\(\int \frac {\text {sech}^6(c+d x)}{a+b \tanh ^2(c+d x)} \, dx\) [114]

Optimal result

Integrand size = 23, antiderivative size = 75 \[ \int \frac {\text {sech}^6(c+d x)}{a+b \tanh ^2(c+d x)} \, dx=\frac {(a+b)^2 \arctan \left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a}}\right )}{\sqrt {a} b^{5/2} d}-\frac {(a+2 b) \tanh (c+d x)}{b^2 d}+\frac {\tanh ^3(c+d x)}{3 b d} \] Output:

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

Mathematica [A] (verified)

Time = 0.47 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.95 \[ \int \frac {\text {sech}^6(c+d x)}{a+b \tanh ^2(c+d x)} \, dx=\frac {(a+b)^2 \arctan \left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a}}\right )}{\sqrt {a} b^{5/2} d}-\frac {\left (3 a+5 b+b \text {sech}^2(c+d x)\right ) \tanh (c+d x)}{3 b^2 d} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.93, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3042, 4158, 300, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4158

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

\(\Big \downarrow \) 300

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 300

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

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 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. \(251\) vs. \(2(65)=130\).

Time = 174.33 (sec) , antiderivative size = 252, normalized size of antiderivative = 3.36

method	result	size
derivativedivides	\(\frac {\frac {2 a \left (a^{2}+2 a b +b^{2}\right ) \left (\frac {\left (-a -\sqrt {\left (a +b \right ) b}-b \right ) \arctan \left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {\left (a +b \right ) b}+a +2 b \right ) a}}\right )}{2 a \sqrt {\left (a +b \right ) b}\, \sqrt {\left (2 \sqrt {\left (a +b \right ) b}+a +2 b \right ) a}}-\frac {\left (a -\sqrt {\left (a +b \right ) b}+b \right ) \operatorname {arctanh}\left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {\left (a +b \right ) b}-a -2 b \right ) a}}\right )}{2 a \sqrt {\left (a +b \right ) b}\, \sqrt {\left (2 \sqrt {\left (a +b \right ) b}-a -2 b \right ) a}}\right )}{b^{2}}+\frac {2 \left (-a -2 b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+2 \left (-2 a -\frac {8 b}{3}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+2 \left (-a -2 b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{b^{2} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right )^{3}}}{d}\)	\(252\)
default	\(\frac {\frac {2 a \left (a^{2}+2 a b +b^{2}\right ) \left (\frac {\left (-a -\sqrt {\left (a +b \right ) b}-b \right ) \arctan \left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {\left (a +b \right ) b}+a +2 b \right ) a}}\right )}{2 a \sqrt {\left (a +b \right ) b}\, \sqrt {\left (2 \sqrt {\left (a +b \right ) b}+a +2 b \right ) a}}-\frac {\left (a -\sqrt {\left (a +b \right ) b}+b \right ) \operatorname {arctanh}\left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {\left (a +b \right ) b}-a -2 b \right ) a}}\right )}{2 a \sqrt {\left (a +b \right ) b}\, \sqrt {\left (2 \sqrt {\left (a +b \right ) b}-a -2 b \right ) a}}\right )}{b^{2}}+\frac {2 \left (-a -2 b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+2 \left (-2 a -\frac {8 b}{3}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+2 \left (-a -2 b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{b^{2} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right )^{3}}}{d}\)	\(252\)
risch	\(\frac {2 \,{\mathrm e}^{4 d x +4 c} a +2 b \,{\mathrm e}^{4 d x +4 c}+4 \,{\mathrm e}^{2 d x +2 c} a +8 \,{\mathrm e}^{2 d x +2 c} b +2 a +\frac {10 b}{3}}{b^{2} d \left ({\mathrm e}^{2 d x +2 c}+1\right )^{3}}-\frac {\ln \left ({\mathrm e}^{2 d x +2 c}+\frac {a \sqrt {-a b}-b \sqrt {-a b}-2 a b}{\left (a +b \right ) \sqrt {-a b}}\right ) a^{2}}{2 \sqrt {-a b}\, d \,b^{2}}-\frac {\ln \left ({\mathrm e}^{2 d x +2 c}+\frac {a \sqrt {-a b}-b \sqrt {-a b}-2 a b}{\left (a +b \right ) \sqrt {-a b}}\right ) a}{\sqrt {-a b}\, d b}-\frac {\ln \left ({\mathrm e}^{2 d x +2 c}+\frac {a \sqrt {-a b}-b \sqrt {-a b}-2 a b}{\left (a +b \right ) \sqrt {-a b}}\right )}{2 \sqrt {-a b}\, d}+\frac {\ln \left ({\mathrm e}^{2 d x +2 c}+\frac {a \sqrt {-a b}-b \sqrt {-a b}+2 a b}{\left (a +b \right ) \sqrt {-a b}}\right ) a^{2}}{2 \sqrt {-a b}\, d \,b^{2}}+\frac {\ln \left ({\mathrm e}^{2 d x +2 c}+\frac {a \sqrt {-a b}-b \sqrt {-a b}+2 a b}{\left (a +b \right ) \sqrt {-a b}}\right ) a}{\sqrt {-a b}\, d b}+\frac {\ln \left ({\mathrm e}^{2 d x +2 c}+\frac {a \sqrt {-a b}-b \sqrt {-a b}+2 a b}{\left (a +b \right ) \sqrt {-a b}}\right )}{2 \sqrt {-a b}\, d}\)	\(433\)

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 864 vs. \(2 (65) = 130\).

Time = 0.12 (sec) , antiderivative size = 2032, normalized size of antiderivative = 27.09 \[ \int \frac {\text {sech}^6(c+d x)}{a+b \tanh ^2(c+d x)} \, dx=\text {Too large to display} \] Input:

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

Output:

[1/6*(12*(a^2*b + a*b^2)*cosh(d*x + c)^4 + 48*(a^2*b + a*b^2)*cosh(d*x + c 
)*sinh(d*x + c)^3 + 12*(a^2*b + a*b^2)*sinh(d*x + c)^4 + 12*a^2*b + 20*a*b 
^2 + 24*(a^2*b + 2*a*b^2)*cosh(d*x + c)^2 + 24*(a^2*b + 2*a*b^2 + 3*(a^2*b 
 + a*b^2)*cosh(d*x + c)^2)*sinh(d*x + c)^2 - 3*((a^2 + 2*a*b + b^2)*cosh(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 + 3*(a^2 + 2*a*b + b^2)*cosh(d*x + c)^4 + 3*(5 
*(a^2 + 2*a*b + b^2)*cosh(d*x + c)^2 + a^2 + 2*a*b + b^2)*sinh(d*x + c)^4 
+ 4*(5*(a^2 + 2*a*b + b^2)*cosh(d*x + c)^3 + 3*(a^2 + 2*a*b + b^2)*cosh(d* 
x + c))*sinh(d*x + c)^3 + 3*(a^2 + 2*a*b + b^2)*cosh(d*x + c)^2 + 3*(5*(a^ 
2 + 2*a*b + b^2)*cosh(d*x + c)^4 + 6*(a^2 + 2*a*b + b^2)*cosh(d*x + c)^2 + 
 a^2 + 2*a*b + b^2)*sinh(d*x + c)^2 + a^2 + 2*a*b + b^2 + 6*((a^2 + 2*a*b 
+ b^2)*cosh(d*x + c)^5 + 2*(a^2 + 2*a*b + b^2)*cosh(d*x + c)^3 + (a^2 + 2* 
a*b + b^2)*cosh(d*x + c))*sinh(d*x + c))*sqrt(-a*b)*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 + b)*cosh(d*x + c)^2 + 2*(a + b)*cosh(d*x + 
 c)*sinh(d*x + c) + (a + b)*sinh(d*x + c)^2 + a - b)*sqrt(-a*b))/((a + b)* 
cosh(d*x + c)^4 + 4*(a + b)*cosh(d*x + c)*sinh(d*x + c)^3 + (a + b)*sin...
 

Sympy [F]

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

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

Output:

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 140 vs. \(2 (65) = 130\).

Time = 0.16 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.87 \[ \int \frac {\text {sech}^6(c+d x)}{a+b \tanh ^2(c+d x)} \, dx=-\frac {2 \, {\left (6 \, {\left (a + 2 \, b\right )} e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, {\left (a + b\right )} e^{\left (-4 \, d x - 4 \, c\right )} + 3 \, a + 5 \, b\right )}}{3 \, {\left (3 \, b^{2} e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, b^{2} e^{\left (-4 \, d x - 4 \, c\right )} + b^{2} e^{\left (-6 \, d x - 6 \, c\right )} + b^{2}\right )} d} - \frac {{\left (a^{2} + 2 \, a b + b^{2}\right )} \arctan \left (\frac {{\left (a + b\right )} e^{\left (-2 \, d x - 2 \, c\right )} + a - b}{2 \, \sqrt {a b}}\right )}{\sqrt {a b} b^{2} d} \] Input:

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

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 135 vs. \(2 (65) = 130\).

Time = 0.31 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.80 \[ \int \frac {\text {sech}^6(c+d x)}{a+b \tanh ^2(c+d x)} \, dx=\frac {\frac {3 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \arctan \left (\frac {a e^{\left (2 \, d x + 2 \, c\right )} + b e^{\left (2 \, d x + 2 \, c\right )} + a - b}{2 \, \sqrt {a b}}\right )}{\sqrt {a b} b^{2}} + \frac {2 \, {\left (3 \, a e^{\left (4 \, d x + 4 \, c\right )} + 3 \, b e^{\left (4 \, d x + 4 \, c\right )} + 6 \, a e^{\left (2 \, d x + 2 \, c\right )} + 12 \, b e^{\left (2 \, d x + 2 \, c\right )} + 3 \, a + 5 \, b\right )}}{b^{2} {\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{3}}}{3 \, d} \] Input:

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

Output:

1/3*(3*(a^2 + 2*a*b + 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)*b^2) + 2*(3*a*e^(4*d*x + 4*c) + 3*b*e^(4 
*d*x + 4*c) + 6*a*e^(2*d*x + 2*c) + 12*b*e^(2*d*x + 2*c) + 3*a + 5*b)/(b^2 
*(e^(2*d*x + 2*c) + 1)^3))/d
 

Mupad [B] (verification not implemented)

Time = 2.89 (sec) , antiderivative size = 252, normalized size of antiderivative = 3.36 \[ \int \frac {\text {sech}^6(c+d x)}{a+b \tanh ^2(c+d x)} \, dx=\frac {4}{b\,d\,\left (2\,{\mathrm {e}}^{2\,c+2\,d\,x}+{\mathrm {e}}^{4\,c+4\,d\,x}+1\right )}-\frac {8}{3\,b\,d\,\left (3\,{\mathrm {e}}^{2\,c+2\,d\,x}+3\,{\mathrm {e}}^{4\,c+4\,d\,x}+{\mathrm {e}}^{6\,c+6\,d\,x}+1\right )}+\frac {2\,\left (a+b\right )}{b^2\,d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}+1\right )}+\frac {\ln \left (-\frac {4\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (a+b\right )}{b^2}-\frac {2\,\left (a+b\right )\,\left (a+b+a\,{\mathrm {e}}^{2\,c+2\,d\,x}-b\,{\mathrm {e}}^{2\,c+2\,d\,x}\right )}{\sqrt {-a}\,b^{5/2}}\right )\,{\left (a+b\right )}^2}{2\,\sqrt {-a}\,b^{5/2}\,d}-\frac {\ln \left (\frac {2\,\left (a+b\right )\,\left (a+b+a\,{\mathrm {e}}^{2\,c+2\,d\,x}-b\,{\mathrm {e}}^{2\,c+2\,d\,x}\right )}{\sqrt {-a}\,b^{5/2}}-\frac {4\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (a+b\right )}{b^2}\right )\,{\left (a+b\right )}^2}{2\,\sqrt {-a}\,b^{5/2}\,d} \] Input:

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

Output:

4/(b*d*(2*exp(2*c + 2*d*x) + exp(4*c + 4*d*x) + 1)) - 8/(3*b*d*(3*exp(2*c 
+ 2*d*x) + 3*exp(4*c + 4*d*x) + exp(6*c + 6*d*x) + 1)) + (2*(a + b))/(b^2* 
d*(exp(2*c + 2*d*x) + 1)) + (log(- (4*exp(2*c + 2*d*x)*(a + b))/b^2 - (2*( 
a + b)*(a + b + a*exp(2*c + 2*d*x) - b*exp(2*c + 2*d*x)))/((-a)^(1/2)*b^(5 
/2)))*(a + b)^2)/(2*(-a)^(1/2)*b^(5/2)*d) - (log((2*(a + b)*(a + b + a*exp 
(2*c + 2*d*x) - b*exp(2*c + 2*d*x)))/((-a)^(1/2)*b^(5/2)) - (4*exp(2*c + 2 
*d*x)*(a + b))/b^2)*(a + b)^2)/(2*(-a)^(1/2)*b^(5/2)*d)
 

Reduce [B] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 1042, normalized size of antiderivative = 13.89 \[ \int \frac {\text {sech}^6(c+d x)}{a+b \tanh ^2(c+d x)} \, dx =\text {Too large to display} \] Input:

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

Output:

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