\(\int \frac {\text {sech}(c+d x)}{(a+b \tanh ^2(c+d x))^3} \, dx\) [127]

Optimal result

Integrand size = 21, antiderivative size = 140 \[ \int \frac {\text {sech}(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=\frac {\left (8 a^2+8 a b+3 b^2\right ) \arctan \left (\frac {\sqrt {a+b} \sinh (c+d x)}{\sqrt {a}}\right )}{8 a^{5/2} (a+b)^{5/2} d}+\frac {b^2 \sinh (c+d x)}{4 a (a+b)^2 d \left (a+(a+b) \sinh ^2(c+d x)\right )^2}+\frac {b (8 a+3 b) \sinh (c+d x)}{8 a^2 (a+b)^2 d \left (a+(a+b) \sinh ^2(c+d x)\right )} \] Output:

1/8*(8*a^2+8*a*b+3*b^2)*arctan((a+b)^(1/2)*sinh(d*x+c)/a^(1/2))/a^(5/2)/(a 
+b)^(5/2)/d+1/4*b^2*sinh(d*x+c)/a/(a+b)^2/d/(a+(a+b)*sinh(d*x+c)^2)^2+1/8* 
b*(8*a+3*b)*sinh(d*x+c)/a^2/(a+b)^2/d/(a+(a+b)*sinh(d*x+c)^2)
 

Mathematica [A] (verified)

Time = 0.69 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.96 \[ \int \frac {\text {sech}(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=\frac {-\frac {\left (8 a^2+8 a b+3 b^2\right ) \arctan \left (\frac {\sqrt {a} \text {csch}(c+d x)}{\sqrt {a+b}}\right )}{(a+b)^{5/2}}+\frac {2 \sqrt {a} b \left (8 a^2-a b-3 b^2+\left (8 a^2+11 a b+3 b^2\right ) \cosh (2 (c+d x))\right ) \sinh (c+d x)}{(a+b)^2 (a-b+(a+b) \cosh (2 (c+d x)))^2}}{8 a^{5/2} d} \] Input:

Integrate[Sech[c + d*x]/(a + b*Tanh[c + d*x]^2)^3,x]
 

Output:

(-(((8*a^2 + 8*a*b + 3*b^2)*ArcTan[(Sqrt[a]*Csch[c + d*x])/Sqrt[a + b]])/( 
a + b)^(5/2)) + (2*Sqrt[a]*b*(8*a^2 - a*b - 3*b^2 + (8*a^2 + 11*a*b + 3*b^ 
2)*Cosh[2*(c + d*x)])*Sinh[c + d*x])/((a + b)^2*(a - b + (a + b)*Cosh[2*(c 
 + d*x)])^2))/(8*a^(5/2)*d)
 

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.10, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3042, 4159, 315, 298, 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.

\(\displaystyle \int \frac {\text {sech}(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sec (i c+i d x)}{\left (a-b \tan (i c+i d x)^2\right )^3}dx\)

\(\Big \downarrow \) 4159

\(\displaystyle \frac {\int \frac {\left (\sinh ^2(c+d x)+1\right )^2}{\left ((a+b) \sinh ^2(c+d x)+a\right )^3}d\sinh (c+d x)}{d}\)

\(\Big \downarrow \) 315

\(\displaystyle \frac {\frac {\int \frac {(4 a+b) \sinh ^2(c+d x)+4 a+3 b}{\left ((a+b) \sinh ^2(c+d x)+a\right )^2}d\sinh (c+d x)}{4 a (a+b)}+\frac {b \sinh (c+d x) \left (\sinh ^2(c+d x)+1\right )}{4 a (a+b) \left ((a+b) \sinh ^2(c+d x)+a\right )^2}}{d}\)

\(\Big \downarrow \) 298

\(\displaystyle \frac {\frac {\frac {\left (8 a^2+8 a b+3 b^2\right ) \int \frac {1}{(a+b) \sinh ^2(c+d x)+a}d\sinh (c+d x)}{2 a (a+b)}+\frac {3 b (2 a+b) \sinh (c+d x)}{2 a (a+b) \left ((a+b) \sinh ^2(c+d x)+a\right )}}{4 a (a+b)}+\frac {b \sinh (c+d x) \left (\sinh ^2(c+d x)+1\right )}{4 a (a+b) \left ((a+b) \sinh ^2(c+d x)+a\right )^2}}{d}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {\frac {\frac {\left (8 a^2+8 a b+3 b^2\right ) \arctan \left (\frac {\sqrt {a+b} \sinh (c+d x)}{\sqrt {a}}\right )}{2 a^{3/2} (a+b)^{3/2}}+\frac {3 b (2 a+b) \sinh (c+d x)}{2 a (a+b) \left ((a+b) \sinh ^2(c+d x)+a\right )}}{4 a (a+b)}+\frac {b \sinh (c+d x) \left (\sinh ^2(c+d x)+1\right )}{4 a (a+b) \left ((a+b) \sinh ^2(c+d x)+a\right )^2}}{d}\)

Input:

Int[Sech[c + d*x]/(a + b*Tanh[c + d*x]^2)^3,x]
 

Output:

((b*Sinh[c + d*x]*(1 + Sinh[c + d*x]^2))/(4*a*(a + b)*(a + (a + b)*Sinh[c 
+ d*x]^2)^2) + (((8*a^2 + 8*a*b + 3*b^2)*ArcTan[(Sqrt[a + b]*Sinh[c + d*x] 
)/Sqrt[a]])/(2*a^(3/2)*(a + b)^(3/2)) + (3*b*(2*a + b)*Sinh[c + d*x])/(2*a 
*(a + b)*(a + (a + b)*Sinh[c + d*x]^2)))/(4*a*(a + b)))/d
 

Defintions of rubi rules used

rule 218

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]
 

rule 298

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[(-( 
b*c - a*d))*x*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] - Simp[(a*d - b*c*( 
2*p + 3))/(2*a*b*(p + 1))   Int[(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, 
 c, d, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/2 + p, 0])
 

rule 315

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim 
p[(a*d - c*b)*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q - 1)/(2*a*b*(p + 1))), 
x] - Simp[1/(2*a*b*(p + 1))   Int[(a + b*x^2)^(p + 1)*(c + d*x^2)^(q - 2)*S 
imp[c*(a*d - c*b*(2*p + 3)) + d*(a*d*(2*(q - 1) + 1) - b*c*(2*(p + q) + 1)) 
*x^2, x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, - 
1] && GtQ[q, 1] && IntBinomialQ[a, b, c, d, 2, p, q, x]
 

rule 3042

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

rule 4159

Int[sec[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^(n_ 
))^(p_.), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Simp[ff/f 
  Subst[Int[ExpandToSum[b*(ff*x)^n + a*(1 - ff^2*x^2)^(n/2), x]^p/(1 - ff^2 
*x^2)^((m + n*p + 1)/2), x], x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f} 
, x] && IntegerQ[(m - 1)/2] && IntegerQ[n/2] && IntegerQ[p]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(393\) vs. \(2(126)=252\).

Time = 29.85 (sec) , antiderivative size = 394, normalized size of antiderivative = 2.81

method	result	size
derivativedivides	\(\frac {\frac {-\frac {b \left (8 a +5 b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{4 a \left (a^{2}+2 a b +b^{2}\right )}-\frac {\left (8 a^{2}+29 a b +12 b^{2}\right ) b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{4 a^{2} \left (a^{2}+2 a b +b^{2}\right )}+\frac {\left (8 a^{2}+29 a b +12 b^{2}\right ) b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{4 a^{2} \left (a^{2}+2 a b +b^{2}\right )}+\frac {b \left (8 a +5 b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a \left (a^{2}+2 a b +b^{2}\right )}}{\left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a +2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +4 b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+a \right )^{2}}+\frac {\left (8 a^{2}+8 a b +3 b^{2}\right ) \left (\frac {\left (\sqrt {\left (a +b \right ) b}+b \right ) \arctan \left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {\left (a +b \right ) b}+a +2 b \right ) a}}\right )}{2 a \sqrt {\left (a +b \right ) b}\, \sqrt {\left (2 \sqrt {\left (a +b \right ) b}+a +2 b \right ) a}}-\frac {\left (\sqrt {\left (a +b \right ) b}-b \right ) \operatorname {arctanh}\left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {\left (a +b \right ) b}-a -2 b \right ) a}}\right )}{2 a \sqrt {\left (a +b \right ) b}\, \sqrt {\left (2 \sqrt {\left (a +b \right ) b}-a -2 b \right ) a}}\right )}{4 a \left (a^{2}+2 a b +b^{2}\right )}}{d}\)	\(394\)
default	\(\frac {\frac {-\frac {b \left (8 a +5 b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{4 a \left (a^{2}+2 a b +b^{2}\right )}-\frac {\left (8 a^{2}+29 a b +12 b^{2}\right ) b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{4 a^{2} \left (a^{2}+2 a b +b^{2}\right )}+\frac {\left (8 a^{2}+29 a b +12 b^{2}\right ) b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{4 a^{2} \left (a^{2}+2 a b +b^{2}\right )}+\frac {b \left (8 a +5 b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a \left (a^{2}+2 a b +b^{2}\right )}}{\left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a +2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +4 b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+a \right )^{2}}+\frac {\left (8 a^{2}+8 a b +3 b^{2}\right ) \left (\frac {\left (\sqrt {\left (a +b \right ) b}+b \right ) \arctan \left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {\left (a +b \right ) b}+a +2 b \right ) a}}\right )}{2 a \sqrt {\left (a +b \right ) b}\, \sqrt {\left (2 \sqrt {\left (a +b \right ) b}+a +2 b \right ) a}}-\frac {\left (\sqrt {\left (a +b \right ) b}-b \right ) \operatorname {arctanh}\left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {\left (a +b \right ) b}-a -2 b \right ) a}}\right )}{2 a \sqrt {\left (a +b \right ) b}\, \sqrt {\left (2 \sqrt {\left (a +b \right ) b}-a -2 b \right ) a}}\right )}{4 a \left (a^{2}+2 a b +b^{2}\right )}}{d}\)	\(394\)
risch	\(\frac {\left (8 \,{\mathrm e}^{6 d x +6 c} a^{2}+11 \,{\mathrm e}^{6 d x +6 c} a b +3 \,{\mathrm e}^{6 d x +6 c} b^{2}+8 \,{\mathrm e}^{4 d x +4 c} a^{2}-13 \,{\mathrm e}^{4 d x +4 c} a b -9 \,{\mathrm e}^{4 d x +4 c} b^{2}-8 \,{\mathrm e}^{2 d x +2 c} a^{2}+13 \,{\mathrm e}^{2 d x +2 c} b a +9 b^{2} {\mathrm e}^{2 d x +2 c}-8 a^{2}-11 a b -3 b^{2}\right ) b \,{\mathrm e}^{d x +c}}{4 \left (a^{2}+2 a b +b^{2}\right ) \left ({\mathrm e}^{4 d x +4 c} a +b \,{\mathrm e}^{4 d x +4 c}+2 \,{\mathrm e}^{2 d x +2 c} a -2 \,{\mathrm e}^{2 d x +2 c} b +a +b \right )^{2} d \,a^{2}}-\frac {\ln \left ({\mathrm e}^{2 d x +2 c}-\frac {2 a \,{\mathrm e}^{d x +c}}{\sqrt {-a^{2}-a b}}-1\right )}{2 \sqrt {-a^{2}-a b}\, \left (a +b \right )^{2} d}-\frac {\ln \left ({\mathrm e}^{2 d x +2 c}-\frac {2 a \,{\mathrm e}^{d x +c}}{\sqrt {-a^{2}-a b}}-1\right ) b}{2 \sqrt {-a^{2}-a b}\, \left (a +b \right )^{2} d a}-\frac {3 \ln \left ({\mathrm e}^{2 d x +2 c}-\frac {2 a \,{\mathrm e}^{d x +c}}{\sqrt {-a^{2}-a b}}-1\right ) b^{2}}{16 \sqrt {-a^{2}-a b}\, \left (a +b \right )^{2} d \,a^{2}}+\frac {\ln \left ({\mathrm e}^{2 d x +2 c}+\frac {2 a \,{\mathrm e}^{d x +c}}{\sqrt {-a^{2}-a b}}-1\right )}{2 \sqrt {-a^{2}-a b}\, \left (a +b \right )^{2} d}+\frac {\ln \left ({\mathrm e}^{2 d x +2 c}+\frac {2 a \,{\mathrm e}^{d x +c}}{\sqrt {-a^{2}-a b}}-1\right ) b}{2 \sqrt {-a^{2}-a b}\, \left (a +b \right )^{2} d a}+\frac {3 \ln \left ({\mathrm e}^{2 d x +2 c}+\frac {2 a \,{\mathrm e}^{d x +c}}{\sqrt {-a^{2}-a b}}-1\right ) b^{2}}{16 \sqrt {-a^{2}-a b}\, \left (a +b \right )^{2} d \,a^{2}}\)	\(569\)

Input:

int(sech(d*x+c)/(a+tanh(d*x+c)^2*b)^3,x,method=_RETURNVERBOSE)
 

Output:

1/d*(2*(-1/8*b*(8*a+5*b)/a/(a^2+2*a*b+b^2)*tanh(1/2*d*x+1/2*c)^7-1/8*(8*a^ 
2+29*a*b+12*b^2)/a^2*b/(a^2+2*a*b+b^2)*tanh(1/2*d*x+1/2*c)^5+1/8*(8*a^2+29 
*a*b+12*b^2)/a^2*b/(a^2+2*a*b+b^2)*tanh(1/2*d*x+1/2*c)^3+1/8*b*(8*a+5*b)/a 
/(a^2+2*a*b+b^2)*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+1/4/a*(8*a^2+8*a*b+3*b^2)/(a 
^2+2*a*b+b^2)*(1/2*(((a+b)*b)^(1/2)+b)/a/((a+b)*b)^(1/2)/((2*((a+b)*b)^(1/ 
2)+a+2*b)*a)^(1/2)*arctan(a*tanh(1/2*d*x+1/2*c)/((2*((a+b)*b)^(1/2)+a+2*b) 
*a)^(1/2))-1/2*(((a+b)*b)^(1/2)-b)/a/((a+b)*b)^(1/2)/((2*((a+b)*b)^(1/2)-a 
-2*b)*a)^(1/2)*arctanh(a*tanh(1/2*d*x+1/2*c)/((2*((a+b)*b)^(1/2)-a-2*b)*a) 
^(1/2))))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 4385 vs. \(2 (126) = 252\).

Time = 0.18 (sec) , antiderivative size = 7917, normalized size of antiderivative = 56.55 \[ \int \frac {\text {sech}(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=\text {Too large to display} \] Input:

integrate(sech(d*x+c)/(a+b*tanh(d*x+c)^2)^3,x, algorithm="fricas")
 

Output:

Too large to include
 

Sympy [F]

\[ \int \frac {\text {sech}(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=\int \frac {\operatorname {sech}{\left (c + d x \right )}}{\left (a + b \tanh ^{2}{\left (c + d x \right )}\right )^{3}}\, dx \] Input:

integrate(sech(d*x+c)/(a+b*tanh(d*x+c)**2)**3,x)
 

Output:

Integral(sech(c + d*x)/(a + b*tanh(c + d*x)**2)**3, x)
 

Maxima [F]

\[ \int \frac {\text {sech}(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=\int { \frac {\operatorname {sech}\left (d x + c\right )}{{\left (b \tanh \left (d x + c\right )^{2} + a\right )}^{3}} \,d x } \] Input:

integrate(sech(d*x+c)/(a+b*tanh(d*x+c)^2)^3,x, algorithm="maxima")
 

Output:

1/4*((8*a^2*b*e^(7*c) + 11*a*b^2*e^(7*c) + 3*b^3*e^(7*c))*e^(7*d*x) + (8*a 
^2*b*e^(5*c) - 13*a*b^2*e^(5*c) - 9*b^3*e^(5*c))*e^(5*d*x) - (8*a^2*b*e^(3 
*c) - 13*a*b^2*e^(3*c) - 9*b^3*e^(3*c))*e^(3*d*x) - (8*a^2*b*e^c + 11*a*b^ 
2*e^c + 3*b^3*e^c)*e^(d*x))/(a^6*d + 4*a^5*b*d + 6*a^4*b^2*d + 4*a^3*b^3*d 
 + a^2*b^4*d + (a^6*d*e^(8*c) + 4*a^5*b*d*e^(8*c) + 6*a^4*b^2*d*e^(8*c) + 
4*a^3*b^3*d*e^(8*c) + a^2*b^4*d*e^(8*c))*e^(8*d*x) + 4*(a^6*d*e^(6*c) + 2* 
a^5*b*d*e^(6*c) - 2*a^3*b^3*d*e^(6*c) - a^2*b^4*d*e^(6*c))*e^(6*d*x) + 2*( 
3*a^6*d*e^(4*c) + 4*a^5*b*d*e^(4*c) + 2*a^4*b^2*d*e^(4*c) + 4*a^3*b^3*d*e^ 
(4*c) + 3*a^2*b^4*d*e^(4*c))*e^(4*d*x) + 4*(a^6*d*e^(2*c) + 2*a^5*b*d*e^(2 
*c) - 2*a^3*b^3*d*e^(2*c) - a^2*b^4*d*e^(2*c))*e^(2*d*x)) + 2*integrate(1/ 
8*((8*a^2*e^(3*c) + 8*a*b*e^(3*c) + 3*b^2*e^(3*c))*e^(3*d*x) + (8*a^2*e^c 
+ 8*a*b*e^c + 3*b^2*e^c)*e^(d*x))/(a^5 + 3*a^4*b + 3*a^3*b^2 + a^2*b^3 + ( 
a^5*e^(4*c) + 3*a^4*b*e^(4*c) + 3*a^3*b^2*e^(4*c) + a^2*b^3*e^(4*c))*e^(4* 
d*x) + 2*(a^5*e^(2*c) + a^4*b*e^(2*c) - a^3*b^2*e^(2*c) - a^2*b^3*e^(2*c)) 
*e^(2*d*x)), x)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {\text {sech}(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(sech(d*x+c)/(a+b*tanh(d*x+c)^2)^3,x, algorithm="giac")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Limit: Max order reached or unable 
to make series expansion Error: Bad Argument Value
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\text {sech}(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=\int \frac {1}{\mathrm {cosh}\left (c+d\,x\right )\,{\left (b\,{\mathrm {tanh}\left (c+d\,x\right )}^2+a\right )}^3} \,d x \] Input:

int(1/(cosh(c + d*x)*(a + b*tanh(c + d*x)^2)^3),x)
 

Output:

int(1/(cosh(c + d*x)*(a + b*tanh(c + d*x)^2)^3), x)
 

Reduce [B] (verification not implemented)

Time = 0.33 (sec) , antiderivative size = 14353, normalized size of antiderivative = 102.52 \[ \int \frac {\text {sech}(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx =\text {Too large to display} \] Input:

int(sech(d*x+c)/(a+b*tanh(d*x+c)^2)^3,x)
 

Output:

(32*e**(8*c + 8*d*x)*sqrt(a)*sqrt(a + b)*atan((e**(c + d*x)*sqrt(a + b) - 
sqrt(b))/sqrt(a))*tanh(c + d*x)**4*a**5*b**2 + 104*e**(8*c + 8*d*x)*sqrt(a 
)*sqrt(a + b)*atan((e**(c + d*x)*sqrt(a + b) - sqrt(b))/sqrt(a))*tanh(c + 
d*x)**4*a**4*b**3 + 132*e**(8*c + 8*d*x)*sqrt(a)*sqrt(a + b)*atan((e**(c + 
 d*x)*sqrt(a + b) - sqrt(b))/sqrt(a))*tanh(c + d*x)**4*a**3*b**4 + 83*e**( 
8*c + 8*d*x)*sqrt(a)*sqrt(a + b)*atan((e**(c + d*x)*sqrt(a + b) - sqrt(b)) 
/sqrt(a))*tanh(c + d*x)**4*a**2*b**5 + 26*e**(8*c + 8*d*x)*sqrt(a)*sqrt(a 
+ b)*atan((e**(c + d*x)*sqrt(a + b) - sqrt(b))/sqrt(a))*tanh(c + d*x)**4*a 
*b**6 + 3*e**(8*c + 8*d*x)*sqrt(a)*sqrt(a + b)*atan((e**(c + d*x)*sqrt(a + 
 b) - sqrt(b))/sqrt(a))*tanh(c + d*x)**4*b**7 + 64*e**(8*c + 8*d*x)*sqrt(a 
)*sqrt(a + b)*atan((e**(c + d*x)*sqrt(a + b) - sqrt(b))/sqrt(a))*tanh(c + 
d*x)**2*a**6*b + 208*e**(8*c + 8*d*x)*sqrt(a)*sqrt(a + b)*atan((e**(c + d* 
x)*sqrt(a + b) - sqrt(b))/sqrt(a))*tanh(c + d*x)**2*a**5*b**2 + 264*e**(8* 
c + 8*d*x)*sqrt(a)*sqrt(a + b)*atan((e**(c + d*x)*sqrt(a + b) - sqrt(b))/s 
qrt(a))*tanh(c + d*x)**2*a**4*b**3 + 166*e**(8*c + 8*d*x)*sqrt(a)*sqrt(a + 
 b)*atan((e**(c + d*x)*sqrt(a + b) - sqrt(b))/sqrt(a))*tanh(c + d*x)**2*a* 
*3*b**4 + 52*e**(8*c + 8*d*x)*sqrt(a)*sqrt(a + b)*atan((e**(c + d*x)*sqrt( 
a + b) - sqrt(b))/sqrt(a))*tanh(c + d*x)**2*a**2*b**5 + 6*e**(8*c + 8*d*x) 
*sqrt(a)*sqrt(a + b)*atan((e**(c + d*x)*sqrt(a + b) - sqrt(b))/sqrt(a))*ta 
nh(c + d*x)**2*a*b**6 + 32*e**(8*c + 8*d*x)*sqrt(a)*sqrt(a + b)*atan((e...