\(\int \frac {\text {sech}^5(c+d x)}{a+b \sinh ^2(c+d x)} \, dx\) [291]

Optimal result

Integrand size = 23, antiderivative size = 138 \[ \int \frac {\text {sech}^5(c+d x)}{a+b \sinh ^2(c+d x)} \, dx=\frac {\left (3 a^2-10 a b+15 b^2\right ) \arctan (\sinh (c+d x))}{8 (a-b)^3 d}-\frac {b^{5/2} \arctan \left (\frac {\sqrt {b} \sinh (c+d x)}{\sqrt {a}}\right )}{\sqrt {a} (a-b)^3 d}+\frac {(3 a-7 b) \text {sech}(c+d x) \tanh (c+d x)}{8 (a-b)^2 d}+\frac {\text {sech}^3(c+d x) \tanh (c+d x)}{4 (a-b) d} \] Output:

1/8*(3*a^2-10*a*b+15*b^2)*arctan(sinh(d*x+c))/(a-b)^3/d-b^(5/2)*arctan(b^( 
1/2)*sinh(d*x+c)/a^(1/2))/a^(1/2)/(a-b)^3/d+1/8*(3*a-7*b)*sech(d*x+c)*tanh 
(d*x+c)/(a-b)^2/d+1/4*sech(d*x+c)^3*tanh(d*x+c)/(a-b)/d
 

Mathematica [A] (verified)

Time = 0.38 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.01 \[ \int \frac {\text {sech}^5(c+d x)}{a+b \sinh ^2(c+d x)} \, dx=\frac {8 b^{5/2} \arctan \left (\frac {\sqrt {a} \text {csch}(c+d x)}{\sqrt {b}}\right )+2 \sqrt {a} \left (3 a^2-10 a b+15 b^2\right ) \arctan \left (\tanh \left (\frac {1}{2} (c+d x)\right )\right )+\sqrt {a} \left (3 a^2-10 a b+7 b^2\right ) \text {sech}(c+d x) \tanh (c+d x)+2 \sqrt {a} (a-b)^2 \text {sech}^3(c+d x) \tanh (c+d x)}{8 \sqrt {a} (a-b)^3 d} \] Input:

Integrate[Sech[c + d*x]^5/(a + b*Sinh[c + d*x]^2),x]
 

Output:

(8*b^(5/2)*ArcTan[(Sqrt[a]*Csch[c + d*x])/Sqrt[b]] + 2*Sqrt[a]*(3*a^2 - 10 
*a*b + 15*b^2)*ArcTan[Tanh[(c + d*x)/2]] + Sqrt[a]*(3*a^2 - 10*a*b + 7*b^2 
)*Sech[c + d*x]*Tanh[c + d*x] + 2*Sqrt[a]*(a - b)^2*Sech[c + d*x]^3*Tanh[c 
 + d*x])/(8*Sqrt[a]*(a - b)^3*d)
 

Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.17, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {3042, 3669, 316, 25, 402, 25, 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.

\(\displaystyle \int \frac {\text {sech}^5(c+d x)}{a+b \sinh ^2(c+d x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\cos (i c+i d x)^5 \left (a-b \sin (i c+i d x)^2\right )}dx\)

\(\Big \downarrow \) 3669

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

\(\Big \downarrow \) 316

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 402

\(\displaystyle \frac {\frac {\frac {(3 a-7 b) \sinh (c+d x)}{2 (a-b) \left (\sinh ^2(c+d x)+1\right )}-\frac {\int -\frac {3 a^2-7 b a+8 b^2+(3 a-7 b) b \sinh ^2(c+d x)}{\left (\sinh ^2(c+d x)+1\right ) \left (b \sinh ^2(c+d x)+a\right )}d\sinh (c+d x)}{2 (a-b)}}{4 (a-b)}+\frac {\sinh (c+d x)}{4 (a-b) \left (\sinh ^2(c+d x)+1\right )^2}}{d}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {\frac {\int \frac {3 a^2-7 b a+8 b^2+(3 a-7 b) b \sinh ^2(c+d x)}{\left (\sinh ^2(c+d x)+1\right ) \left (b \sinh ^2(c+d x)+a\right )}d\sinh (c+d x)}{2 (a-b)}+\frac {(3 a-7 b) \sinh (c+d x)}{2 (a-b) \left (\sinh ^2(c+d x)+1\right )}}{4 (a-b)}+\frac {\sinh (c+d x)}{4 (a-b) \left (\sinh ^2(c+d x)+1\right )^2}}{d}\)

\(\Big \downarrow \) 397

\(\displaystyle \frac {\frac {\frac {\frac {\left (3 a^2-10 a b+15 b^2\right ) \int \frac {1}{\sinh ^2(c+d x)+1}d\sinh (c+d x)}{a-b}-\frac {8 b^3 \int \frac {1}{b \sinh ^2(c+d x)+a}d\sinh (c+d x)}{a-b}}{2 (a-b)}+\frac {(3 a-7 b) \sinh (c+d x)}{2 (a-b) \left (\sinh ^2(c+d x)+1\right )}}{4 (a-b)}+\frac {\sinh (c+d x)}{4 (a-b) \left (\sinh ^2(c+d x)+1\right )^2}}{d}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {\frac {\frac {\frac {\left (3 a^2-10 a b+15 b^2\right ) \arctan (\sinh (c+d x))}{a-b}-\frac {8 b^3 \int \frac {1}{b \sinh ^2(c+d x)+a}d\sinh (c+d x)}{a-b}}{2 (a-b)}+\frac {(3 a-7 b) \sinh (c+d x)}{2 (a-b) \left (\sinh ^2(c+d x)+1\right )}}{4 (a-b)}+\frac {\sinh (c+d x)}{4 (a-b) \left (\sinh ^2(c+d x)+1\right )^2}}{d}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {\frac {\frac {\frac {\left (3 a^2-10 a b+15 b^2\right ) \arctan (\sinh (c+d x))}{a-b}-\frac {8 b^{5/2} \arctan \left (\frac {\sqrt {b} \sinh (c+d x)}{\sqrt {a}}\right )}{\sqrt {a} (a-b)}}{2 (a-b)}+\frac {(3 a-7 b) \sinh (c+d x)}{2 (a-b) \left (\sinh ^2(c+d x)+1\right )}}{4 (a-b)}+\frac {\sinh (c+d x)}{4 (a-b) \left (\sinh ^2(c+d x)+1\right )^2}}{d}\)

Input:

Int[Sech[c + d*x]^5/(a + b*Sinh[c + d*x]^2),x]
 

Output:

(Sinh[c + d*x]/(4*(a - b)*(1 + Sinh[c + d*x]^2)^2) + ((((3*a^2 - 10*a*b + 
15*b^2)*ArcTan[Sinh[c + d*x]])/(a - b) - (8*b^(5/2)*ArcTan[(Sqrt[b]*Sinh[c 
 + d*x])/Sqrt[a]])/(Sqrt[a]*(a - b)))/(2*(a - b)) + ((3*a - 7*b)*Sinh[c + 
d*x])/(2*(a - b)*(1 + Sinh[c + d*x]^2)))/(4*(a - b)))/d
 

Defintions of rubi rules used

rule 25

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

rule 216

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])
 

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 316

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]
 

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 402

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]
 

rule 3042

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

rule 3669

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]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(351\) vs. \(2(124)=248\).

Time = 0.25 (sec) , antiderivative size = 352, normalized size of antiderivative = 2.55

\[\frac {\frac {\frac {2 \left (\left (-\frac {5}{8} a^{2}+\frac {7}{4} a b -\frac {9}{8} b^{2}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}+\left (\frac {3}{8} a^{2}-\frac {1}{4} a b -\frac {1}{8} b^{2}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+\left (-\frac {3}{8} a^{2}+\frac {1}{4} a b +\frac {1}{8} b^{2}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+\left (\frac {5}{8} a^{2}-\frac {7}{4} a b +\frac {9}{8} b^{2}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right )^{4}}+\frac {\left (3 a^{2}-10 a b +15 b^{2}\right ) \arctan \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}}{\left (a -b \right )^{3}}-\frac {2 b^{3} a \left (-\frac {\left (a +\sqrt {-b \left (a -b \right )}-b \right ) \operatorname {arctanh}\left (\frac {\tanh \left (\frac {d x}{2}+\frac {c}{2}\right ) a}{\sqrt {\left (2 \sqrt {-b \left (a -b \right )}+a -2 b \right ) a}}\right )}{2 a \sqrt {-b \left (a -b \right )}\, \sqrt {\left (2 \sqrt {-b \left (a -b \right )}+a -2 b \right ) a}}+\frac {\left (-a +\sqrt {-b \left (a -b \right )}+b \right ) \arctan \left (\frac {\tanh \left (\frac {d x}{2}+\frac {c}{2}\right ) a}{\sqrt {\left (2 \sqrt {-b \left (a -b \right )}-a +2 b \right ) a}}\right )}{2 a \sqrt {-b \left (a -b \right )}\, \sqrt {\left (2 \sqrt {-b \left (a -b \right )}-a +2 b \right ) a}}\right )}{\left (a -b \right )^{3}}}{d}\]

Input:

int(sech(d*x+c)^5/(a+b*sinh(d*x+c)^2),x)
 

Output:

1/d*(2/(a-b)^3*(((-5/8*a^2+7/4*a*b-9/8*b^2)*tanh(1/2*d*x+1/2*c)^7+(3/8*a^2 
-1/4*a*b-1/8*b^2)*tanh(1/2*d*x+1/2*c)^5+(-3/8*a^2+1/4*a*b+1/8*b^2)*tanh(1/ 
2*d*x+1/2*c)^3+(5/8*a^2-7/4*a*b+9/8*b^2)*tanh(1/2*d*x+1/2*c))/(tanh(1/2*d* 
x+1/2*c)^2+1)^4+1/8*(3*a^2-10*a*b+15*b^2)*arctan(tanh(1/2*d*x+1/2*c)))-2*b 
^3/(a-b)^3*a*(-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(tanh(1/2*d*x+1/2*c)*a/((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(tanh(1/2*d*x+1/2*c)*a/((2*(-b*(a-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. 2827 vs. \(2 (124) = 248\).

Time = 0.20 (sec) , antiderivative size = 5500, normalized size of antiderivative = 39.86 \[ \int \frac {\text {sech}^5(c+d x)}{a+b \sinh ^2(c+d x)} \, dx=\text {Too large to display} \] Input:

integrate(sech(d*x+c)^5/(a+b*sinh(d*x+c)^2),x, algorithm="fricas")
 

Output:

Too large to include
 

Sympy [F]

\[ \int \frac {\text {sech}^5(c+d x)}{a+b \sinh ^2(c+d x)} \, dx=\int \frac {\operatorname {sech}^{5}{\left (c + d x \right )}}{a + b \sinh ^{2}{\left (c + d x \right )}}\, dx \] Input:

integrate(sech(d*x+c)**5/(a+b*sinh(d*x+c)**2),x)
 

Output:

Integral(sech(c + d*x)**5/(a + b*sinh(c + d*x)**2), x)
 

Maxima [F]

\[ \int \frac {\text {sech}^5(c+d x)}{a+b \sinh ^2(c+d x)} \, dx=\int { \frac {\operatorname {sech}\left (d x + c\right )^{5}}{b \sinh \left (d x + c\right )^{2} + a} \,d x } \] Input:

integrate(sech(d*x+c)^5/(a+b*sinh(d*x+c)^2),x, algorithm="maxima")
 

Output:

1/4*(3*a^2*e^c - 10*a*b*e^c + 15*b^2*e^c)*arctan(e^(d*x + c))*e^(-c)/(a^3* 
d - 3*a^2*b*d + 3*a*b^2*d - b^3*d) + 1/4*((3*a*e^(7*c) - 7*b*e^(7*c))*e^(7 
*d*x) + (11*a*e^(5*c) - 15*b*e^(5*c))*e^(5*d*x) - (11*a*e^(3*c) - 15*b*e^( 
3*c))*e^(3*d*x) - (3*a*e^c - 7*b*e^c)*e^(d*x))/(a^2*d - 2*a*b*d + b^2*d + 
(a^2*d*e^(8*c) - 2*a*b*d*e^(8*c) + b^2*d*e^(8*c))*e^(8*d*x) + 4*(a^2*d*e^( 
6*c) - 2*a*b*d*e^(6*c) + b^2*d*e^(6*c))*e^(6*d*x) + 6*(a^2*d*e^(4*c) - 2*a 
*b*d*e^(4*c) + b^2*d*e^(4*c))*e^(4*d*x) + 4*(a^2*d*e^(2*c) - 2*a*b*d*e^(2* 
c) + b^2*d*e^(2*c))*e^(2*d*x)) - 32*integrate(1/16*(b^3*e^(3*d*x + 3*c) + 
b^3*e^(d*x + c))/(a^3*b - 3*a^2*b^2 + 3*a*b^3 - b^4 + (a^3*b*e^(4*c) - 3*a 
^2*b^2*e^(4*c) + 3*a*b^3*e^(4*c) - b^4*e^(4*c))*e^(4*d*x) + 2*(2*a^4*e^(2* 
c) - 7*a^3*b*e^(2*c) + 9*a^2*b^2*e^(2*c) - 5*a*b^3*e^(2*c) + b^4*e^(2*c))* 
e^(2*d*x)), x)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {\text {sech}^5(c+d x)}{a+b \sinh ^2(c+d x)} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(sech(d*x+c)^5/(a+b*sinh(d*x+c)^2),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 [B] (verification not implemented)

Time = 12.96 (sec) , antiderivative size = 6237, normalized size of antiderivative = 45.20 \[ \int \frac {\text {sech}^5(c+d x)}{a+b \sinh ^2(c+d x)} \, dx=\text {Too large to display} \] Input:

int(1/(cosh(c + d*x)^5*(a + b*sinh(c + d*x)^2)),x)
 

Output:

(4*exp(c + d*x))/((a*d - b*d)*(4*exp(2*c + 2*d*x) + 6*exp(4*c + 4*d*x) + 4 
*exp(6*c + 6*d*x) + exp(8*c + 8*d*x) + 1)) - (6*exp(c + d*x))/((a*d - b*d) 
*(3*exp(2*c + 2*d*x) + 3*exp(4*c + 4*d*x) + exp(6*c + 6*d*x) + 1)) + (atan 
((exp(d*x)*exp(c)*(243*a^12*(a^6*d^2 + b^6*d^2 - 6*a*b^5*d^2 - 6*a^5*b*d^2 
 + 15*a^2*b^4*d^2 - 20*a^3*b^3*d^2 + 15*a^4*b^2*d^2)^(1/2) + 3840*b^12*(a^ 
6*d^2 + b^6*d^2 - 6*a*b^5*d^2 - 6*a^5*b*d^2 + 15*a^2*b^4*d^2 - 20*a^3*b^3* 
d^2 + 15*a^4*b^2*d^2)^(1/2) - 110560*a*b^11*(a^6*d^2 + b^6*d^2 - 6*a*b^5*d 
^2 - 6*a^5*b*d^2 + 15*a^2*b^4*d^2 - 20*a^3*b^3*d^2 + 15*a^4*b^2*d^2)^(1/2) 
 - 4050*a^11*b*(a^6*d^2 + b^6*d^2 - 6*a*b^5*d^2 - 6*a^5*b*d^2 + 15*a^2*b^4 
*d^2 - 20*a^3*b^3*d^2 + 15*a^4*b^2*d^2)^(1/2) + 976143*a^2*b^10*(a^6*d^2 + 
 b^6*d^2 - 6*a*b^5*d^2 - 6*a^5*b*d^2 + 15*a^2*b^4*d^2 - 20*a^3*b^3*d^2 + 1 
5*a^4*b^2*d^2)^(1/2) - 2740050*a^3*b^9*(a^6*d^2 + b^6*d^2 - 6*a*b^5*d^2 - 
6*a^5*b*d^2 + 15*a^2*b^4*d^2 - 20*a^3*b^3*d^2 + 15*a^4*b^2*d^2)^(1/2) + 42 
52775*a^4*b^8*(a^6*d^2 + b^6*d^2 - 6*a*b^5*d^2 - 6*a^5*b*d^2 + 15*a^2*b^4* 
d^2 - 20*a^3*b^3*d^2 + 15*a^4*b^2*d^2)^(1/2) - 4316760*a^5*b^7*(a^6*d^2 + 
b^6*d^2 - 6*a*b^5*d^2 - 6*a^5*b*d^2 + 15*a^2*b^4*d^2 - 20*a^3*b^3*d^2 + 15 
*a^4*b^2*d^2)^(1/2) + 3087390*a^6*b^6*(a^6*d^2 + b^6*d^2 - 6*a*b^5*d^2 - 6 
*a^5*b*d^2 + 15*a^2*b^4*d^2 - 20*a^3*b^3*d^2 + 15*a^4*b^2*d^2)^(1/2) - 160 
8364*a^7*b^5*(a^6*d^2 + b^6*d^2 - 6*a*b^5*d^2 - 6*a^5*b*d^2 + 15*a^2*b^4*d 
^2 - 20*a^3*b^3*d^2 + 15*a^4*b^2*d^2)^(1/2) + 615750*a^8*b^4*(a^6*d^2 +...
 

Reduce [B] (verification not implemented)

Time = 0.47 (sec) , antiderivative size = 2728, normalized size of antiderivative = 19.77 \[ \int \frac {\text {sech}^5(c+d x)}{a+b \sinh ^2(c+d x)} \, dx =\text {Too large to display} \] Input:

int(sech(d*x+c)^5/(a+b*sinh(d*x+c)^2),x)
 

Output:

(3*e**(8*c + 8*d*x)*atan(e**(c + d*x))*a**3 - 10*e**(8*c + 8*d*x)*atan(e** 
(c + d*x))*a**2*b + 15*e**(8*c + 8*d*x)*atan(e**(c + d*x))*a*b**2 + 12*e** 
(6*c + 6*d*x)*atan(e**(c + d*x))*a**3 - 40*e**(6*c + 6*d*x)*atan(e**(c + d 
*x))*a**2*b + 60*e**(6*c + 6*d*x)*atan(e**(c + d*x))*a*b**2 + 18*e**(4*c + 
 4*d*x)*atan(e**(c + d*x))*a**3 - 60*e**(4*c + 4*d*x)*atan(e**(c + d*x))*a 
**2*b + 90*e**(4*c + 4*d*x)*atan(e**(c + d*x))*a*b**2 + 12*e**(2*c + 2*d*x 
)*atan(e**(c + d*x))*a**3 - 40*e**(2*c + 2*d*x)*atan(e**(c + d*x))*a**2*b 
+ 60*e**(2*c + 2*d*x)*atan(e**(c + d*x))*a*b**2 + 3*atan(e**(c + d*x))*a** 
3 - 10*atan(e**(c + d*x))*a**2*b + 15*atan(e**(c + d*x))*a*b**2 + 4*e**(8* 
c + 8*d*x)*sqrt(b)*sqrt(a)*sqrt(a - b)*sqrt(2*sqrt(a)*sqrt(a - b) + 2*a - 
b)*atan((e**(c + d*x)*b)/(sqrt(b)*sqrt(2*sqrt(a)*sqrt(a - b) + 2*a - b)))* 
b + 16*e**(6*c + 6*d*x)*sqrt(b)*sqrt(a)*sqrt(a - b)*sqrt(2*sqrt(a)*sqrt(a 
- b) + 2*a - b)*atan((e**(c + d*x)*b)/(sqrt(b)*sqrt(2*sqrt(a)*sqrt(a - b) 
+ 2*a - b)))*b + 24*e**(4*c + 4*d*x)*sqrt(b)*sqrt(a)*sqrt(a - b)*sqrt(2*sq 
rt(a)*sqrt(a - b) + 2*a - b)*atan((e**(c + d*x)*b)/(sqrt(b)*sqrt(2*sqrt(a) 
*sqrt(a - b) + 2*a - b)))*b + 16*e**(2*c + 2*d*x)*sqrt(b)*sqrt(a)*sqrt(a - 
 b)*sqrt(2*sqrt(a)*sqrt(a - b) + 2*a - b)*atan((e**(c + d*x)*b)/(sqrt(b)*s 
qrt(2*sqrt(a)*sqrt(a - b) + 2*a - b)))*b + 4*sqrt(b)*sqrt(a)*sqrt(a - b)*s 
qrt(2*sqrt(a)*sqrt(a - b) + 2*a - b)*atan((e**(c + d*x)*b)/(sqrt(b)*sqrt(2 
*sqrt(a)*sqrt(a - b) + 2*a - b)))*b - 4*e**(8*c + 8*d*x)*sqrt(b)*sqrt(2...