\(\int \frac {\cosh ^3(c+d x)}{a+b \text {csch}(c+d x)} \, dx\) [29]

Optimal result

Integrand size = 21, antiderivative size = 85 \[ \int \frac {\cosh ^3(c+d x)}{a+b \text {csch}(c+d x)} \, dx=-\frac {b \left (a^2+b^2\right ) \log (b+a \sinh (c+d x))}{a^4 d}+\frac {\left (a^2+b^2\right ) \sinh (c+d x)}{a^3 d}-\frac {b \sinh ^2(c+d x)}{2 a^2 d}+\frac {\sinh ^3(c+d x)}{3 a d} \] Output:

-b*(a^2+b^2)*ln(b+a*sinh(d*x+c))/a^4/d+(a^2+b^2)*sinh(d*x+c)/a^3/d-1/2*b*s 
inh(d*x+c)^2/a^2/d+1/3*sinh(d*x+c)^3/a/d
 

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.88 \[ \int \frac {\cosh ^3(c+d x)}{a+b \text {csch}(c+d x)} \, dx=\frac {-6 b \left (a^2+b^2\right ) \log (b+a \sinh (c+d x))+6 a \left (a^2+b^2\right ) \sinh (c+d x)-3 a^2 b \sinh ^2(c+d x)+2 a^3 \sinh ^3(c+d x)}{6 a^4 d} \] Input:

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

Output:

(-6*b*(a^2 + b^2)*Log[b + a*Sinh[c + d*x]] + 6*a*(a^2 + b^2)*Sinh[c + d*x] 
 - 3*a^2*b*Sinh[c + d*x]^2 + 2*a^3*Sinh[c + d*x]^3)/(6*a^4*d)
 

Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.88, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.524, Rules used = {3042, 4360, 26, 26, 3042, 26, 3316, 26, 27, 522, 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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\cos (i c+i d x)^3}{a+i b \csc (i c+i d x)}dx\)

\(\Big \downarrow \) 4360

\(\displaystyle \int \frac {i \sinh (c+d x) \cosh ^3(c+d x)}{i a \sinh (c+d x)+i b}dx\)

\(\Big \downarrow \) 26

\(\displaystyle i \int -\frac {i \cosh ^3(c+d x) \sinh (c+d x)}{b+a \sinh (c+d x)}dx\)

\(\Big \downarrow \) 26

\(\displaystyle \int \frac {\sinh (c+d x) \cosh ^3(c+d x)}{a \sinh (c+d x)+b}dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 26

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

\(\Big \downarrow \) 3316

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

\(\Big \downarrow \) 26

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 522

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

(-(b*(a^2 + b^2)*Log[b + a*Sinh[c + d*x]]) + a*(a^2 + b^2)*Sinh[c + d*x] - 
 (a^2*b*Sinh[c + d*x]^2)/2 + (a^3*Sinh[c + d*x]^3)/3)/(a^4*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 522

Int[((e_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_. 
), x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*(c + d*x)^n*(a + b*x^2)^p, x], 
x] /; FreeQ[{a, b, c, d, e, m, n}, x] && IGtQ[p, 0]
 

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 3316

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_ 
.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[1/(b^p* 
f)   Subst[Int[(a + x)^m*(c + (d/b)*x)^n*(b^2 - x^2)^((p - 1)/2), x], x, b* 
Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IntegerQ[(p - 1) 
/2] && NeQ[a^2 - b^2, 0]
 

rule 4360

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + 
(a_))^(m_.), x_Symbol] :> Int[(g*Cos[e + f*x])^p*((b + a*Sin[e + f*x])^m/Si 
n[e + f*x]^m), x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(243\) vs. \(2(81)=162\).

Time = 17.67 (sec) , antiderivative size = 244, normalized size of antiderivative = 2.87

method	result	size
risch	\(\frac {x b}{a^{2}}+\frac {x \,b^{3}}{a^{4}}+\frac {{\mathrm e}^{3 d x +3 c}}{24 d a}-\frac {b \,{\mathrm e}^{2 d x +2 c}}{8 a^{2} d}+\frac {3 \,{\mathrm e}^{d x +c}}{8 a d}+\frac {{\mathrm e}^{d x +c} b^{2}}{2 a^{3} d}-\frac {3 \,{\mathrm e}^{-d x -c}}{8 d a}-\frac {{\mathrm e}^{-d x -c} b^{2}}{2 a^{3} d}-\frac {b \,{\mathrm e}^{-2 d x -2 c}}{8 a^{2} d}-\frac {{\mathrm e}^{-3 d x -3 c}}{24 d a}+\frac {2 b c}{a^{2} d}+\frac {2 b^{3} c}{a^{4} d}-\frac {b \ln \left ({\mathrm e}^{2 d x +2 c}+\frac {2 b \,{\mathrm e}^{d x +c}}{a}-1\right )}{a^{2} d}-\frac {b^{3} \ln \left ({\mathrm e}^{2 d x +2 c}+\frac {2 b \,{\mathrm e}^{d x +c}}{a}-1\right )}{a^{4} d}\)	\(244\)
derivativedivides	\(\frac {-\frac {1}{3 a \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {a +b}{2 a^{2} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {2 a^{2}+a b +2 b^{2}}{2 a^{3} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\left (a^{2}+b^{2}\right ) b \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{4}}+\frac {2 b \left (-\frac {a^{2}}{2}-\frac {b^{2}}{2}\right ) \ln \left (-\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +2 a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+b \right )}{a^{4}}-\frac {1}{3 a \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {-a +b}{2 a^{2} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {2 a^{2}-a b +2 b^{2}}{2 a^{3} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {\left (a^{2}+b^{2}\right ) b \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{4}}}{d}\)	\(245\)
default	\(\frac {-\frac {1}{3 a \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {a +b}{2 a^{2} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {2 a^{2}+a b +2 b^{2}}{2 a^{3} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\left (a^{2}+b^{2}\right ) b \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{4}}+\frac {2 b \left (-\frac {a^{2}}{2}-\frac {b^{2}}{2}\right ) \ln \left (-\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +2 a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+b \right )}{a^{4}}-\frac {1}{3 a \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {-a +b}{2 a^{2} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {2 a^{2}-a b +2 b^{2}}{2 a^{3} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {\left (a^{2}+b^{2}\right ) b \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{4}}}{d}\)	\(245\)

Input:

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

Output:

x*b/a^2+x/a^4*b^3+1/24/d/a*exp(3*d*x+3*c)-1/8*b/a^2/d*exp(2*d*x+2*c)+3/8/a 
/d*exp(d*x+c)+1/2/a^3/d*exp(d*x+c)*b^2-3/8/d/a*exp(-d*x-c)-1/2/a^3/d*exp(- 
d*x-c)*b^2-1/8*b/a^2/d*exp(-2*d*x-2*c)-1/24/d/a*exp(-3*d*x-3*c)+2*b/a^2/d* 
c+2*b^3/a^4/d*c-b/a^2/d*ln(exp(2*d*x+2*c)+2/a*b*exp(d*x+c)-1)-b^3/a^4/d*ln 
(exp(2*d*x+2*c)+2/a*b*exp(d*x+c)-1)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 652 vs. \(2 (81) = 162\).

Time = 0.08 (sec) , antiderivative size = 652, normalized size of antiderivative = 7.67 \[ \int \frac {\cosh ^3(c+d x)}{a+b \text {csch}(c+d x)} \, dx =\text {Too large to display} \] Input:

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

Output:

1/24*(a^3*cosh(d*x + c)^6 + a^3*sinh(d*x + c)^6 - 3*a^2*b*cosh(d*x + c)^5 
+ 24*(a^2*b + b^3)*d*x*cosh(d*x + c)^3 + 3*(2*a^3*cosh(d*x + c) - a^2*b)*s 
inh(d*x + c)^5 + 3*(3*a^3 + 4*a*b^2)*cosh(d*x + c)^4 + 3*(5*a^3*cosh(d*x + 
 c)^2 - 5*a^2*b*cosh(d*x + c) + 3*a^3 + 4*a*b^2)*sinh(d*x + c)^4 - 3*a^2*b 
*cosh(d*x + c) + 2*(10*a^3*cosh(d*x + c)^3 - 15*a^2*b*cosh(d*x + c)^2 + 12 
*(a^2*b + b^3)*d*x + 6*(3*a^3 + 4*a*b^2)*cosh(d*x + c))*sinh(d*x + c)^3 - 
a^3 - 3*(3*a^3 + 4*a*b^2)*cosh(d*x + c)^2 + 3*(5*a^3*cosh(d*x + c)^4 - 10* 
a^2*b*cosh(d*x + c)^3 + 24*(a^2*b + b^3)*d*x*cosh(d*x + c) - 3*a^3 - 4*a*b 
^2 + 6*(3*a^3 + 4*a*b^2)*cosh(d*x + c)^2)*sinh(d*x + c)^2 - 24*((a^2*b + b 
^3)*cosh(d*x + c)^3 + 3*(a^2*b + b^3)*cosh(d*x + c)^2*sinh(d*x + c) + 3*(a 
^2*b + b^3)*cosh(d*x + c)*sinh(d*x + c)^2 + (a^2*b + b^3)*sinh(d*x + c)^3) 
*log(2*(a*sinh(d*x + c) + b)/(cosh(d*x + c) - sinh(d*x + c))) + 3*(2*a^3*c 
osh(d*x + c)^5 - 5*a^2*b*cosh(d*x + c)^4 + 24*(a^2*b + b^3)*d*x*cosh(d*x + 
 c)^2 + 4*(3*a^3 + 4*a*b^2)*cosh(d*x + c)^3 - a^2*b - 2*(3*a^3 + 4*a*b^2)* 
cosh(d*x + c))*sinh(d*x + c))/(a^4*d*cosh(d*x + c)^3 + 3*a^4*d*cosh(d*x + 
c)^2*sinh(d*x + c) + 3*a^4*d*cosh(d*x + c)*sinh(d*x + c)^2 + a^4*d*sinh(d* 
x + c)^3)
 

Sympy [F]

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

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

Output:

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 183 vs. \(2 (81) = 162\).

Time = 0.05 (sec) , antiderivative size = 183, normalized size of antiderivative = 2.15 \[ \int \frac {\cosh ^3(c+d x)}{a+b \text {csch}(c+d x)} \, dx=-\frac {{\left (3 \, a b e^{\left (-d x - c\right )} - a^{2} - 3 \, {\left (3 \, a^{2} + 4 \, b^{2}\right )} e^{\left (-2 \, d x - 2 \, c\right )}\right )} e^{\left (3 \, d x + 3 \, c\right )}}{24 \, a^{3} d} - \frac {{\left (a^{2} b + b^{3}\right )} {\left (d x + c\right )}}{a^{4} d} - \frac {3 \, a b e^{\left (-2 \, d x - 2 \, c\right )} + a^{2} e^{\left (-3 \, d x - 3 \, c\right )} + 3 \, {\left (3 \, a^{2} + 4 \, b^{2}\right )} e^{\left (-d x - c\right )}}{24 \, a^{3} d} - \frac {{\left (a^{2} b + b^{3}\right )} \log \left (-2 \, b e^{\left (-d x - c\right )} + a e^{\left (-2 \, d x - 2 \, c\right )} - a\right )}{a^{4} d} \] Input:

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

Output:

-1/24*(3*a*b*e^(-d*x - c) - a^2 - 3*(3*a^2 + 4*b^2)*e^(-2*d*x - 2*c))*e^(3 
*d*x + 3*c)/(a^3*d) - (a^2*b + b^3)*(d*x + c)/(a^4*d) - 1/24*(3*a*b*e^(-2* 
d*x - 2*c) + a^2*e^(-3*d*x - 3*c) + 3*(3*a^2 + 4*b^2)*e^(-d*x - c))/(a^3*d 
) - (a^2*b + b^3)*log(-2*b*e^(-d*x - c) + a*e^(-2*d*x - 2*c) - a)/(a^4*d)
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.71 \[ \int \frac {\cosh ^3(c+d x)}{a+b \text {csch}(c+d x)} \, dx=\frac {\frac {a^{2} {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{3} - 3 \, a b {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{2} + 12 \, a^{2} {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )} + 12 \, b^{2} {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}}{a^{3}} - \frac {24 \, {\left (a^{2} b + b^{3}\right )} \log \left ({\left | a {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )} + 2 \, b \right |}\right )}{a^{4}}}{24 \, d} \] Input:

integrate(cosh(d*x+c)^3/(a+b*csch(d*x+c)),x, algorithm="giac")
 

Output:

1/24*((a^2*(e^(d*x + c) - e^(-d*x - c))^3 - 3*a*b*(e^(d*x + c) - e^(-d*x - 
 c))^2 + 12*a^2*(e^(d*x + c) - e^(-d*x - c)) + 12*b^2*(e^(d*x + c) - e^(-d 
*x - c)))/a^3 - 24*(a^2*b + b^3)*log(abs(a*(e^(d*x + c) - e^(-d*x - c)) + 
2*b))/a^4)/d
 

Mupad [B] (verification not implemented)

Time = 2.74 (sec) , antiderivative size = 180, normalized size of antiderivative = 2.12 \[ \int \frac {\cosh ^3(c+d x)}{a+b \text {csch}(c+d x)} \, dx=\frac {x\,\left (a^2\,b+b^3\right )}{a^4}-\frac {{\mathrm {e}}^{-3\,c-3\,d\,x}}{24\,a\,d}+\frac {{\mathrm {e}}^{3\,c+3\,d\,x}}{24\,a\,d}-\frac {b\,{\mathrm {e}}^{-2\,c-2\,d\,x}}{8\,a^2\,d}-\frac {b\,{\mathrm {e}}^{2\,c+2\,d\,x}}{8\,a^2\,d}-\frac {{\mathrm {e}}^{-c-d\,x}\,\left (3\,a^2+4\,b^2\right )}{8\,a^3\,d}-\frac {\ln \left (2\,b\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c-a+a\,{\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}\right )\,\left (a^2\,b+b^3\right )}{a^4\,d}+\frac {{\mathrm {e}}^{c+d\,x}\,\left (3\,a^2+4\,b^2\right )}{8\,a^3\,d} \] Input:

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

Output:

(x*(a^2*b + b^3))/a^4 - exp(- 3*c - 3*d*x)/(24*a*d) + exp(3*c + 3*d*x)/(24 
*a*d) - (b*exp(- 2*c - 2*d*x))/(8*a^2*d) - (b*exp(2*c + 2*d*x))/(8*a^2*d) 
- (exp(- c - d*x)*(3*a^2 + 4*b^2))/(8*a^3*d) - (log(2*b*exp(d*x)*exp(c) - 
a + a*exp(2*c)*exp(2*d*x))*(a^2*b + b^3))/(a^4*d) + (exp(c + d*x)*(3*a^2 + 
 4*b^2))/(8*a^3*d)
 

Reduce [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 251, normalized size of antiderivative = 2.95 \[ \int \frac {\cosh ^3(c+d x)}{a+b \text {csch}(c+d x)} \, dx=\frac {e^{6 d x +6 c} a^{3}-3 e^{5 d x +5 c} a^{2} b +9 e^{4 d x +4 c} a^{3}+12 e^{4 d x +4 c} a \,b^{2}-24 e^{3 d x +3 c} \mathrm {log}\left (e^{2 d x +2 c} a +2 e^{d x +c} b -a \right ) a^{2} b -24 e^{3 d x +3 c} \mathrm {log}\left (e^{2 d x +2 c} a +2 e^{d x +c} b -a \right ) b^{3}+24 e^{3 d x +3 c} a^{2} b d x +24 e^{3 d x +3 c} b^{3} d x -9 e^{2 d x +2 c} a^{3}-12 e^{2 d x +2 c} a \,b^{2}-3 e^{d x +c} a^{2} b -a^{3}}{24 e^{3 d x +3 c} a^{4} d} \] Input:

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

Output:

(e**(6*c + 6*d*x)*a**3 - 3*e**(5*c + 5*d*x)*a**2*b + 9*e**(4*c + 4*d*x)*a* 
*3 + 12*e**(4*c + 4*d*x)*a*b**2 - 24*e**(3*c + 3*d*x)*log(e**(2*c + 2*d*x) 
*a + 2*e**(c + d*x)*b - a)*a**2*b - 24*e**(3*c + 3*d*x)*log(e**(2*c + 2*d* 
x)*a + 2*e**(c + d*x)*b - a)*b**3 + 24*e**(3*c + 3*d*x)*a**2*b*d*x + 24*e* 
*(3*c + 3*d*x)*b**3*d*x - 9*e**(2*c + 2*d*x)*a**3 - 12*e**(2*c + 2*d*x)*a* 
b**2 - 3*e**(c + d*x)*a**2*b - a**3)/(24*e**(3*c + 3*d*x)*a**4*d)