Integrand size = 23, antiderivative size = 85 \[ \int \frac {\tanh ^2(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^2} \, dx=\frac {x}{a^2}-\frac {(a+2 b) \text {arctanh}\left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a+b}}\right )}{2 a^2 \sqrt {b} \sqrt {a+b} d}-\frac {\tanh (c+d x)}{2 a d \left (a+b-b \tanh ^2(c+d x)\right )} \]
x/a^2-1/2*(a+2*b)*arctanh(b^(1/2)*tanh(d*x+c)/(a+b)^(1/2))/a^2/d/b^(1/2)/( a+b)^(1/2)-1/2*tanh(d*x+c)/a/d/(a+b-b*tanh(d*x+c)^2)
Leaf count is larger than twice the leaf count of optimal. \(326\) vs. \(2(85)=170\).
Time = 4.77 (sec) , antiderivative size = 326, normalized size of antiderivative = 3.84 \[ \int \frac {\tanh ^2(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^2} \, dx=\frac {(a+2 b+a \cosh (2 (c+d x)))^2 \text {sech}^4(c+d x) \left (\frac {16 x}{a^2}-\frac {(a+2 b) \text {arctanh}\left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a+b}}\right )}{b^{3/2} (a+b)^{3/2} d}+\frac {\left (a^3-6 a^2 b-24 a b^2-16 b^3\right ) \text {arctanh}\left (\frac {\text {sech}(d x) (\cosh (2 c)-\sinh (2 c)) ((a+2 b) \sinh (d x)-a \sinh (2 c+d x))}{2 \sqrt {a+b} \sqrt {b (\cosh (c)-\sinh (c))^4}}\right ) (\cosh (2 c)-\sinh (2 c))}{a^2 b (a+b)^{3/2} d \sqrt {b (\cosh (c)-\sinh (c))^4}}+\frac {\left (a^2+8 a b+8 b^2\right ) \text {sech}(2 c) ((a+2 b) \sinh (2 c)-a \sinh (2 d x))}{a^2 b (a+b) d (a+2 b+a \cosh (2 (c+d x)))}+\frac {a \sinh (2 (c+d x))}{b (a+b) d (a+2 b+a \cosh (2 (c+d x)))}\right )}{64 \left (a+b \text {sech}^2(c+d x)\right )^2} \]
((a + 2*b + a*Cosh[2*(c + d*x)])^2*Sech[c + d*x]^4*((16*x)/a^2 - ((a + 2*b )*ArcTanh[(Sqrt[b]*Tanh[c + d*x])/Sqrt[a + b]])/(b^(3/2)*(a + b)^(3/2)*d) + ((a^3 - 6*a^2*b - 24*a*b^2 - 16*b^3)*ArcTanh[(Sech[d*x]*(Cosh[2*c] - Sin h[2*c])*((a + 2*b)*Sinh[d*x] - a*Sinh[2*c + d*x]))/(2*Sqrt[a + b]*Sqrt[b*( Cosh[c] - Sinh[c])^4])]*(Cosh[2*c] - Sinh[2*c]))/(a^2*b*(a + b)^(3/2)*d*Sq rt[b*(Cosh[c] - Sinh[c])^4]) + ((a^2 + 8*a*b + 8*b^2)*Sech[2*c]*((a + 2*b) *Sinh[2*c] - a*Sinh[2*d*x]))/(a^2*b*(a + b)*d*(a + 2*b + a*Cosh[2*(c + d*x )])) + (a*Sinh[2*(c + d*x)])/(b*(a + b)*d*(a + 2*b + a*Cosh[2*(c + d*x)])) ))/(64*(a + b*Sech[c + d*x]^2)^2)
Time = 0.35 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.14, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {3042, 25, 4629, 25, 2075, 373, 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.
| \(\displaystyle \int \frac {\tanh ^2(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int -\frac {\tan (i c+i d x)^2}{\left (a+b \sec (i c+i d x)^2\right )^2}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\int \frac {\tan (i c+i d x)^2}{\left (b \sec (i c+i d x)^2+a\right )^2}dx\) |
| \(\Big \downarrow \) 4629 |
| \(\displaystyle -\frac {\int -\frac {\tanh ^2(c+d x)}{\left (1-\tanh ^2(c+d x)\right ) \left (a+b \left (1-\tanh ^2(c+d x)\right )\right )^2}d\tanh (c+d x)}{d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {\tanh ^2(c+d x)}{\left (1-\tanh ^2(c+d x)\right ) \left (a+b \left (1-\tanh ^2(c+d x)\right )\right )^2}d\tanh (c+d x)}{d}\) |
| \(\Big \downarrow \) 2075 |
| \(\displaystyle \frac {\int \frac {\tanh ^2(c+d x)}{\left (1-\tanh ^2(c+d x)\right ) \left (-b \tanh ^2(c+d x)+a+b\right )^2}d\tanh (c+d x)}{d}\) |
| \(\Big \downarrow \) 373 |
| \(\displaystyle -\frac {\frac {\tanh (c+d x)}{2 a \left (a-b \tanh ^2(c+d x)+b\right )}-\frac {\int \frac {\tanh ^2(c+d x)+1}{\left (1-\tanh ^2(c+d x)\right ) \left (-b \tanh ^2(c+d x)+a+b\right )}d\tanh (c+d x)}{2 a}}{d}\) |
| \(\Big \downarrow \) 397 |
| \(\displaystyle -\frac {\frac {\tanh (c+d x)}{2 a \left (a-b \tanh ^2(c+d x)+b\right )}-\frac {\frac {2 \int \frac {1}{1-\tanh ^2(c+d x)}d\tanh (c+d x)}{a}-\frac {(a+2 b) \int \frac {1}{-b \tanh ^2(c+d x)+a+b}d\tanh (c+d x)}{a}}{2 a}}{d}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle -\frac {\frac {\tanh (c+d x)}{2 a \left (a-b \tanh ^2(c+d x)+b\right )}-\frac {\frac {2 \text {arctanh}(\tanh (c+d x))}{a}-\frac {(a+2 b) \int \frac {1}{-b \tanh ^2(c+d x)+a+b}d\tanh (c+d x)}{a}}{2 a}}{d}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle -\frac {\frac {\tanh (c+d x)}{2 a \left (a-b \tanh ^2(c+d x)+b\right )}-\frac {\frac {2 \text {arctanh}(\tanh (c+d x))}{a}-\frac {(a+2 b) \text {arctanh}\left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a+b}}\right )}{a \sqrt {b} \sqrt {a+b}}}{2 a}}{d}\) |
-((-1/2*((2*ArcTanh[Tanh[c + d*x]])/a - ((a + 2*b)*ArcTanh[(Sqrt[b]*Tanh[c + d*x])/Sqrt[a + b]])/(a*Sqrt[b]*Sqrt[a + b]))/a + Tanh[c + d*x]/(2*a*(a + b - b*Tanh[c + d*x]^2)))/d)
3.2.51.3.1 Defintions of rubi rules used
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[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ ), x_Symbol] :> Simp[e*(e*x)^(m - 1)*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(2*(b*c - a*d)*(p + 1))), x] - Simp[e^2/(2*(b*c - a*d)*(p + 1)) Int[(e *x)^(m - 2)*(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[c*(m - 1) + d*(m + 2*p + 2*q + 3)*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && GtQ[m, 1] && LeQ[m, 3] && IntBinomialQ[a, b, c, d, e, m, 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[(u_)^(p_.)*(v_)^(q_.)*((e_.)*(x_))^(m_.), x_Symbol] :> Int[(e*x)^m*Expa ndToSum[u, x]^p*ExpandToSum[v, x]^q, x] /; FreeQ[{e, m, p, q}, x] && Binomi alQ[{u, v}, x] && EqQ[BinomialDegree[u, x] - BinomialDegree[v, x], 0] && ! BinomialMatchQ[{u, v}, x]
Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*((d_.)*tan[(e_.) + (f _.)*(x_)])^(m_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Sim p[ff/f Subst[Int[(d*ff*x)^m*((a + b*(1 + ff^2*x^2)^(n/2))^p/(1 + ff^2*x^2 )), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, d, e, f, m, p}, x] && Inte gerQ[n/2] && (IntegerQ[m/2] || EqQ[n, 2])
Leaf count of result is larger than twice the leaf count of optimal. \(235\) vs. \(2(73)=146\).
Time = 6.53 (sec) , antiderivative size = 236, normalized size of antiderivative = 2.78
| method | result | size |
| derivativedivides | \(\frac {\frac {\ln \left (1+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{2}}-\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{2}}+\frac {\frac {2 \left (-\frac {\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} a}{2}-\frac {\tanh \left (\frac {d x}{2}+\frac {c}{2}\right ) a}{2}\right )}{\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a +\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} b +2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b}+\left (a +2 b \right ) \left (-\frac {\ln \left (\sqrt {a +b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {b}+\sqrt {a +b}\right )}{4 \sqrt {b}\, \sqrt {a +b}}+\frac {\ln \left (\sqrt {a +b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {b}+\sqrt {a +b}\right )}{4 \sqrt {b}\, \sqrt {a +b}}\right )}{a^{2}}}{d}\) | \(236\) |
| default | \(\frac {\frac {\ln \left (1+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{2}}-\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{2}}+\frac {\frac {2 \left (-\frac {\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} a}{2}-\frac {\tanh \left (\frac {d x}{2}+\frac {c}{2}\right ) a}{2}\right )}{\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a +\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} b +2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b}+\left (a +2 b \right ) \left (-\frac {\ln \left (\sqrt {a +b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {b}+\sqrt {a +b}\right )}{4 \sqrt {b}\, \sqrt {a +b}}+\frac {\ln \left (\sqrt {a +b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {b}+\sqrt {a +b}\right )}{4 \sqrt {b}\, \sqrt {a +b}}\right )}{a^{2}}}{d}\) | \(236\) |
| risch | \(\frac {x}{a^{2}}+\frac {{\mathrm e}^{2 d x +2 c} a +2 b \,{\mathrm e}^{2 d x +2 c}+a}{d \,a^{2} \left (a \,{\mathrm e}^{4 d x +4 c}+2 \,{\mathrm e}^{2 d x +2 c} a +4 b \,{\mathrm e}^{2 d x +2 c}+a \right )}+\frac {\ln \left ({\mathrm e}^{2 d x +2 c}+\frac {a \sqrt {a b +b^{2}}+2 b \sqrt {a b +b^{2}}+2 a b +2 b^{2}}{a \sqrt {a b +b^{2}}}\right )}{4 \sqrt {a b +b^{2}}\, d a}+\frac {\ln \left ({\mathrm e}^{2 d x +2 c}+\frac {a \sqrt {a b +b^{2}}+2 b \sqrt {a b +b^{2}}+2 a b +2 b^{2}}{a \sqrt {a b +b^{2}}}\right ) b}{2 \sqrt {a b +b^{2}}\, d \,a^{2}}-\frac {\ln \left ({\mathrm e}^{2 d x +2 c}+\frac {a \sqrt {a b +b^{2}}+2 b \sqrt {a b +b^{2}}-2 a b -2 b^{2}}{a \sqrt {a b +b^{2}}}\right )}{4 \sqrt {a b +b^{2}}\, d a}-\frac {\ln \left ({\mathrm e}^{2 d x +2 c}+\frac {a \sqrt {a b +b^{2}}+2 b \sqrt {a b +b^{2}}-2 a b -2 b^{2}}{a \sqrt {a b +b^{2}}}\right ) b}{2 \sqrt {a b +b^{2}}\, d \,a^{2}}\) | \(376\) |
1/d*(1/a^2*ln(1+tanh(1/2*d*x+1/2*c))-1/a^2*ln(tanh(1/2*d*x+1/2*c)-1)+2/a^2 *((-1/2*tanh(1/2*d*x+1/2*c)^3*a-1/2*tanh(1/2*d*x+1/2*c)*a)/(tanh(1/2*d*x+1 /2*c)^4*a+tanh(1/2*d*x+1/2*c)^4*b+2*tanh(1/2*d*x+1/2*c)^2*a-2*tanh(1/2*d*x +1/2*c)^2*b+a+b)+1/2*(a+2*b)*(-1/4/b^(1/2)/(a+b)^(1/2)*ln((a+b)^(1/2)*tanh (1/2*d*x+1/2*c)^2+2*tanh(1/2*d*x+1/2*c)*b^(1/2)+(a+b)^(1/2))+1/4/b^(1/2)/( a+b)^(1/2)*ln((a+b)^(1/2)*tanh(1/2*d*x+1/2*c)^2-2*tanh(1/2*d*x+1/2*c)*b^(1 /2)+(a+b)^(1/2)))))
Leaf count of result is larger than twice the leaf count of optimal. 803 vs. \(2 (76) = 152\).
Time = 0.30 (sec) , antiderivative size = 1846, normalized size of antiderivative = 21.72 \[ \int \frac {\tanh ^2(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^2} \, dx=\text {Too large to display} \]
[1/4*(4*(a^2*b + a*b^2)*d*x*cosh(d*x + c)^4 + 16*(a^2*b + a*b^2)*d*x*cosh( d*x + c)*sinh(d*x + c)^3 + 4*(a^2*b + a*b^2)*d*x*sinh(d*x + c)^4 + 4*a^2*b + 4*a*b^2 + 4*(a^2*b + a*b^2)*d*x + 4*(a^2*b + 3*a*b^2 + 2*b^3 + 2*(a^2*b + 3*a*b^2 + 2*b^3)*d*x)*cosh(d*x + c)^2 + 4*(6*(a^2*b + a*b^2)*d*x*cosh(d *x + c)^2 + a^2*b + 3*a*b^2 + 2*b^3 + 2*(a^2*b + 3*a*b^2 + 2*b^3)*d*x)*sin h(d*x + c)^2 + ((a^2 + 2*a*b)*cosh(d*x + c)^4 + 4*(a^2 + 2*a*b)*cosh(d*x + c)*sinh(d*x + c)^3 + (a^2 + 2*a*b)*sinh(d*x + c)^4 + 2*(a^2 + 4*a*b + 4*b ^2)*cosh(d*x + c)^2 + 2*(3*(a^2 + 2*a*b)*cosh(d*x + c)^2 + a^2 + 4*a*b + 4 *b^2)*sinh(d*x + c)^2 + a^2 + 2*a*b + 4*((a^2 + 2*a*b)*cosh(d*x + c)^3 + ( a^2 + 4*a*b + 4*b^2)*cosh(d*x + c))*sinh(d*x + c))*sqrt(a*b + b^2)*log((a^ 2*cosh(d*x + c)^4 + 4*a^2*cosh(d*x + c)*sinh(d*x + c)^3 + a^2*sinh(d*x + c )^4 + 2*(a^2 + 2*a*b)*cosh(d*x + c)^2 + 2*(3*a^2*cosh(d*x + c)^2 + a^2 + 2 *a*b)*sinh(d*x + c)^2 + a^2 + 8*a*b + 8*b^2 + 4*(a^2*cosh(d*x + c)^3 + (a^ 2 + 2*a*b)*cosh(d*x + c))*sinh(d*x + c) + 4*(a*cosh(d*x + c)^2 + 2*a*cosh( d*x + c)*sinh(d*x + c) + a*sinh(d*x + c)^2 + a + 2*b)*sqrt(a*b + b^2))/(a* cosh(d*x + c)^4 + 4*a*cosh(d*x + c)*sinh(d*x + c)^3 + a*sinh(d*x + c)^4 + 2*(a + 2*b)*cosh(d*x + c)^2 + 2*(3*a*cosh(d*x + c)^2 + a + 2*b)*sinh(d*x + c)^2 + 4*(a*cosh(d*x + c)^3 + (a + 2*b)*cosh(d*x + c))*sinh(d*x + c) + a) ) + 8*(2*(a^2*b + a*b^2)*d*x*cosh(d*x + c)^3 + (a^2*b + 3*a*b^2 + 2*b^3 + 2*(a^2*b + 3*a*b^2 + 2*b^3)*d*x)*cosh(d*x + c))*sinh(d*x + c))/((a^4*b ...
\[ \int \frac {\tanh ^2(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^2} \, dx=\int \frac {\tanh ^{2}{\left (c + d x \right )}}{\left (a + b \operatorname {sech}^{2}{\left (c + d x \right )}\right )^{2}}\, dx \]
Leaf count of result is larger than twice the leaf count of optimal. 597 vs. \(2 (76) = 152\).
Time = 0.35 (sec) , antiderivative size = 597, normalized size of antiderivative = 7.02 \[ \int \frac {\tanh ^2(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^2} \, dx=-\frac {{\left (a^{2} + 6 \, a b + 4 \, 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 )}{16 \, {\left (a^{3} + a^{2} b\right )} \sqrt {{\left (a + b\right )} b} d} + \frac {{\left (a^{2} + 6 \, a b + 4 \, 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 )}{16 \, {\left (a^{3} + a^{2} b\right )} \sqrt {{\left (a + b\right )} b} d} + \frac {a^{2} + 2 \, a b + {\left (a^{2} + 8 \, a b + 8 \, b^{2}\right )} e^{\left (2 \, d x + 2 \, c\right )}}{4 \, {\left (a^{4} + a^{3} b + {\left (a^{4} + a^{3} b\right )} e^{\left (4 \, d x + 4 \, c\right )} + 2 \, {\left (a^{4} + 3 \, a^{3} b + 2 \, a^{2} b^{2}\right )} e^{\left (2 \, d x + 2 \, c\right )}\right )} d} - \frac {a^{2} + 2 \, a b + {\left (a^{2} + 8 \, a b + 8 \, b^{2}\right )} e^{\left (-2 \, d x - 2 \, c\right )}}{4 \, {\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 {{\left (a + 2 \, b\right )} e^{\left (-2 \, d x - 2 \, c\right )} + a}{2 \, {\left (a^{3} + a^{2} b + 2 \, {\left (a^{3} + 3 \, a^{2} b + 2 \, a b^{2}\right )} e^{\left (-2 \, d x - 2 \, c\right )} + {\left (a^{3} + a^{2} b\right )} e^{\left (-4 \, d x - 4 \, c\right )}\right )} d} + \frac {\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 \, \sqrt {{\left (a + b\right )} b} {\left (a + b\right )} d} + \frac {\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 )}{4 \, a^{2} d} - \frac {\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 )}{4 \, a^{2} d} \]
-1/16*(a^2 + 6*a*b + 4*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) + 1/16*(a^2 + 6*a*b + 4*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 + 2*a*b + (a^2 + 8*a*b + 8*b^2)*e^(2*d*x + 2*c))/((a^4 + a^3*b + (a^4 + a^3*b)*e^(4*d*x + 4*c) + 2*(a^4 + 3*a^3*b + 2*a^2*b^2)*e^(2*d*x + 2*c))*d) - 1/4*(a^2 + 2*a*b + (a ^2 + 8*a*b + 8*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*((a + 2*b)*e^(-2*d*x - 2*c) + a)/((a^3 + a^2*b + 2*(a^3 + 3*a^2*b + 2*a*b^2)*e ^(-2*d*x - 2*c) + (a^3 + a^2*b)*e^(-4*d*x - 4*c))*d) + 1/8*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)))/(sqrt((a + b)*b)*(a + b)*d) + 1/4*log(a*e^(4*d*x + 4*c) + 2*(a + 2*b)*e^(2*d*x + 2*c) + a)/(a^2*d) - 1/4*log(2*(a + 2*b)*e^(-2*d*x - 2*c) + a*e^(-4*d*x - 4*c) + a)/(a^2*d)
\[ \int \frac {\tanh ^2(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^2} \, dx=\int { \frac {\tanh \left (d x + c\right )^{2}}{{\left (b \operatorname {sech}\left (d x + c\right )^{2} + a\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {\tanh ^2(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^2} \, dx=\int \frac {{\mathrm {cosh}\left (c+d\,x\right )}^2\,\left ({\mathrm {cosh}\left (c+d\,x\right )}^2-1\right )}{{\left (a\,{\mathrm {cosh}\left (c+d\,x\right )}^2+b\right )}^2} \,d x \]