Integrand size = 23, antiderivative size = 217 \[ \int \frac {\text {sech}^3(c+d x)}{\left (a+b \sinh ^2(c+d x)\right )^3} \, dx=\frac {(a-7 b) \arctan (\sinh (c+d x))}{2 (a-b)^4 d}+\frac {b^{3/2} \left (35 a^2-14 a b+3 b^2\right ) \arctan \left (\frac {\sqrt {b} \sinh (c+d x)}{\sqrt {a}}\right )}{8 a^{5/2} (a-b)^4 d}+\frac {b (2 a+b) \sinh (c+d x)}{4 a (a-b)^2 d \left (a+b \sinh ^2(c+d x)\right )^2}+\frac {(4 a-b) b (a+3 b) \sinh (c+d x)}{8 a^2 (a-b)^3 d \left (a+b \sinh ^2(c+d x)\right )}+\frac {\text {sech}(c+d x) \tanh (c+d x)}{2 (a-b) d \left (a+b \sinh ^2(c+d x)\right )^2} \]
1/2*(a-7*b)*arctan(sinh(d*x+c))/(a-b)^4/d+1/8*b^(3/2)*(35*a^2-14*a*b+3*b^2 )*arctan(sinh(d*x+c)*b^(1/2)/a^(1/2))/a^(5/2)/(a-b)^4/d+1/4*b*(2*a+b)*sinh (d*x+c)/a/(a-b)^2/d/(a+b*sinh(d*x+c)^2)^2+1/8*(4*a-b)*b*(a+3*b)*sinh(d*x+c )/a^2/(a-b)^3/d/(a+b*sinh(d*x+c)^2)+1/2*sech(d*x+c)*tanh(d*x+c)/(a-b)/d/(a +b*sinh(d*x+c)^2)^2
Time = 1.12 (sec) , antiderivative size = 232, normalized size of antiderivative = 1.07 \[ \int \frac {\text {sech}^3(c+d x)}{\left (a+b \sinh ^2(c+d x)\right )^3} \, dx=\frac {-35 a^2 b^{3/2} \arctan \left (\frac {\sqrt {a} \text {csch}(c+d x)}{\sqrt {b}}\right )+14 a b^{5/2} \arctan \left (\frac {\sqrt {a} \text {csch}(c+d x)}{\sqrt {b}}\right )-3 b^{7/2} \arctan \left (\frac {\sqrt {a} \text {csch}(c+d x)}{\sqrt {b}}\right )+8 a^{7/2} \arctan \left (\tanh \left (\frac {1}{2} (c+d x)\right )\right )-56 a^{5/2} b \arctan \left (\tanh \left (\frac {1}{2} (c+d x)\right )\right )+\frac {2 \sqrt {a} (a-b) b^2 \left (26 a^2-21 a b+3 b^2+(11 a-3 b) b \cosh (2 (c+d x))\right ) \sinh (c+d x)}{(2 a-b+b \cosh (2 (c+d x)))^2}+4 a^{5/2} (a-b) \text {sech}(c+d x) \tanh (c+d x)}{8 a^{5/2} (a-b)^4 d} \]
(-35*a^2*b^(3/2)*ArcTan[(Sqrt[a]*Csch[c + d*x])/Sqrt[b]] + 14*a*b^(5/2)*Ar cTan[(Sqrt[a]*Csch[c + d*x])/Sqrt[b]] - 3*b^(7/2)*ArcTan[(Sqrt[a]*Csch[c + d*x])/Sqrt[b]] + 8*a^(7/2)*ArcTan[Tanh[(c + d*x)/2]] - 56*a^(5/2)*b*ArcTa n[Tanh[(c + d*x)/2]] + (2*Sqrt[a]*(a - b)*b^2*(26*a^2 - 21*a*b + 3*b^2 + ( 11*a - 3*b)*b*Cosh[2*(c + d*x)])*Sinh[c + d*x])/(2*a - b + b*Cosh[2*(c + d *x)])^2 + 4*a^(5/2)*(a - b)*Sech[c + d*x]*Tanh[c + d*x])/(8*a^(5/2)*(a - b )^4*d)
Time = 0.46 (sec) , antiderivative size = 252, normalized size of antiderivative = 1.16, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.435, Rules used = {3042, 3669, 316, 25, 402, 27, 402, 397, 216, 218}
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.
| \(\displaystyle \int \frac {\text {sech}^3(c+d x)}{\left (a+b \sinh ^2(c+d x)\right )^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\cos (i c+i d x)^3 \left (a-b \sin (i c+i d x)^2\right )^3}dx\) |
| \(\Big \downarrow \) 3669 |
| \(\displaystyle \frac {\int \frac {1}{\left (\sinh ^2(c+d x)+1\right )^2 \left (b \sinh ^2(c+d x)+a\right )^3}d\sinh (c+d x)}{d}\) |
| \(\Big \downarrow \) 316 |
| \(\displaystyle \frac {\frac {\sinh (c+d x)}{2 (a-b) \left (\sinh ^2(c+d x)+1\right ) \left (a+b \sinh ^2(c+d x)\right )^2}-\frac {\int -\frac {5 b \sinh ^2(c+d x)+a-2 b}{\left (\sinh ^2(c+d x)+1\right ) \left (b \sinh ^2(c+d x)+a\right )^3}d\sinh (c+d x)}{2 (a-b)}}{d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\int \frac {5 b \sinh ^2(c+d x)+a-2 b}{\left (\sinh ^2(c+d x)+1\right ) \left (b \sinh ^2(c+d x)+a\right )^3}d\sinh (c+d x)}{2 (a-b)}+\frac {\sinh (c+d x)}{2 (a-b) \left (\sinh ^2(c+d x)+1\right ) \left (a+b \sinh ^2(c+d x)\right )^2}}{d}\) |
| \(\Big \downarrow \) 402 |
| \(\displaystyle \frac {\frac {\frac {\int \frac {2 \left (2 a^2-8 b a+3 b^2+3 b (2 a+b) \sinh ^2(c+d x)\right )}{\left (\sinh ^2(c+d x)+1\right ) \left (b \sinh ^2(c+d x)+a\right )^2}d\sinh (c+d x)}{4 a (a-b)}+\frac {b (2 a+b) \sinh (c+d x)}{2 a (a-b) \left (a+b \sinh ^2(c+d x)\right )^2}}{2 (a-b)}+\frac {\sinh (c+d x)}{2 (a-b) \left (\sinh ^2(c+d x)+1\right ) \left (a+b \sinh ^2(c+d x)\right )^2}}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\frac {\int \frac {2 a^2-8 b a+3 b^2+3 b (2 a+b) \sinh ^2(c+d x)}{\left (\sinh ^2(c+d x)+1\right ) \left (b \sinh ^2(c+d x)+a\right )^2}d\sinh (c+d x)}{2 a (a-b)}+\frac {b (2 a+b) \sinh (c+d x)}{2 a (a-b) \left (a+b \sinh ^2(c+d x)\right )^2}}{2 (a-b)}+\frac {\sinh (c+d x)}{2 (a-b) \left (\sinh ^2(c+d x)+1\right ) \left (a+b \sinh ^2(c+d x)\right )^2}}{d}\) |
| \(\Big \downarrow \) 402 |
| \(\displaystyle \frac {\frac {\frac {\frac {\int \frac {4 a^3-24 b a^2+11 b^2 a-3 b^3+(4 a-b) b (a+3 b) \sinh ^2(c+d x)}{\left (\sinh ^2(c+d x)+1\right ) \left (b \sinh ^2(c+d x)+a\right )}d\sinh (c+d x)}{2 a (a-b)}+\frac {b (4 a-b) (a+3 b) \sinh (c+d x)}{2 a (a-b) \left (a+b \sinh ^2(c+d x)\right )}}{2 a (a-b)}+\frac {b (2 a+b) \sinh (c+d x)}{2 a (a-b) \left (a+b \sinh ^2(c+d x)\right )^2}}{2 (a-b)}+\frac {\sinh (c+d x)}{2 (a-b) \left (\sinh ^2(c+d x)+1\right ) \left (a+b \sinh ^2(c+d x)\right )^2}}{d}\) |
| \(\Big \downarrow \) 397 |
| \(\displaystyle \frac {\frac {\frac {\frac {\frac {b^2 \left (35 a^2-14 a b+3 b^2\right ) \int \frac {1}{b \sinh ^2(c+d x)+a}d\sinh (c+d x)}{a-b}+\frac {4 a^2 (a-7 b) \int \frac {1}{\sinh ^2(c+d x)+1}d\sinh (c+d x)}{a-b}}{2 a (a-b)}+\frac {b (4 a-b) (a+3 b) \sinh (c+d x)}{2 a (a-b) \left (a+b \sinh ^2(c+d x)\right )}}{2 a (a-b)}+\frac {b (2 a+b) \sinh (c+d x)}{2 a (a-b) \left (a+b \sinh ^2(c+d x)\right )^2}}{2 (a-b)}+\frac {\sinh (c+d x)}{2 (a-b) \left (\sinh ^2(c+d x)+1\right ) \left (a+b \sinh ^2(c+d x)\right )^2}}{d}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {\frac {\frac {\frac {\frac {b^2 \left (35 a^2-14 a b+3 b^2\right ) \int \frac {1}{b \sinh ^2(c+d x)+a}d\sinh (c+d x)}{a-b}+\frac {4 a^2 (a-7 b) \arctan (\sinh (c+d x))}{a-b}}{2 a (a-b)}+\frac {b (4 a-b) (a+3 b) \sinh (c+d x)}{2 a (a-b) \left (a+b \sinh ^2(c+d x)\right )}}{2 a (a-b)}+\frac {b (2 a+b) \sinh (c+d x)}{2 a (a-b) \left (a+b \sinh ^2(c+d x)\right )^2}}{2 (a-b)}+\frac {\sinh (c+d x)}{2 (a-b) \left (\sinh ^2(c+d x)+1\right ) \left (a+b \sinh ^2(c+d x)\right )^2}}{d}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {\frac {\frac {\frac {\frac {b^{3/2} \left (35 a^2-14 a b+3 b^2\right ) \arctan \left (\frac {\sqrt {b} \sinh (c+d x)}{\sqrt {a}}\right )}{\sqrt {a} (a-b)}+\frac {4 a^2 (a-7 b) \arctan (\sinh (c+d x))}{a-b}}{2 a (a-b)}+\frac {b (4 a-b) (a+3 b) \sinh (c+d x)}{2 a (a-b) \left (a+b \sinh ^2(c+d x)\right )}}{2 a (a-b)}+\frac {b (2 a+b) \sinh (c+d x)}{2 a (a-b) \left (a+b \sinh ^2(c+d x)\right )^2}}{2 (a-b)}+\frac {\sinh (c+d x)}{2 (a-b) \left (\sinh ^2(c+d x)+1\right ) \left (a+b \sinh ^2(c+d x)\right )^2}}{d}\) |
(Sinh[c + d*x]/(2*(a - b)*(1 + Sinh[c + d*x]^2)*(a + b*Sinh[c + d*x]^2)^2) + ((b*(2*a + b)*Sinh[c + d*x])/(2*a*(a - b)*(a + b*Sinh[c + d*x]^2)^2) + (((4*a^2*(a - 7*b)*ArcTan[Sinh[c + d*x]])/(a - b) + (b^(3/2)*(35*a^2 - 14* a*b + 3*b^2)*ArcTan[(Sqrt[b]*Sinh[c + d*x])/Sqrt[a]])/(Sqrt[a]*(a - b)))/( 2*a*(a - b)) + ((4*a - b)*b*(a + 3*b)*Sinh[c + d*x])/(2*a*(a - b)*(a + b*S inh[c + d*x]^2)))/(2*a*(a - b)))/(2*(a - b)))/d
3.4.47.3.1 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[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim p[(-b)*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(2*a*(p + 1)*(b*c - a*d)) ), x] + Simp[1/(2*a*(p + 1)*(b*c - a*d)) Int[(a + b*x^2)^(p + 1)*(c + d*x ^2)^q*Simp[b*c + 2*(p + 1)*(b*c - a*d) + d*b*(2*(p + q + 2) + 1)*x^2, x], x ], x] /; FreeQ[{a, b, c, d, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && ! ( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomialQ[a, b, c, d, 2, p, q, x]
Int[((e_) + (f_.)*(x_)^2)/(((a_) + (b_.)*(x_)^2)*((c_) + (d_.)*(x_)^2)), x_ Symbol] :> Simp[(b*e - a*f)/(b*c - a*d) Int[1/(a + b*x^2), x], x] - Simp[ (d*e - c*f)/(b*c - a*d) Int[1/(c + d*x^2), x], x] /; FreeQ[{a, b, c, d, e , f}, x]
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_.)*((e_) + (f_.)*(x _)^2), x_Symbol] :> Simp[(-(b*e - a*f))*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^ (q + 1)/(a*2*(b*c - a*d)*(p + 1))), x] + Simp[1/(a*2*(b*c - a*d)*(p + 1)) Int[(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[c*(b*e - a*f) + e*2*(b*c - a*d) *(p + 1) + d*(b*e - a*f)*(2*(p + q + 2) + 1)*x^2, x], x], x] /; FreeQ[{a, b , c, d, e, f, q}, x] && LtQ[p, -1]
Int[cos[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^ (p_.), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Simp[ff/f S ubst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b*ff^2*x^2)^p, x], x, Sin[e + f*x] /ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]
Leaf count of result is larger than twice the leaf count of optimal. \(476\) vs. \(2(199)=398\).
Time = 0.17 (sec) , antiderivative size = 477, normalized size of antiderivative = 2.20
\[\frac {\frac {2 b^{2} \left (\frac {-\frac {\left (13 a^{2}-18 a b +5 b^{2}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{8 a}+\frac {\left (39 a^{3}-98 a^{2} b +71 a \,b^{2}-12 b^{3}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{8 a^{2}}-\frac {\left (39 a^{3}-98 a^{2} b +71 a \,b^{2}-12 b^{3}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{8 a^{2}}+\frac {\left (13 a^{2}-18 a b +5 b^{2}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 a}}{\left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a -2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +4 b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+a \right )^{2}}+\frac {\left (35 a^{2}-14 a b +3 b^{2}\right ) \left (-\frac {\left (a +\sqrt {-b \left (a -b \right )}-b \right ) \operatorname {arctanh}\left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {-b \left (a -b \right )}+a -2 b \right ) a}}\right )}{2 a \sqrt {-b \left (a -b \right )}\, \sqrt {\left (2 \sqrt {-b \left (a -b \right )}+a -2 b \right ) a}}+\frac {\left (-a +\sqrt {-b \left (a -b \right )}+b \right ) \arctan \left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {-b \left (a -b \right )}-a +2 b \right ) a}}\right )}{2 a \sqrt {-b \left (a -b \right )}\, \sqrt {\left (2 \sqrt {-b \left (a -b \right )}-a +2 b \right ) a}}\right )}{8 a}\right )}{\left (a -b \right )^{4}}+\frac {\frac {2 \left (\left (-\frac {a}{2}+\frac {b}{2}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+\left (\frac {a}{2}-\frac {b}{2}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right )^{2}}+\left (a -7 b \right ) \arctan \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (a -b \right )^{4}}}{d}\]
1/d*(2*b^2/(a-b)^4*((-1/8*(13*a^2-18*a*b+5*b^2)/a*tanh(1/2*d*x+1/2*c)^7+1/ 8*(39*a^3-98*a^2*b+71*a*b^2-12*b^3)/a^2*tanh(1/2*d*x+1/2*c)^5-1/8*(39*a^3- 98*a^2*b+71*a*b^2-12*b^3)/a^2*tanh(1/2*d*x+1/2*c)^3+1/8*(13*a^2-18*a*b+5*b ^2)/a*tanh(1/2*d*x+1/2*c))/(tanh(1/2*d*x+1/2*c)^4*a-2*tanh(1/2*d*x+1/2*c)^ 2*a+4*b*tanh(1/2*d*x+1/2*c)^2+a)^2+1/8/a*(35*a^2-14*a*b+3*b^2)*(-1/2*(a+(- b*(a-b))^(1/2)-b)/a/(-b*(a-b))^(1/2)/((2*(-b*(a-b))^(1/2)+a-2*b)*a)^(1/2)* arctanh(a*tanh(1/2*d*x+1/2*c)/((2*(-b*(a-b))^(1/2)+a-2*b)*a)^(1/2))+1/2*(- a+(-b*(a-b))^(1/2)+b)/a/(-b*(a-b))^(1/2)/((2*(-b*(a-b))^(1/2)-a+2*b)*a)^(1 /2)*arctan(a*tanh(1/2*d*x+1/2*c)/((2*(-b*(a-b))^(1/2)-a+2*b)*a)^(1/2))))+2 /(a-b)^4*(((-1/2*a+1/2*b)*tanh(1/2*d*x+1/2*c)^3+(1/2*a-1/2*b)*tanh(1/2*d*x +1/2*c))/(tanh(1/2*d*x+1/2*c)^2+1)^2+1/2*(a-7*b)*arctan(tanh(1/2*d*x+1/2*c ))))
Leaf count of result is larger than twice the leaf count of optimal. 10237 vs. \(2 (199) = 398\).
Time = 0.52 (sec) , antiderivative size = 18765, normalized size of antiderivative = 86.47 \[ \int \frac {\text {sech}^3(c+d x)}{\left (a+b \sinh ^2(c+d x)\right )^3} \, dx=\text {Too large to display} \]
Timed out. \[ \int \frac {\text {sech}^3(c+d x)}{\left (a+b \sinh ^2(c+d x)\right )^3} \, dx=\text {Timed out} \]
\[ \int \frac {\text {sech}^3(c+d x)}{\left (a+b \sinh ^2(c+d x)\right )^3} \, dx=\int { \frac {\operatorname {sech}\left (d x + c\right )^{3}}{{\left (b \sinh \left (d x + c\right )^{2} + a\right )}^{3}} \,d x } \]
(a*e^c - 7*b*e^c)*arctan(e^(d*x + c))*e^(-c)/(a^4*d - 4*a^3*b*d + 6*a^2*b^ 2*d - 4*a*b^3*d + b^4*d) + 1/4*((4*a^2*b^2*e^(11*c) + 11*a*b^3*e^(11*c) - 3*b^4*e^(11*c))*e^(11*d*x) + (32*a^3*b*e^(9*c) + 32*a^2*b^2*e^(9*c) - 31*a *b^3*e^(9*c) + 3*b^4*e^(9*c))*e^(9*d*x) + 2*(32*a^4*e^(7*c) - 48*a^3*b*e^( 7*c) + 46*a^2*b^2*e^(7*c) - 21*a*b^3*e^(7*c) + 3*b^4*e^(7*c))*e^(7*d*x) - 2*(32*a^4*e^(5*c) - 48*a^3*b*e^(5*c) + 46*a^2*b^2*e^(5*c) - 21*a*b^3*e^(5* c) + 3*b^4*e^(5*c))*e^(5*d*x) - (32*a^3*b*e^(3*c) + 32*a^2*b^2*e^(3*c) - 3 1*a*b^3*e^(3*c) + 3*b^4*e^(3*c))*e^(3*d*x) - (4*a^2*b^2*e^c + 11*a*b^3*e^c - 3*b^4*e^c)*e^(d*x))/(a^5*b^2*d - 3*a^4*b^3*d + 3*a^3*b^4*d - a^2*b^5*d + (a^5*b^2*d*e^(12*c) - 3*a^4*b^3*d*e^(12*c) + 3*a^3*b^4*d*e^(12*c) - a^2* b^5*d*e^(12*c))*e^(12*d*x) + 2*(4*a^6*b*d*e^(10*c) - 13*a^5*b^2*d*e^(10*c) + 15*a^4*b^3*d*e^(10*c) - 7*a^3*b^4*d*e^(10*c) + a^2*b^5*d*e^(10*c))*e^(1 0*d*x) + (16*a^7*d*e^(8*c) - 48*a^6*b*d*e^(8*c) + 47*a^5*b^2*d*e^(8*c) - 1 3*a^4*b^3*d*e^(8*c) - 3*a^3*b^4*d*e^(8*c) + a^2*b^5*d*e^(8*c))*e^(8*d*x) + 4*(8*a^7*d*e^(6*c) - 28*a^6*b*d*e^(6*c) + 37*a^5*b^2*d*e^(6*c) - 23*a^4*b ^3*d*e^(6*c) + 7*a^3*b^4*d*e^(6*c) - a^2*b^5*d*e^(6*c))*e^(6*d*x) + (16*a^ 7*d*e^(4*c) - 48*a^6*b*d*e^(4*c) + 47*a^5*b^2*d*e^(4*c) - 13*a^4*b^3*d*e^( 4*c) - 3*a^3*b^4*d*e^(4*c) + a^2*b^5*d*e^(4*c))*e^(4*d*x) + 2*(4*a^6*b*d*e ^(2*c) - 13*a^5*b^2*d*e^(2*c) + 15*a^4*b^3*d*e^(2*c) - 7*a^3*b^4*d*e^(2*c) + a^2*b^5*d*e^(2*c))*e^(2*d*x)) + 8*integrate(1/32*((35*a^2*b^2*e^(3*c...
\[ \int \frac {\text {sech}^3(c+d x)}{\left (a+b \sinh ^2(c+d x)\right )^3} \, dx=\int { \frac {\operatorname {sech}\left (d x + c\right )^{3}}{{\left (b \sinh \left (d x + c\right )^{2} + a\right )}^{3}} \,d x } \]
Timed out. \[ \int \frac {\text {sech}^3(c+d x)}{\left (a+b \sinh ^2(c+d x)\right )^3} \, dx=\int \frac {1}{{\mathrm {cosh}\left (c+d\,x\right )}^3\,{\left (b\,{\mathrm {sinh}\left (c+d\,x\right )}^2+a\right )}^3} \,d x \]