\(\int \frac {\text {csch}^3(c+d x)}{(a+b \tanh ^2(c+d x))^2} \, dx\) [39]

Optimal result

Integrand size = 23, antiderivative size = 141 \[ \int \frac {\text {csch}^3(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^2} \, dx=\frac {(a+4 b) \text {arctanh}(\cosh (c+d x))}{2 a^3 d}-\frac {\sqrt {b} (3 a+4 b) \text {arctanh}\left (\frac {\sqrt {b} \text {sech}(c+d x)}{\sqrt {a+b}}\right )}{2 a^3 \sqrt {a+b} d}-\frac {\coth (c+d x) \text {csch}(c+d x)}{2 a d \left (a+b-b \text {sech}^2(c+d x)\right )}-\frac {b \text {sech}(c+d x)}{a^2 d \left (a+b-b \text {sech}^2(c+d x)\right )} \] Output:

1/2*(a+4*b)*arctanh(cosh(d*x+c))/a^3/d-1/2*b^(1/2)*(3*a+4*b)*arctanh(b^(1/ 
2)*sech(d*x+c)/(a+b)^(1/2))/a^3/(a+b)^(1/2)/d-1/2*coth(d*x+c)*csch(d*x+c)/ 
a/d/(a+b-b*sech(d*x+c)^2)-b*sech(d*x+c)/a^2/d/(a+b-b*sech(d*x+c)^2)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 3.53 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.57 \[ \int \frac {\text {csch}^3(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^2} \, dx=-\frac {\frac {4 i \sqrt {b} (3 a+4 b) \arctan \left (\frac {-i \sqrt {a+b}-\sqrt {a} \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {b}}\right )}{\sqrt {a+b}}+\frac {4 i \sqrt {b} (3 a+4 b) \arctan \left (\frac {-i \sqrt {a+b}+\sqrt {a} \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {b}}\right )}{\sqrt {a+b}}+\frac {8 a b \cosh (c+d x)}{a-b+(a+b) \cosh (2 (c+d x))}+a \text {csch}^2\left (\frac {1}{2} (c+d x)\right )-4 (a+4 b) \log \left (\cosh \left (\frac {1}{2} (c+d x)\right )\right )+4 (a+4 b) \log \left (\sinh \left (\frac {1}{2} (c+d x)\right )\right )+a \text {sech}^2\left (\frac {1}{2} (c+d x)\right )}{8 a^3 d} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.65 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.06, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {3042, 26, 4147, 373, 402, 27, 397, 219, 221}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 26

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

\(\Big \downarrow \) 4147

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

\(\Big \downarrow \) 373

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

\(\Big \downarrow \) 402

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 397

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

\(\Big \downarrow \) 219

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

-((Sech[c + d*x]/(2*a*(1 - Sech[c + d*x]^2)*(a + b - b*Sech[c + d*x]^2)) - 
 ((((a + 4*b)*ArcTanh[Sech[c + d*x]])/a - (Sqrt[b]*(3*a + 4*b)*ArcTanh[(Sq 
rt[b]*Sech[c + d*x])/Sqrt[a + b]])/(a*Sqrt[a + b]))/a - (2*b*Sech[c + d*x] 
)/(a*(a + b - b*Sech[c + d*x]^2)))/(2*a))/d)
 

Defintions of rubi rules used

rule 26

Int[(Complex[0, a_])*(Fx_), x_Symbol] :> Simp[(Complex[Identity[0], a])   I 
nt[Fx, x], x] /; FreeQ[a, x] && EqQ[a^2, 1]
 

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 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 221

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x 
/Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
 

rule 373

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ 
), x_Symbol] :> Simp[e*(e*x)^(m - 1)*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 
1)/(2*(b*c - a*d)*(p + 1))), x] - Simp[e^2/(2*(b*c - a*d)*(p + 1))   Int[(e 
*x)^(m - 2)*(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[c*(m - 1) + d*(m + 2*p + 
 2*q + 3)*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 
 0] && LtQ[p, -1] && GtQ[m, 1] && LeQ[m, 3] && IntBinomialQ[a, b, c, d, e, 
m, 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 4147

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

Maple [A] (verified)

Time = 5.12 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.33

method	result	size
derivativedivides	\(\frac {\frac {\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{8 a^{2}}+\frac {2 b \left (\frac {\left (-\frac {a}{2}-b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-\frac {a}{2}}{\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a +2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +4 b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+a}-\frac {\left (3 a +4 b \right ) \operatorname {arctanh}\left (\frac {2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 a +4 b}{4 \sqrt {a b +b^{2}}}\right )}{4 \sqrt {a b +b^{2}}}\right )}{a^{3}}-\frac {1}{8 a^{2} \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\frac {\left (-2 a -8 b \right ) \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 a^{3}}}{d}\)	\(187\)
default	\(\frac {\frac {\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{8 a^{2}}+\frac {2 b \left (\frac {\left (-\frac {a}{2}-b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-\frac {a}{2}}{\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a +2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +4 b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+a}-\frac {\left (3 a +4 b \right ) \operatorname {arctanh}\left (\frac {2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 a +4 b}{4 \sqrt {a b +b^{2}}}\right )}{4 \sqrt {a b +b^{2}}}\right )}{a^{3}}-\frac {1}{8 a^{2} \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\frac {\left (-2 a -8 b \right ) \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 a^{3}}}{d}\)	\(187\)
risch	\(-\frac {{\mathrm e}^{d x +c} \left ({\mathrm e}^{6 d x +6 c} a +2 \,{\mathrm e}^{6 d x +6 c} b +3 \,{\mathrm e}^{4 d x +4 c} a -2 b \,{\mathrm e}^{4 d x +4 c}+3 \,{\mathrm e}^{2 d x +2 c} a -2 \,{\mathrm e}^{2 d x +2 c} b +a +2 b \right )}{d \,a^{2} \left ({\mathrm e}^{2 d x +2 c}-1\right )^{2} \left ({\mathrm e}^{4 d x +4 c} a +b \,{\mathrm e}^{4 d x +4 c}+2 \,{\mathrm e}^{2 d x +2 c} a -2 \,{\mathrm e}^{2 d x +2 c} b +a +b \right )}-\frac {\ln \left ({\mathrm e}^{d x +c}-1\right )}{2 a^{2} d}-\frac {2 \ln \left ({\mathrm e}^{d x +c}-1\right ) b}{d \,a^{3}}+\frac {\ln \left ({\mathrm e}^{d x +c}+1\right )}{2 a^{2} d}+\frac {2 \ln \left ({\mathrm e}^{d x +c}+1\right ) b}{d \,a^{3}}+\frac {3 \sqrt {\left (a +b \right ) b}\, \ln \left ({\mathrm e}^{2 d x +2 c}-\frac {2 \sqrt {\left (a +b \right ) b}\, {\mathrm e}^{d x +c}}{a +b}+1\right )}{4 \left (a +b \right ) d \,a^{2}}+\frac {\sqrt {\left (a +b \right ) b}\, \ln \left ({\mathrm e}^{2 d x +2 c}-\frac {2 \sqrt {\left (a +b \right ) b}\, {\mathrm e}^{d x +c}}{a +b}+1\right ) b}{\left (a +b \right ) d \,a^{3}}-\frac {3 \sqrt {\left (a +b \right ) b}\, \ln \left ({\mathrm e}^{2 d x +2 c}+\frac {2 \sqrt {\left (a +b \right ) b}\, {\mathrm e}^{d x +c}}{a +b}+1\right )}{4 \left (a +b \right ) d \,a^{2}}-\frac {\sqrt {\left (a +b \right ) b}\, \ln \left ({\mathrm e}^{2 d x +2 c}+\frac {2 \sqrt {\left (a +b \right ) b}\, {\mathrm e}^{d x +c}}{a +b}+1\right ) b}{\left (a +b \right ) d \,a^{3}}\)	\(435\)

Input:

int(csch(d*x+c)^3/(a+tanh(d*x+c)^2*b)^2,x,method=_RETURNVERBOSE)
 

Output:

1/d*(1/8*tanh(1/2*d*x+1/2*c)^2/a^2+2*b/a^3*(((-1/2*a-b)*tanh(1/2*d*x+1/2*c 
)^2-1/2*a)/(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)-1/4*(3*a+4*b)/(a*b+b^2)^(1/2)*arctanh(1/4*(2*tanh(1/2*d*x 
+1/2*c)^2*a+2*a+4*b)/(a*b+b^2)^(1/2)))-1/8/a^2/tanh(1/2*d*x+1/2*c)^2+1/4/a 
^3*(-2*a-8*b)*ln(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. 3368 vs. \(2 (132) = 264\).

Time = 0.18 (sec) , antiderivative size = 6335, normalized size of antiderivative = 44.93 \[ \int \frac {\text {csch}^3(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^2} \, dx=\text {Too large to display} \] Input:

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

Output:

Too large to include
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F(-2)]

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

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.34 (sec) , antiderivative size = 2493, normalized size of antiderivative = 17.68 \[ \int \frac {\text {csch}^3(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^2} \, dx =\text {Too large to display} \] Input:

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

Output:

(3*e**(8*c + 8*d*x)*sqrt(b)*sqrt(a + b)*log(e**(2*c + 2*d*x)*sqrt(a + b) + 
 sqrt(a + b) - 2*e**(c + d*x)*sqrt(b))*a**2 + 7*e**(8*c + 8*d*x)*sqrt(b)*s 
qrt(a + b)*log(e**(2*c + 2*d*x)*sqrt(a + b) + sqrt(a + b) - 2*e**(c + d*x) 
*sqrt(b))*a*b + 4*e**(8*c + 8*d*x)*sqrt(b)*sqrt(a + b)*log(e**(2*c + 2*d*x 
)*sqrt(a + b) + sqrt(a + b) - 2*e**(c + d*x)*sqrt(b))*b**2 - 3*e**(8*c + 8 
*d*x)*sqrt(b)*sqrt(a + b)*log(e**(2*c + 2*d*x)*sqrt(a + b) + sqrt(a + b) + 
 2*e**(c + d*x)*sqrt(b))*a**2 - 7*e**(8*c + 8*d*x)*sqrt(b)*sqrt(a + b)*log 
(e**(2*c + 2*d*x)*sqrt(a + b) + sqrt(a + b) + 2*e**(c + d*x)*sqrt(b))*a*b 
- 4*e**(8*c + 8*d*x)*sqrt(b)*sqrt(a + b)*log(e**(2*c + 2*d*x)*sqrt(a + b) 
+ sqrt(a + b) + 2*e**(c + d*x)*sqrt(b))*b**2 - 12*e**(6*c + 6*d*x)*sqrt(b) 
*sqrt(a + b)*log(e**(2*c + 2*d*x)*sqrt(a + b) + sqrt(a + b) - 2*e**(c + d* 
x)*sqrt(b))*a*b - 16*e**(6*c + 6*d*x)*sqrt(b)*sqrt(a + b)*log(e**(2*c + 2* 
d*x)*sqrt(a + b) + sqrt(a + b) - 2*e**(c + d*x)*sqrt(b))*b**2 + 12*e**(6*c 
 + 6*d*x)*sqrt(b)*sqrt(a + b)*log(e**(2*c + 2*d*x)*sqrt(a + b) + sqrt(a + 
b) + 2*e**(c + d*x)*sqrt(b))*a*b + 16*e**(6*c + 6*d*x)*sqrt(b)*sqrt(a + b) 
*log(e**(2*c + 2*d*x)*sqrt(a + b) + sqrt(a + b) + 2*e**(c + d*x)*sqrt(b))* 
b**2 - 6*e**(4*c + 4*d*x)*sqrt(b)*sqrt(a + b)*log(e**(2*c + 2*d*x)*sqrt(a 
+ b) + sqrt(a + b) - 2*e**(c + d*x)*sqrt(b))*a**2 + 10*e**(4*c + 4*d*x)*sq 
rt(b)*sqrt(a + b)*log(e**(2*c + 2*d*x)*sqrt(a + b) + sqrt(a + b) - 2*e**(c 
 + d*x)*sqrt(b))*a*b + 24*e**(4*c + 4*d*x)*sqrt(b)*sqrt(a + b)*log(e**(...