\(\int \frac {1}{(a+b \cosh (c+d x))^3} \, dx\) [69]

Optimal result

Integrand size = 12, antiderivative size = 133 \[ \int \frac {1}{(a+b \cosh (c+d x))^3} \, dx=\frac {\left (2 a^2+b^2\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{5/2} (a+b)^{5/2} d}-\frac {b \sinh (c+d x)}{2 \left (a^2-b^2\right ) d (a+b \cosh (c+d x))^2}-\frac {3 a b \sinh (c+d x)}{2 \left (a^2-b^2\right )^2 d (a+b \cosh (c+d x))} \] Output:

(2*a^2+b^2)*arctanh((a-b)^(1/2)*tanh(1/2*d*x+1/2*c)/(a+b)^(1/2))/(a-b)^(5/ 
2)/(a+b)^(5/2)/d-1/2*b*sinh(d*x+c)/(a^2-b^2)/d/(a+b*cosh(d*x+c))^2-3/2*a*b 
*sinh(d*x+c)/(a^2-b^2)^2/d/(a+b*cosh(d*x+c))
 

Mathematica [A] (verified)

Time = 0.31 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.85 \[ \int \frac {1}{(a+b \cosh (c+d x))^3} \, dx=\frac {-\frac {2 \left (2 a^2+b^2\right ) \arctan \left (\frac {(a-b) \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )}{\left (-a^2+b^2\right )^{5/2}}+\frac {b \left (-4 a^2+b^2-3 a b \cosh (c+d x)\right ) \sinh (c+d x)}{(a-b)^2 (a+b)^2 (a+b \cosh (c+d x))^2}}{2 d} \] Input:

Integrate[(a + b*Cosh[c + d*x])^(-3),x]
 

Output:

((-2*(2*a^2 + b^2)*ArcTan[((a - b)*Tanh[(c + d*x)/2])/Sqrt[-a^2 + b^2]])/( 
-a^2 + b^2)^(5/2) + (b*(-4*a^2 + b^2 - 3*a*b*Cosh[c + d*x])*Sinh[c + d*x]) 
/((a - b)^2*(a + b)^2*(a + b*Cosh[c + d*x])^2))/(2*d)
 

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.20, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.833, Rules used = {3042, 3143, 25, 3042, 3233, 25, 27, 3042, 3138, 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.

\(\displaystyle \int \frac {1}{(a+b \cosh (c+d x))^3} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3143

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3233

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {b \sinh (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \cosh (c+d x))^2}+\frac {-\frac {3 a b \sinh (c+d x)}{d \left (a^2-b^2\right ) (a+b \cosh (c+d x))}+\frac {\left (2 a^2+b^2\right ) \int \frac {1}{a+b \sin \left (i c+i d x+\frac {\pi }{2}\right )}dx}{a^2-b^2}}{2 \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 3138

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

\(\Big \downarrow \) 218

\(\displaystyle \frac {\frac {2 \left (2 a^2+b^2\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{d \sqrt {a-b} \sqrt {a+b} \left (a^2-b^2\right )}-\frac {3 a b \sinh (c+d x)}{d \left (a^2-b^2\right ) (a+b \cosh (c+d x))}}{2 \left (a^2-b^2\right )}-\frac {b \sinh (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \cosh (c+d x))^2}\)

Input:

Int[(a + b*Cosh[c + d*x])^(-3),x]
 

Output:

-1/2*(b*Sinh[c + d*x])/((a^2 - b^2)*d*(a + b*Cosh[c + d*x])^2) + ((2*(2*a^ 
2 + b^2)*ArcTanh[(Sqrt[a - b]*Tanh[(c + d*x)/2])/Sqrt[a + b]])/(Sqrt[a - b 
]*Sqrt[a + b]*(a^2 - b^2)*d) - (3*a*b*Sinh[c + d*x])/((a^2 - b^2)*d*(a + b 
*Cosh[c + d*x])))/(2*(a^2 - b^2))
 

Defintions of rubi rules used

rule 25

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

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, 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 3042

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

rule 3138

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{ 
e = FreeFactors[Tan[(c + d*x)/2], x]}, Simp[2*(e/d)   Subst[Int[1/(a + b + 
(a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] 
 && NeQ[a^2 - b^2, 0]
 

rule 3143

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos 
[c + d*x]*((a + b*Sin[c + d*x])^(n + 1)/(d*(n + 1)*(a^2 - b^2))), x] + Simp 
[1/((n + 1)*(a^2 - b^2))   Int[(a + b*Sin[c + d*x])^(n + 1)*Simp[a*(n + 1) 
- b*(n + 2)*Sin[c + d*x], x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - 
 b^2, 0] && LtQ[n, -1] && IntegerQ[2*n]
 

rule 3233

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
(f_.)*(x_)]), x_Symbol] :> Simp[(-(b*c - a*d))*Cos[e + f*x]*((a + b*Sin[e + 
 f*x])^(m + 1)/(f*(m + 1)*(a^2 - b^2))), x] + Simp[1/((m + 1)*(a^2 - b^2)) 
  Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[(a*c - b*d)*(m + 1) - (b*c - a*d)*( 
m + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c 
- a*d, 0] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && IntegerQ[2*m]
 

Maple [A] (verified)

Time = 0.63 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.40

method	result	size
derivativedivides	\(\frac {-\frac {2 \left (-\frac {\left (4 a +b \right ) b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{2 \left (a -b \right ) \left (a^{2}+2 a b +b^{2}\right )}+\frac {b \left (4 a -b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a +b \right ) \left (a^{2}-2 a b +b^{2}\right )}\right )}{\left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-a -b \right )^{2}}+\frac {\left (2 a^{2}+b^{2}\right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}}{d}\)	\(186\)
default	\(\frac {-\frac {2 \left (-\frac {\left (4 a +b \right ) b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{2 \left (a -b \right ) \left (a^{2}+2 a b +b^{2}\right )}+\frac {b \left (4 a -b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a +b \right ) \left (a^{2}-2 a b +b^{2}\right )}\right )}{\left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-a -b \right )^{2}}+\frac {\left (2 a^{2}+b^{2}\right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}}{d}\)	\(186\)
risch	\(\frac {2 a^{2} b \,{\mathrm e}^{3 d x +3 c}+b^{3} {\mathrm e}^{3 d x +3 c}+6 \,{\mathrm e}^{2 d x +2 c} a^{3}+3 a \,b^{2} {\mathrm e}^{2 d x +2 c}+10 \,{\mathrm e}^{d x +c} a^{2} b -b^{3} {\mathrm e}^{d x +c}+3 b^{2} a}{d \left (a^{2}-b^{2}\right )^{2} \left ({\mathrm e}^{2 d x +2 c} b +2 \,{\mathrm e}^{d x +c} a +b \right )^{2}}+\frac {\ln \left ({\mathrm e}^{d x +c}+\frac {a \sqrt {a^{2}-b^{2}}-a^{2}+b^{2}}{b \sqrt {a^{2}-b^{2}}}\right ) a^{2}}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d}+\frac {\ln \left ({\mathrm e}^{d x +c}+\frac {a \sqrt {a^{2}-b^{2}}-a^{2}+b^{2}}{b \sqrt {a^{2}-b^{2}}}\right ) b^{2}}{2 \sqrt {a^{2}-b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d}-\frac {\ln \left ({\mathrm e}^{d x +c}+\frac {a \sqrt {a^{2}-b^{2}}+a^{2}-b^{2}}{b \sqrt {a^{2}-b^{2}}}\right ) a^{2}}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d}-\frac {\ln \left ({\mathrm e}^{d x +c}+\frac {a \sqrt {a^{2}-b^{2}}+a^{2}-b^{2}}{b \sqrt {a^{2}-b^{2}}}\right ) b^{2}}{2 \sqrt {a^{2}-b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d}\)	\(431\)

Input:

int(1/(a+b*cosh(d*x+c))^3,x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1239 vs. \(2 (120) = 240\).

Time = 0.13 (sec) , antiderivative size = 2591, normalized size of antiderivative = 19.48 \[ \int \frac {1}{(a+b \cosh (c+d x))^3} \, dx=\text {Too large to display} \] Input:

integrate(1/(a+b*cosh(d*x+c))^3,x, algorithm="fricas")
 

Output:

[1/2*(6*a^3*b^2 - 6*a*b^4 + 2*(2*a^4*b - a^2*b^3 - b^5)*cosh(d*x + c)^3 + 
2*(2*a^4*b - a^2*b^3 - b^5)*sinh(d*x + c)^3 + 6*(2*a^5 - a^3*b^2 - a*b^4)* 
cosh(d*x + c)^2 + 6*(2*a^5 - a^3*b^2 - a*b^4 + (2*a^4*b - a^2*b^3 - b^5)*c 
osh(d*x + c))*sinh(d*x + c)^2 + ((2*a^2*b^2 + b^4)*cosh(d*x + c)^4 + (2*a^ 
2*b^2 + b^4)*sinh(d*x + c)^4 + 2*a^2*b^2 + b^4 + 4*(2*a^3*b + a*b^3)*cosh( 
d*x + c)^3 + 4*(2*a^3*b + a*b^3 + (2*a^2*b^2 + b^4)*cosh(d*x + c))*sinh(d* 
x + c)^3 + 2*(4*a^4 + 4*a^2*b^2 + b^4)*cosh(d*x + c)^2 + 2*(4*a^4 + 4*a^2* 
b^2 + b^4 + 3*(2*a^2*b^2 + b^4)*cosh(d*x + c)^2 + 6*(2*a^3*b + a*b^3)*cosh 
(d*x + c))*sinh(d*x + c)^2 + 4*(2*a^3*b + a*b^3)*cosh(d*x + c) + 4*(2*a^3* 
b + a*b^3 + (2*a^2*b^2 + b^4)*cosh(d*x + c)^3 + 3*(2*a^3*b + a*b^3)*cosh(d 
*x + c)^2 + (4*a^4 + 4*a^2*b^2 + b^4)*cosh(d*x + c))*sinh(d*x + c))*sqrt(a 
^2 - b^2)*log((b^2*cosh(d*x + c)^2 + b^2*sinh(d*x + c)^2 + 2*a*b*cosh(d*x 
+ c) + 2*a^2 - b^2 + 2*(b^2*cosh(d*x + c) + a*b)*sinh(d*x + c) - 2*sqrt(a^ 
2 - b^2)*(b*cosh(d*x + c) + b*sinh(d*x + c) + a))/(b*cosh(d*x + c)^2 + b*s 
inh(d*x + c)^2 + 2*a*cosh(d*x + c) + 2*(b*cosh(d*x + c) + a)*sinh(d*x + c) 
 + b)) + 2*(10*a^4*b - 11*a^2*b^3 + b^5)*cosh(d*x + c) + 2*(10*a^4*b - 11* 
a^2*b^3 + b^5 + 3*(2*a^4*b - a^2*b^3 - b^5)*cosh(d*x + c)^2 + 6*(2*a^5 - a 
^3*b^2 - a*b^4)*cosh(d*x + c))*sinh(d*x + c))/((a^6*b^2 - 3*a^4*b^4 + 3*a^ 
2*b^6 - b^8)*d*cosh(d*x + c)^4 + (a^6*b^2 - 3*a^4*b^4 + 3*a^2*b^6 - b^8)*d 
*sinh(d*x + c)^4 + 4*(a^7*b - 3*a^5*b^3 + 3*a^3*b^5 - a*b^7)*d*cosh(d*x...
 

Sympy [F(-1)]

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

integrate(1/(a+b*cosh(d*x+c))**3,x)
 

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{(a+b \cosh (c+d x))^3} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(1/(a+b*cosh(d*x+c))^3,x, algorithm="maxima")
 

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(4*a^2-4*b^2>0)', see `assume?` f 
or more de
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.47 \[ \int \frac {1}{(a+b \cosh (c+d x))^3} \, dx=\frac {\frac {{\left (2 \, a^{2} + b^{2}\right )} \arctan \left (\frac {b e^{\left (d x + c\right )} + a}{\sqrt {-a^{2} + b^{2}}}\right )}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {-a^{2} + b^{2}}} + \frac {2 \, a^{2} b e^{\left (3 \, d x + 3 \, c\right )} + b^{3} e^{\left (3 \, d x + 3 \, c\right )} + 6 \, a^{3} e^{\left (2 \, d x + 2 \, c\right )} + 3 \, a b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 10 \, a^{2} b e^{\left (d x + c\right )} - b^{3} e^{\left (d x + c\right )} + 3 \, a b^{2}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} {\left (b e^{\left (2 \, d x + 2 \, c\right )} + 2 \, a e^{\left (d x + c\right )} + b\right )}^{2}}}{d} \] Input:

integrate(1/(a+b*cosh(d*x+c))^3,x, algorithm="giac")
 

Output:

((2*a^2 + b^2)*arctan((b*e^(d*x + c) + a)/sqrt(-a^2 + b^2))/((a^4 - 2*a^2* 
b^2 + b^4)*sqrt(-a^2 + b^2)) + (2*a^2*b*e^(3*d*x + 3*c) + b^3*e^(3*d*x + 3 
*c) + 6*a^3*e^(2*d*x + 2*c) + 3*a*b^2*e^(2*d*x + 2*c) + 10*a^2*b*e^(d*x + 
c) - b^3*e^(d*x + c) + 3*a*b^2)/((a^4 - 2*a^2*b^2 + b^4)*(b*e^(2*d*x + 2*c 
) + 2*a*e^(d*x + c) + b)^2))/d
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 1035, normalized size of antiderivative = 7.78 \[ \int \frac {1}{(a+b \cosh (c+d x))^3} \, dx =\text {Too large to display} \] Input:

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

Output:

( - 8*e**(4*c + 4*d*x)*sqrt( - a**2 + b**2)*atan((e**(c + d*x)*b + a)/sqrt 
( - a**2 + b**2))*a**3*b**2 - 4*e**(4*c + 4*d*x)*sqrt( - a**2 + b**2)*atan 
((e**(c + d*x)*b + a)/sqrt( - a**2 + b**2))*a*b**4 - 32*e**(3*c + 3*d*x)*s 
qrt( - a**2 + b**2)*atan((e**(c + d*x)*b + a)/sqrt( - a**2 + b**2))*a**4*b 
 - 16*e**(3*c + 3*d*x)*sqrt( - a**2 + b**2)*atan((e**(c + d*x)*b + a)/sqrt 
( - a**2 + b**2))*a**2*b**3 - 32*e**(2*c + 2*d*x)*sqrt( - a**2 + b**2)*ata 
n((e**(c + d*x)*b + a)/sqrt( - a**2 + b**2))*a**5 - 32*e**(2*c + 2*d*x)*sq 
rt( - a**2 + b**2)*atan((e**(c + d*x)*b + a)/sqrt( - a**2 + b**2))*a**3*b* 
*2 - 8*e**(2*c + 2*d*x)*sqrt( - a**2 + b**2)*atan((e**(c + d*x)*b + a)/sqr 
t( - a**2 + b**2))*a*b**4 - 32*e**(c + d*x)*sqrt( - a**2 + b**2)*atan((e** 
(c + d*x)*b + a)/sqrt( - a**2 + b**2))*a**4*b - 16*e**(c + d*x)*sqrt( - a* 
*2 + b**2)*atan((e**(c + d*x)*b + a)/sqrt( - a**2 + b**2))*a**2*b**3 - 8*s 
qrt( - a**2 + b**2)*atan((e**(c + d*x)*b + a)/sqrt( - a**2 + b**2))*a**3*b 
**2 - 4*sqrt( - a**2 + b**2)*atan((e**(c + d*x)*b + a)/sqrt( - a**2 + b**2 
))*a*b**4 - 2*e**(4*c + 4*d*x)*a**4*b**2 + e**(4*c + 4*d*x)*a**2*b**4 + e* 
*(4*c + 4*d*x)*b**6 + 16*e**(2*c + 2*d*x)*a**6 - 12*e**(2*c + 2*d*x)*a**4* 
b**2 - 6*e**(2*c + 2*d*x)*a**2*b**4 + 2*e**(2*c + 2*d*x)*b**6 + 32*e**(c + 
 d*x)*a**5*b - 40*e**(c + d*x)*a**3*b**3 + 8*e**(c + d*x)*a*b**5 + 10*a**4 
*b**2 - 11*a**2*b**4 + b**6)/(4*a*d*(e**(4*c + 4*d*x)*a**6*b**2 - 3*e**(4* 
c + 4*d*x)*a**4*b**4 + 3*e**(4*c + 4*d*x)*a**2*b**6 - e**(4*c + 4*d*x)*...