\(\int \frac {\text {csch}^2(c+d x)}{(a+b \tanh ^2(c+d x))^2} \, dx\) [38]

Optimal result

Integrand size = 23, antiderivative size = 81 \[ \int \frac {\text {csch}^2(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^2} \, dx=-\frac {3 \sqrt {b} \arctan \left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a}}\right )}{2 a^{5/2} d}-\frac {\coth (c+d x)}{a^2 d}-\frac {b \tanh (c+d x)}{2 a^2 d \left (a+b \tanh ^2(c+d x)\right )} \] Output:

-3/2*b^(1/2)*arctan(b^(1/2)*tanh(d*x+c)/a^(1/2))/a^(5/2)/d-coth(d*x+c)/a^2 
/d-1/2*b*tanh(d*x+c)/a^2/d/(a+b*tanh(d*x+c)^2)
 

Mathematica [A] (verified)

Time = 0.70 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.06 \[ \int \frac {\text {csch}^2(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^2} \, dx=\frac {-3 \sqrt {b} \arctan \left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a}}\right )-2 \sqrt {a} \coth (c+d x)-\frac {\sqrt {a} b \sinh (2 (c+d x))}{a-b+(a+b) \cosh (2 (c+d x))}}{2 a^{5/2} d} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3042, 25, 4146, 253, 264, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 4146

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

\(\Big \downarrow \) 253

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

\(\Big \downarrow \) 264

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

\(\Big \downarrow \) 218

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

Input:

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

Output:

((3*(-((Sqrt[b]*ArcTan[(Sqrt[b]*Tanh[c + d*x])/Sqrt[a]])/a^(3/2)) - Coth[c 
 + d*x]/a))/(2*a) + Coth[c + d*x]/(2*a*(a + 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 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 253

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-(c*x 
)^(m + 1))*((a + b*x^2)^(p + 1)/(2*a*c*(p + 1))), x] + Simp[(m + 2*p + 3)/( 
2*a*(p + 1))   Int[(c*x)^m*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, m 
}, x] && LtQ[p, -1] && IntBinomialQ[a, b, c, 2, m, p, x]
 

rule 264

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^( 
m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + 2*p + 3)/(a*c 
^2*(m + 1)))   Int[(c*x)^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, p 
}, x] && LtQ[m, -1] && IntBinomialQ[a, b, c, 2, m, p, x]
 

rule 3042

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

rule 4146

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(251\) vs. \(2(69)=138\).

Time = 3.36 (sec) , antiderivative size = 252, normalized size of antiderivative = 3.11

method	result	size
risch	\(-\frac {2 \,{\mathrm e}^{4 d x +4 c} a^{2}+3 \,{\mathrm e}^{4 d x +4 c} a b +3 \,{\mathrm e}^{4 d x +4 c} b^{2}+4 \,{\mathrm e}^{2 d x +2 c} a^{2}-6 b^{2} {\mathrm e}^{2 d x +2 c}+2 a^{2}+5 a b +3 b^{2}}{d \,a^{2} \left ({\mathrm e}^{2 d x +2 c}-1\right ) \left (a +b \right ) \left ({\mathrm e}^{4 d x +4 c} a +b \,{\mathrm e}^{4 d x +4 c}+2 \,{\mathrm e}^{2 d x +2 c} a -2 \,{\mathrm e}^{2 d x +2 c} b +a +b \right )}+\frac {3 \sqrt {-a b}\, \ln \left ({\mathrm e}^{2 d x +2 c}-\frac {2 \sqrt {-a b}-a +b}{a +b}\right )}{4 a^{3} d}-\frac {3 \sqrt {-a b}\, \ln \left ({\mathrm e}^{2 d x +2 c}+\frac {2 \sqrt {-a b}+a -b}{a +b}\right )}{4 a^{3} d}\)	\(252\)
derivativedivides	\(\frac {-\frac {\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a^{2}}+\frac {4 b \left (\frac {-\frac {\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{4}-\frac {\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{4}}{\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a +2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +4 b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+a}+\frac {3 a \left (\frac {\left (a +\sqrt {\left (a +b \right ) b}+b \right ) \arctan \left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {\left (a +b \right ) b}+a +2 b \right ) a}}\right )}{2 a \sqrt {\left (a +b \right ) b}\, \sqrt {\left (2 \sqrt {\left (a +b \right ) b}+a +2 b \right ) a}}-\frac {\left (-a +\sqrt {\left (a +b \right ) b}-b \right ) \operatorname {arctanh}\left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {\left (a +b \right ) b}-a -2 b \right ) a}}\right )}{2 a \sqrt {\left (a +b \right ) b}\, \sqrt {\left (2 \sqrt {\left (a +b \right ) b}-a -2 b \right ) a}}\right )}{4}\right )}{a^{2}}-\frac {1}{2 a^{2} \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}}{d}\)	\(266\)
default	\(\frac {-\frac {\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a^{2}}+\frac {4 b \left (\frac {-\frac {\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{4}-\frac {\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{4}}{\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a +2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +4 b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+a}+\frac {3 a \left (\frac {\left (a +\sqrt {\left (a +b \right ) b}+b \right ) \arctan \left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {\left (a +b \right ) b}+a +2 b \right ) a}}\right )}{2 a \sqrt {\left (a +b \right ) b}\, \sqrt {\left (2 \sqrt {\left (a +b \right ) b}+a +2 b \right ) a}}-\frac {\left (-a +\sqrt {\left (a +b \right ) b}-b \right ) \operatorname {arctanh}\left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {\left (a +b \right ) b}-a -2 b \right ) a}}\right )}{2 a \sqrt {\left (a +b \right ) b}\, \sqrt {\left (2 \sqrt {\left (a +b \right ) b}-a -2 b \right ) a}}\right )}{4}\right )}{a^{2}}-\frac {1}{2 a^{2} \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}}{d}\)	\(266\)

Input:

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

Output:

-(2*exp(4*d*x+4*c)*a^2+3*exp(4*d*x+4*c)*a*b+3*exp(4*d*x+4*c)*b^2+4*exp(2*d 
*x+2*c)*a^2-6*b^2*exp(2*d*x+2*c)+2*a^2+5*a*b+3*b^2)/d/a^2/(exp(2*d*x+2*c)- 
1)/(a+b)/(exp(4*d*x+4*c)*a+b*exp(4*d*x+4*c)+2*exp(2*d*x+2*c)*a-2*exp(2*d*x 
+2*c)*b+a+b)+3/4/a^3*(-a*b)^(1/2)/d*ln(exp(2*d*x+2*c)-(2*(-a*b)^(1/2)-a+b) 
/(a+b))-3/4/a^3*(-a*b)^(1/2)/d*ln(exp(2*d*x+2*c)+(2*(-a*b)^(1/2)+a-b)/(a+b 
))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1120 vs. \(2 (69) = 138\).

Time = 0.14 (sec) , antiderivative size = 2562, normalized size of antiderivative = 31.63 \[ \int \frac {\text {csch}^2(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^2} \, dx=\text {Too large to display} \] Input:

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

Output:

[-1/4*(4*(2*a^2 + 3*a*b + 3*b^2)*cosh(d*x + c)^4 + 16*(2*a^2 + 3*a*b + 3*b 
^2)*cosh(d*x + c)*sinh(d*x + c)^3 + 4*(2*a^2 + 3*a*b + 3*b^2)*sinh(d*x + c 
)^4 + 8*(2*a^2 - 3*b^2)*cosh(d*x + c)^2 + 8*(3*(2*a^2 + 3*a*b + 3*b^2)*cos 
h(d*x + c)^2 + 2*a^2 - 3*b^2)*sinh(d*x + c)^2 - 3*((a^2 + 2*a*b + b^2)*cos 
h(d*x + c)^6 + 6*(a^2 + 2*a*b + b^2)*cosh(d*x + c)*sinh(d*x + c)^5 + (a^2 
+ 2*a*b + b^2)*sinh(d*x + c)^6 + (a^2 - 2*a*b - 3*b^2)*cosh(d*x + c)^4 + ( 
15*(a^2 + 2*a*b + b^2)*cosh(d*x + c)^2 + a^2 - 2*a*b - 3*b^2)*sinh(d*x + c 
)^4 + 4*(5*(a^2 + 2*a*b + b^2)*cosh(d*x + c)^3 + (a^2 - 2*a*b - 3*b^2)*cos 
h(d*x + c))*sinh(d*x + c)^3 - (a^2 - 2*a*b - 3*b^2)*cosh(d*x + c)^2 + (15* 
(a^2 + 2*a*b + b^2)*cosh(d*x + c)^4 + 6*(a^2 - 2*a*b - 3*b^2)*cosh(d*x + c 
)^2 - a^2 + 2*a*b + 3*b^2)*sinh(d*x + c)^2 - a^2 - 2*a*b - b^2 + 2*(3*(a^2 
 + 2*a*b + b^2)*cosh(d*x + c)^5 + 2*(a^2 - 2*a*b - 3*b^2)*cosh(d*x + c)^3 
- (a^2 - 2*a*b - 3*b^2)*cosh(d*x + c))*sinh(d*x + c))*sqrt(-b/a)*log(((a^2 
 + 2*a*b + b^2)*cosh(d*x + c)^4 + 4*(a^2 + 2*a*b + b^2)*cosh(d*x + c)*sinh 
(d*x + c)^3 + (a^2 + 2*a*b + b^2)*sinh(d*x + c)^4 + 2*(a^2 - b^2)*cosh(d*x 
 + c)^2 + 2*(3*(a^2 + 2*a*b + b^2)*cosh(d*x + c)^2 + a^2 - b^2)*sinh(d*x + 
 c)^2 + a^2 - 6*a*b + b^2 + 4*((a^2 + 2*a*b + b^2)*cosh(d*x + c)^3 + (a^2 
- b^2)*cosh(d*x + c))*sinh(d*x + c) - 4*((a^2 + a*b)*cosh(d*x + c)^2 + 2*( 
a^2 + a*b)*cosh(d*x + c)*sinh(d*x + c) + (a^2 + a*b)*sinh(d*x + c)^2 + a^2 
 - a*b)*sqrt(-b/a))/((a + b)*cosh(d*x + c)^4 + 4*(a + b)*cosh(d*x + c)*...
 

Sympy [F]

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

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

Output:

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 212 vs. \(2 (69) = 138\).

Time = 0.19 (sec) , antiderivative size = 212, normalized size of antiderivative = 2.62 \[ \int \frac {\text {csch}^2(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^2} \, dx=-\frac {2 \, a^{2} + 5 \, a b + 3 \, b^{2} + 2 \, {\left (2 \, a^{2} - 3 \, b^{2}\right )} e^{\left (-2 \, d x - 2 \, c\right )} + {\left (2 \, a^{2} + 3 \, a b + 3 \, b^{2}\right )} e^{\left (-4 \, d x - 4 \, c\right )}}{{\left (a^{4} + 2 \, a^{3} b + a^{2} b^{2} + {\left (a^{4} - 2 \, a^{3} b - 3 \, a^{2} b^{2}\right )} e^{\left (-2 \, d x - 2 \, c\right )} - {\left (a^{4} - 2 \, a^{3} b - 3 \, a^{2} b^{2}\right )} e^{\left (-4 \, d x - 4 \, c\right )} - {\left (a^{4} + 2 \, a^{3} b + a^{2} b^{2}\right )} e^{\left (-6 \, d x - 6 \, c\right )}\right )} d} + \frac {3 \, b \arctan \left (\frac {{\left (a + b\right )} e^{\left (-2 \, d x - 2 \, c\right )} + a - b}{2 \, \sqrt {a b}}\right )}{2 \, \sqrt {a b} a^{2} d} \] Input:

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

Output:

-(2*a^2 + 5*a*b + 3*b^2 + 2*(2*a^2 - 3*b^2)*e^(-2*d*x - 2*c) + (2*a^2 + 3* 
a*b + 3*b^2)*e^(-4*d*x - 4*c))/((a^4 + 2*a^3*b + a^2*b^2 + (a^4 - 2*a^3*b 
- 3*a^2*b^2)*e^(-2*d*x - 2*c) - (a^4 - 2*a^3*b - 3*a^2*b^2)*e^(-4*d*x - 4* 
c) - (a^4 + 2*a^3*b + a^2*b^2)*e^(-6*d*x - 6*c))*d) + 3/2*b*arctan(1/2*((a 
 + b)*e^(-2*d*x - 2*c) + a - b)/sqrt(a*b))/(sqrt(a*b)*a^2*d)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 227 vs. \(2 (69) = 138\).

Time = 0.34 (sec) , antiderivative size = 227, normalized size of antiderivative = 2.80 \[ \int \frac {\text {csch}^2(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^2} \, dx=-\frac {\frac {3 \, b \arctan \left (\frac {a e^{\left (2 \, d x + 2 \, c\right )} + b e^{\left (2 \, d x + 2 \, c\right )} + a - b}{2 \, \sqrt {a b}}\right )}{\sqrt {a b} a^{2}} + \frac {2 \, {\left (2 \, a^{2} e^{\left (4 \, d x + 4 \, c\right )} + 3 \, a b e^{\left (4 \, d x + 4 \, c\right )} + 3 \, b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 4 \, a^{2} e^{\left (2 \, d x + 2 \, c\right )} - 6 \, b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 2 \, a^{2} + 5 \, a b + 3 \, b^{2}\right )}}{{\left (a^{3} + a^{2} b\right )} {\left (a e^{\left (6 \, d x + 6 \, c\right )} + b e^{\left (6 \, d x + 6 \, c\right )} + a e^{\left (4 \, d x + 4 \, c\right )} - 3 \, b e^{\left (4 \, d x + 4 \, c\right )} - a e^{\left (2 \, d x + 2 \, c\right )} + 3 \, b e^{\left (2 \, d x + 2 \, c\right )} - a - b\right )}}}{2 \, d} \] Input:

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

Output:

-1/2*(3*b*arctan(1/2*(a*e^(2*d*x + 2*c) + b*e^(2*d*x + 2*c) + a - b)/sqrt( 
a*b))/(sqrt(a*b)*a^2) + 2*(2*a^2*e^(4*d*x + 4*c) + 3*a*b*e^(4*d*x + 4*c) + 
 3*b^2*e^(4*d*x + 4*c) + 4*a^2*e^(2*d*x + 2*c) - 6*b^2*e^(2*d*x + 2*c) + 2 
*a^2 + 5*a*b + 3*b^2)/((a^3 + a^2*b)*(a*e^(6*d*x + 6*c) + b*e^(6*d*x + 6*c 
) + a*e^(4*d*x + 4*c) - 3*b*e^(4*d*x + 4*c) - a*e^(2*d*x + 2*c) + 3*b*e^(2 
*d*x + 2*c) - a - b)))/d
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.31 (sec) , antiderivative size = 1177, normalized size of antiderivative = 14.53 \[ \int \frac {\text {csch}^2(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^2} \, dx =\text {Too large to display} \] Input:

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

Output:

( - 3*e**(6*c + 6*d*x)*sqrt(b)*sqrt(a)*atan((e**(c + d*x)*sqrt(a + b) - sq 
rt(b))/sqrt(a))*a**2 + 6*e**(6*c + 6*d*x)*sqrt(b)*sqrt(a)*atan((e**(c + d* 
x)*sqrt(a + b) - sqrt(b))/sqrt(a))*a*b + 9*e**(6*c + 6*d*x)*sqrt(b)*sqrt(a 
)*atan((e**(c + d*x)*sqrt(a + b) - sqrt(b))/sqrt(a))*b**2 - 3*e**(4*c + 4* 
d*x)*sqrt(b)*sqrt(a)*atan((e**(c + d*x)*sqrt(a + b) - sqrt(b))/sqrt(a))*a* 
*2 + 18*e**(4*c + 4*d*x)*sqrt(b)*sqrt(a)*atan((e**(c + d*x)*sqrt(a + b) - 
sqrt(b))/sqrt(a))*a*b - 27*e**(4*c + 4*d*x)*sqrt(b)*sqrt(a)*atan((e**(c + 
d*x)*sqrt(a + b) - sqrt(b))/sqrt(a))*b**2 + 3*e**(2*c + 2*d*x)*sqrt(b)*sqr 
t(a)*atan((e**(c + d*x)*sqrt(a + b) - sqrt(b))/sqrt(a))*a**2 - 18*e**(2*c 
+ 2*d*x)*sqrt(b)*sqrt(a)*atan((e**(c + d*x)*sqrt(a + b) - sqrt(b))/sqrt(a) 
)*a*b + 27*e**(2*c + 2*d*x)*sqrt(b)*sqrt(a)*atan((e**(c + d*x)*sqrt(a + b) 
 - sqrt(b))/sqrt(a))*b**2 + 3*sqrt(b)*sqrt(a)*atan((e**(c + d*x)*sqrt(a + 
b) - sqrt(b))/sqrt(a))*a**2 - 6*sqrt(b)*sqrt(a)*atan((e**(c + d*x)*sqrt(a 
+ b) - sqrt(b))/sqrt(a))*a*b - 9*sqrt(b)*sqrt(a)*atan((e**(c + d*x)*sqrt(a 
 + b) - sqrt(b))/sqrt(a))*b**2 + 3*e**(6*c + 6*d*x)*sqrt(b)*sqrt(a)*atan(( 
e**(c + d*x)*sqrt(a + b) + sqrt(b))/sqrt(a))*a**2 - 6*e**(6*c + 6*d*x)*sqr 
t(b)*sqrt(a)*atan((e**(c + d*x)*sqrt(a + b) + sqrt(b))/sqrt(a))*a*b - 9*e* 
*(6*c + 6*d*x)*sqrt(b)*sqrt(a)*atan((e**(c + d*x)*sqrt(a + b) + sqrt(b))/s 
qrt(a))*b**2 + 3*e**(4*c + 4*d*x)*sqrt(b)*sqrt(a)*atan((e**(c + d*x)*sqrt( 
a + b) + sqrt(b))/sqrt(a))*a**2 - 18*e**(4*c + 4*d*x)*sqrt(b)*sqrt(a)*a...