\(\int \frac {\tanh ^2(c+d x)}{(a+b \text {sech}^2(c+d x))^2} \, dx\) [151]

Optimal result

Integrand size = 23, antiderivative size = 85 \[ \int \frac {\tanh ^2(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^2} \, dx=\frac {x}{a^2}-\frac {(a+2 b) \text {arctanh}\left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a+b}}\right )}{2 a^2 \sqrt {b} \sqrt {a+b} d}-\frac {\tanh (c+d x)}{2 a d \left (a+b-b \tanh ^2(c+d x)\right )} \] Output:

x/a^2-1/2*(a+2*b)*arctanh(b^(1/2)*tanh(d*x+c)/(a+b)^(1/2))/a^2/b^(1/2)/(a+ 
b)^(1/2)/d-1/2*tanh(d*x+c)/a/d/(a+b-b*tanh(d*x+c)^2)
                                                                                    
                                                                                    
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(326\) vs. \(2(85)=170\).

Time = 2.88 (sec) , antiderivative size = 326, normalized size of antiderivative = 3.84 \[ \int \frac {\tanh ^2(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^2} \, dx=\frac {(a+2 b+a \cosh (2 (c+d x)))^2 \text {sech}^4(c+d x) \left (\frac {16 x}{a^2}-\frac {(a+2 b) \text {arctanh}\left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a+b}}\right )}{b^{3/2} (a+b)^{3/2} d}+\frac {\left (a^3-6 a^2 b-24 a b^2-16 b^3\right ) \text {arctanh}\left (\frac {\text {sech}(d x) (\cosh (2 c)-\sinh (2 c)) ((a+2 b) \sinh (d x)-a \sinh (2 c+d x))}{2 \sqrt {a+b} \sqrt {b (\cosh (c)-\sinh (c))^4}}\right ) (\cosh (2 c)-\sinh (2 c))}{a^2 b (a+b)^{3/2} d \sqrt {b (\cosh (c)-\sinh (c))^4}}+\frac {\left (a^2+8 a b+8 b^2\right ) \text {sech}(2 c) ((a+2 b) \sinh (2 c)-a \sinh (2 d x))}{a^2 b (a+b) d (a+2 b+a \cosh (2 (c+d x)))}+\frac {a \sinh (2 (c+d x))}{b (a+b) d (a+2 b+a \cosh (2 (c+d x)))}\right )}{64 \left (a+b \text {sech}^2(c+d x)\right )^2} \] Input:

Integrate[Tanh[c + d*x]^2/(a + b*Sech[c + d*x]^2)^2,x]
 

Output:

((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)
 

Rubi [A] (verified)

Time = 0.39 (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

\(\displaystyle -\frac {\frac {\tanh (c+d x)}{2 a \left (a-b \tanh ^2(c+d x)+b\right )}-\frac {\frac {2 \text {arctanh}(\tanh (c+d x))}{a}-\frac {(a+2 b) \text {arctanh}\left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a+b}}\right )}{a \sqrt {b} \sqrt {a+b}}}{2 a}}{d}\)

Input:

Int[Tanh[c + d*x]^2/(a + b*Sech[c + d*x]^2)^2,x]
 

Output:

-((-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)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 219

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])
 

rule 221

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]
 

rule 373

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]
 

rule 397

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]
 

rule 2075

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]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4629

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])
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(235\) vs. \(2(73)=146\).

Time = 6.67 (sec) , antiderivative size = 236, normalized size of antiderivative = 2.78

method	result	size
derivativedivides	\(\frac {\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{2}}-\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{2}}+\frac {\frac {2 \left (-\frac {\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} a}{2}-\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{2}\right )}{\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a +\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} b +2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b}+\left (a +2 b \right ) \left (-\frac {\ln \left (\sqrt {a +b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {b}+\sqrt {a +b}\right )}{4 \sqrt {b}\, \sqrt {a +b}}+\frac {\ln \left (\sqrt {a +b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {b}+\sqrt {a +b}\right )}{4 \sqrt {b}\, \sqrt {a +b}}\right )}{a^{2}}}{d}\)	\(236\)
default	\(\frac {\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{2}}-\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{2}}+\frac {\frac {2 \left (-\frac {\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} a}{2}-\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{2}\right )}{\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a +\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} b +2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b}+\left (a +2 b \right ) \left (-\frac {\ln \left (\sqrt {a +b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {b}+\sqrt {a +b}\right )}{4 \sqrt {b}\, \sqrt {a +b}}+\frac {\ln \left (\sqrt {a +b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {b}+\sqrt {a +b}\right )}{4 \sqrt {b}\, \sqrt {a +b}}\right )}{a^{2}}}{d}\)	\(236\)
risch	\(\frac {x}{a^{2}}+\frac {a \,{\mathrm e}^{2 d x +2 c}+2 b \,{\mathrm e}^{2 d x +2 c}+a}{a^{2} d \left ({\mathrm e}^{4 d x +4 c} a +2 a \,{\mathrm e}^{2 d x +2 c}+4 b \,{\mathrm e}^{2 d x +2 c}+a \right )}+\frac {\ln \left ({\mathrm e}^{2 d x +2 c}+\frac {a \sqrt {a b +b^{2}}+2 b \sqrt {a b +b^{2}}+2 a b +2 b^{2}}{a \sqrt {a b +b^{2}}}\right )}{4 \sqrt {a b +b^{2}}\, d a}+\frac {\ln \left ({\mathrm e}^{2 d x +2 c}+\frac {a \sqrt {a b +b^{2}}+2 b \sqrt {a b +b^{2}}+2 a b +2 b^{2}}{a \sqrt {a b +b^{2}}}\right ) b}{2 \sqrt {a b +b^{2}}\, d \,a^{2}}-\frac {\ln \left ({\mathrm e}^{2 d x +2 c}+\frac {a \sqrt {a b +b^{2}}+2 b \sqrt {a b +b^{2}}-2 a b -2 b^{2}}{a \sqrt {a b +b^{2}}}\right )}{4 \sqrt {a b +b^{2}}\, d a}-\frac {\ln \left ({\mathrm e}^{2 d x +2 c}+\frac {a \sqrt {a b +b^{2}}+2 b \sqrt {a b +b^{2}}-2 a b -2 b^{2}}{a \sqrt {a b +b^{2}}}\right ) b}{2 \sqrt {a b +b^{2}}\, d \,a^{2}}\)	\(376\)

Input:

int(tanh(d*x+c)^2/(a+b*sech(d*x+c)^2)^2,x,method=_RETURNVERBOSE)
 

Output:

1/d*(1/a^2*ln(tanh(1/2*d*x+1/2*c)+1)-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*a*tanh(1/2*d*x+1/2*c))/(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)))))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 803 vs. \(2 (76) = 152\).

Time = 0.35 (sec) , antiderivative size = 1846, normalized size of antiderivative = 21.72 \[ \int \frac {\tanh ^2(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^2} \, dx=\text {Too large to display} \] Input:

integrate(tanh(d*x+c)^2/(a+b*sech(d*x+c)^2)^2,x, algorithm="fricas")
 

Output:

[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 ...
 

Sympy [F]

\[ \int \frac {\tanh ^2(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^2} \, dx=\int \frac {\tanh ^{2}{\left (c + d x \right )}}{\left (a + b \operatorname {sech}^{2}{\left (c + d x \right )}\right )^{2}}\, dx \] Input:

integrate(tanh(d*x+c)**2/(a+b*sech(d*x+c)**2)**2,x)
 

Output:

Integral(tanh(c + d*x)**2/(a + b*sech(c + d*x)**2)**2, x)
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 597 vs. \(2 (76) = 152\).

Time = 0.19 (sec) , antiderivative size = 597, normalized size of antiderivative = 7.02 \[ \int \frac {\tanh ^2(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^2} \, dx =\text {Too large to display} \] Input:

integrate(tanh(d*x+c)^2/(a+b*sech(d*x+c)^2)^2,x, algorithm="maxima")
 

Output:

-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)
 

Giac [A] (verification not implemented)

Time = 0.58 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.61 \[ \int \frac {\tanh ^2(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^2} \, dx=-\frac {\frac {{\left (a + 2 \, b\right )} \arctan \left (\frac {a e^{\left (2 \, d x + 2 \, c\right )} + a + 2 \, b}{2 \, \sqrt {-a b - b^{2}}}\right )}{\sqrt {-a b - b^{2}} a^{2}} - \frac {2 \, {\left (d x + c\right )}}{a^{2}} - \frac {2 \, {\left (a e^{\left (2 \, d x + 2 \, c\right )} + 2 \, b e^{\left (2 \, d x + 2 \, c\right )} + a\right )}}{{\left (a e^{\left (4 \, d x + 4 \, c\right )} + 2 \, a e^{\left (2 \, d x + 2 \, c\right )} + 4 \, b e^{\left (2 \, d x + 2 \, c\right )} + a\right )} a^{2}}}{2 \, d} \] Input:

integrate(tanh(d*x+c)^2/(a+b*sech(d*x+c)^2)^2,x, algorithm="giac")
 

Output:

-1/2*((a + 2*b)*arctan(1/2*(a*e^(2*d*x + 2*c) + a + 2*b)/sqrt(-a*b - b^2)) 
/(sqrt(-a*b - b^2)*a^2) - 2*(d*x + c)/a^2 - 2*(a*e^(2*d*x + 2*c) + 2*b*e^( 
2*d*x + 2*c) + a)/((a*e^(4*d*x + 4*c) + 2*a*e^(2*d*x + 2*c) + 4*b*e^(2*d*x 
 + 2*c) + a)*a^2))/d
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\tanh ^2(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^2} \, dx=\int \frac {{\mathrm {cosh}\left (c+d\,x\right )}^2\,\left ({\mathrm {cosh}\left (c+d\,x\right )}^2-1\right )}{{\left (a\,{\mathrm {cosh}\left (c+d\,x\right )}^2+b\right )}^2} \,d x \] Input:

int(tanh(c + d*x)^2/(a + b/cosh(c + d*x)^2)^2,x)
 

Output:

int((cosh(c + d*x)^2*(cosh(c + d*x)^2 - 1))/(b + a*cosh(c + d*x)^2)^2, x)
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 1199, normalized size of antiderivative = 14.11 \[ \int \frac {\tanh ^2(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^2} \, dx =\text {Too large to display} \] Input:

int(tanh(d*x+c)^2/(a+b*sech(d*x+c)^2)^2,x)
 

Output:

( - e**(4*c + 4*d*x)*sqrt(b)*sqrt(a + b)*log( - sqrt(2*sqrt(b)*sqrt(a + b) 
 - a - 2*b) + e**(c + d*x)*sqrt(a))*a**2 - 2*e**(4*c + 4*d*x)*sqrt(b)*sqrt 
(a + b)*log( - sqrt(2*sqrt(b)*sqrt(a + b) - a - 2*b) + e**(c + d*x)*sqrt(a 
))*a*b - e**(4*c + 4*d*x)*sqrt(b)*sqrt(a + b)*log(sqrt(2*sqrt(b)*sqrt(a + 
b) - a - 2*b) + e**(c + d*x)*sqrt(a))*a**2 - 2*e**(4*c + 4*d*x)*sqrt(b)*sq 
rt(a + b)*log(sqrt(2*sqrt(b)*sqrt(a + b) - a - 2*b) + e**(c + d*x)*sqrt(a) 
)*a*b + e**(4*c + 4*d*x)*sqrt(b)*sqrt(a + b)*log(2*sqrt(b)*sqrt(a + b) + e 
**(2*c + 2*d*x)*a + a + 2*b)*a**2 + 2*e**(4*c + 4*d*x)*sqrt(b)*sqrt(a + b) 
*log(2*sqrt(b)*sqrt(a + b) + e**(2*c + 2*d*x)*a + a + 2*b)*a*b - 2*e**(2*c 
 + 2*d*x)*sqrt(b)*sqrt(a + b)*log( - sqrt(2*sqrt(b)*sqrt(a + b) - a - 2*b) 
 + e**(c + d*x)*sqrt(a))*a**2 - 8*e**(2*c + 2*d*x)*sqrt(b)*sqrt(a + b)*log 
( - sqrt(2*sqrt(b)*sqrt(a + b) - a - 2*b) + e**(c + d*x)*sqrt(a))*a*b - 8* 
e**(2*c + 2*d*x)*sqrt(b)*sqrt(a + b)*log( - sqrt(2*sqrt(b)*sqrt(a + b) - a 
 - 2*b) + e**(c + d*x)*sqrt(a))*b**2 - 2*e**(2*c + 2*d*x)*sqrt(b)*sqrt(a + 
 b)*log(sqrt(2*sqrt(b)*sqrt(a + b) - a - 2*b) + e**(c + d*x)*sqrt(a))*a**2 
 - 8*e**(2*c + 2*d*x)*sqrt(b)*sqrt(a + b)*log(sqrt(2*sqrt(b)*sqrt(a + b) - 
 a - 2*b) + e**(c + d*x)*sqrt(a))*a*b - 8*e**(2*c + 2*d*x)*sqrt(b)*sqrt(a 
+ b)*log(sqrt(2*sqrt(b)*sqrt(a + b) - a - 2*b) + e**(c + d*x)*sqrt(a))*b** 
2 + 2*e**(2*c + 2*d*x)*sqrt(b)*sqrt(a + b)*log(2*sqrt(b)*sqrt(a + b) + e** 
(2*c + 2*d*x)*a + a + 2*b)*a**2 + 8*e**(2*c + 2*d*x)*sqrt(b)*sqrt(a + b...