Integrand size = 23, antiderivative size = 144 \[ \int \frac {\cosh ^2(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^2} \, dx=\frac {(a-4 b) x}{2 a^3}+\frac {b^{3/2} (5 a+4 b) \text {arctanh}\left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a+b}}\right )}{2 a^3 (a+b)^{3/2} d}+\frac {\cosh (c+d x) \sinh (c+d x)}{2 a d \left (a+b-b \tanh ^2(c+d x)\right )}+\frac {b (a+2 b) \tanh (c+d x)}{2 a^2 (a+b) d \left (a+b-b \tanh ^2(c+d x)\right )} \]
1/2*(a-4*b)*x/a^3+1/2*b^(3/2)*(5*a+4*b)*arctanh(b^(1/2)*tanh(d*x+c)/(a+b)^ (1/2))/a^3/(a+b)^(3/2)/d+1/2*cosh(d*x+c)*sinh(d*x+c)/a/d/(a+b-b*tanh(d*x+c )^2)+1/2*b*(a+2*b)*tanh(d*x+c)/a^2/(a+b)/d/(a+b-b*tanh(d*x+c)^2)
Time = 1.58 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.72 \[ \int \frac {\cosh ^2(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^2} \, dx=\frac {2 (a-4 b) (c+d x)+\frac {2 b^{3/2} (5 a+4 b) \text {arctanh}\left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a+b}}\right )}{(a+b)^{3/2}}+\left (a+\frac {2 a b^2}{(a+b) (a+2 b+a \cosh (2 (c+d x)))}\right ) \sinh (2 (c+d x))}{4 a^3 d} \]
(2*(a - 4*b)*(c + d*x) + (2*b^(3/2)*(5*a + 4*b)*ArcTanh[(Sqrt[b]*Tanh[c + d*x])/Sqrt[a + b]])/(a + b)^(3/2) + (a + (2*a*b^2)/((a + b)*(a + 2*b + a*C osh[2*(c + d*x)])))*Sinh[2*(c + d*x)])/(4*a^3*d)
Time = 0.40 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.15, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {3042, 4634, 316, 402, 27, 397, 219, 221}
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.
| \(\displaystyle \int \frac {\cosh ^2(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sec (i c+i d x)^2 \left (a+b \sec (i c+i d x)^2\right )^2}dx\) |
| \(\Big \downarrow \) 4634 |
| \(\displaystyle \frac {\int \frac {1}{\left (1-\tanh ^2(c+d x)\right )^2 \left (-b \tanh ^2(c+d x)+a+b\right )^2}d\tanh (c+d x)}{d}\) |
| \(\Big \downarrow \) 316 |
| \(\displaystyle \frac {\frac {\int \frac {-3 b \tanh ^2(c+d x)+a-b}{\left (1-\tanh ^2(c+d x)\right ) \left (-b \tanh ^2(c+d x)+a+b\right )^2}d\tanh (c+d x)}{2 a}+\frac {\tanh (c+d x)}{2 a \left (1-\tanh ^2(c+d x)\right ) \left (a-b \tanh ^2(c+d x)+b\right )}}{d}\) |
| \(\Big \downarrow \) 402 |
| \(\displaystyle \frac {\frac {\frac {b (a+2 b) \tanh (c+d x)}{a (a+b) \left (a-b \tanh ^2(c+d x)+b\right )}-\frac {\int -\frac {2 \left (a^2-2 b a-2 b^2-b (a+2 b) \tanh ^2(c+d x)\right )}{\left (1-\tanh ^2(c+d x)\right ) \left (-b \tanh ^2(c+d x)+a+b\right )}d\tanh (c+d x)}{2 a (a+b)}}{2 a}+\frac {\tanh (c+d x)}{2 a \left (1-\tanh ^2(c+d x)\right ) \left (a-b \tanh ^2(c+d x)+b\right )}}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\frac {\int \frac {a^2-2 b a-2 b^2-b (a+2 b) \tanh ^2(c+d x)}{\left (1-\tanh ^2(c+d x)\right ) \left (-b \tanh ^2(c+d x)+a+b\right )}d\tanh (c+d x)}{a (a+b)}+\frac {b (a+2 b) \tanh (c+d x)}{a (a+b) \left (a-b \tanh ^2(c+d x)+b\right )}}{2 a}+\frac {\tanh (c+d x)}{2 a \left (1-\tanh ^2(c+d x)\right ) \left (a-b \tanh ^2(c+d x)+b\right )}}{d}\) |
| \(\Big \downarrow \) 397 |
| \(\displaystyle \frac {\frac {\frac {\frac {b^2 (5 a+4 b) \int \frac {1}{-b \tanh ^2(c+d x)+a+b}d\tanh (c+d x)}{a}+\frac {(a-4 b) (a+b) \int \frac {1}{1-\tanh ^2(c+d x)}d\tanh (c+d x)}{a}}{a (a+b)}+\frac {b (a+2 b) \tanh (c+d x)}{a (a+b) \left (a-b \tanh ^2(c+d x)+b\right )}}{2 a}+\frac {\tanh (c+d x)}{2 a \left (1-\tanh ^2(c+d x)\right ) \left (a-b \tanh ^2(c+d x)+b\right )}}{d}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {\frac {\frac {b^2 (5 a+4 b) \int \frac {1}{-b \tanh ^2(c+d x)+a+b}d\tanh (c+d x)}{a}+\frac {(a-4 b) (a+b) \text {arctanh}(\tanh (c+d x))}{a}}{a (a+b)}+\frac {b (a+2 b) \tanh (c+d x)}{a (a+b) \left (a-b \tanh ^2(c+d x)+b\right )}}{2 a}+\frac {\tanh (c+d x)}{2 a \left (1-\tanh ^2(c+d x)\right ) \left (a-b \tanh ^2(c+d x)+b\right )}}{d}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {\frac {\frac {\frac {b^{3/2} (5 a+4 b) \text {arctanh}\left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a+b}}\right )}{a \sqrt {a+b}}+\frac {(a-4 b) (a+b) \text {arctanh}(\tanh (c+d x))}{a}}{a (a+b)}+\frac {b (a+2 b) \tanh (c+d x)}{a (a+b) \left (a-b \tanh ^2(c+d x)+b\right )}}{2 a}+\frac {\tanh (c+d x)}{2 a \left (1-\tanh ^2(c+d x)\right ) \left (a-b \tanh ^2(c+d x)+b\right )}}{d}\) |
(Tanh[c + d*x]/(2*a*(1 - Tanh[c + d*x]^2)*(a + b - b*Tanh[c + d*x]^2)) + ( (((a - 4*b)*(a + b)*ArcTanh[Tanh[c + d*x]])/a + (b^(3/2)*(5*a + 4*b)*ArcTa nh[(Sqrt[b]*Tanh[c + d*x])/Sqrt[a + b]])/(a*Sqrt[a + b]))/(a*(a + b)) + (b *(a + 2*b)*Tanh[c + d*x])/(a*(a + b)*(a + b - b*Tanh[c + d*x]^2)))/(2*a))/ d
3.1.84.3.1 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]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[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[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_) )^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[ff/f Subst[Int[(1 + ff^2*x^2)^(m/2 - 1)*ExpandToSum[a + b*(1 + ff^2*x^2)^(n/2), x]^p, x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ [m/2] && IntegerQ[n/2]
Leaf count of result is larger than twice the leaf count of optimal. \(337\) vs. \(2(128)=256\).
Time = 1.26 (sec) , antiderivative size = 338, normalized size of antiderivative = 2.35
| method | result | size |
| risch | \(\frac {x}{2 a^{2}}-\frac {2 x b}{a^{3}}+\frac {{\mathrm e}^{2 d x +2 c}}{8 a^{2} d}-\frac {{\mathrm e}^{-2 d x -2 c}}{8 a^{2} d}-\frac {b^{2} \left ({\mathrm e}^{2 d x +2 c} a +2 b \,{\mathrm e}^{2 d x +2 c}+a \right )}{a^{3} \left (a +b \right ) d \left (a \,{\mathrm e}^{4 d x +4 c}+2 \,{\mathrm e}^{2 d x +2 c} a +4 b \,{\mathrm e}^{2 d x +2 c}+a \right )}+\frac {5 \sqrt {\left (a +b \right ) b}\, b \ln \left ({\mathrm e}^{2 d x +2 c}-\frac {2 \sqrt {\left (a +b \right ) b}-a -2 b}{a}\right )}{4 \left (a +b \right )^{2} d \,a^{2}}+\frac {\sqrt {\left (a +b \right ) b}\, b^{2} \ln \left ({\mathrm e}^{2 d x +2 c}-\frac {2 \sqrt {\left (a +b \right ) b}-a -2 b}{a}\right )}{\left (a +b \right )^{2} d \,a^{3}}-\frac {5 \sqrt {\left (a +b \right ) b}\, b \ln \left ({\mathrm e}^{2 d x +2 c}+\frac {2 \sqrt {\left (a +b \right ) b}+a +2 b}{a}\right )}{4 \left (a +b \right )^{2} d \,a^{2}}-\frac {\sqrt {\left (a +b \right ) b}\, b^{2} \ln \left ({\mathrm e}^{2 d x +2 c}+\frac {2 \sqrt {\left (a +b \right ) b}+a +2 b}{a}\right )}{\left (a +b \right )^{2} d \,a^{3}}\) | \(338\) |
| derivativedivides | \(\frac {-\frac {1}{2 a^{2} \left (1+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {1}{2 a^{2} \left (1+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+\frac {\left (a -4 b \right ) \ln \left (1+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a^{3}}-\frac {2 b^{2} \left (\frac {-\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{2 \left (a +b \right )}-\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a +b \right )}}{\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a +\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} b +2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b}+\frac {\left (5 a +4 b \right ) \left (-\frac {\ln \left (\sqrt {a +b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {b}+\sqrt {a +b}\right )}{4 \sqrt {b}\, \sqrt {a +b}}+\frac {\ln \left (\sqrt {a +b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {b}+\sqrt {a +b}\right )}{4 \sqrt {b}\, \sqrt {a +b}}\right )}{2 a +2 b}\right )}{a^{3}}+\frac {1}{2 a^{2} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {1}{2 a^{2} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\left (-a +4 b \right ) \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 a^{3}}}{d}\) | \(341\) |
| default | \(\frac {-\frac {1}{2 a^{2} \left (1+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {1}{2 a^{2} \left (1+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+\frac {\left (a -4 b \right ) \ln \left (1+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a^{3}}-\frac {2 b^{2} \left (\frac {-\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{2 \left (a +b \right )}-\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a +b \right )}}{\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a +\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} b +2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b}+\frac {\left (5 a +4 b \right ) \left (-\frac {\ln \left (\sqrt {a +b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {b}+\sqrt {a +b}\right )}{4 \sqrt {b}\, \sqrt {a +b}}+\frac {\ln \left (\sqrt {a +b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {b}+\sqrt {a +b}\right )}{4 \sqrt {b}\, \sqrt {a +b}}\right )}{2 a +2 b}\right )}{a^{3}}+\frac {1}{2 a^{2} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {1}{2 a^{2} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\left (-a +4 b \right ) \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 a^{3}}}{d}\) | \(341\) |
1/2*x/a^2-2*x/a^3*b+1/8/a^2/d*exp(2*d*x+2*c)-1/8/a^2/d*exp(-2*d*x-2*c)-b^2 *(exp(2*d*x+2*c)*a+2*b*exp(2*d*x+2*c)+a)/a^3/(a+b)/d/(a*exp(4*d*x+4*c)+2*e xp(2*d*x+2*c)*a+4*b*exp(2*d*x+2*c)+a)+5/4*((a+b)*b)^(1/2)/(a+b)^2*b/d/a^2* ln(exp(2*d*x+2*c)-(2*((a+b)*b)^(1/2)-a-2*b)/a)+((a+b)*b)^(1/2)/(a+b)^2*b^2 /d/a^3*ln(exp(2*d*x+2*c)-(2*((a+b)*b)^(1/2)-a-2*b)/a)-5/4*((a+b)*b)^(1/2)/ (a+b)^2*b/d/a^2*ln(exp(2*d*x+2*c)+(2*((a+b)*b)^(1/2)+a+2*b)/a)-((a+b)*b)^( 1/2)/(a+b)^2*b^2/d/a^3*ln(exp(2*d*x+2*c)+(2*((a+b)*b)^(1/2)+a+2*b)/a)
Leaf count of result is larger than twice the leaf count of optimal. 1731 vs. \(2 (134) = 268\).
Time = 0.32 (sec) , antiderivative size = 3739, normalized size of antiderivative = 25.97 \[ \int \frac {\cosh ^2(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^2} \, dx=\text {Too large to display} \]
[1/8*((a^3 + a^2*b)*cosh(d*x + c)^8 + 8*(a^3 + a^2*b)*cosh(d*x + c)*sinh(d *x + c)^7 + (a^3 + a^2*b)*sinh(d*x + c)^8 + 2*(a^3 + 3*a^2*b + 2*a*b^2 + 2 *(a^3 - 3*a^2*b - 4*a*b^2)*d*x)*cosh(d*x + c)^6 + 2*(a^3 + 3*a^2*b + 2*a*b ^2 + 2*(a^3 - 3*a^2*b - 4*a*b^2)*d*x + 14*(a^3 + a^2*b)*cosh(d*x + c)^2)*s inh(d*x + c)^6 + 4*(14*(a^3 + a^2*b)*cosh(d*x + c)^3 + 3*(a^3 + 3*a^2*b + 2*a*b^2 + 2*(a^3 - 3*a^2*b - 4*a*b^2)*d*x)*cosh(d*x + c))*sinh(d*x + c)^5 - 8*(a*b^2 + 2*b^3 - (a^3 - a^2*b - 10*a*b^2 - 8*b^3)*d*x)*cosh(d*x + c)^4 + 2*(35*(a^3 + a^2*b)*cosh(d*x + c)^4 - 4*a*b^2 - 8*b^3 + 4*(a^3 - a^2*b - 10*a*b^2 - 8*b^3)*d*x + 15*(a^3 + 3*a^2*b + 2*a*b^2 + 2*(a^3 - 3*a^2*b - 4*a*b^2)*d*x)*cosh(d*x + c)^2)*sinh(d*x + c)^4 + 8*(7*(a^3 + a^2*b)*cosh( d*x + c)^5 + 5*(a^3 + 3*a^2*b + 2*a*b^2 + 2*(a^3 - 3*a^2*b - 4*a*b^2)*d*x) *cosh(d*x + c)^3 - 4*(a*b^2 + 2*b^3 - (a^3 - a^2*b - 10*a*b^2 - 8*b^3)*d*x )*cosh(d*x + c))*sinh(d*x + c)^3 - a^3 - a^2*b - 2*(a^3 + 3*a^2*b + 6*a*b^ 2 - 2*(a^3 - 3*a^2*b - 4*a*b^2)*d*x)*cosh(d*x + c)^2 + 2*(14*(a^3 + a^2*b) *cosh(d*x + c)^6 + 15*(a^3 + 3*a^2*b + 2*a*b^2 + 2*(a^3 - 3*a^2*b - 4*a*b^ 2)*d*x)*cosh(d*x + c)^4 - a^3 - 3*a^2*b - 6*a*b^2 + 2*(a^3 - 3*a^2*b - 4*a *b^2)*d*x - 24*(a*b^2 + 2*b^3 - (a^3 - a^2*b - 10*a*b^2 - 8*b^3)*d*x)*cosh (d*x + c)^2)*sinh(d*x + c)^2 + 2*((5*a^2*b + 4*a*b^2)*cosh(d*x + c)^6 + 6* (5*a^2*b + 4*a*b^2)*cosh(d*x + c)*sinh(d*x + c)^5 + (5*a^2*b + 4*a*b^2)*si nh(d*x + c)^6 + 2*(5*a^2*b + 14*a*b^2 + 8*b^3)*cosh(d*x + c)^4 + (10*a^...
\[ \int \frac {\cosh ^2(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^2} \, dx=\int \frac {\cosh ^{2}{\left (c + d x \right )}}{\left (a + b \operatorname {sech}^{2}{\left (c + d x \right )}\right )^{2}}\, dx \]
Leaf count of result is larger than twice the leaf count of optimal. 696 vs. \(2 (134) = 268\).
Time = 0.31 (sec) , antiderivative size = 696, normalized size of antiderivative = 4.83 \[ \int \frac {\cosh ^2(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^2} \, dx=\frac {{\left (3 \, a^{2} b + 12 \, a b^{2} + 8 \, b^{3}\right )} \log \left (\frac {a e^{\left (2 \, d x + 2 \, c\right )} + a + 2 \, b - 2 \, \sqrt {{\left (a + b\right )} b}}{a e^{\left (2 \, d x + 2 \, c\right )} + a + 2 \, b + 2 \, \sqrt {{\left (a + b\right )} b}}\right )}{16 \, {\left (a^{4} + a^{3} b\right )} \sqrt {{\left (a + b\right )} b} d} - \frac {{\left (3 \, a^{2} b + 12 \, a b^{2} + 8 \, b^{3}\right )} \log \left (\frac {a e^{\left (-2 \, d x - 2 \, c\right )} + a + 2 \, b - 2 \, \sqrt {{\left (a + b\right )} b}}{a e^{\left (-2 \, d x - 2 \, c\right )} + a + 2 \, b + 2 \, \sqrt {{\left (a + b\right )} b}}\right )}{16 \, {\left (a^{4} + a^{3} b\right )} \sqrt {{\left (a + b\right )} b} d} + \frac {{\left (3 \, a b + 2 \, b^{2}\right )} \log \left (\frac {a e^{\left (-2 \, d x - 2 \, c\right )} + a + 2 \, b - 2 \, \sqrt {{\left (a + b\right )} b}}{a e^{\left (-2 \, d x - 2 \, c\right )} + a + 2 \, b + 2 \, \sqrt {{\left (a + b\right )} b}}\right )}{8 \, {\left (a^{3} + a^{2} b\right )} \sqrt {{\left (a + b\right )} b} d} - \frac {a^{2} b + 2 \, a b^{2} + {\left (a^{2} b + 8 \, a b^{2} + 8 \, b^{3}\right )} e^{\left (2 \, d x + 2 \, c\right )}}{4 \, {\left (a^{5} + a^{4} b + {\left (a^{5} + a^{4} b\right )} e^{\left (4 \, d x + 4 \, c\right )} + 2 \, {\left (a^{5} + 3 \, a^{4} b + 2 \, a^{3} b^{2}\right )} e^{\left (2 \, d x + 2 \, c\right )}\right )} d} + \frac {a^{2} b + 2 \, a b^{2} + {\left (a^{2} b + 8 \, a b^{2} + 8 \, b^{3}\right )} e^{\left (-2 \, d x - 2 \, c\right )}}{4 \, {\left (a^{5} + a^{4} b + 2 \, {\left (a^{5} + 3 \, a^{4} b + 2 \, a^{3} b^{2}\right )} e^{\left (-2 \, d x - 2 \, c\right )} + {\left (a^{5} + a^{4} b\right )} e^{\left (-4 \, d x - 4 \, c\right )}\right )} d} - \frac {a b + {\left (a b + 2 \, b^{2}\right )} e^{\left (-2 \, d x - 2 \, c\right )}}{2 \, {\left (a^{4} + a^{3} b + 2 \, {\left (a^{4} + 3 \, a^{3} b + 2 \, a^{2} b^{2}\right )} e^{\left (-2 \, d x - 2 \, c\right )} + {\left (a^{4} + a^{3} b\right )} e^{\left (-4 \, d x - 4 \, c\right )}\right )} d} + \frac {d x + c}{2 \, a^{2} d} + \frac {e^{\left (2 \, d x + 2 \, c\right )}}{8 \, a^{2} d} - \frac {e^{\left (-2 \, d x - 2 \, c\right )}}{8 \, a^{2} d} - \frac {b \log \left (a e^{\left (4 \, d x + 4 \, c\right )} + 2 \, {\left (a + 2 \, b\right )} e^{\left (2 \, d x + 2 \, c\right )} + a\right )}{2 \, a^{3} d} + \frac {b \log \left (2 \, {\left (a + 2 \, b\right )} e^{\left (-2 \, d x - 2 \, c\right )} + a e^{\left (-4 \, d x - 4 \, c\right )} + a\right )}{2 \, a^{3} d} \]
1/16*(3*a^2*b + 12*a*b^2 + 8*b^3)*log((a*e^(2*d*x + 2*c) + a + 2*b - 2*sqr t((a + b)*b))/(a*e^(2*d*x + 2*c) + a + 2*b + 2*sqrt((a + b)*b)))/((a^4 + a ^3*b)*sqrt((a + b)*b)*d) - 1/16*(3*a^2*b + 12*a*b^2 + 8*b^3)*log((a*e^(-2* d*x - 2*c) + a + 2*b - 2*sqrt((a + b)*b))/(a*e^(-2*d*x - 2*c) + a + 2*b + 2*sqrt((a + b)*b)))/((a^4 + a^3*b)*sqrt((a + b)*b)*d) + 1/8*(3*a*b + 2*b^2 )*log((a*e^(-2*d*x - 2*c) + a + 2*b - 2*sqrt((a + b)*b))/(a*e^(-2*d*x - 2* c) + a + 2*b + 2*sqrt((a + b)*b)))/((a^3 + a^2*b)*sqrt((a + b)*b)*d) - 1/4 *(a^2*b + 2*a*b^2 + (a^2*b + 8*a*b^2 + 8*b^3)*e^(2*d*x + 2*c))/((a^5 + a^4 *b + (a^5 + a^4*b)*e^(4*d*x + 4*c) + 2*(a^5 + 3*a^4*b + 2*a^3*b^2)*e^(2*d* x + 2*c))*d) + 1/4*(a^2*b + 2*a*b^2 + (a^2*b + 8*a*b^2 + 8*b^3)*e^(-2*d*x - 2*c))/((a^5 + a^4*b + 2*(a^5 + 3*a^4*b + 2*a^3*b^2)*e^(-2*d*x - 2*c) + ( a^5 + a^4*b)*e^(-4*d*x - 4*c))*d) - 1/2*(a*b + (a*b + 2*b^2)*e^(-2*d*x - 2 *c))/((a^4 + a^3*b + 2*(a^4 + 3*a^3*b + 2*a^2*b^2)*e^(-2*d*x - 2*c) + (a^4 + a^3*b)*e^(-4*d*x - 4*c))*d) + 1/2*(d*x + c)/(a^2*d) + 1/8*e^(2*d*x + 2* c)/(a^2*d) - 1/8*e^(-2*d*x - 2*c)/(a^2*d) - 1/2*b*log(a*e^(4*d*x + 4*c) + 2*(a + 2*b)*e^(2*d*x + 2*c) + a)/(a^3*d) + 1/2*b*log(2*(a + 2*b)*e^(-2*d*x - 2*c) + a*e^(-4*d*x - 4*c) + a)/(a^3*d)
\[ \int \frac {\cosh ^2(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^2} \, dx=\int { \frac {\cosh \left (d x + c\right )^{2}}{{\left (b \operatorname {sech}\left (d x + c\right )^{2} + a\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {\cosh ^2(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^2} \, dx=\int \frac {{\mathrm {cosh}\left (c+d\,x\right )}^2}{{\left (a+\frac {b}{{\mathrm {cosh}\left (c+d\,x\right )}^2}\right )}^2} \,d x \]