\(\int x^m \cosh ^4(a+b \log (c x^n)) \, dx\) [246]

Optimal result

Integrand size = 17, antiderivative size = 266 \[ \int x^m \cosh ^4\left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {24 b^4 n^4 x^{1+m}}{(1+m) \left ((1+m)^2-16 b^2 n^2\right ) \left ((1+m)^2-4 b^2 n^2\right )}-\frac {12 b^2 (1+m) n^2 x^{1+m} \cosh ^2\left (a+b \log \left (c x^n\right )\right )}{\left ((1+m)^2-16 b^2 n^2\right ) \left ((1+m)^2-4 b^2 n^2\right )}+\frac {(1+m) x^{1+m} \cosh ^4\left (a+b \log \left (c x^n\right )\right )}{(1+m)^2-16 b^2 n^2}+\frac {24 b^3 n^3 x^{1+m} \cosh \left (a+b \log \left (c x^n\right )\right ) \sinh \left (a+b \log \left (c x^n\right )\right )}{\left ((1+m)^2-16 b^2 n^2\right ) \left ((1+m)^2-4 b^2 n^2\right )}-\frac {4 b n x^{1+m} \cosh ^3\left (a+b \log \left (c x^n\right )\right ) \sinh \left (a+b \log \left (c x^n\right )\right )}{(1+m)^2-16 b^2 n^2} \] Output:

24*b^4*n^4*x^(1+m)/(1+m)/((1+m)^2-16*b^2*n^2)/((1+m)^2-4*b^2*n^2)-12*b^2*( 
1+m)*n^2*x^(1+m)*cosh(a+b*ln(c*x^n))^2/((1+m)^2-16*b^2*n^2)/((1+m)^2-4*b^2 
*n^2)+(1+m)*x^(1+m)*cosh(a+b*ln(c*x^n))^4/((1+m)^2-16*b^2*n^2)+24*b^3*n^3* 
x^(1+m)*cosh(a+b*ln(c*x^n))*sinh(a+b*ln(c*x^n))/((1+m)^2-16*b^2*n^2)/((1+m 
)^2-4*b^2*n^2)-4*b*n*x^(1+m)*cosh(a+b*ln(c*x^n))^3*sinh(a+b*ln(c*x^n))/((1 
+m)^2-16*b^2*n^2)
 

Mathematica [A] (verified)

Time = 2.40 (sec) , antiderivative size = 311, normalized size of antiderivative = 1.17 \[ \int x^m \cosh ^4\left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {1}{8} x^{1+m} \left (\frac {3}{1+m}+\frac {4 \sinh (2 b n \log (x)) \left (-2 b n \cosh \left (2 \left (a-b n \log (x)+b \log \left (c x^n\right )\right )\right )+(1+m) \sinh \left (2 \left (a-b n \log (x)+b \log \left (c x^n\right )\right )\right )\right )}{(1+m-2 b n) (1+m+2 b n)}+\frac {4 \cosh (2 b n \log (x)) \left ((1+m) \cosh \left (2 \left (a-b n \log (x)+b \log \left (c x^n\right )\right )\right )-2 b n \sinh \left (2 \left (a-b n \log (x)+b \log \left (c x^n\right )\right )\right )\right )}{(1+m-2 b n) (1+m+2 b n)}+\frac {\sinh (4 b n \log (x)) \left (-4 b n \cosh \left (4 \left (a-b n \log (x)+b \log \left (c x^n\right )\right )\right )+(1+m) \sinh \left (4 \left (a-b n \log (x)+b \log \left (c x^n\right )\right )\right )\right )}{(1+m-4 b n) (1+m+4 b n)}+\frac {\cosh (4 b n \log (x)) \left ((1+m) \cosh \left (4 \left (a-b n \log (x)+b \log \left (c x^n\right )\right )\right )-4 b n \sinh \left (4 \left (a-b n \log (x)+b \log \left (c x^n\right )\right )\right )\right )}{(1+m-4 b n) (1+m+4 b n)}\right ) \] Input:

Integrate[x^m*Cosh[a + b*Log[c*x^n]]^4,x]
 

Output:

(x^(1 + m)*(3/(1 + m) + (4*Sinh[2*b*n*Log[x]]*(-2*b*n*Cosh[2*(a - b*n*Log[ 
x] + b*Log[c*x^n])] + (1 + m)*Sinh[2*(a - b*n*Log[x] + b*Log[c*x^n])]))/(( 
1 + m - 2*b*n)*(1 + m + 2*b*n)) + (4*Cosh[2*b*n*Log[x]]*((1 + m)*Cosh[2*(a 
 - b*n*Log[x] + b*Log[c*x^n])] - 2*b*n*Sinh[2*(a - b*n*Log[x] + b*Log[c*x^ 
n])]))/((1 + m - 2*b*n)*(1 + m + 2*b*n)) + (Sinh[4*b*n*Log[x]]*(-4*b*n*Cos 
h[4*(a - b*n*Log[x] + b*Log[c*x^n])] + (1 + m)*Sinh[4*(a - b*n*Log[x] + b* 
Log[c*x^n])]))/((1 + m - 4*b*n)*(1 + m + 4*b*n)) + (Cosh[4*b*n*Log[x]]*((1 
 + m)*Cosh[4*(a - b*n*Log[x] + b*Log[c*x^n])] - 4*b*n*Sinh[4*(a - b*n*Log[ 
x] + b*Log[c*x^n])]))/((1 + m - 4*b*n)*(1 + m + 4*b*n))))/8
 

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 236, normalized size of antiderivative = 0.89, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {6056, 6056, 15}

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.

\(\displaystyle \int x^m \cosh ^4\left (a+b \log \left (c x^n\right )\right ) \, dx\)

\(\Big \downarrow \) 6056

\(\displaystyle -\frac {12 b^2 n^2 \int x^m \cosh ^2\left (a+b \log \left (c x^n\right )\right )dx}{(m+1)^2-16 b^2 n^2}+\frac {(m+1) x^{m+1} \cosh ^4\left (a+b \log \left (c x^n\right )\right )}{-16 b^2 n^2+m^2+2 m+1}-\frac {4 b n x^{m+1} \sinh \left (a+b \log \left (c x^n\right )\right ) \cosh ^3\left (a+b \log \left (c x^n\right )\right )}{(m+1)^2-16 b^2 n^2}\)

\(\Big \downarrow \) 6056

\(\displaystyle -\frac {12 b^2 n^2 \left (-\frac {2 b^2 n^2 \int x^mdx}{(m+1)^2-4 b^2 n^2}+\frac {(m+1) x^{m+1} \cosh ^2\left (a+b \log \left (c x^n\right )\right )}{-4 b^2 n^2+m^2+2 m+1}-\frac {2 b n x^{m+1} \sinh \left (a+b \log \left (c x^n\right )\right ) \cosh \left (a+b \log \left (c x^n\right )\right )}{(m+1)^2-4 b^2 n^2}\right )}{(m+1)^2-16 b^2 n^2}+\frac {(m+1) x^{m+1} \cosh ^4\left (a+b \log \left (c x^n\right )\right )}{-16 b^2 n^2+m^2+2 m+1}-\frac {4 b n x^{m+1} \sinh \left (a+b \log \left (c x^n\right )\right ) \cosh ^3\left (a+b \log \left (c x^n\right )\right )}{(m+1)^2-16 b^2 n^2}\)

\(\Big \downarrow \) 15

\(\displaystyle \frac {(m+1) x^{m+1} \cosh ^4\left (a+b \log \left (c x^n\right )\right )}{-16 b^2 n^2+m^2+2 m+1}-\frac {12 b^2 n^2 \left (\frac {(m+1) x^{m+1} \cosh ^2\left (a+b \log \left (c x^n\right )\right )}{-4 b^2 n^2+m^2+2 m+1}-\frac {2 b n x^{m+1} \sinh \left (a+b \log \left (c x^n\right )\right ) \cosh \left (a+b \log \left (c x^n\right )\right )}{(m+1)^2-4 b^2 n^2}-\frac {2 b^2 n^2 x^{m+1}}{(m+1) \left ((m+1)^2-4 b^2 n^2\right )}\right )}{(m+1)^2-16 b^2 n^2}-\frac {4 b n x^{m+1} \sinh \left (a+b \log \left (c x^n\right )\right ) \cosh ^3\left (a+b \log \left (c x^n\right )\right )}{(m+1)^2-16 b^2 n^2}\)

Input:

Int[x^m*Cosh[a + b*Log[c*x^n]]^4,x]
 

Output:

((1 + m)*x^(1 + m)*Cosh[a + b*Log[c*x^n]]^4)/(1 + 2*m + m^2 - 16*b^2*n^2) 
- (4*b*n*x^(1 + m)*Cosh[a + b*Log[c*x^n]]^3*Sinh[a + b*Log[c*x^n]])/((1 + 
m)^2 - 16*b^2*n^2) - (12*b^2*n^2*((-2*b^2*n^2*x^(1 + m))/((1 + m)*((1 + m) 
^2 - 4*b^2*n^2)) + ((1 + m)*x^(1 + m)*Cosh[a + b*Log[c*x^n]]^2)/(1 + 2*m + 
 m^2 - 4*b^2*n^2) - (2*b*n*x^(1 + m)*Cosh[a + b*Log[c*x^n]]*Sinh[a + b*Log 
[c*x^n]])/((1 + m)^2 - 4*b^2*n^2)))/((1 + m)^2 - 16*b^2*n^2)
 

Defintions of rubi rules used

rule 15

Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ 
{a, m}, x] && NeQ[m, -1]
 

rule 6056

Int[Cosh[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(d_.)]^(p_)*((e_.)*(x_))^(m_ 
.), x_Symbol] :> Simp[(-(m + 1))*(e*x)^(m + 1)*(Cosh[d*(a + b*Log[c*x^n])]^ 
p/(b^2*d^2*e*n^2*p^2 - e*(m + 1)^2)), x] + (Simp[b*d*n*p*(e*x)^(m + 1)*Sinh 
[d*(a + b*Log[c*x^n])]*(Cosh[d*(a + b*Log[c*x^n])]^(p - 1)/(b^2*d^2*e*n^2*p 
^2 - e*(m + 1)^2)), x] + Simp[b^2*d^2*n^2*p*((p - 1)/(b^2*d^2*n^2*p^2 - (m 
+ 1)^2))   Int[(e*x)^m*Cosh[d*(a + b*Log[c*x^n])]^(p - 2), x], x]) /; FreeQ 
[{a, b, c, d, e, m, n}, x] && IGtQ[p, 1] && NeQ[b^2*d^2*n^2*p^2 - (m + 1)^2 
, 0]
 

Maple [F]

\[\int x^{m} {\cosh \left (a +b \ln \left (c \,x^{n}\right )\right )}^{4}d x\]

Input:

int(x^m*cosh(a+b*ln(c*x^n))^4,x)
 

Output:

int(x^m*cosh(a+b*ln(c*x^n))^4,x)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1123 vs. \(2 (283) = 566\).

Time = 0.13 (sec) , antiderivative size = 1123, normalized size of antiderivative = 4.22 \[ \int x^m \cosh ^4\left (a+b \log \left (c x^n\right )\right ) \, dx=\text {Too large to display} \] Input:

integrate(x^m*cosh(a+b*log(c*x^n))^4,x, algorithm="fricas")
 

Output:

1/8*((m^4 + 4*m^3 - 4*(b^2*m^2 + 2*b^2*m + b^2)*n^2 + 6*m^2 + 4*m + 1)*x*c 
osh(b*n*log(x) + b*log(c) + a)^4*cosh(m*log(x)) + 4*(m^4 + 4*m^3 - 16*(b^2 
*m^2 + 2*b^2*m + b^2)*n^2 + 6*m^2 + 4*m + 1)*x*cosh(b*n*log(x) + b*log(c) 
+ a)^2*cosh(m*log(x)) + ((m^4 + 4*m^3 - 4*(b^2*m^2 + 2*b^2*m + b^2)*n^2 + 
6*m^2 + 4*m + 1)*x*cosh(m*log(x)) + (m^4 + 4*m^3 - 4*(b^2*m^2 + 2*b^2*m + 
b^2)*n^2 + 6*m^2 + 4*m + 1)*x*sinh(m*log(x)))*sinh(b*n*log(x) + b*log(c) + 
 a)^4 + 16*((4*(b^3*m + b^3)*n^3 - (b*m^3 + 3*b*m^2 + 3*b*m + b)*n)*x*cosh 
(b*n*log(x) + b*log(c) + a)*cosh(m*log(x)) + (4*(b^3*m + b^3)*n^3 - (b*m^3 
 + 3*b*m^2 + 3*b*m + b)*n)*x*cosh(b*n*log(x) + b*log(c) + a)*sinh(m*log(x) 
))*sinh(b*n*log(x) + b*log(c) + a)^3 + 3*(64*b^4*n^4 + m^4 + 4*m^3 - 20*(b 
^2*m^2 + 2*b^2*m + b^2)*n^2 + 6*m^2 + 4*m + 1)*x*cosh(m*log(x)) + 2*(3*(m^ 
4 + 4*m^3 - 4*(b^2*m^2 + 2*b^2*m + b^2)*n^2 + 6*m^2 + 4*m + 1)*x*cosh(b*n* 
log(x) + b*log(c) + a)^2*cosh(m*log(x)) + 2*(m^4 + 4*m^3 - 16*(b^2*m^2 + 2 
*b^2*m + b^2)*n^2 + 6*m^2 + 4*m + 1)*x*cosh(m*log(x)) + (3*(m^4 + 4*m^3 - 
4*(b^2*m^2 + 2*b^2*m + b^2)*n^2 + 6*m^2 + 4*m + 1)*x*cosh(b*n*log(x) + b*l 
og(c) + a)^2 + 2*(m^4 + 4*m^3 - 16*(b^2*m^2 + 2*b^2*m + b^2)*n^2 + 6*m^2 + 
 4*m + 1)*x)*sinh(m*log(x)))*sinh(b*n*log(x) + b*log(c) + a)^2 + 16*((4*(b 
^3*m + b^3)*n^3 - (b*m^3 + 3*b*m^2 + 3*b*m + b)*n)*x*cosh(b*n*log(x) + b*l 
og(c) + a)^3*cosh(m*log(x)) + (16*(b^3*m + b^3)*n^3 - (b*m^3 + 3*b*m^2 + 3 
*b*m + b)*n)*x*cosh(b*n*log(x) + b*log(c) + a)*cosh(m*log(x)) + ((4*(b^...
 

Sympy [F(-1)]

Timed out. \[ \int x^m \cosh ^4\left (a+b \log \left (c x^n\right )\right ) \, dx=\text {Timed out} \] Input:

integrate(x**m*cosh(a+b*ln(c*x**n))**4,x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 161, normalized size of antiderivative = 0.61 \[ \int x^m \cosh ^4\left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {c^{4 \, b} x e^{\left (4 \, b \log \left (x^{n}\right ) + m \log \left (x\right ) + 4 \, a\right )}}{16 \, {\left (4 \, b n + m + 1\right )}} + \frac {c^{2 \, b} x e^{\left (2 \, b \log \left (x^{n}\right ) + m \log \left (x\right ) + 2 \, a\right )}}{4 \, {\left (2 \, b n + m + 1\right )}} - \frac {x e^{\left (-2 \, b \log \left (x^{n}\right ) + m \log \left (x\right ) - 2 \, a\right )}}{4 \, {\left (2 \, b c^{2 \, b} n - c^{2 \, b} {\left (m + 1\right )}\right )}} - \frac {x e^{\left (-4 \, b \log \left (x^{n}\right ) + m \log \left (x\right ) - 4 \, a\right )}}{16 \, {\left (4 \, b c^{4 \, b} n - c^{4 \, b} {\left (m + 1\right )}\right )}} + \frac {3 \, x^{m + 1}}{8 \, {\left (m + 1\right )}} \] Input:

integrate(x^m*cosh(a+b*log(c*x^n))^4,x, algorithm="maxima")
 

Output:

1/16*c^(4*b)*x*e^(4*b*log(x^n) + m*log(x) + 4*a)/(4*b*n + m + 1) + 1/4*c^( 
2*b)*x*e^(2*b*log(x^n) + m*log(x) + 2*a)/(2*b*n + m + 1) - 1/4*x*e^(-2*b*l 
og(x^n) + m*log(x) - 2*a)/(2*b*c^(2*b)*n - c^(2*b)*(m + 1)) - 1/16*x*e^(-4 
*b*log(x^n) + m*log(x) - 4*a)/(4*b*c^(4*b)*n - c^(4*b)*(m + 1)) + 3/8*x^(m 
 + 1)/(m + 1)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 6880 vs. \(2 (283) = 566\).

Time = 0.20 (sec) , antiderivative size = 6880, normalized size of antiderivative = 25.86 \[ \int x^m \cosh ^4\left (a+b \log \left (c x^n\right )\right ) \, dx=\text {Too large to display} \] Input:

integrate(x^m*cosh(a+b*log(c*x^n))^4,x, algorithm="giac")
 

Output:

b^3*c^(4*b)*m*n^3*x*x^(4*b*n)*x^m*e^(4*a)/(64*b^4*m*n^4 + 64*b^4*n^4 - 20* 
b^2*m^3*n^2 - 60*b^2*m^2*n^2 + m^5 - 60*b^2*m*n^2 + 5*m^4 - 20*b^2*n^2 + 1 
0*m^3 + 10*m^2 + 5*m + 1) + 8*b^3*c^(2*b)*m*n^3*x*x^(2*b*n)*x^m*e^(2*a)/(6 
4*b^4*m*n^4 + 64*b^4*n^4 - 20*b^2*m^3*n^2 - 60*b^2*m^2*n^2 + m^5 - 60*b^2* 
m*n^2 + 5*m^4 - 20*b^2*n^2 + 10*m^3 + 10*m^2 + 5*m + 1) - 1/4*b^2*c^(4*b)* 
m^2*n^2*x*x^(4*b*n)*x^m*e^(4*a)/(64*b^4*m*n^4 + 64*b^4*n^4 - 20*b^2*m^3*n^ 
2 - 60*b^2*m^2*n^2 + m^5 - 60*b^2*m*n^2 + 5*m^4 - 20*b^2*n^2 + 10*m^3 + 10 
*m^2 + 5*m + 1) + b^3*c^(4*b)*n^3*x*x^(4*b*n)*x^m*e^(4*a)/(64*b^4*m*n^4 + 
64*b^4*n^4 - 20*b^2*m^3*n^2 - 60*b^2*m^2*n^2 + m^5 - 60*b^2*m*n^2 + 5*m^4 
- 20*b^2*n^2 + 10*m^3 + 10*m^2 + 5*m + 1) - 4*b^2*c^(2*b)*m^2*n^2*x*x^(2*b 
*n)*x^m*e^(2*a)/(64*b^4*m*n^4 + 64*b^4*n^4 - 20*b^2*m^3*n^2 - 60*b^2*m^2*n 
^2 + m^5 - 60*b^2*m*n^2 + 5*m^4 - 20*b^2*n^2 + 10*m^3 + 10*m^2 + 5*m + 1) 
+ 8*b^3*c^(2*b)*n^3*x*x^(2*b*n)*x^m*e^(2*a)/(64*b^4*m*n^4 + 64*b^4*n^4 - 2 
0*b^2*m^3*n^2 - 60*b^2*m^2*n^2 + m^5 - 60*b^2*m*n^2 + 5*m^4 - 20*b^2*n^2 + 
 10*m^3 + 10*m^2 + 5*m + 1) + 24*b^4*n^4*x*x^m/(64*b^4*m*n^4 + 64*b^4*n^4 
- 20*b^2*m^3*n^2 - 60*b^2*m^2*n^2 + m^5 - 60*b^2*m*n^2 + 5*m^4 - 20*b^2*n^ 
2 + 10*m^3 + 10*m^2 + 5*m + 1) - 1/4*b*c^(4*b)*m^3*n*x*x^(4*b*n)*x^m*e^(4* 
a)/(64*b^4*m*n^4 + 64*b^4*n^4 - 20*b^2*m^3*n^2 - 60*b^2*m^2*n^2 + m^5 - 60 
*b^2*m*n^2 + 5*m^4 - 20*b^2*n^2 + 10*m^3 + 10*m^2 + 5*m + 1) - 1/2*b^2*c^( 
4*b)*m*n^2*x*x^(4*b*n)*x^m*e^(4*a)/(64*b^4*m*n^4 + 64*b^4*n^4 - 20*b^2*...
 

Mupad [B] (verification not implemented)

Time = 2.16 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.50 \[ \int x^m \cosh ^4\left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {3\,x\,x^m}{8\,m+8}+\frac {x\,x^m\,{\mathrm {e}}^{-2\,a}}{{\left (c\,x^n\right )}^{2\,b}\,\left (4\,m-8\,b\,n+4\right )}+\frac {x\,x^m\,{\mathrm {e}}^{2\,a}\,{\left (c\,x^n\right )}^{2\,b}}{4\,m+8\,b\,n+4}+\frac {x\,x^m\,{\mathrm {e}}^{-4\,a}}{{\left (c\,x^n\right )}^{4\,b}\,\left (16\,m-64\,b\,n+16\right )}+\frac {x\,x^m\,{\mathrm {e}}^{4\,a}\,{\left (c\,x^n\right )}^{4\,b}}{16\,m+64\,b\,n+16} \] Input:

int(x^m*cosh(a + b*log(c*x^n))^4,x)
 

Output:

(3*x*x^m)/(8*m + 8) + (x*x^m*exp(-2*a))/((c*x^n)^(2*b)*(4*m - 8*b*n + 4)) 
+ (x*x^m*exp(2*a)*(c*x^n)^(2*b))/(4*m + 8*b*n + 4) + (x*x^m*exp(-4*a))/((c 
*x^n)^(4*b)*(16*m - 64*b*n + 16)) + (x*x^m*exp(4*a)*(c*x^n)^(4*b))/(16*m + 
 64*b*n + 16)
 

Reduce [B] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 1329, normalized size of antiderivative = 5.00 \[ \int x^m \cosh ^4\left (a+b \log \left (c x^n\right )\right ) \, dx =\text {Too large to display} \] Input:

int(x^m*cosh(a+b*log(c*x^n))^4,x)
 

Output:

(x**m*x*(16*x**(8*b*n)*e**(8*a)*c**(8*b)*b**3*m*n**3 + 16*x**(8*b*n)*e**(8 
*a)*c**(8*b)*b**3*n**3 - 4*x**(8*b*n)*e**(8*a)*c**(8*b)*b**2*m**2*n**2 - 8 
*x**(8*b*n)*e**(8*a)*c**(8*b)*b**2*m*n**2 - 4*x**(8*b*n)*e**(8*a)*c**(8*b) 
*b**2*n**2 - 4*x**(8*b*n)*e**(8*a)*c**(8*b)*b*m**3*n - 12*x**(8*b*n)*e**(8 
*a)*c**(8*b)*b*m**2*n - 12*x**(8*b*n)*e**(8*a)*c**(8*b)*b*m*n - 4*x**(8*b* 
n)*e**(8*a)*c**(8*b)*b*n + x**(8*b*n)*e**(8*a)*c**(8*b)*m**4 + 4*x**(8*b*n 
)*e**(8*a)*c**(8*b)*m**3 + 6*x**(8*b*n)*e**(8*a)*c**(8*b)*m**2 + 4*x**(8*b 
*n)*e**(8*a)*c**(8*b)*m + x**(8*b*n)*e**(8*a)*c**(8*b) + 128*x**(6*b*n)*e* 
*(6*a)*c**(6*b)*b**3*m*n**3 + 128*x**(6*b*n)*e**(6*a)*c**(6*b)*b**3*n**3 - 
 64*x**(6*b*n)*e**(6*a)*c**(6*b)*b**2*m**2*n**2 - 128*x**(6*b*n)*e**(6*a)* 
c**(6*b)*b**2*m*n**2 - 64*x**(6*b*n)*e**(6*a)*c**(6*b)*b**2*n**2 - 8*x**(6 
*b*n)*e**(6*a)*c**(6*b)*b*m**3*n - 24*x**(6*b*n)*e**(6*a)*c**(6*b)*b*m**2* 
n - 24*x**(6*b*n)*e**(6*a)*c**(6*b)*b*m*n - 8*x**(6*b*n)*e**(6*a)*c**(6*b) 
*b*n + 4*x**(6*b*n)*e**(6*a)*c**(6*b)*m**4 + 16*x**(6*b*n)*e**(6*a)*c**(6* 
b)*m**3 + 24*x**(6*b*n)*e**(6*a)*c**(6*b)*m**2 + 16*x**(6*b*n)*e**(6*a)*c* 
*(6*b)*m + 4*x**(6*b*n)*e**(6*a)*c**(6*b) + 384*x**(4*b*n)*e**(4*a)*c**(4* 
b)*b**4*n**4 - 120*x**(4*b*n)*e**(4*a)*c**(4*b)*b**2*m**2*n**2 - 240*x**(4 
*b*n)*e**(4*a)*c**(4*b)*b**2*m*n**2 - 120*x**(4*b*n)*e**(4*a)*c**(4*b)*b** 
2*n**2 + 6*x**(4*b*n)*e**(4*a)*c**(4*b)*m**4 + 24*x**(4*b*n)*e**(4*a)*c**( 
4*b)*m**3 + 36*x**(4*b*n)*e**(4*a)*c**(4*b)*m**2 + 24*x**(4*b*n)*e**(4*...