\(\int \frac {\sinh ^3(c+d x)}{(a+b \text {sech}^2(c+d x))^2} \, dx\) [34]

Optimal result

Integrand size = 23, antiderivative size = 114 \[ \int \frac {\sinh ^3(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^2} \, dx=\frac {\sqrt {b} (3 a+5 b) \arctan \left (\frac {\sqrt {a} \cosh (c+d x)}{\sqrt {b}}\right )}{2 a^{7/2} d}-\frac {(a+2 b) \cosh (c+d x)}{a^3 d}+\frac {\cosh ^3(c+d x)}{3 a^2 d}-\frac {b (a+b) \cosh (c+d x)}{2 a^3 d \left (b+a \cosh ^2(c+d x)\right )} \] Output:

1/2*b^(1/2)*(3*a+5*b)*arctan(a^(1/2)*cosh(d*x+c)/b^(1/2))/a^(7/2)/d-(a+2*b 
)*cosh(d*x+c)/a^3/d+1/3*cosh(d*x+c)^3/a^2/d-1/2*b*(a+b)*cosh(d*x+c)/a^3/d/ 
(b+a*cosh(d*x+c)^2)
                                                                                    
                                                                                    
 

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 3.90 (sec) , antiderivative size = 861, normalized size of antiderivative = 7.55 \[ \int \frac {\sinh ^3(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^2} \, dx =\text {Too large to display} \] Input:

Integrate[Sinh[c + d*x]^3/(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*((9*a^3*ArcTan[((Sqrt[a 
] - I*Sqrt[a + b]*Sqrt[(Cosh[c] - Sinh[c])^2])*Sinh[c]*Tanh[(d*x)/2] + Cos 
h[c]*(Sqrt[a] - I*Sqrt[a + b]*Sqrt[(Cosh[c] - Sinh[c])^2]*Tanh[(d*x)/2]))/ 
Sqrt[b]])/b^(3/2) + 576*a*Sqrt[b]*ArcTan[((Sqrt[a] - I*Sqrt[a + b]*Sqrt[(C 
osh[c] - Sinh[c])^2])*Sinh[c]*Tanh[(d*x)/2] + Cosh[c]*(Sqrt[a] - I*Sqrt[a 
+ b]*Sqrt[(Cosh[c] - Sinh[c])^2]*Tanh[(d*x)/2]))/Sqrt[b]] + 960*b^(3/2)*Ar 
cTan[((Sqrt[a] - I*Sqrt[a + b]*Sqrt[(Cosh[c] - Sinh[c])^2])*Sinh[c]*Tanh[( 
d*x)/2] + Cosh[c]*(Sqrt[a] - I*Sqrt[a + b]*Sqrt[(Cosh[c] - Sinh[c])^2]*Tan 
h[(d*x)/2]))/Sqrt[b]] + (9*a^3*ArcTan[((Sqrt[a] + I*Sqrt[a + b]*Sqrt[(Cosh 
[c] - Sinh[c])^2])*Sinh[c]*Tanh[(d*x)/2] + Cosh[c]*(Sqrt[a] + I*Sqrt[a + b 
]*Sqrt[(Cosh[c] - Sinh[c])^2]*Tanh[(d*x)/2]))/Sqrt[b]])/b^(3/2) + 576*a*Sq 
rt[b]*ArcTan[((Sqrt[a] + I*Sqrt[a + b]*Sqrt[(Cosh[c] - Sinh[c])^2])*Sinh[c 
]*Tanh[(d*x)/2] + Cosh[c]*(Sqrt[a] + I*Sqrt[a + b]*Sqrt[(Cosh[c] - Sinh[c] 
)^2]*Tanh[(d*x)/2]))/Sqrt[b]] + 960*b^(3/2)*ArcTan[((Sqrt[a] + I*Sqrt[a + 
b]*Sqrt[(Cosh[c] - Sinh[c])^2])*Sinh[c]*Tanh[(d*x)/2] + Cosh[c]*(Sqrt[a] + 
 I*Sqrt[a + b]*Sqrt[(Cosh[c] - Sinh[c])^2]*Tanh[(d*x)/2]))/Sqrt[b]] - (9*a 
^3*ArcTan[(Sqrt[a] - I*Sqrt[a + b]*Tanh[(c + d*x)/2])/Sqrt[b]])/b^(3/2) - 
(9*a^3*ArcTan[(Sqrt[a] + I*Sqrt[a + b]*Tanh[(c + d*x)/2])/Sqrt[b]])/b^(3/2 
) - 96*Sqrt[a]*(3*a + 8*b)*Cosh[c]*Cosh[d*x] + 32*a^(3/2)*Cosh[3*c]*Cosh[3 
*d*x] - (384*a^(3/2)*b*Cosh[c + d*x])/(a + 2*b + a*Cosh[2*(c + d*x)]) -...
 

Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.94, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3042, 26, 4621, 360, 1467, 2009}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 26

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

\(\Big \downarrow \) 4621

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

\(\Big \downarrow \) 360

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

\(\Big \downarrow \) 1467

\(\displaystyle -\frac {\frac {b (a+b) \cosh (c+d x)}{2 a^3 \left (a \cosh ^2(c+d x)+b\right )}-\frac {\int \left (2 a \cosh ^2(c+d x)-2 (a+2 b)+\frac {5 b^2+3 a b}{a \cosh ^2(c+d x)+b}\right )d\cosh (c+d x)}{2 a^3}}{d}\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {\frac {b (a+b) \cosh (c+d x)}{2 a^3 \left (a \cosh ^2(c+d x)+b\right )}-\frac {\frac {\sqrt {b} (3 a+5 b) \arctan \left (\frac {\sqrt {a} \cosh (c+d x)}{\sqrt {b}}\right )}{\sqrt {a}}-2 (a+2 b) \cosh (c+d x)+\frac {2}{3} a \cosh ^3(c+d x)}{2 a^3}}{d}\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 26

Int[(Complex[0, a_])*(Fx_), x_Symbol] :> Simp[(Complex[Identity[0], a])   I 
nt[Fx, x], x] /; FreeQ[a, x] && EqQ[a^2, 1]
 

rule 360

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] : 
> Simp[(-a)^(m/2 - 1)*(b*c - a*d)*x*((a + b*x^2)^(p + 1)/(2*b^(m/2 + 1)*(p 
+ 1))), x] + Simp[1/(2*b^(m/2 + 1)*(p + 1))   Int[(a + b*x^2)^(p + 1)*Expan 
dToSum[2*b*(p + 1)*x^2*Together[(b^(m/2)*x^(m - 2)*(c + d*x^2) - (-a)^(m/2 
- 1)*(b*c - a*d))/(a + b*x^2)] - (-a)^(m/2 - 1)*(b*c - a*d), x], x], x] /; 
FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && IGtQ[m/2, 0] & 
& (IntegerQ[p] || EqQ[m + 2*p + 1, 0])
 

rule 1467

Int[((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), 
 x_Symbol] :> Int[ExpandIntegrand[(d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], 
x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e 
 + a*e^2, 0] && IGtQ[p, 0] && IGtQ[q, -2]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

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

rule 4621

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(263\) vs. \(2(100)=200\).

Time = 307.18 (sec) , antiderivative size = 264, normalized size of antiderivative = 2.32

method	result	size
derivativedivides	\(\frac {\frac {1}{3 a^{2} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {1}{2 a^{2} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {a +4 b}{2 a^{3} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {1}{3 a^{2} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {1}{2 a^{2} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {-a -4 b}{2 a^{3} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {4 b \left (\frac {\left (-\frac {a}{4}+\frac {b}{4}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-\frac {a}{4}-\frac {b}{4}}{\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a +\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} b +2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b}+\frac {\left (3 a +5 b \right ) \arctan \left (\frac {2 \left (a +b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+2 a -2 b}{4 \sqrt {a b}}\right )}{8 \sqrt {a b}}\right )}{a^{3}}}{d}\)	\(264\)
default	\(\frac {\frac {1}{3 a^{2} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {1}{2 a^{2} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {a +4 b}{2 a^{3} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {1}{3 a^{2} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {1}{2 a^{2} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {-a -4 b}{2 a^{3} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {4 b \left (\frac {\left (-\frac {a}{4}+\frac {b}{4}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-\frac {a}{4}-\frac {b}{4}}{\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a +\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} b +2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b}+\frac {\left (3 a +5 b \right ) \arctan \left (\frac {2 \left (a +b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+2 a -2 b}{4 \sqrt {a b}}\right )}{8 \sqrt {a b}}\right )}{a^{3}}}{d}\)	\(264\)
risch	\(\frac {{\mathrm e}^{3 d x +3 c}}{24 a^{2} d}-\frac {3 \,{\mathrm e}^{d x +c}}{8 a^{2} d}-\frac {{\mathrm e}^{d x +c} b}{a^{3} d}-\frac {3 \,{\mathrm e}^{-d x -c}}{8 a^{2} d}-\frac {{\mathrm e}^{-d x -c} b}{a^{3} d}+\frac {{\mathrm e}^{-3 d x -3 c}}{24 a^{2} d}-\frac {b \left (a +b \right ) {\mathrm e}^{d x +c} \left ({\mathrm e}^{2 d x +2 c}+1\right )}{a^{3} d \left ({\mathrm e}^{4 d x +4 c} a +2 a \,{\mathrm e}^{2 d x +2 c}+4 b \,{\mathrm e}^{2 d x +2 c}+a \right )}+\frac {3 \sqrt {-a b}\, \ln \left ({\mathrm e}^{2 d x +2 c}+\frac {2 \sqrt {-a b}\, {\mathrm e}^{d x +c}}{a}+1\right )}{4 a^{3} d}+\frac {5 \sqrt {-a b}\, \ln \left ({\mathrm e}^{2 d x +2 c}+\frac {2 \sqrt {-a b}\, {\mathrm e}^{d x +c}}{a}+1\right ) b}{4 a^{4} d}-\frac {3 \sqrt {-a b}\, \ln \left ({\mathrm e}^{2 d x +2 c}-\frac {2 \sqrt {-a b}\, {\mathrm e}^{d x +c}}{a}+1\right )}{4 a^{3} d}-\frac {5 \sqrt {-a b}\, \ln \left ({\mathrm e}^{2 d x +2 c}-\frac {2 \sqrt {-a b}\, {\mathrm e}^{d x +c}}{a}+1\right ) b}{4 a^{4} d}\)	\(342\)

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2021 vs. \(2 (100) = 200\).

Time = 0.32 (sec) , antiderivative size = 3804, normalized size of antiderivative = 33.37 \[ \int \frac {\sinh ^3(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^2} \, dx=\text {Too large to display} \] Input:

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

Output:

Too large to include
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

1/24*(a^2*e^(10*d*x + 10*c) + a^2 - (7*a^2*e^(8*c) + 20*a*b*e^(8*c))*e^(8* 
d*x) - 2*(13*a^2*e^(6*c) + 66*a*b*e^(6*c) + 60*b^2*e^(6*c))*e^(6*d*x) - 2* 
(13*a^2*e^(4*c) + 66*a*b*e^(4*c) + 60*b^2*e^(4*c))*e^(4*d*x) - (7*a^2*e^(2 
*c) + 20*a*b*e^(2*c))*e^(2*d*x))/(a^4*d*e^(7*d*x + 7*c) + a^4*d*e^(3*d*x + 
 3*c) + 2*(a^4*d*e^(5*c) + 2*a^3*b*d*e^(5*c))*e^(5*d*x)) + 1/8*integrate(8 
*((3*a*b*e^(3*c) + 5*b^2*e^(3*c))*e^(3*d*x) - (3*a*b*e^c + 5*b^2*e^c)*e^(d 
*x))/(a^4*e^(4*d*x + 4*c) + a^4 + 2*(a^4*e^(2*c) + 2*a^3*b*e^(2*c))*e^(2*d 
*x)), x)
 

Giac [F(-2)]

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

integrate(sinh(d*x+c)^3/(a+b*sech(d*x+c)^2)^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 [F(-1)]

Timed out. \[ \int \frac {\sinh ^3(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^2} \, dx=\int \frac {{\mathrm {cosh}\left (c+d\,x\right )}^4\,{\mathrm {sinh}\left (c+d\,x\right )}^3}{{\left (a\,{\mathrm {cosh}\left (c+d\,x\right )}^2+b\right )}^2} \,d x \] Input:

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

Output:

int((cosh(c + d*x)^4*sinh(c + d*x)^3)/(b + a*cosh(c + d*x)^2)^2, x)
 

Reduce [B] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 2965, normalized size of antiderivative = 26.01 \[ \int \frac {\sinh ^3(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^2} \, dx =\text {Too large to display} \] Input:

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

Output:

(36*e**(7*c + 7*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)))*a**2 + 60*e**(7*c + 7*d*x)*sqrt(b)*sqrt(a)*sqrt(a + b)*sqrt(2*sq 
rt(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*b + 72*e**(5*c + 5*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)))*a**2 + 264*e**(5*c + 5*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)))*a*b + 240*e**(5* 
c + 5*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)))* 
b**2 + 36*e**(3*c + 3*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)))*a**2 + 60*e**(3*c + 3*d*x)*sqrt(b)*sqrt(a)*sqrt(a + b)*sqr 
t(2*sqrt(b)*sqrt(a + b) + a + 2*b)*atan((e**(c + d*x)*a)/(sqrt(a)*sqrt(2*s 
qrt(b)*sqrt(a + b) + a + 2*b)))*a*b - 36*e**(7*c + 7*d*x)*sqrt(a)*sqrt(2*s 
qrt(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*b - 60*e**(7*c + 7*d*x)*sqrt(a)*sqrt(2*sqr 
t(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*b**2 - 72*e**(5*c + 5*d*x)*sqrt(a)*sqrt(2*sq...