\(\int \frac {\text {sech}^7(c+d x)}{(a+b \tanh ^2(c+d x))^2} \, dx\) [124]

Optimal result

Integrand size = 23, antiderivative size = 155 \[ \int \frac {\text {sech}^7(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^2} \, dx=\frac {(4 a+5 b) \arctan (\sinh (c+d x))}{2 b^3 d}-\frac {(4 a-b) (a+b)^{3/2} \arctan \left (\frac {\sqrt {a+b} \sinh (c+d x)}{\sqrt {a}}\right )}{2 a^{3/2} b^3 d}+\frac {(a+b) (2 a+b) \sinh (c+d x)}{2 a b^2 d \left (a+(a+b) \sinh ^2(c+d x)\right )}-\frac {\text {sech}(c+d x) \tanh (c+d x)}{2 b d \left (a+(a+b) \sinh ^2(c+d x)\right )} \] Output:

1/2*(4*a+5*b)*arctan(sinh(d*x+c))/b^3/d-1/2*(4*a-b)*(a+b)^(3/2)*arctan((a+ 
b)^(1/2)*sinh(d*x+c)/a^(1/2))/a^(3/2)/b^3/d+1/2*(a+b)*(2*a+b)*sinh(d*x+c)/ 
a/b^2/d/(a+(a+b)*sinh(d*x+c)^2)-1/2*sech(d*x+c)*tanh(d*x+c)/b/d/(a+(a+b)*s 
inh(d*x+c)^2)
 

Mathematica [A] (verified)

Time = 0.94 (sec) , antiderivative size = 265, normalized size of antiderivative = 1.71 \[ \int \frac {\text {sech}^7(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^2} \, dx=\frac {2 \sqrt {a} b (a+b)^{5/2} \sinh (c+d x)+(a-b) \left ((4 a-b) (a+b)^2 \arctan \left (\frac {\sqrt {a} \text {csch}(c+d x)}{\sqrt {a+b}}\right )+2 a^{3/2} \sqrt {a+b} (4 a+5 b) \arctan \left (\tanh \left (\frac {1}{2} (c+d x)\right )\right )+a^{3/2} b \sqrt {a+b} \text {sech}(c+d x) \tanh (c+d x)\right )+(a+b) \cosh (2 (c+d x)) \left ((4 a-b) (a+b)^2 \arctan \left (\frac {\sqrt {a} \text {csch}(c+d x)}{\sqrt {a+b}}\right )+2 a^{3/2} \sqrt {a+b} (4 a+5 b) \arctan \left (\tanh \left (\frac {1}{2} (c+d x)\right )\right )+a^{3/2} b \sqrt {a+b} \text {sech}(c+d x) \tanh (c+d x)\right )}{2 a^{3/2} b^3 \sqrt {a+b} d (a-b+(a+b) \cosh (2 (c+d x)))} \] Input:

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

Output:

(2*Sqrt[a]*b*(a + b)^(5/2)*Sinh[c + d*x] + (a - b)*((4*a - b)*(a + b)^2*Ar 
cTan[(Sqrt[a]*Csch[c + d*x])/Sqrt[a + b]] + 2*a^(3/2)*Sqrt[a + b]*(4*a + 5 
*b)*ArcTan[Tanh[(c + d*x)/2]] + a^(3/2)*b*Sqrt[a + b]*Sech[c + d*x]*Tanh[c 
 + d*x]) + (a + b)*Cosh[2*(c + d*x)]*((4*a - b)*(a + b)^2*ArcTan[(Sqrt[a]* 
Csch[c + d*x])/Sqrt[a + b]] + 2*a^(3/2)*Sqrt[a + b]*(4*a + 5*b)*ArcTan[Tan 
h[(c + d*x)/2]] + a^(3/2)*b*Sqrt[a + b]*Sech[c + d*x]*Tanh[c + d*x]))/(2*a 
^(3/2)*b^3*Sqrt[a + b]*d*(a - b + (a + b)*Cosh[2*(c + d*x)]))
 

Rubi [A] (verified)

Time = 0.71 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.05, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {3042, 4159, 316, 402, 27, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4159

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

\(\Big \downarrow \) 316

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

\(\Big \downarrow \) 402

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 397

\(\displaystyle \frac {\frac {\frac {(a+b) (2 a+b) \sinh (c+d x)}{a b \left ((a+b) \sinh ^2(c+d x)+a\right )}-\frac {\frac {(4 a-b) (a+b)^2 \int \frac {1}{(a+b) \sinh ^2(c+d x)+a}d\sinh (c+d x)}{b}-\frac {a (4 a+5 b) \int \frac {1}{\sinh ^2(c+d x)+1}d\sinh (c+d x)}{b}}{a b}}{2 b}-\frac {\sinh (c+d x)}{2 b \left (\sinh ^2(c+d x)+1\right ) \left ((a+b) \sinh ^2(c+d x)+a\right )}}{d}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {\frac {\frac {(a+b) (2 a+b) \sinh (c+d x)}{a b \left ((a+b) \sinh ^2(c+d x)+a\right )}-\frac {\frac {(4 a-b) (a+b)^2 \int \frac {1}{(a+b) \sinh ^2(c+d x)+a}d\sinh (c+d x)}{b}-\frac {a (4 a+5 b) \arctan (\sinh (c+d x))}{b}}{a b}}{2 b}-\frac {\sinh (c+d x)}{2 b \left (\sinh ^2(c+d x)+1\right ) \left ((a+b) \sinh ^2(c+d x)+a\right )}}{d}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {\frac {\frac {(a+b) (2 a+b) \sinh (c+d x)}{a b \left ((a+b) \sinh ^2(c+d x)+a\right )}-\frac {\frac {(4 a-b) (a+b)^{3/2} \arctan \left (\frac {\sqrt {a+b} \sinh (c+d x)}{\sqrt {a}}\right )}{\sqrt {a} b}-\frac {a (4 a+5 b) \arctan (\sinh (c+d x))}{b}}{a b}}{2 b}-\frac {\sinh (c+d x)}{2 b \left (\sinh ^2(c+d x)+1\right ) \left ((a+b) \sinh ^2(c+d x)+a\right )}}{d}\)

Input:

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

Output:

(-1/2*Sinh[c + d*x]/(b*(1 + Sinh[c + d*x]^2)*(a + (a + b)*Sinh[c + d*x]^2) 
) + (-((-((a*(4*a + 5*b)*ArcTan[Sinh[c + d*x]])/b) + ((4*a - b)*(a + b)^(3 
/2)*ArcTan[(Sqrt[a + b]*Sinh[c + d*x])/Sqrt[a]])/(Sqrt[a]*b))/(a*b)) + ((a 
 + b)*(2*a + b)*Sinh[c + d*x])/(a*b*(a + (a + b)*Sinh[c + d*x]^2)))/(2*b)) 
/d
 

Defintions of rubi rules used

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 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. \(350\) vs. \(2(139)=278\).

Time = 0.62 (sec) , antiderivative size = 351, normalized size of antiderivative = 2.26

\[\frac {-\frac {2 \left (\frac {\frac {b \left (a^{2}+2 a b +b^{2}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{2 a}-\frac {b \left (a^{2}+2 a b +b^{2}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a}}{\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a +2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +4 b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+a}+\frac {\left (4 a^{3}+7 a^{2} b +2 b^{2} a -b^{3}\right ) \left (\frac {\left (\sqrt {\left (a +b \right ) b}+b \right ) \arctan \left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {\left (a +b \right ) b}+a +2 b \right ) a}}\right )}{2 a \sqrt {\left (a +b \right ) b}\, \sqrt {\left (2 \sqrt {\left (a +b \right ) b}+a +2 b \right ) a}}-\frac {\left (\sqrt {\left (a +b \right ) b}-b \right ) \operatorname {arctanh}\left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {\left (a +b \right ) b}-a -2 b \right ) a}}\right )}{2 a \sqrt {\left (a +b \right ) b}\, \sqrt {\left (2 \sqrt {\left (a +b \right ) b}-a -2 b \right ) a}}\right )}{2}\right )}{b^{3}}+\frac {\frac {2 \left (-\frac {b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{2}+\frac {b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{2}\right )}{\left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right )^{2}}+\left (5 b +4 a \right ) \arctan \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{3}}}{d}\]

Input:

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

Output:

1/d*(-2/b^3*((1/2*b*(a^2+2*a*b+b^2)/a*tanh(1/2*d*x+1/2*c)^3-1/2*b*(a^2+2*a 
*b+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)+1/2*(4*a^3+7*a^2*b+2*a*b^2-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)*arct 
anh(a*tanh(1/2*d*x+1/2*c)/((2*((a+b)*b)^(1/2)-a-2*b)*a)^(1/2))))+2/b^3*((- 
1/2*b*tanh(1/2*d*x+1/2*c)^3+1/2*b*tanh(1/2*d*x+1/2*c))/(tanh(1/2*d*x+1/2*c 
)^2+1)^2+1/2*(5*b+4*a)*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. 3490 vs. \(2 (139) = 278\).

Time = 0.18 (sec) , antiderivative size = 6396, normalized size of antiderivative = 41.26 \[ \int \frac {\text {sech}^7(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^2} \, dx=\text {Too large to display} \] Input:

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

Output:

Too large to include
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F(-2)]

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

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.36 (sec) , antiderivative size = 2051, normalized size of antiderivative = 13.23 \[ \int \frac {\text {sech}^7(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^2} \, dx =\text {Too large to display} \] Input:

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

Output:

(8*e**(8*c + 8*d*x)*atan(e**(c + d*x))*a**4 + 18*e**(8*c + 8*d*x)*atan(e** 
(c + d*x))*a**3*b + 10*e**(8*c + 8*d*x)*atan(e**(c + d*x))*a**2*b**2 + 32* 
e**(6*c + 6*d*x)*atan(e**(c + d*x))*a**4 + 40*e**(6*c + 6*d*x)*atan(e**(c 
+ d*x))*a**3*b + 48*e**(4*c + 4*d*x)*atan(e**(c + d*x))*a**4 + 44*e**(4*c 
+ 4*d*x)*atan(e**(c + d*x))*a**3*b - 20*e**(4*c + 4*d*x)*atan(e**(c + d*x) 
)*a**2*b**2 + 32*e**(2*c + 2*d*x)*atan(e**(c + d*x))*a**4 + 40*e**(2*c + 2 
*d*x)*atan(e**(c + d*x))*a**3*b + 8*atan(e**(c + d*x))*a**4 + 18*atan(e**( 
c + d*x))*a**3*b + 10*atan(e**(c + d*x))*a**2*b**2 - 4*e**(8*c + 8*d*x)*sq 
rt(a)*sqrt(a + b)*atan((e**(c + d*x)*sqrt(a + b) - sqrt(b))/sqrt(a))*a**3 
- 7*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*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**2 + e**(8*c + 8*d*x)*sqrt( 
a)*sqrt(a + b)*atan((e**(c + d*x)*sqrt(a + b) - sqrt(b))/sqrt(a))*b**3 - 1 
6*e**(6*c + 6*d*x)*sqrt(a)*sqrt(a + b)*atan((e**(c + d*x)*sqrt(a + b) - sq 
rt(b))/sqrt(a))*a**3 - 12*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 + 4*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**2 - 
24*e**(4*c + 4*d*x)*sqrt(a)*sqrt(a + b)*atan((e**(c + d*x)*sqrt(a + b) - s 
qrt(b))/sqrt(a))*a**3 - 10*e**(4*c + 4*d*x)*sqrt(a)*sqrt(a + b)*atan((e**( 
c + d*x)*sqrt(a + b) - sqrt(b))/sqrt(a))*a**2*b + 12*e**(4*c + 4*d*x)*s...