrm e}^{4 d x +4 c} b -b \right )^{2}}+\frac {16 \ln \left ({\mathrm e}^{2 d x +2 c}+\frac {2 \left (4 a^{2}+b^{2}\right )^{\frac {5}{2}} a -64 a^{6}-48 a^{4} b^{2}-12 a^{2} b^{4}-b^{6}}{\left (4 a^{2}+b^{2}\right )^{\frac {5}{2}} b}\right ) a^{2}}{\left (4 a^{2}+b^{2}\right )^{\frac {5}{2}} d}-\frac {2 \ln \left ({\mathrm e}^{2 d x +2 c}+\frac {2 \left (4 a^{2}+b^{2}\right )^{\frac {5}{2}} a -64 a^{6}-48 a^{4} b^{2}-12 a^{2} b^{4}-b^{6}}{\left (4 a^{2}+b^{2}\right )^{\frac {5}{2}} b}\right ) b^{2}}{\left (4 a^{2}+b^{2}\right )^{\frac {5}{2}} d}-\frac {16 \ln \left ({\mathrm e}^{2 d x +2 c}+\frac {2 \left (4 a^{2}+b^{2}\right )^{\frac {5}{2}} a +64 a^{6}+48 a^{4} b^{2}+12 a^{2} b^{4}+b^{6}}{\left (4 a^{2}+b^{2}\right )^{\frac {5}{2}} b}\right ) a^{2}}{\left (4 a^{2}+b^{2}\right )^{\frac {5}{2}} d}+\frac {2 \ln \left ({\mathrm e}^{2 d x +2 c}+\frac {2 \left (4 a^{2}+b^{2}\right )^{\frac {5}{2}} a +64 a^{6}+48 a^{4} b^{2}+12 a^{2} b^{4}+b^{6}}{\left (4 a^{2}+b^{2}\right )^{\frac {5}{2}} b}\right ) b^{2}}{\left (4 a^{2}+b^{2}\right )^{\frac {5}{2}} d}\) | \(481\) |
| derivativedivides | \(\frac {-\frac {2 \left (-\frac {b^{2} \left (10 a^{2}+b^{2}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{a \left (16 a^{4}+8 a^{2} b^{2}+b^{4}\right )}+\frac {\left (32 a^{4}-14 a^{2} b^{2}-b^{4}\right ) b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{a^{2} \left (16 a^{4}+8 a^{2} b^{2}+b^{4}\right )}+\frac {\left (58 a^{2}+b^{2}\right ) b^{2} \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{a \left (16 a^{4}+8 a^{2} b^{2}+b^{4}\right )}-\frac {2 b \left (32 a^{4}+18 a^{2} b^{2}+b^{4}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{a^{2} \left (16 a^{4}+8 a^{2} b^{2}+b^{4}\right )}+\frac {\left (58 a^{2}+b^{2}\right ) b^{2} \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{a \left (16 a^{4}+8 a^{2} b^{2}+b^{4}\right )}+\frac {\left (32 a^{4}-14 a^{2} b^{2}-b^{4}\right ) b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{a^{2} \left (16 a^{4}+8 a^{2} b^{2}+b^{4}\right )}-\frac {b^{2} \left (10 a^{2}+b^{2}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{a \left (16 a^{4}+8 a^{2} b^{2}+b^{4}\right )}\right )}{\left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a +2 b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}-2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )^{2}}-\frac {4 \left (8 a^{2}-b^{2}\right ) a \left (\frac {\left (-4 a^{2}-b^{2}\right ) \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +\sqrt {4 a^{2}+b^{2}}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}{2 \left (4 a^{2}+b^{2}\right )^{\frac {3}{2}} a}+\frac {\ln \left (-\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +\sqrt {4 a^{2}+b^{2}}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-a \right )}{2 \sqrt {4 a^{2}+b^{2}}\, a}\right )}{16 a^{4}+8 a^{2} b^{2}+b^{4}}}{d}\) | \(588\) |
| default | \(\frac {-\frac {2 \left (-\frac {b^{2} \left (10 a^{2}+b^{2}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{a \left (16 a^{4}+8 a^{2} b^{2}+b^{4}\right )}+\frac {\left (32 a^{4}-14 a^{2} b^{2}-b^{4}\right ) b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{a^{2} \left (16 a^{4}+8 a^{2} b^{2}+b^{4}\right )}+\frac {\left (58 a^{2}+b^{2}\right ) b^{2} \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{a \left (16 a^{4}+8 a^{2} b^{2}+b^{4}\right )}-\frac {2 b \left (32 a^{4}+18 a^{2} b^{2}+b^{4}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{a^{2} \left (16 a^{4}+8 a^{2} b^{2}+b^{4}\right )}+\frac {\left (58 a^{2}+b^{2}\right ) b^{2} \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{a \left (16 a^{4}+8 a^{2} b^{2}+b^{4}\right )}+\frac {\left (32 a^{4}-14 a^{2} b^{2}-b^{4}\right ) b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{a^{2} \left (16 a^{4}+8 a^{2} b^{2}+b^{4}\right )}-\frac {b^{2} \left (10 a^{2}+b^{2}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{a \left (16 a^{4}+8 a^{2} b^{2}+b^{4}\right )}\right )}{\left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a +2 b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}-2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )^{2}}-\frac {4 \left (8 a^{2}-b^{2}\right ) a \left (\frac {\left (-4 a^{2}-b^{2}\right ) \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +\sqrt {4 a^{2}+b^{2}}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}{2 \left (4 a^{2}+b^{2}\right )^{\frac {3}{2}} a}+\frac {\ln \left (-\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +\sqrt {4 a^{2}+b^{2}}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-a \right )}{2 \sqrt {4 a^{2}+b^{2}}\, a}\right )}{16 a^{4}+8 a^{2} b^{2}+b^{4}}}{d}\) | \(588\) |
4*(8*a^2*b*exp(6*d*x+6*c)-b^3*exp(6*d*x+6*c)+48*a^3*exp(4*d*x+4*c)-6*a*b^2 *exp(4*d*x+4*c)-40*a^2*b*exp(2*d*x+2*c)-b^3*exp(2*d*x+2*c)+6*a*b^2)/d/(4*a ^2+b^2)^2/(4*a*exp(2*d*x+2*c)+exp(4*d*x+4*c)*b-b)^2+16/(4*a^2+b^2)^(5/2)/d *ln(exp(2*d*x+2*c)+(2*(4*a^2+b^2)^(5/2)*a-64*a^6-48*a^4*b^2-12*a^2*b^4-b^6 )/(4*a^2+b^2)^(5/2)/b)*a^2-2/(4*a^2+b^2)^(5/2)/d*ln(exp(2*d*x+2*c)+(2*(4*a ^2+b^2)^(5/2)*a-64*a^6-48*a^4*b^2-12*a^2*b^4-b^6)/(4*a^2+b^2)^(5/2)/b)*b^2 -16/(4*a^2+b^2)^(5/2)/d*ln(exp(2*d*x+2*c)+(2*(4*a^2+b^2)^(5/2)*a+64*a^6+48 *a^4*b^2+12*a^2*b^4+b^6)/(4*a^2+b^2)^(5/2)/b)*a^2+2/(4*a^2+b^2)^(5/2)/d*ln (exp(2*d*x+2*c)+(2*(4*a^2+b^2)^(5/2)*a+64*a^6+48*a^4*b^2+12*a^2*b^4+b^6)/( 4*a^2+b^2)^(5/2)/b)*b^2
Leaf count of result is larger than twice the leaf count of optimal. 2439 vs. \(2 (142) = 284\).
Time = 0.29 (sec) , antiderivative size = 2439, normalized size of antiderivative = 17.06 \[ \int \frac {1}{(a+b \cosh (c+d x) \sinh (c+d x))^3} \, dx=\text {Too large to display} \]
2*(2*(32*a^4*b + 4*a^2*b^3 - b^5)*cosh(d*x + c)^6 + 12*(32*a^4*b + 4*a^2*b ^3 - b^5)*cosh(d*x + c)*sinh(d*x + c)^5 + 2*(32*a^4*b + 4*a^2*b^3 - b^5)*s inh(d*x + c)^6 + 48*a^3*b^2 + 12*a*b^4 + 12*(32*a^5 + 4*a^3*b^2 - a*b^4)*c osh(d*x + c)^4 + 6*(64*a^5 + 8*a^3*b^2 - 2*a*b^4 + 5*(32*a^4*b + 4*a^2*b^3 - b^5)*cosh(d*x + c)^2)*sinh(d*x + c)^4 + 8*(5*(32*a^4*b + 4*a^2*b^3 - b^ 5)*cosh(d*x + c)^3 + 6*(32*a^5 + 4*a^3*b^2 - a*b^4)*cosh(d*x + c))*sinh(d* x + c)^3 - 2*(160*a^4*b + 44*a^2*b^3 + b^5)*cosh(d*x + c)^2 - 2*(160*a^4*b + 44*a^2*b^3 + b^5 - 15*(32*a^4*b + 4*a^2*b^3 - b^5)*cosh(d*x + c)^4 - 36 *(32*a^5 + 4*a^3*b^2 - a*b^4)*cosh(d*x + c)^2)*sinh(d*x + c)^2 - ((8*a^2*b ^2 - b^4)*cosh(d*x + c)^8 + 8*(8*a^2*b^2 - b^4)*cosh(d*x + c)*sinh(d*x + c )^7 + (8*a^2*b^2 - b^4)*sinh(d*x + c)^8 + 8*(8*a^3*b - a*b^3)*cosh(d*x + c )^6 + 4*(16*a^3*b - 2*a*b^3 + 7*(8*a^2*b^2 - b^4)*cosh(d*x + c)^2)*sinh(d* x + c)^6 + 8*(7*(8*a^2*b^2 - b^4)*cosh(d*x + c)^3 + 6*(8*a^3*b - a*b^3)*co sh(d*x + c))*sinh(d*x + c)^5 + 2*(64*a^4 - 16*a^2*b^2 + b^4)*cosh(d*x + c) ^4 + 2*(35*(8*a^2*b^2 - b^4)*cosh(d*x + c)^4 + 64*a^4 - 16*a^2*b^2 + b^4 + 60*(8*a^3*b - a*b^3)*cosh(d*x + c)^2)*sinh(d*x + c)^4 + 8*a^2*b^2 - b^4 + 8*(7*(8*a^2*b^2 - b^4)*cosh(d*x + c)^5 + 20*(8*a^3*b - a*b^3)*cosh(d*x + c)^3 + (64*a^4 - 16*a^2*b^2 + b^4)*cosh(d*x + c))*sinh(d*x + c)^3 - 8*(8*a ^3*b - a*b^3)*cosh(d*x + c)^2 + 4*(7*(8*a^2*b^2 - b^4)*cosh(d*x + c)^6 + 3 0*(8*a^3*b - a*b^3)*cosh(d*x + c)^4 - 16*a^3*b + 2*a*b^3 + 3*(64*a^4 - ...
Timed out. \[ \int \frac {1}{(a+b \cosh (c+d x) \sinh (c+d x))^3} \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 327 vs. \(2 (142) = 284\).
Time = 0.30 (sec) , antiderivative size = 327, normalized size of antiderivative = 2.29 \[ \int \frac {1}{(a+b \cosh (c+d x) \sinh (c+d x))^3} \, dx=\frac {2 \, {\left (8 \, a^{2} - b^{2}\right )} \log \left (\frac {b e^{\left (-2 \, d x - 2 \, c\right )} - 2 \, a - \sqrt {4 \, a^{2} + b^{2}}}{b e^{\left (-2 \, d x - 2 \, c\right )} - 2 \, a + \sqrt {4 \, a^{2} + b^{2}}}\right )}{{\left (16 \, a^{4} + 8 \, a^{2} b^{2} + b^{4}\right )} \sqrt {4 \, a^{2} + b^{2}} d} - \frac {4 \, {\left (6 \, a b^{2} + {\left (40 \, a^{2} b + b^{3}\right )} e^{\left (-2 \, d x - 2 \, c\right )} + 6 \, {\left (8 \, a^{3} - a b^{2}\right )} e^{\left (-4 \, d x - 4 \, c\right )} - {\left (8 \, a^{2} b - b^{3}\right )} e^{\left (-6 \, d x - 6 \, c\right )}\right )}}{{\left (16 \, a^{4} b^{2} + 8 \, a^{2} b^{4} + b^{6} + 8 \, {\left (16 \, a^{5} b + 8 \, a^{3} b^{3} + a b^{5}\right )} e^{\left (-2 \, d x - 2 \, c\right )} + 2 \, {\left (128 \, a^{6} + 48 \, a^{4} b^{2} - b^{6}\right )} e^{\left (-4 \, d x - 4 \, c\right )} - 8 \, {\left (16 \, a^{5} b + 8 \, a^{3} b^{3} + a b^{5}\right )} e^{\left (-6 \, d x - 6 \, c\right )} + {\left (16 \, a^{4} b^{2} + 8 \, a^{2} b^{4} + b^{6}\right )} e^{\left (-8 \, d x - 8 \, c\right )}\right )} d} \]
2*(8*a^2 - b^2)*log((b*e^(-2*d*x - 2*c) - 2*a - sqrt(4*a^2 + b^2))/(b*e^(- 2*d*x - 2*c) - 2*a + sqrt(4*a^2 + b^2)))/((16*a^4 + 8*a^2*b^2 + b^4)*sqrt( 4*a^2 + b^2)*d) - 4*(6*a*b^2 + (40*a^2*b + b^3)*e^(-2*d*x - 2*c) + 6*(8*a^ 3 - a*b^2)*e^(-4*d*x - 4*c) - (8*a^2*b - b^3)*e^(-6*d*x - 6*c))/((16*a^4*b ^2 + 8*a^2*b^4 + b^6 + 8*(16*a^5*b + 8*a^3*b^3 + a*b^5)*e^(-2*d*x - 2*c) + 2*(128*a^6 + 48*a^4*b^2 - b^6)*e^(-4*d*x - 4*c) - 8*(16*a^5*b + 8*a^3*b^3 + a*b^5)*e^(-6*d*x - 6*c) + (16*a^4*b^2 + 8*a^2*b^4 + b^6)*e^(-8*d*x - 8* c))*d)
Time = 0.60 (sec) , antiderivative size = 256, normalized size of antiderivative = 1.79 \[ \int \frac {1}{(a+b \cosh (c+d x) \sinh (c+d x))^3} \, dx=\frac {2 \, {\left (\frac {{\left (8 \, a^{2} - b^{2}\right )} \log \left (\frac {{\left | 2 \, b e^{\left (2 \, d x + 2 \, c\right )} + 4 \, a - 2 \, \sqrt {4 \, a^{2} + b^{2}} \right |}}{{\left | 2 \, b e^{\left (2 \, d x + 2 \, c\right )} + 4 \, a + 2 \, \sqrt {4 \, a^{2} + b^{2}} \right |}}\right )}{{\left (16 \, a^{4} + 8 \, a^{2} b^{2} + b^{4}\right )} \sqrt {4 \, a^{2} + b^{2}}} + \frac {2 \, {\left (8 \, a^{2} b e^{\left (6 \, d x + 6 \, c\right )} - b^{3} e^{\left (6 \, d x + 6 \, c\right )} + 48 \, a^{3} e^{\left (4 \, d x + 4 \, c\right )} - 6 \, a b^{2} e^{\left (4 \, d x + 4 \, c\right )} - 40 \, a^{2} b e^{\left (2 \, d x + 2 \, c\right )} - b^{3} e^{\left (2 \, d x + 2 \, c\right )} + 6 \, a b^{2}\right )}}{{\left (16 \, a^{4} + 8 \, a^{2} b^{2} + b^{4}\right )} {\left (b e^{\left (4 \, d x + 4 \, c\right )} + 4 \, a e^{\left (2 \, d x + 2 \, c\right )} - b\right )}^{2}}\right )}}{d} \]
2*((8*a^2 - b^2)*log(abs(2*b*e^(2*d*x + 2*c) + 4*a - 2*sqrt(4*a^2 + b^2))/ abs(2*b*e^(2*d*x + 2*c) + 4*a + 2*sqrt(4*a^2 + b^2)))/((16*a^4 + 8*a^2*b^2 + b^4)*sqrt(4*a^2 + b^2)) + 2*(8*a^2*b*e^(6*d*x + 6*c) - b^3*e^(6*d*x + 6 *c) + 48*a^3*e^(4*d*x + 4*c) - 6*a*b^2*e^(4*d*x + 4*c) - 40*a^2*b*e^(2*d*x + 2*c) - b^3*e^(2*d*x + 2*c) + 6*a*b^2)/((16*a^4 + 8*a^2*b^2 + b^4)*(b*e^ (4*d*x + 4*c) + 4*a*e^(2*d*x + 2*c) - b)^2))/d
Timed out. \[ \int \frac {1}{(a+b \cosh (c+d x) \sinh (c+d x))^3} \, dx=\int \frac {1}{{\left (a+b\,\mathrm {cosh}\left (c+d\,x\right )\,\mathrm {sinh}\left (c+d\,x\right )\right )}^3} \,d x \]