\(\int \frac {\text {sech}^5(c+d x)}{a+b \text {sech}^2(c+d x)} \, dx\) [81]

Optimal result
Mathematica [B] (verified)
Rubi [A] (verified)
Maple [B] (verified)
Fricas [B] (verification not implemented)
Sympy [F]
Maxima [F]
Giac [F(-2)]
Mupad [B] (verification not implemented)
Reduce [B] (verification not implemented)

Optimal result

Integrand size = 23, antiderivative size = 86 \[ \int \frac {\text {sech}^5(c+d x)}{a+b \text {sech}^2(c+d x)} \, dx=-\frac {(2 a-b) \arctan (\sinh (c+d x))}{2 b^2 d}+\frac {a^{3/2} \arctan \left (\frac {\sqrt {a} \sinh (c+d x)}{\sqrt {a+b}}\right )}{b^2 \sqrt {a+b} d}+\frac {\text {sech}(c+d x) \tanh (c+d x)}{2 b d} \] Output:

-1/2*(2*a-b)*arctan(sinh(d*x+c))/b^2/d+a^(3/2)*arctan(a^(1/2)*sinh(d*x+c)/ 
(a+b)^(1/2))/b^2/(a+b)^(1/2)/d+1/2*sech(d*x+c)*tanh(d*x+c)/b/d
                                                                                    
                                                                                    
 

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(213\) vs. \(2(86)=172\).

Time = 1.43 (sec) , antiderivative size = 213, normalized size of antiderivative = 2.48 \[ \int \frac {\text {sech}^5(c+d x)}{a+b \text {sech}^2(c+d x)} \, dx=\frac {\cosh (c) (a+2 b+a \cosh (2 (c+d x))) \text {sech}^2(c+d x) \left (b \sqrt {a+b} \text {sech}^2(c) \text {sech}^2(c+d x) \sqrt {(\cosh (c)-\sinh (c))^2} \sinh (d x)+2 a^{3/2} \arctan \left (\frac {\sqrt {a+b} \text {csch}(c+d x) \sqrt {(\cosh (c)-\sinh (c))^2} (\cosh (c)+\sinh (c))}{\sqrt {a}}\right ) (-1+\tanh (c))-\sqrt {a+b} \text {sech}(c) \sqrt {(\cosh (c)-\sinh (c))^2} \left (2 (2 a-b) \arctan \left (\tanh \left (\frac {1}{2} (c+d x)\right )\right )-b \text {sech}(c+d x) \tanh (c)\right )\right )}{4 b^2 \sqrt {a+b} d \left (a+b \text {sech}^2(c+d x)\right ) \sqrt {(\cosh (c)-\sinh (c))^2}} \] Input:

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

Output:

(Cosh[c]*(a + 2*b + a*Cosh[2*(c + d*x)])*Sech[c + d*x]^2*(b*Sqrt[a + b]*Se 
ch[c]^2*Sech[c + d*x]^2*Sqrt[(Cosh[c] - Sinh[c])^2]*Sinh[d*x] + 2*a^(3/2)* 
ArcTan[(Sqrt[a + b]*Csch[c + d*x]*Sqrt[(Cosh[c] - Sinh[c])^2]*(Cosh[c] + S 
inh[c]))/Sqrt[a]]*(-1 + Tanh[c]) - Sqrt[a + b]*Sech[c]*Sqrt[(Cosh[c] - Sin 
h[c])^2]*(2*(2*a - b)*ArcTan[Tanh[(c + d*x)/2]] - b*Sech[c + d*x]*Tanh[c]) 
))/(4*b^2*Sqrt[a + b]*d*(a + b*Sech[c + d*x]^2)*Sqrt[(Cosh[c] - Sinh[c])^2 
])
 

Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.08, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3042, 4635, 316, 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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sec (i c+i d x)^5}{a+b \sec (i c+i d x)^2}dx\)

\(\Big \downarrow \) 4635

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

\(\Big \downarrow \) 316

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

\(\Big \downarrow \) 397

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

\(\Big \downarrow \) 216

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

\(\Big \downarrow \) 218

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

Input:

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

Output:

(-1/2*(((2*a - b)*ArcTan[Sinh[c + d*x]])/b - (2*a^(3/2)*ArcTan[(Sqrt[a]*Si 
nh[c + d*x])/Sqrt[a + b]])/(b*Sqrt[a + b]))/b + Sinh[c + d*x]/(2*b*(1 + Si 
nh[c + d*x]^2)))/d
 

Defintions of rubi rules used

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 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4635
Int[sec[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_ 
))^(p_), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Simp[ff/f 
 Subst[Int[ExpandToSum[b + 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] && In 
tegerQ[(m - 1)/2] && IntegerQ[n/2] && IntegerQ[p]
 
Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(158\) vs. \(2(74)=148\).

Time = 1.06 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.85

method result size
derivativedivides \(\frac {-\frac {2 \left (\frac {\frac {b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{2}-\frac {b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{2}}{\left (1+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}+\frac {\left (2 a -b \right ) \arctan \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}\right )}{b^{2}}+\frac {2 a^{2} \left (\frac {\arctan \left (\frac {2 \sqrt {a +b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-2 \sqrt {b}}{2 \sqrt {a}}\right )}{2 \sqrt {a +b}\, \sqrt {a}}+\frac {\arctan \left (\frac {2 \sqrt {a +b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+2 \sqrt {b}}{2 \sqrt {a}}\right )}{2 \sqrt {a +b}\, \sqrt {a}}\right )}{b^{2}}}{d}\) \(159\)
default \(\frac {-\frac {2 \left (\frac {\frac {b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{2}-\frac {b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{2}}{\left (1+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}+\frac {\left (2 a -b \right ) \arctan \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}\right )}{b^{2}}+\frac {2 a^{2} \left (\frac {\arctan \left (\frac {2 \sqrt {a +b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-2 \sqrt {b}}{2 \sqrt {a}}\right )}{2 \sqrt {a +b}\, \sqrt {a}}+\frac {\arctan \left (\frac {2 \sqrt {a +b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+2 \sqrt {b}}{2 \sqrt {a}}\right )}{2 \sqrt {a +b}\, \sqrt {a}}\right )}{b^{2}}}{d}\) \(159\)
risch \(\frac {{\mathrm e}^{d x +c} \left ({\mathrm e}^{2 d x +2 c}-1\right )}{d b \left ({\mathrm e}^{2 d x +2 c}+1\right )^{2}}+\frac {i \ln \left ({\mathrm e}^{d x +c}-i\right ) a}{d \,b^{2}}-\frac {i \ln \left ({\mathrm e}^{d x +c}-i\right )}{2 d b}-\frac {i \ln \left ({\mathrm e}^{d x +c}+i\right ) a}{d \,b^{2}}+\frac {i \ln \left ({\mathrm e}^{d x +c}+i\right )}{2 d b}+\frac {\sqrt {-\left (a +b \right ) a}\, a \ln \left ({\mathrm e}^{2 d x +2 c}+\frac {2 \sqrt {-\left (a +b \right ) a}\, {\mathrm e}^{d x +c}}{a}-1\right )}{2 \left (a +b \right ) d \,b^{2}}-\frac {\sqrt {-\left (a +b \right ) a}\, a \ln \left ({\mathrm e}^{2 d x +2 c}-\frac {2 \sqrt {-\left (a +b \right ) a}\, {\mathrm e}^{d x +c}}{a}-1\right )}{2 \left (a +b \right ) d \,b^{2}}\) \(223\)

Input:

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

Output:

1/d*(-2/b^2*((1/2*b*tanh(1/2*d*x+1/2*c)^3-1/2*b*tanh(1/2*d*x+1/2*c))/(1+ta 
nh(1/2*d*x+1/2*c)^2)^2+1/2*(2*a-b)*arctan(tanh(1/2*d*x+1/2*c)))+2/b^2*a^2* 
(1/2/(a+b)^(1/2)/a^(1/2)*arctan(1/2*(2*(a+b)^(1/2)*tanh(1/2*d*x+1/2*c)-2*b 
^(1/2))/a^(1/2))+1/2/(a+b)^(1/2)/a^(1/2)*arctan(1/2*(2*(a+b)^(1/2)*tanh(1/ 
2*d*x+1/2*c)+2*b^(1/2))/a^(1/2))))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 706 vs. \(2 (74) = 148\).

Time = 0.26 (sec) , antiderivative size = 1518, normalized size of antiderivative = 17.65 \[ \int \frac {\text {sech}^5(c+d x)}{a+b \text {sech}^2(c+d x)} \, dx=\text {Too large to display} \] Input:

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

Output:

[1/2*(2*b*cosh(d*x + c)^3 + 6*b*cosh(d*x + c)*sinh(d*x + c)^2 + 2*b*sinh(d 
*x + c)^3 + (a*cosh(d*x + c)^4 + 4*a*cosh(d*x + c)*sinh(d*x + c)^3 + a*sin 
h(d*x + c)^4 + 2*a*cosh(d*x + c)^2 + 2*(3*a*cosh(d*x + c)^2 + a)*sinh(d*x 
+ c)^2 + 4*(a*cosh(d*x + c)^3 + a*cosh(d*x + c))*sinh(d*x + c) + a)*sqrt(- 
a/(a + b))*log((a*cosh(d*x + c)^4 + 4*a*cosh(d*x + c)*sinh(d*x + c)^3 + a* 
sinh(d*x + c)^4 - 2*(3*a + 2*b)*cosh(d*x + c)^2 + 2*(3*a*cosh(d*x + c)^2 - 
 3*a - 2*b)*sinh(d*x + c)^2 + 4*(a*cosh(d*x + c)^3 - (3*a + 2*b)*cosh(d*x 
+ c))*sinh(d*x + c) + 4*((a + b)*cosh(d*x + c)^3 + 3*(a + b)*cosh(d*x + c) 
*sinh(d*x + c)^2 + (a + b)*sinh(d*x + c)^3 - (a + b)*cosh(d*x + c) + (3*(a 
 + b)*cosh(d*x + c)^2 - a - b)*sinh(d*x + c))*sqrt(-a/(a + b)) + a)/(a*cos 
h(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)) - 
 2*((2*a - b)*cosh(d*x + c)^4 + 4*(2*a - b)*cosh(d*x + c)*sinh(d*x + c)^3 
+ (2*a - b)*sinh(d*x + c)^4 + 2*(2*a - b)*cosh(d*x + c)^2 + 2*(3*(2*a - b) 
*cosh(d*x + c)^2 + 2*a - b)*sinh(d*x + c)^2 + 4*((2*a - b)*cosh(d*x + c)^3 
 + (2*a - b)*cosh(d*x + c))*sinh(d*x + c) + 2*a - b)*arctan(cosh(d*x + c) 
+ sinh(d*x + c)) - 2*b*cosh(d*x + c) + 2*(3*b*cosh(d*x + c)^2 - b)*sinh(d* 
x + c))/(b^2*d*cosh(d*x + c)^4 + 4*b^2*d*cosh(d*x + c)*sinh(d*x + c)^3 + b 
^2*d*sinh(d*x + c)^4 + 2*b^2*d*cosh(d*x + c)^2 + b^2*d + 2*(3*b^2*d*cos...
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

(e^(3*d*x + 3*c) - e^(d*x + c))/(b*d*e^(4*d*x + 4*c) + 2*b*d*e^(2*d*x + 2* 
c) + b*d) - (2*a*e^c - b*e^c)*arctan(e^(d*x + c))*e^(-c)/(b^2*d) + 32*inte 
grate(1/16*(a^2*e^(3*d*x + 3*c) + a^2*e^(d*x + c))/(a*b^2*e^(4*d*x + 4*c) 
+ a*b^2 + 2*(a*b^2*e^(2*c) + 2*b^3*e^(2*c))*e^(2*d*x)), x)
 

Giac [F(-2)]

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

integrate(sech(d*x+c)^5/(a+b*sech(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 = 4.43 (sec) , antiderivative size = 946, normalized size of antiderivative = 11.00 \[ \int \frac {\text {sech}^5(c+d x)}{a+b \text {sech}^2(c+d x)} \, dx =\text {Too large to display} \] Input:

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

Output:

((a^3)^(1/2)*(2*atan((exp(d*x)*exp(c)*((64*(6*b^3*d*(a^3)^(3/2) + 2*b^6*d* 
(a^3)^(1/2) + 6*a*b^2*d*(a^3)^(3/2) - 4*a*b^5*d*(a^3)^(1/2) - 6*a^2*b^4*d* 
(a^3)^(1/2)))/(a^4*b^4*(a + b)*(a*b + b^2)*(b^5*d^2 + a*b^4*d^2)^(1/2)*(b^ 
4*d^2*(a + b))^(1/2)*(3*a^3 - 3*a*b^2 + b^3)) - (32*(3*a^5*(b^5*d^2 + a*b^ 
4*d^2)^(1/2) + a^2*b^3*(b^5*d^2 + a*b^4*d^2)^(1/2) - 3*a^3*b^2*(b^5*d^2 + 
a*b^4*d^2)^(1/2)))/(a^2*b^6*d*(a + b)^2*(a*b + b^2)*(b^5*d^2 + a*b^4*d^2)^ 
(1/2)*(a^3)^(1/2)*(3*a^3 - 3*a*b^2 + b^3))) + (32*exp(3*c)*exp(3*d*x)*(3*a 
^5*(b^5*d^2 + a*b^4*d^2)^(1/2) + a^2*b^3*(b^5*d^2 + a*b^4*d^2)^(1/2) - 3*a 
^3*b^2*(b^5*d^2 + a*b^4*d^2)^(1/2)))/(a^2*b^6*d*(a + b)^2*(a*b + b^2)*(b^5 
*d^2 + a*b^4*d^2)^(1/2)*(a^3)^(1/2)*(3*a^3 - 3*a*b^2 + b^3)))*((a^2*b^7*(b 
^5*d^2 + a*b^4*d^2)^(1/2))/64 + (a^3*b^6*(b^5*d^2 + a*b^4*d^2)^(1/2))/32 + 
 (a^4*b^5*(b^5*d^2 + a*b^4*d^2)^(1/2))/64)) + 2*atan((a^2*exp(d*x)*exp(c)* 
(b^4*d^2*(a + b))^(1/2))/(2*b^2*d*(a + b)*(a^3)^(1/2)))))/(2*(b^5*d^2 + a* 
b^4*d^2)^(1/2)) - (atan((exp(d*x)*exp(c)*(18*a^7*(b^4*d^2)^(1/2) - b^7*(b^ 
4*d^2)^(1/2) - 21*a^2*b^5*(b^4*d^2)^(1/2) + 12*a^3*b^4*(b^4*d^2)^(1/2) + 3 
0*a^4*b^3*(b^4*d^2)^(1/2) - 36*a^5*b^2*(b^4*d^2)^(1/2) + 8*a*b^6*(b^4*d^2) 
^(1/2) - 9*a^6*b*(b^4*d^2)^(1/2)))/(b^8*d*(4*a^2 - 4*a*b + b^2)^(1/2) + 9* 
a^2*b^6*d*(4*a^2 - 4*a*b + b^2)^(1/2) + 6*a^3*b^5*d*(4*a^2 - 4*a*b + b^2)^ 
(1/2) - 18*a^4*b^4*d*(4*a^2 - 4*a*b + b^2)^(1/2) + 9*a^6*b^2*d*(4*a^2 - 4* 
a*b + b^2)^(1/2) - 6*a*b^7*d*(4*a^2 - 4*a*b + b^2)^(1/2)))*(4*a^2 - 4*a...
 

Reduce [B] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 1850, normalized size of antiderivative = 21.51 \[ \int \frac {\text {sech}^5(c+d x)}{a+b \text {sech}^2(c+d x)} \, dx =\text {Too large to display} \] Input:

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

Output:

( - 4*e**(4*c + 4*d*x)*atan(e**(c + d*x))*a**2 - 2*e**(4*c + 4*d*x)*atan(e 
**(c + d*x))*a*b + 2*e**(4*c + 4*d*x)*atan(e**(c + d*x))*b**2 - 8*e**(2*c 
+ 2*d*x)*atan(e**(c + d*x))*a**2 - 4*e**(2*c + 2*d*x)*atan(e**(c + d*x))*a 
*b + 4*e**(2*c + 2*d*x)*atan(e**(c + d*x))*b**2 - 4*atan(e**(c + d*x))*a** 
2 - 2*atan(e**(c + d*x))*a*b + 2*atan(e**(c + d*x))*b**2 - 2*e**(4*c + 4*d 
*x)*sqrt(b)*sqrt(a)*sqrt(a + b)*sqrt(2*sqrt(b)*sqrt(a + b) + a + 2*b)*atan 
((e**(c + d*x)*a)/(sqrt(a)*sqrt(2*sqrt(b)*sqrt(a + b) + a + 2*b))) - 4*e** 
(2*c + 2*d*x)*sqrt(b)*sqrt(a)*sqrt(a + b)*sqrt(2*sqrt(b)*sqrt(a + b) + a + 
 2*b)*atan((e**(c + d*x)*a)/(sqrt(a)*sqrt(2*sqrt(b)*sqrt(a + b) + a + 2*b) 
)) - 2*sqrt(b)*sqrt(a)*sqrt(a + b)*sqrt(2*sqrt(b)*sqrt(a + b) + a + 2*b)*a 
tan((e**(c + d*x)*a)/(sqrt(a)*sqrt(2*sqrt(b)*sqrt(a + b) + a + 2*b))) + 2* 
e**(4*c + 4*d*x)*sqrt(a)*sqrt(2*sqrt(b)*sqrt(a + b) + a + 2*b)*atan((e**(c 
 + d*x)*a)/(sqrt(a)*sqrt(2*sqrt(b)*sqrt(a + b) + a + 2*b)))*a + 2*e**(4*c 
+ 4*d*x)*sqrt(a)*sqrt(2*sqrt(b)*sqrt(a + b) + a + 2*b)*atan((e**(c + d*x)* 
a)/(sqrt(a)*sqrt(2*sqrt(b)*sqrt(a + b) + a + 2*b)))*b + 4*e**(2*c + 2*d*x) 
*sqrt(a)*sqrt(2*sqrt(b)*sqrt(a + b) + a + 2*b)*atan((e**(c + d*x)*a)/(sqrt 
(a)*sqrt(2*sqrt(b)*sqrt(a + b) + a + 2*b)))*a + 4*e**(2*c + 2*d*x)*sqrt(a) 
*sqrt(2*sqrt(b)*sqrt(a + b) + a + 2*b)*atan((e**(c + d*x)*a)/(sqrt(a)*sqrt 
(2*sqrt(b)*sqrt(a + b) + a + 2*b)))*b + 2*sqrt(a)*sqrt(2*sqrt(b)*sqrt(a + 
b) + a + 2*b)*atan((e**(c + d*x)*a)/(sqrt(a)*sqrt(2*sqrt(b)*sqrt(a + b)...