\(\int \frac {\text {sech}^6(c+d x)}{(a+b \tanh ^2(c+d x))^3} \, dx\) [132]

Optimal result

Integrand size = 23, antiderivative size = 132 \[ \int \frac {\text {sech}^6(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=\frac {\left (3 a^2-2 a b+3 b^2\right ) \arctan \left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a}}\right )}{8 a^{5/2} b^{5/2} d}+\frac {(a+b)^2 \tanh (c+d x)}{4 a b^2 d \left (a+b \tanh ^2(c+d x)\right )^2}-\frac {(5 a-3 b) (a+b) \tanh (c+d x)}{8 a^2 b^2 d \left (a+b \tanh ^2(c+d x)\right )} \] Output:

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

Mathematica [A] (verified)

Time = 1.16 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.97 \[ \int \frac {\text {sech}^6(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=\frac {\left (3 a^2-2 a b+3 b^2\right ) \arctan \left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a}}\right )-\frac {\sqrt {a} \sqrt {b} (a+b) \left (3 a^2-10 a b+3 b^2+3 \left (a^2-b^2\right ) \cosh (2 (c+d x))\right ) \sinh (2 (c+d x))}{(a-b+(a+b) \cosh (2 (c+d x)))^2}}{8 a^{5/2} b^{5/2} d} \] Input:

Integrate[Sech[c + d*x]^6/(a + b*Tanh[c + d*x]^2)^3,x]
 

Output:

((3*a^2 - 2*a*b + 3*b^2)*ArcTan[(Sqrt[b]*Tanh[c + d*x])/Sqrt[a]] - (Sqrt[a 
]*Sqrt[b]*(a + b)*(3*a^2 - 10*a*b + 3*b^2 + 3*(a^2 - b^2)*Cosh[2*(c + d*x) 
])*Sinh[2*(c + d*x)])/(a - b + (a + b)*Cosh[2*(c + d*x)])^2)/(8*a^(5/2)*b^ 
(5/2)*d)
 

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 143, 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, 4158, 315, 25, 298, 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.

\(\displaystyle \int \frac {\text {sech}^6(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sec (i c+i d x)^6}{\left (a-b \tan (i c+i d x)^2\right )^3}dx\)

\(\Big \downarrow \) 4158

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

\(\Big \downarrow \) 315

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 298

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

\(\Big \downarrow \) 218

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

Input:

Int[Sech[c + d*x]^6/(a + b*Tanh[c + d*x]^2)^3,x]
 

Output:

(((a + b)*Tanh[c + d*x]*(1 - Tanh[c + d*x]^2))/(4*a*b*(a + b*Tanh[c + d*x] 
^2)^2) - (((2 - (3*a)/b - (3*b)/a)*ArcTan[(Sqrt[b]*Tanh[c + d*x])/Sqrt[a]] 
)/(2*Sqrt[a]*Sqrt[b]) + (3*(a/b - b/a)*Tanh[c + d*x])/(2*(a + b*Tanh[c + d 
*x]^2)))/(4*a*b))/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 298

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[(-( 
b*c - a*d))*x*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] - Simp[(a*d - b*c*( 
2*p + 3))/(2*a*b*(p + 1))   Int[(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, 
 c, d, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/2 + p, 0])
 

rule 315

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim 
p[(a*d - c*b)*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q - 1)/(2*a*b*(p + 1))), 
x] - Simp[1/(2*a*b*(p + 1))   Int[(a + b*x^2)^(p + 1)*(c + d*x^2)^(q - 2)*S 
imp[c*(a*d - c*b*(2*p + 3)) + d*(a*d*(2*(q - 1) + 1) - b*c*(2*(p + q) + 1)) 
*x^2, x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, - 
1] && GtQ[q, 1] && IntBinomialQ[a, b, c, d, 2, p, q, x]
 

rule 3042

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

rule 4158

Int[sec[(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[ff/(c^(m - 1)*f)   Subst[Int[(c^2 + ff^2*x^2)^(m/2 - 1)*(a + b*(ff*x)^n)^ 
p, x], x, c*(Tan[e + f*x]/ff)], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && I 
ntegerQ[m/2] && (IntegersQ[n, p] || IGtQ[m, 0] || IGtQ[p, 0] || EqQ[n^2, 4] 
 || EqQ[n^2, 16])
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(375\) vs. \(2(118)=236\).

Time = 0.64 (sec) , antiderivative size = 376, normalized size of antiderivative = 2.85

\[\frac {-\frac {2 \left (\frac {\left (3 a^{2}-2 a b -5 b^{2}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{8 b^{2} a}+\frac {\left (9 a^{3}+14 a^{2} b -7 b^{2} a -12 b^{3}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{8 a^{2} b^{2}}+\frac {\left (9 a^{3}+14 a^{2} b -7 b^{2} a -12 b^{3}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{8 a^{2} b^{2}}+\frac {\left (3 a^{2}-2 a b -5 b^{2}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 b^{2} a}\right )}{\left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a +2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +4 b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+a \right )^{2}}-\frac {\left (3 a^{2}-2 a b +3 b^{2}\right ) \left (\frac {\left (a +\sqrt {\left (a +b \right ) b}+b \right ) \arctan \left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {\left (a +b \right ) b}+a +2 b \right ) a}}\right )}{2 a \sqrt {\left (a +b \right ) b}\, \sqrt {\left (2 \sqrt {\left (a +b \right ) b}+a +2 b \right ) a}}-\frac {\left (-a +\sqrt {\left (a +b \right ) b}-b \right ) \operatorname {arctanh}\left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {\left (a +b \right ) b}-a -2 b \right ) a}}\right )}{2 a \sqrt {\left (a +b \right ) b}\, \sqrt {\left (2 \sqrt {\left (a +b \right ) b}-a -2 b \right ) a}}\right )}{4 a \,b^{2}}}{d}\]

Input:

int(sech(d*x+c)^6/(a+tanh(d*x+c)^2*b)^3,x)
 

Output:

1/d*(-2*(1/8*(3*a^2-2*a*b-5*b^2)/b^2/a*tanh(1/2*d*x+1/2*c)^7+1/8*(9*a^3+14 
*a^2*b-7*a*b^2-12*b^3)/a^2/b^2*tanh(1/2*d*x+1/2*c)^5+1/8*(9*a^3+14*a^2*b-7 
*a*b^2-12*b^3)/a^2/b^2*tanh(1/2*d*x+1/2*c)^3+1/8*(3*a^2-2*a*b-5*b^2)/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)^2-1/4/a*(3*a^2-2*a*b+3*b^2)/b^2*(1/2*(a+((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+((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))))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2464 vs. \(2 (118) = 236\).

Time = 0.20 (sec) , antiderivative size = 5233, normalized size of antiderivative = 39.64 \[ \int \frac {\text {sech}^6(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=\text {Too large to display} \] Input:

integrate(sech(d*x+c)^6/(a+b*tanh(d*x+c)^2)^3,x, algorithm="fricas")
 

Output:

Too large to include
 

Sympy [F]

\[ \int \frac {\text {sech}^6(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=\int \frac {\operatorname {sech}^{6}{\left (c + d x \right )}}{\left (a + b \tanh ^{2}{\left (c + d x \right )}\right )^{3}}\, dx \] Input:

integrate(sech(d*x+c)**6/(a+b*tanh(d*x+c)**2)**3,x)
 

Output:

Integral(sech(c + d*x)**6/(a + b*tanh(c + d*x)**2)**3, x)
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 332 vs. \(2 (118) = 236\).

Time = 0.33 (sec) , antiderivative size = 332, normalized size of antiderivative = 2.52 \[ \int \frac {\text {sech}^6(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=-\frac {3 \, a^{3} + 3 \, a^{2} b - 3 \, a b^{2} - 3 \, b^{3} + {\left (9 \, a^{3} - 13 \, a^{2} b - 13 \, a b^{2} + 9 \, b^{3}\right )} e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, {\left (3 \, a^{3} - 5 \, a^{2} b + 5 \, a b^{2} - 3 \, b^{3}\right )} e^{\left (-4 \, d x - 4 \, c\right )} + {\left (3 \, a^{3} + a^{2} b + a b^{2} + 3 \, b^{3}\right )} e^{\left (-6 \, d x - 6 \, c\right )}}{4 \, {\left (a^{4} b^{2} + 2 \, a^{3} b^{3} + a^{2} b^{4} + 4 \, {\left (a^{4} b^{2} - a^{2} b^{4}\right )} e^{\left (-2 \, d x - 2 \, c\right )} + 2 \, {\left (3 \, a^{4} b^{2} - 2 \, a^{3} b^{3} + 3 \, a^{2} b^{4}\right )} e^{\left (-4 \, d x - 4 \, c\right )} + 4 \, {\left (a^{4} b^{2} - a^{2} b^{4}\right )} e^{\left (-6 \, d x - 6 \, c\right )} + {\left (a^{4} b^{2} + 2 \, a^{3} b^{3} + a^{2} b^{4}\right )} e^{\left (-8 \, d x - 8 \, c\right )}\right )} d} - \frac {{\left (3 \, a^{2} - 2 \, a b + 3 \, b^{2}\right )} \arctan \left (\frac {{\left (a + b\right )} e^{\left (-2 \, d x - 2 \, c\right )} + a - b}{2 \, \sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{2} b^{2} d} \] Input:

integrate(sech(d*x+c)^6/(a+b*tanh(d*x+c)^2)^3,x, algorithm="maxima")
 

Output:

-1/4*(3*a^3 + 3*a^2*b - 3*a*b^2 - 3*b^3 + (9*a^3 - 13*a^2*b - 13*a*b^2 + 9 
*b^3)*e^(-2*d*x - 2*c) + 3*(3*a^3 - 5*a^2*b + 5*a*b^2 - 3*b^3)*e^(-4*d*x - 
 4*c) + (3*a^3 + a^2*b + a*b^2 + 3*b^3)*e^(-6*d*x - 6*c))/((a^4*b^2 + 2*a^ 
3*b^3 + a^2*b^4 + 4*(a^4*b^2 - a^2*b^4)*e^(-2*d*x - 2*c) + 2*(3*a^4*b^2 - 
2*a^3*b^3 + 3*a^2*b^4)*e^(-4*d*x - 4*c) + 4*(a^4*b^2 - a^2*b^4)*e^(-6*d*x 
- 6*c) + (a^4*b^2 + 2*a^3*b^3 + a^2*b^4)*e^(-8*d*x - 8*c))*d) - 1/8*(3*a^2 
 - 2*a*b + 3*b^2)*arctan(1/2*((a + b)*e^(-2*d*x - 2*c) + a - b)/sqrt(a*b)) 
/(sqrt(a*b)*a^2*b^2*d)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 322 vs. \(2 (118) = 236\).

Time = 0.58 (sec) , antiderivative size = 322, normalized size of antiderivative = 2.44 \[ \int \frac {\text {sech}^6(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=\frac {\frac {{\left (3 \, a^{2} - 2 \, a b + 3 \, b^{2}\right )} \arctan \left (\frac {a e^{\left (2 \, d x + 2 \, c\right )} + b e^{\left (2 \, d x + 2 \, c\right )} + a - b}{2 \, \sqrt {a b}}\right )}{\sqrt {a b} a^{2} b^{2}} + \frac {2 \, {\left (3 \, a^{3} e^{\left (6 \, d x + 6 \, c\right )} + a^{2} b e^{\left (6 \, d x + 6 \, c\right )} + a b^{2} e^{\left (6 \, d x + 6 \, c\right )} + 3 \, b^{3} e^{\left (6 \, d x + 6 \, c\right )} + 9 \, a^{3} e^{\left (4 \, d x + 4 \, c\right )} - 15 \, a^{2} b e^{\left (4 \, d x + 4 \, c\right )} + 15 \, a b^{2} e^{\left (4 \, d x + 4 \, c\right )} - 9 \, b^{3} e^{\left (4 \, d x + 4 \, c\right )} + 9 \, a^{3} e^{\left (2 \, d x + 2 \, c\right )} - 13 \, a^{2} b e^{\left (2 \, d x + 2 \, c\right )} - 13 \, a b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 9 \, b^{3} e^{\left (2 \, d x + 2 \, c\right )} + 3 \, a^{3} + 3 \, a^{2} b - 3 \, a b^{2} - 3 \, b^{3}\right )}}{{\left (a e^{\left (4 \, d x + 4 \, c\right )} + b e^{\left (4 \, d x + 4 \, c\right )} + 2 \, a e^{\left (2 \, d x + 2 \, c\right )} - 2 \, b e^{\left (2 \, d x + 2 \, c\right )} + a + b\right )}^{2} a^{2} b^{2}}}{8 \, d} \] Input:

integrate(sech(d*x+c)^6/(a+b*tanh(d*x+c)^2)^3,x, algorithm="giac")
 

Output:

1/8*((3*a^2 - 2*a*b + 3*b^2)*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*b^2) + 2*(3*a^3*e^(6*d*x + 6*c) + 
a^2*b*e^(6*d*x + 6*c) + a*b^2*e^(6*d*x + 6*c) + 3*b^3*e^(6*d*x + 6*c) + 9* 
a^3*e^(4*d*x + 4*c) - 15*a^2*b*e^(4*d*x + 4*c) + 15*a*b^2*e^(4*d*x + 4*c) 
- 9*b^3*e^(4*d*x + 4*c) + 9*a^3*e^(2*d*x + 2*c) - 13*a^2*b*e^(2*d*x + 2*c) 
 - 13*a*b^2*e^(2*d*x + 2*c) + 9*b^3*e^(2*d*x + 2*c) + 3*a^3 + 3*a^2*b - 3* 
a*b^2 - 3*b^3)/((a*e^(4*d*x + 4*c) + b*e^(4*d*x + 4*c) + 2*a*e^(2*d*x + 2* 
c) - 2*b*e^(2*d*x + 2*c) + a + b)^2*a^2*b^2))/d
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\text {sech}^6(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=\int \frac {1}{{\mathrm {cosh}\left (c+d\,x\right )}^6\,{\left (b\,{\mathrm {tanh}\left (c+d\,x\right )}^2+a\right )}^3} \,d x \] Input:

int(1/(cosh(c + d*x)^6*(a + b*tanh(c + d*x)^2)^3),x)
 

Output:

int(1/(cosh(c + d*x)^6*(a + b*tanh(c + d*x)^2)^3), x)
                                                                                    
                                                                                    
 

Reduce [B] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 2976, normalized size of antiderivative = 22.55 \[ \int \frac {\text {sech}^6(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx =\text {Too large to display} \] Input:

int(sech(d*x+c)^6/(a+b*tanh(d*x+c)^2)^3,x)
 

Output:

(6*e**(8*c + 8*d*x)*sqrt(b)*sqrt(a)*atan((e**(c + d*x)*sqrt(a + b) - sqrt( 
b))/sqrt(a))*a**5 + 2*e**(8*c + 8*d*x)*sqrt(b)*sqrt(a)*atan((e**(c + d*x)* 
sqrt(a + b) - sqrt(b))/sqrt(a))*a**4*b - 4*e**(8*c + 8*d*x)*sqrt(b)*sqrt(a 
)*atan((e**(c + d*x)*sqrt(a + b) - sqrt(b))/sqrt(a))*a**3*b**2 + 4*e**(8*c 
 + 8*d*x)*sqrt(b)*sqrt(a)*atan((e**(c + d*x)*sqrt(a + b) - sqrt(b))/sqrt(a 
))*a**2*b**3 - 2*e**(8*c + 8*d*x)*sqrt(b)*sqrt(a)*atan((e**(c + d*x)*sqrt( 
a + b) - sqrt(b))/sqrt(a))*a*b**4 - 6*e**(8*c + 8*d*x)*sqrt(b)*sqrt(a)*ata 
n((e**(c + d*x)*sqrt(a + b) - sqrt(b))/sqrt(a))*b**5 + 24*e**(6*c + 6*d*x) 
*sqrt(b)*sqrt(a)*atan((e**(c + d*x)*sqrt(a + b) - sqrt(b))/sqrt(a))*a**5 - 
 40*e**(6*c + 6*d*x)*sqrt(b)*sqrt(a)*atan((e**(c + d*x)*sqrt(a + b) - sqrt 
(b))/sqrt(a))*a**4*b + 16*e**(6*c + 6*d*x)*sqrt(b)*sqrt(a)*atan((e**(c + d 
*x)*sqrt(a + b) - sqrt(b))/sqrt(a))*a**3*b**2 + 16*e**(6*c + 6*d*x)*sqrt(b 
)*sqrt(a)*atan((e**(c + d*x)*sqrt(a + b) - sqrt(b))/sqrt(a))*a**2*b**3 - 4 
0*e**(6*c + 6*d*x)*sqrt(b)*sqrt(a)*atan((e**(c + d*x)*sqrt(a + b) - sqrt(b 
))/sqrt(a))*a*b**4 + 24*e**(6*c + 6*d*x)*sqrt(b)*sqrt(a)*atan((e**(c + d*x 
)*sqrt(a + b) - sqrt(b))/sqrt(a))*b**5 + 36*e**(4*c + 4*d*x)*sqrt(b)*sqrt( 
a)*atan((e**(c + d*x)*sqrt(a + b) - sqrt(b))/sqrt(a))*a**5 - 84*e**(4*c + 
4*d*x)*sqrt(b)*sqrt(a)*atan((e**(c + d*x)*sqrt(a + b) - sqrt(b))/sqrt(a))* 
a**4*b + 136*e**(4*c + 4*d*x)*sqrt(b)*sqrt(a)*atan((e**(c + d*x)*sqrt(a + 
b) - sqrt(b))/sqrt(a))*a**3*b**2 - 136*e**(4*c + 4*d*x)*sqrt(b)*sqrt(a)...