\(\int \frac {\sinh ^3(c+d x)}{(a+b \tanh ^2(c+d x))^2} \, dx\) [34]

Optimal result
Mathematica [C] (verified)
Rubi [A] (verified)
Maple [B] (verified)
Fricas [B] (verification not implemented)
Sympy [F]
Maxima [F]
Giac [F(-2)]
Mupad [F(-1)]
Reduce [B] (verification not implemented)

Optimal result

Integrand size = 23, antiderivative size = 124 \[ \int \frac {\sinh ^3(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^2} \, dx=\frac {(3 a-2 b) \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \text {sech}(c+d x)}{\sqrt {a+b}}\right )}{2 (a+b)^{7/2} d}-\frac {(a-b) \cosh (c+d x)}{(a+b)^3 d}+\frac {\cosh ^3(c+d x)}{3 (a+b)^2 d}+\frac {a b \text {sech}(c+d x)}{2 (a+b)^3 d \left (a+b-b \text {sech}^2(c+d x)\right )} \] Output:

1/2*(3*a-2*b)*b^(1/2)*arctanh(b^(1/2)*sech(d*x+c)/(a+b)^(1/2))/(a+b)^(7/2) 
/d-(a-b)*cosh(d*x+c)/(a+b)^3/d+1/3*cosh(d*x+c)^3/(a+b)^2/d+1/2*a*b*sech(d* 
x+c)/(a+b)^3/d/(a+b-b*sech(d*x+c)^2)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 1.38 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.29 \[ \int \frac {\sinh ^3(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^2} \, dx=\frac {\frac {6 i (3 a-2 b) \sqrt {b} \left (\arctan \left (\frac {-i \sqrt {a+b}-\sqrt {a} \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {b}}\right )+\arctan \left (\frac {-i \sqrt {a+b}+\sqrt {a} \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {b}}\right )\right )}{(a+b)^{7/2}}+\frac {3 \cosh (c+d x) \left (5 b+a \left (-3+\frac {4 b}{a-b+(a+b) \cosh (2 (c+d x))}\right )\right )}{(a+b)^3}+\frac {\cosh (3 (c+d x))}{(a+b)^2}}{12 d} \] Input:

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

Output:

(((6*I)*(3*a - 2*b)*Sqrt[b]*(ArcTan[((-I)*Sqrt[a + b] - Sqrt[a]*Tanh[(c + 
d*x)/2])/Sqrt[b]] + ArcTan[((-I)*Sqrt[a + b] + Sqrt[a]*Tanh[(c + d*x)/2])/ 
Sqrt[b]]))/(a + b)^(7/2) + (3*Cosh[c + d*x]*(5*b + a*(-3 + (4*b)/(a - b + 
(a + b)*Cosh[2*(c + d*x)]))))/(a + b)^3 + Cosh[3*(c + d*x)]/(a + b)^2)/(12 
*d)
 

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.02, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3042, 26, 4147, 25, 361, 1584, 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.

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

\(\Big \downarrow \) 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

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

\(\Big \downarrow \) 4147

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 361

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

\(\Big \downarrow \) 1584

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\frac {a b \text {sech}(c+d x)}{2 (a+b)^3 \left (a-b \text {sech}^2(c+d x)+b\right )}-\frac {1}{2} b \left (-\frac {(3 a-2 b) \text {arctanh}\left (\frac {\sqrt {b} \text {sech}(c+d x)}{\sqrt {a+b}}\right )}{\sqrt {b} (a+b)^{7/2}}-\frac {2 \cosh ^3(c+d x)}{3 b (a+b)^2}+\frac {2 (a-b) \cosh (c+d x)}{b (a+b)^3}\right )}{d}\)

Input:

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

Output:

(-1/2*(b*(-(((3*a - 2*b)*ArcTanh[(Sqrt[b]*Sech[c + d*x])/Sqrt[a + b]])/(Sq 
rt[b]*(a + b)^(7/2))) + (2*(a - b)*Cosh[c + d*x])/(b*(a + b)^3) - (2*Cosh[ 
c + d*x]^3)/(3*b*(a + b)^2))) + (a*b*Sech[c + d*x])/(2*(a + b)^3*(a + b - 
b*Sech[c + d*x]^2)))/d
 

Defintions of rubi rules used

rule 25
Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 361
Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] : 
> Simp[(-a)^(m/2 - 1)*(b*c - a*d)*x*((a + b*x^2)^(p + 1)/(2*b^(m/2 + 1)*(p 
+ 1))), x] + Simp[1/(2*b^(m/2 + 1)*(p + 1))   Int[x^m*(a + b*x^2)^(p + 1)*E 
xpandToSum[2*b*(p + 1)*Together[(b^(m/2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c 
- a*d)*x^(-m + 2))/(a + b*x^2)] - ((-a)^(m/2 - 1)*(b*c - a*d))/x^m, x], x], 
 x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && ILtQ[m/ 
2, 0] && (IntegerQ[p] || EqQ[m + 2*p + 1, 0])
 

rule 1584
Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + ( 
c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(f*x)^m*(d + e*x^2)^q* 
(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] && NeQ[ 
b^2 - 4*a*c, 0] && IGtQ[p, 0] && IGtQ[q, -2]
 

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 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 [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(266\) vs. \(2(110)=220\).

Time = 27.62 (sec) , antiderivative size = 267, normalized size of antiderivative = 2.15

method result size
derivativedivides \(\frac {\frac {1}{3 \left (a +b \right )^{2} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {1}{2 \left (a +b \right )^{2} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {a -3 b}{2 \left (a +b \right )^{3} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {1}{3 \left (a +b \right )^{2} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {1}{2 \left (a +b \right )^{2} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {-a +3 b}{2 \left (a +b \right )^{3} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {4 b \left (\frac {\left (-\frac {a}{4}-\frac {b}{2}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-\frac {a}{4}}{\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a +2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +4 b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+a}-\frac {\left (3 a -2 b \right ) \operatorname {arctanh}\left (\frac {2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 a +4 b}{4 \sqrt {a b +b^{2}}}\right )}{8 \sqrt {a b +b^{2}}}\right )}{\left (a +b \right )^{3}}}{d}\) \(267\)
default \(\frac {\frac {1}{3 \left (a +b \right )^{2} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {1}{2 \left (a +b \right )^{2} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {a -3 b}{2 \left (a +b \right )^{3} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {1}{3 \left (a +b \right )^{2} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {1}{2 \left (a +b \right )^{2} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {-a +3 b}{2 \left (a +b \right )^{3} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {4 b \left (\frac {\left (-\frac {a}{4}-\frac {b}{2}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-\frac {a}{4}}{\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a +2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +4 b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+a}-\frac {\left (3 a -2 b \right ) \operatorname {arctanh}\left (\frac {2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 a +4 b}{4 \sqrt {a b +b^{2}}}\right )}{8 \sqrt {a b +b^{2}}}\right )}{\left (a +b \right )^{3}}}{d}\) \(267\)
risch \(\frac {{\mathrm e}^{3 d x +3 c}}{24 \left (a^{2}+2 a b +b^{2}\right ) d}-\frac {3 \,{\mathrm e}^{d x +c} a}{8 \left (a +b \right ) \left (a^{2}+2 a b +b^{2}\right ) d}+\frac {5 \,{\mathrm e}^{d x +c} b}{8 \left (a +b \right ) \left (a^{2}+2 a b +b^{2}\right ) d}-\frac {3 \,{\mathrm e}^{-d x -c} a}{8 \left (a^{3}+3 a^{2} b +3 b^{2} a +b^{3}\right ) d}+\frac {5 \,{\mathrm e}^{-d x -c} b}{8 \left (a^{3}+3 a^{2} b +3 b^{2} a +b^{3}\right ) d}+\frac {{\mathrm e}^{-3 d x -3 c}}{24 \left (a^{2}+2 a b +b^{2}\right ) d}+\frac {\left ({\mathrm e}^{2 d x +2 c}+1\right ) a b \,{\mathrm e}^{d x +c}}{\left ({\mathrm e}^{4 d x +4 c} a +b \,{\mathrm e}^{4 d x +4 c}+2 \,{\mathrm e}^{2 d x +2 c} a -2 \,{\mathrm e}^{2 d x +2 c} b +a +b \right ) d \left (a +b \right ) \left (a^{2}+2 a b +b^{2}\right )}+\frac {3 \sqrt {\left (a +b \right ) b}\, \ln \left ({\mathrm e}^{2 d x +2 c}+\frac {2 \sqrt {\left (a +b \right ) b}\, {\mathrm e}^{d x +c}}{a +b}+1\right ) a}{4 \left (a +b \right )^{4} d}-\frac {\sqrt {\left (a +b \right ) b}\, \ln \left ({\mathrm e}^{2 d x +2 c}+\frac {2 \sqrt {\left (a +b \right ) b}\, {\mathrm e}^{d x +c}}{a +b}+1\right ) b}{2 \left (a +b \right )^{4} d}-\frac {3 \sqrt {\left (a +b \right ) b}\, \ln \left ({\mathrm e}^{2 d x +2 c}-\frac {2 \sqrt {\left (a +b \right ) b}\, {\mathrm e}^{d x +c}}{a +b}+1\right ) a}{4 \left (a +b \right )^{4} d}+\frac {\sqrt {\left (a +b \right ) b}\, \ln \left ({\mathrm e}^{2 d x +2 c}-\frac {2 \sqrt {\left (a +b \right ) b}\, {\mathrm e}^{d x +c}}{a +b}+1\right ) b}{2 \left (a +b \right )^{4} d}\) \(480\)

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2650 vs. \(2 (113) = 226\).

Time = 0.19 (sec) , antiderivative size = 5025, normalized size of antiderivative = 40.52 \[ \int \frac {\sinh ^3(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^2} \, dx=\text {Too large to display} \] Input:

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

Output:

Too large to include
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

1/24*(a^2 + 2*a*b + b^2 + (a^2*e^(10*c) + 2*a*b*e^(10*c) + b^2*e^(10*c))*e 
^(10*d*x) - (7*a^2*e^(8*c) - 6*a*b*e^(8*c) - 13*b^2*e^(8*c))*e^(8*d*x) - 2 
*(13*a^2*e^(6*c) - 40*a*b*e^(6*c) + 7*b^2*e^(6*c))*e^(6*d*x) - 2*(13*a^2*e 
^(4*c) - 40*a*b*e^(4*c) + 7*b^2*e^(4*c))*e^(4*d*x) - (7*a^2*e^(2*c) - 6*a* 
b*e^(2*c) - 13*b^2*e^(2*c))*e^(2*d*x))/((a^4*d*e^(7*c) + 4*a^3*b*d*e^(7*c) 
 + 6*a^2*b^2*d*e^(7*c) + 4*a*b^3*d*e^(7*c) + b^4*d*e^(7*c))*e^(7*d*x) + 2* 
(a^4*d*e^(5*c) + 2*a^3*b*d*e^(5*c) - 2*a*b^3*d*e^(5*c) - b^4*d*e^(5*c))*e^ 
(5*d*x) + (a^4*d*e^(3*c) + 4*a^3*b*d*e^(3*c) + 6*a^2*b^2*d*e^(3*c) + 4*a*b 
^3*d*e^(3*c) + b^4*d*e^(3*c))*e^(3*d*x)) - 1/8*integrate(8*((3*a*b*e^(3*c) 
 - 2*b^2*e^(3*c))*e^(3*d*x) - (3*a*b*e^c - 2*b^2*e^c)*e^(d*x))/(a^4 + 4*a^ 
3*b + 6*a^2*b^2 + 4*a*b^3 + b^4 + (a^4*e^(4*c) + 4*a^3*b*e^(4*c) + 6*a^2*b 
^2*e^(4*c) + 4*a*b^3*e^(4*c) + b^4*e^(4*c))*e^(4*d*x) + 2*(a^4*e^(2*c) + 2 
*a^3*b*e^(2*c) - 2*a*b^3*e^(2*c) - b^4*e^(2*c))*e^(2*d*x)), x)
 

Giac [F(-2)]

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

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

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

Output:

int(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 = 1523, normalized size of antiderivative = 12.28 \[ \int \frac {\sinh ^3(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^2} \, dx =\text {Too large to display} \] Input:

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

Output:

( - 18*e**(7*c + 7*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 - 6*e**(7*c + 7*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 + 12*e**(7*c + 7*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 + 18*e**(7 
*c + 7*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 + 6*e**(7*c + 7*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 - 12*e**(7*c + 7*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 - 36*e**(5*c + 5*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 + 60*e**(5*c + 5*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 - 24* 
e**(5*c + 5*d*x)*sqrt(b)*sqrt(a + b)*log(e**(2*c + 2*d*x)*sqrt(a + b) + sq 
rt(a + b) - 2*e**(c + d*x)*sqrt(b))*b**2 + 36*e**(5*c + 5*d*x)*sqrt(b)*sqr 
t(a + b)*log(e**(2*c + 2*d*x)*sqrt(a + b) + sqrt(a + b) + 2*e**(c + d*x)*s 
qrt(b))*a**2 - 60*e**(5*c + 5*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 + 24*e**(5*c + 5 
*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 - 18*e**(3*c + 3*d*x)*sqrt(b)*sqrt(a + b)...