\(\int \frac {\text {sech}^7(c+d x)}{(a+b \tanh ^2(c+d x))^3} \, dx\) [133]

Optimal result

Integrand size = 23, antiderivative size = 156 \[ \int \frac {\text {sech}^7(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=-\frac {\arctan (\sinh (c+d x))}{b^3 d}+\frac {\sqrt {a+b} \left (8 a^2-4 a b+3 b^2\right ) \arctan \left (\frac {\sqrt {a+b} \sinh (c+d x)}{\sqrt {a}}\right )}{8 a^{5/2} b^3 d}+\frac {(a+b) \sinh (c+d x)}{4 a b d \left (a+(a+b) \sinh ^2(c+d x)\right )^2}-\frac {(4 a-3 b) (a+b) \sinh (c+d x)}{8 a^2 b^2 d \left (a+(a+b) \sinh ^2(c+d x)\right )} \] Output:

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 2.11 (sec) , antiderivative size = 317, normalized size of antiderivative = 2.03 \[ \int \frac {\text {sech}^7(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=-\frac {\frac {2 \sqrt {a+b} \left (8 a^2-4 a b+3 b^2\right ) \arctan \left (\frac {\sqrt {a} \text {csch}(c+d x)}{\sqrt {a+b}}\right )}{a^{5/2}}+\frac {2 \left (8 a^3+4 a^2 b-a b^2+3 b^3\right ) \arctan \left (\frac {\sqrt {a} \text {csch}(c+d x)}{\sqrt {a+b}}\right )}{a^{5/2} \sqrt {a+b}}+64 \arctan \left (\tanh \left (\frac {1}{2} (c+d x)\right )\right )+\frac {i \sqrt {a+b} \left (8 a^2-4 a b+3 b^2\right ) \log (a-b+(a+b) \cosh (2 (c+d x)))}{a^{5/2}}-\frac {i \left (8 a^3+4 a^2 b-a b^2+3 b^3\right ) \log (a-b+(a+b) \cosh (2 (c+d x)))}{a^{5/2} \sqrt {a+b}}-\frac {32 b^2 (a+b) \sinh (c+d x)}{a (a-b+(a+b) \cosh (2 (c+d x)))^2}+\frac {8 b \left (4 a^2+a b-3 b^2\right ) \sinh (c+d x)}{a^2 (a-b+(a+b) \cosh (2 (c+d x)))}}{32 b^3 d} \] Input:

Integrate[Sech[c + d*x]^7/(a + b*Tanh[c + d*x]^2)^3,x]
 

Output:

-1/32*((2*Sqrt[a + b]*(8*a^2 - 4*a*b + 3*b^2)*ArcTan[(Sqrt[a]*Csch[c + d*x 
])/Sqrt[a + b]])/a^(5/2) + (2*(8*a^3 + 4*a^2*b - a*b^2 + 3*b^3)*ArcTan[(Sq 
rt[a]*Csch[c + d*x])/Sqrt[a + b]])/(a^(5/2)*Sqrt[a + b]) + 64*ArcTan[Tanh[ 
(c + d*x)/2]] + (I*Sqrt[a + b]*(8*a^2 - 4*a*b + 3*b^2)*Log[a - b + (a + b) 
*Cosh[2*(c + d*x)]])/a^(5/2) - (I*(8*a^3 + 4*a^2*b - a*b^2 + 3*b^3)*Log[a 
- b + (a + b)*Cosh[2*(c + d*x)]])/(a^(5/2)*Sqrt[a + b]) - (32*b^2*(a + b)* 
Sinh[c + d*x])/(a*(a - b + (a + b)*Cosh[2*(c + d*x)])^2) + (8*b*(4*a^2 + a 
*b - 3*b^2)*Sinh[c + d*x])/(a^2*(a - b + (a + b)*Cosh[2*(c + d*x)])))/(b^3 
*d)
 

Rubi [A] (verified)

Time = 0.71 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.09, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3042, 4159, 316, 402, 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.

\(\displaystyle \int \frac {\text {sech}^7(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sec (i c+i d x)^7}{\left (a-b \tan (i c+i d x)^2\right )^3}dx\)

\(\Big \downarrow \) 4159

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

\(\Big \downarrow \) 316

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

\(\Big \downarrow \) 402

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

\(\Big \downarrow \) 397

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

\(\Big \downarrow \) 216

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

\(\Big \downarrow \) 218

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

Input:

Int[Sech[c + d*x]^7/(a + b*Tanh[c + d*x]^2)^3,x]
 

Output:

(((a + b)*Sinh[c + d*x])/(4*a*b*(a + (a + b)*Sinh[c + d*x]^2)^2) - (-1/2*( 
(-8*a^2*ArcTan[Sinh[c + d*x]])/b + (Sqrt[a + b]*(8*a^2 - 4*a*b + 3*b^2)*Ar 
cTan[(Sqrt[a + b]*Sinh[c + d*x])/Sqrt[a]])/(Sqrt[a]*b))/(a*b) + ((4*a - 3* 
b)*(a + b)*Sinh[c + d*x])/(2*a*b*(a + (a + b)*Sinh[c + d*x]^2)))/(4*a*b))/ 
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 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 4159

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(388\) vs. \(2(142)=284\).

Time = 0.63 (sec) , antiderivative size = 389, normalized size of antiderivative = 2.49

\[\frac {\frac {\frac {2 \left (\frac {b \left (4 a^{2}-a b -5 b^{2}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{8 a}+\frac {\left (4 a^{3}+23 a^{2} b +7 b^{2} a -12 b^{3}\right ) b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{8 a^{2}}-\frac {\left (4 a^{3}+23 a^{2} b +7 b^{2} a -12 b^{3}\right ) b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{8 a^{2}}-\frac {b \left (4 a^{2}-a b -5 b^{2}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 a}\right )}{\left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a +2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +4 b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+a \right )^{2}}+\frac {\left (8 a^{3}+4 a^{2} b -b^{2} a +3 b^{3}\right ) \left (\frac {\left (\sqrt {\left (a +b \right ) b}+b \right ) \arctan \left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {\left (a +b \right ) b}+a +2 b \right ) a}}\right )}{2 a \sqrt {\left (a +b \right ) b}\, \sqrt {\left (2 \sqrt {\left (a +b \right ) b}+a +2 b \right ) a}}-\frac {\left (\sqrt {\left (a +b \right ) b}-b \right ) \operatorname {arctanh}\left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {\left (a +b \right ) b}-a -2 b \right ) a}}\right )}{2 a \sqrt {\left (a +b \right ) b}\, \sqrt {\left (2 \sqrt {\left (a +b \right ) b}-a -2 b \right ) a}}\right )}{4 a}}{b^{3}}-\frac {2 \arctan \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{3}}}{d}\]

Input:

int(sech(d*x+c)^7/(a+tanh(d*x+c)^2*b)^3,x)
 

Output:

1/d*(2/b^3*((1/8*b*(4*a^2-a*b-5*b^2)/a*tanh(1/2*d*x+1/2*c)^7+1/8*(4*a^3+23 
*a^2*b+7*a*b^2-12*b^3)/a^2*b*tanh(1/2*d*x+1/2*c)^5-1/8*(4*a^3+23*a^2*b+7*a 
*b^2-12*b^3)/a^2*b*tanh(1/2*d*x+1/2*c)^3-1/8*b*(4*a^2-a*b-5*b^2)/a*tanh(1/ 
2*d*x+1/2*c))/(tanh(1/2*d*x+1/2*c)^4*a+2*tanh(1/2*d*x+1/2*c)^2*a+4*b*tanh( 
1/2*d*x+1/2*c)^2+a)^2+1/8/a*(8*a^3+4*a^2*b-a*b^2+3*b^3)*(1/2*(((a+b)*b)^(1 
/2)+b)/a/((a+b)*b)^(1/2)/((2*((a+b)*b)^(1/2)+a+2*b)*a)^(1/2)*arctan(a*tanh 
(1/2*d*x+1/2*c)/((2*((a+b)*b)^(1/2)+a+2*b)*a)^(1/2))-1/2*(((a+b)*b)^(1/2)- 
b)/a/((a+b)*b)^(1/2)/((2*((a+b)*b)^(1/2)-a-2*b)*a)^(1/2)*arctanh(a*tanh(1/ 
2*d*x+1/2*c)/((2*((a+b)*b)^(1/2)-a-2*b)*a)^(1/2))))-2/b^3*arctan(tanh(1/2* 
d*x+1/2*c)))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 4457 vs. \(2 (142) = 284\).

Time = 0.26 (sec) , antiderivative size = 8070, normalized size of antiderivative = 51.73 \[ \int \frac {\text {sech}^7(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=\text {Too large to display} \] Input:

integrate(sech(d*x+c)^7/(a+b*tanh(d*x+c)^2)^3,x, algorithm="fricas")
 

Output:

Too large to include
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\text {sech}^7(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=\text {Timed out} \] Input:

integrate(sech(d*x+c)**7/(a+b*tanh(d*x+c)**2)**3,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {\text {sech}^7(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=\int { \frac {\operatorname {sech}\left (d x + c\right )^{7}}{{\left (b \tanh \left (d x + c\right )^{2} + a\right )}^{3}} \,d x } \] Input:

integrate(sech(d*x+c)^7/(a+b*tanh(d*x+c)^2)^3,x, algorithm="maxima")
 

Output:

-1/4*((4*a^3*e^(7*c) + 5*a^2*b*e^(7*c) - 2*a*b^2*e^(7*c) - 3*b^3*e^(7*c))* 
e^(7*d*x) + (4*a^3*e^(5*c) - 19*a^2*b*e^(5*c) - 14*a*b^2*e^(5*c) + 9*b^3*e 
^(5*c))*e^(5*d*x) - (4*a^3*e^(3*c) - 19*a^2*b*e^(3*c) - 14*a*b^2*e^(3*c) + 
 9*b^3*e^(3*c))*e^(3*d*x) - (4*a^3*e^c + 5*a^2*b*e^c - 2*a*b^2*e^c - 3*b^3 
*e^c)*e^(d*x))/(a^4*b^2*d + 2*a^3*b^3*d + a^2*b^4*d + (a^4*b^2*d*e^(8*c) + 
 2*a^3*b^3*d*e^(8*c) + a^2*b^4*d*e^(8*c))*e^(8*d*x) + 4*(a^4*b^2*d*e^(6*c) 
 - a^2*b^4*d*e^(6*c))*e^(6*d*x) + 2*(3*a^4*b^2*d*e^(4*c) - 2*a^3*b^3*d*e^( 
4*c) + 3*a^2*b^4*d*e^(4*c))*e^(4*d*x) + 4*(a^4*b^2*d*e^(2*c) - a^2*b^4*d*e 
^(2*c))*e^(2*d*x)) - 2*arctan(e^(d*x + c))/(b^3*d) + 128*integrate(1/512*( 
(8*a^3*e^(3*c) + 4*a^2*b*e^(3*c) - a*b^2*e^(3*c) + 3*b^3*e^(3*c))*e^(3*d*x 
) + (8*a^3*e^c + 4*a^2*b*e^c - a*b^2*e^c + 3*b^3*e^c)*e^(d*x))/(a^3*b^3 + 
a^2*b^4 + (a^3*b^3*e^(4*c) + a^2*b^4*e^(4*c))*e^(4*d*x) + 2*(a^3*b^3*e^(2* 
c) - a^2*b^4*e^(2*c))*e^(2*d*x)), x)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {\text {sech}^7(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(sech(d*x+c)^7/(a+b*tanh(d*x+c)^2)^3,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. \[ \int \frac {\text {sech}^7(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=\int \frac {1}{{\mathrm {cosh}\left (c+d\,x\right )}^7\,{\left (b\,{\mathrm {tanh}\left (c+d\,x\right )}^2+a\right )}^3} \,d x \] Input:

int(1/(cosh(c + d*x)^7*(a + b*tanh(c + d*x)^2)^3),x)
 

Output:

int(1/(cosh(c + d*x)^7*(a + b*tanh(c + d*x)^2)^3), x)
                                                                                    
                                                                                    
 

Reduce [B] (verification not implemented)

Time = 0.33 (sec) , antiderivative size = 2818, normalized size of antiderivative = 18.06 \[ \int \frac {\text {sech}^7(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx =\text {Too large to display} \] Input:

int(sech(d*x+c)^7/(a+b*tanh(d*x+c)^2)^3,x)
 

Output:

( - 16*e**(8*c + 8*d*x)*atan(e**(c + d*x))*a**5 - 32*e**(8*c + 8*d*x)*atan 
(e**(c + d*x))*a**4*b - 16*e**(8*c + 8*d*x)*atan(e**(c + d*x))*a**3*b**2 - 
 64*e**(6*c + 6*d*x)*atan(e**(c + d*x))*a**5 + 64*e**(6*c + 6*d*x)*atan(e* 
*(c + d*x))*a**3*b**2 - 96*e**(4*c + 4*d*x)*atan(e**(c + d*x))*a**5 + 64*e 
**(4*c + 4*d*x)*atan(e**(c + d*x))*a**4*b - 96*e**(4*c + 4*d*x)*atan(e**(c 
 + d*x))*a**3*b**2 - 64*e**(2*c + 2*d*x)*atan(e**(c + d*x))*a**5 + 64*e**( 
2*c + 2*d*x)*atan(e**(c + d*x))*a**3*b**2 - 16*atan(e**(c + d*x))*a**5 - 3 
2*atan(e**(c + d*x))*a**4*b - 16*atan(e**(c + d*x))*a**3*b**2 + 8*e**(8*c 
+ 8*d*x)*sqrt(a)*sqrt(a + b)*atan((e**(c + d*x)*sqrt(a + b) - sqrt(b))/sqr 
t(a))*a**4 + 12*e**(8*c + 8*d*x)*sqrt(a)*sqrt(a + b)*atan((e**(c + d*x)*sq 
rt(a + b) - sqrt(b))/sqrt(a))*a**3*b + 3*e**(8*c + 8*d*x)*sqrt(a)*sqrt(a + 
 b)*atan((e**(c + d*x)*sqrt(a + b) - sqrt(b))/sqrt(a))*a**2*b**2 + 2*e**(8 
*c + 8*d*x)*sqrt(a)*sqrt(a + b)*atan((e**(c + d*x)*sqrt(a + b) - sqrt(b))/ 
sqrt(a))*a*b**3 + 3*e**(8*c + 8*d*x)*sqrt(a)*sqrt(a + b)*atan((e**(c + d*x 
)*sqrt(a + b) - sqrt(b))/sqrt(a))*b**4 + 32*e**(6*c + 6*d*x)*sqrt(a)*sqrt( 
a + b)*atan((e**(c + d*x)*sqrt(a + b) - sqrt(b))/sqrt(a))*a**4 - 16*e**(6* 
c + 6*d*x)*sqrt(a)*sqrt(a + b)*atan((e**(c + d*x)*sqrt(a + b) - sqrt(b))/s 
qrt(a))*a**3*b - 20*e**(6*c + 6*d*x)*sqrt(a)*sqrt(a + b)*atan((e**(c + d*x 
)*sqrt(a + b) - sqrt(b))/sqrt(a))*a**2*b**2 + 16*e**(6*c + 6*d*x)*sqrt(a)* 
sqrt(a + b)*atan((e**(c + d*x)*sqrt(a + b) - sqrt(b))/sqrt(a))*a*b**3 -...