\(\int x^m \sinh ^3(a+b \log (c x^n)) \, dx\) [272]

Optimal result

Integrand size = 17, antiderivative size = 203 \[ \int x^m \sinh ^3\left (a+b \log \left (c x^n\right )\right ) \, dx=-\frac {6 b^3 n^3 x^{1+m} \cosh \left (a+b \log \left (c x^n\right )\right )}{\left ((1+m)^2-9 b^2 n^2\right ) \left ((1+m)^2-b^2 n^2\right )}+\frac {6 b^2 (1+m) n^2 x^{1+m} \sinh \left (a+b \log \left (c x^n\right )\right )}{\left ((1+m)^2-9 b^2 n^2\right ) \left ((1+m)^2-b^2 n^2\right )}-\frac {3 b n x^{1+m} \cosh \left (a+b \log \left (c x^n\right )\right ) \sinh ^2\left (a+b \log \left (c x^n\right )\right )}{(1+m)^2-9 b^2 n^2}+\frac {(1+m) x^{1+m} \sinh ^3\left (a+b \log \left (c x^n\right )\right )}{(1+m)^2-9 b^2 n^2} \] Output:

-6*b^3*n^3*x^(1+m)*cosh(a+b*ln(c*x^n))/((1+m)^2-9*b^2*n^2)/((1+m)^2-b^2*n^ 
2)+6*b^2*(1+m)*n^2*x^(1+m)*sinh(a+b*ln(c*x^n))/((1+m)^2-9*b^2*n^2)/((1+m)^ 
2-b^2*n^2)-3*b*n*x^(1+m)*cosh(a+b*ln(c*x^n))*sinh(a+b*ln(c*x^n))^2/((1+m)^ 
2-9*b^2*n^2)+(1+m)*x^(1+m)*sinh(a+b*ln(c*x^n))^3/((1+m)^2-9*b^2*n^2)
 

Mathematica [A] (verified)

Time = 0.92 (sec) , antiderivative size = 292, normalized size of antiderivative = 1.44 \[ \int x^m \sinh ^3\left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {1}{4} x^{1+m} \left (-\frac {3 \cosh (b n \log (x)) \left (-b n \cosh \left (a-b n \log (x)+b \log \left (c x^n\right )\right )+(1+m) \sinh \left (a-b n \log (x)+b \log \left (c x^n\right )\right )\right )}{(1+m-b n) (1+m+b n)}-\frac {3 \sinh (b n \log (x)) \left ((1+m) \cosh \left (a-b n \log (x)+b \log \left (c x^n\right )\right )-b n \sinh \left (a-b n \log (x)+b \log \left (c x^n\right )\right )\right )}{(1+m-b n) (1+m+b n)}+\frac {\cosh (3 b n \log (x)) \left (-3 b n \cosh \left (3 \left (a-b n \log (x)+b \log \left (c x^n\right )\right )\right )+(1+m) \sinh \left (3 \left (a-b n \log (x)+b \log \left (c x^n\right )\right )\right )\right )}{(1+m-3 b n) (1+m+3 b n)}+\frac {\sinh (3 b n \log (x)) \left ((1+m) \cosh \left (3 \left (a-b n \log (x)+b \log \left (c x^n\right )\right )\right )-3 b n \sinh \left (3 \left (a-b n \log (x)+b \log \left (c x^n\right )\right )\right )\right )}{(1+m-3 b n) (1+m+3 b n)}\right ) \] Input:

Integrate[x^m*Sinh[a + b*Log[c*x^n]]^3,x]
 

Output:

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

Rubi [A] (verified)

Time = 0.41 (sec) , antiderivative size = 188, normalized size of antiderivative = 0.93, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {6055, 6053}

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.

\(\displaystyle \int x^m \sinh ^3\left (a+b \log \left (c x^n\right )\right ) \, dx\)

\(\Big \downarrow \) 6055

\(\displaystyle \frac {6 b^2 n^2 \int x^m \sinh \left (a+b \log \left (c x^n\right )\right )dx}{(m+1)^2-9 b^2 n^2}+\frac {(m+1) x^{m+1} \sinh ^3\left (a+b \log \left (c x^n\right )\right )}{-9 b^2 n^2+m^2+2 m+1}-\frac {3 b n x^{m+1} \sinh ^2\left (a+b \log \left (c x^n\right )\right ) \cosh \left (a+b \log \left (c x^n\right )\right )}{(m+1)^2-9 b^2 n^2}\)

\(\Big \downarrow \) 6053

\(\displaystyle \frac {(m+1) x^{m+1} \sinh ^3\left (a+b \log \left (c x^n\right )\right )}{-9 b^2 n^2+m^2+2 m+1}-\frac {3 b n x^{m+1} \sinh ^2\left (a+b \log \left (c x^n\right )\right ) \cosh \left (a+b \log \left (c x^n\right )\right )}{(m+1)^2-9 b^2 n^2}+\frac {6 b^2 n^2 \left (\frac {(m+1) x^{m+1} \sinh \left (a+b \log \left (c x^n\right )\right )}{(-b n+m+1) (b n+m+1)}-\frac {b n x^{m+1} \cosh \left (a+b \log \left (c x^n\right )\right )}{(m+1)^2-b^2 n^2}\right )}{(m+1)^2-9 b^2 n^2}\)

Input:

Int[x^m*Sinh[a + b*Log[c*x^n]]^3,x]
 

Output:

(-3*b*n*x^(1 + m)*Cosh[a + b*Log[c*x^n]]*Sinh[a + b*Log[c*x^n]]^2)/((1 + m 
)^2 - 9*b^2*n^2) + ((1 + m)*x^(1 + m)*Sinh[a + b*Log[c*x^n]]^3)/(1 + 2*m + 
 m^2 - 9*b^2*n^2) + (6*b^2*n^2*(-((b*n*x^(1 + m)*Cosh[a + b*Log[c*x^n]])/( 
(1 + m)^2 - b^2*n^2)) + ((1 + m)*x^(1 + m)*Sinh[a + b*Log[c*x^n]])/((1 + m 
 - b*n)*(1 + m + b*n))))/((1 + m)^2 - 9*b^2*n^2)
 

Defintions of rubi rules used

rule 6053

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

rule 6055

Int[((e_.)*(x_))^(m_.)*Sinh[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(d_.)]^(p 
_), x_Symbol] :> Simp[(-(m + 1))*(e*x)^(m + 1)*(Sinh[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)*Cosh 
[d*(a + b*Log[c*x^n])]*(Sinh[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*Sinh[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]

\[\int x^{m} {\sinh \left (a +b \ln \left (c \,x^{n}\right )\right )}^{3}d x\]

Input:

int(x^m*sinh(a+b*ln(c*x^n))^3,x)
 

Output:

int(x^m*sinh(a+b*ln(c*x^n))^3,x)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 585 vs. \(2 (214) = 428\).

Time = 0.12 (sec) , antiderivative size = 585, normalized size of antiderivative = 2.88 \[ \int x^m \sinh ^3\left (a+b \log \left (c x^n\right )\right ) \, dx =\text {Too large to display} \] Input:

integrate(x^m*sinh(a+b*log(c*x^n))^3,x, algorithm="fricas")
 

Output:

1/4*(3*(b^3*n^3 - (b*m^2 + 2*b*m + b)*n)*x*cosh(b*n*log(x) + b*log(c) + a) 
^3*cosh(m*log(x)) - 3*(9*b^3*n^3 - (b*m^2 + 2*b*m + b)*n)*x*cosh(b*n*log(x 
) + b*log(c) + a)*cosh(m*log(x)) + ((m^3 - (b^2*m + b^2)*n^2 + 3*m^2 + 3*m 
 + 1)*x*cosh(m*log(x)) + (m^3 - (b^2*m + b^2)*n^2 + 3*m^2 + 3*m + 1)*x*sin 
h(m*log(x)))*sinh(b*n*log(x) + b*log(c) + a)^3 + 9*((b^3*n^3 - (b*m^2 + 2* 
b*m + b)*n)*x*cosh(b*n*log(x) + b*log(c) + a)*cosh(m*log(x)) + (b^3*n^3 - 
(b*m^2 + 2*b*m + b)*n)*x*cosh(b*n*log(x) + b*log(c) + a)*sinh(m*log(x)))*s 
inh(b*n*log(x) + b*log(c) + a)^2 + 3*((m^3 - (b^2*m + b^2)*n^2 + 3*m^2 + 3 
*m + 1)*x*cosh(b*n*log(x) + b*log(c) + a)^2*cosh(m*log(x)) - (m^3 - 9*(b^2 
*m + b^2)*n^2 + 3*m^2 + 3*m + 1)*x*cosh(m*log(x)) + ((m^3 - (b^2*m + b^2)* 
n^2 + 3*m^2 + 3*m + 1)*x*cosh(b*n*log(x) + b*log(c) + a)^2 - (m^3 - 9*(b^2 
*m + b^2)*n^2 + 3*m^2 + 3*m + 1)*x)*sinh(m*log(x)))*sinh(b*n*log(x) + b*lo 
g(c) + a) + 3*((b^3*n^3 - (b*m^2 + 2*b*m + b)*n)*x*cosh(b*n*log(x) + b*log 
(c) + a)^3 - (9*b^3*n^3 - (b*m^2 + 2*b*m + b)*n)*x*cosh(b*n*log(x) + b*log 
(c) + a))*sinh(m*log(x)))/(9*b^4*n^4 + m^4 + 4*m^3 - 10*(b^2*m^2 + 2*b^2*m 
 + b^2)*n^2 + 6*m^2 + 4*m + 1)
 

Sympy [F(-1)]

Timed out. \[ \int x^m \sinh ^3\left (a+b \log \left (c x^n\right )\right ) \, dx=\text {Timed out} \] Input:

integrate(x**m*sinh(a+b*ln(c*x**n))**3,x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.68 \[ \int x^m \sinh ^3\left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {c^{3 \, b} x e^{\left (3 \, b \log \left (x^{n}\right ) + m \log \left (x\right ) + 3 \, a\right )}}{8 \, {\left (3 \, b n + m + 1\right )}} - \frac {3 \, c^{b} x e^{\left (b \log \left (x^{n}\right ) + m \log \left (x\right ) + a\right )}}{8 \, {\left (b n + m + 1\right )}} - \frac {3 \, x e^{\left (-b \log \left (x^{n}\right ) + m \log \left (x\right ) - a\right )}}{8 \, {\left (b c^{b} n - c^{b} {\left (m + 1\right )}\right )}} + \frac {x e^{\left (-3 \, b \log \left (x^{n}\right ) + m \log \left (x\right ) - 3 \, a\right )}}{8 \, {\left (3 \, b c^{3 \, b} n - c^{3 \, b} {\left (m + 1\right )}\right )}} \] Input:

integrate(x^m*sinh(a+b*log(c*x^n))^3,x, algorithm="maxima")
 

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 3225 vs. \(2 (214) = 428\).

Time = 0.19 (sec) , antiderivative size = 3225, normalized size of antiderivative = 15.89 \[ \int x^m \sinh ^3\left (a+b \log \left (c x^n\right )\right ) \, dx=\text {Too large to display} \] Input:

integrate(x^m*sinh(a+b*log(c*x^n))^3,x, algorithm="giac")
 

Output:

3/8*b^3*c^(3*b)*n^3*x*x^(3*b*n)*x^m*e^(3*a)/(9*b^4*n^4 - 10*b^2*m^2*n^2 - 
20*b^2*m*n^2 + m^4 - 10*b^2*n^2 + 4*m^3 + 6*m^2 + 4*m + 1) - 27/8*b^3*c^b* 
n^3*x*x^(b*n)*x^m*e^a/(9*b^4*n^4 - 10*b^2*m^2*n^2 - 20*b^2*m*n^2 + m^4 - 1 
0*b^2*n^2 + 4*m^3 + 6*m^2 + 4*m + 1) - 1/8*b^2*c^(3*b)*m*n^2*x*x^(3*b*n)*x 
^m*e^(3*a)/(9*b^4*n^4 - 10*b^2*m^2*n^2 - 20*b^2*m*n^2 + m^4 - 10*b^2*n^2 + 
 4*m^3 + 6*m^2 + 4*m + 1) + 27/8*b^2*c^b*m*n^2*x*x^(b*n)*x^m*e^a/(9*b^4*n^ 
4 - 10*b^2*m^2*n^2 - 20*b^2*m*n^2 + m^4 - 10*b^2*n^2 + 4*m^3 + 6*m^2 + 4*m 
 + 1) - 3/8*b*c^(3*b)*m^2*n*x*x^(3*b*n)*x^m*e^(3*a)/(9*b^4*n^4 - 10*b^2*m^ 
2*n^2 - 20*b^2*m*n^2 + m^4 - 10*b^2*n^2 + 4*m^3 + 6*m^2 + 4*m + 1) - 1/8*b 
^2*c^(3*b)*n^2*x*x^(3*b*n)*x^m*e^(3*a)/(9*b^4*n^4 - 10*b^2*m^2*n^2 - 20*b^ 
2*m*n^2 + m^4 - 10*b^2*n^2 + 4*m^3 + 6*m^2 + 4*m + 1) + 3/8*b*c^b*m^2*n*x* 
x^(b*n)*x^m*e^a/(9*b^4*n^4 - 10*b^2*m^2*n^2 - 20*b^2*m*n^2 + m^4 - 10*b^2* 
n^2 + 4*m^3 + 6*m^2 + 4*m + 1) + 27/8*b^2*c^b*n^2*x*x^(b*n)*x^m*e^a/(9*b^4 
*n^4 - 10*b^2*m^2*n^2 - 20*b^2*m*n^2 + m^4 - 10*b^2*n^2 + 4*m^3 + 6*m^2 + 
4*m + 1) + 1/8*c^(3*b)*m^3*x*x^(3*b*n)*x^m*e^(3*a)/(9*b^4*n^4 - 10*b^2*m^2 
*n^2 - 20*b^2*m*n^2 + m^4 - 10*b^2*n^2 + 4*m^3 + 6*m^2 + 4*m + 1) - 3/4*b* 
c^(3*b)*m*n*x*x^(3*b*n)*x^m*e^(3*a)/(9*b^4*n^4 - 10*b^2*m^2*n^2 - 20*b^2*m 
*n^2 + m^4 - 10*b^2*n^2 + 4*m^3 + 6*m^2 + 4*m + 1) - 3/8*c^b*m^3*x*x^(b*n) 
*x^m*e^a/(9*b^4*n^4 - 10*b^2*m^2*n^2 - 20*b^2*m*n^2 + m^4 - 10*b^2*n^2 + 4 
*m^3 + 6*m^2 + 4*m + 1) + 3/4*b*c^b*m*n*x*x^(b*n)*x^m*e^a/(9*b^4*n^4 - ...
 

Mupad [B] (verification not implemented)

Time = 1.79 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.58 \[ \int x^m \sinh ^3\left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {3\,x\,x^m\,{\mathrm {e}}^{-a}}{{\left (c\,x^n\right )}^b\,\left (8\,m-8\,b\,n+8\right )}-\frac {x\,x^m\,{\mathrm {e}}^{-3\,a}}{{\left (c\,x^n\right )}^{3\,b}\,\left (8\,m-24\,b\,n+8\right )}+\frac {x\,x^m\,{\mathrm {e}}^{3\,a}\,{\left (c\,x^n\right )}^{3\,b}}{8\,m+24\,b\,n+8}-\frac {3\,x\,x^m\,{\mathrm {e}}^a\,{\left (c\,x^n\right )}^b}{8\,m+8\,b\,n+8} \] Input:

int(x^m*sinh(a + b*log(c*x^n))^3,x)
 

Output:

(3*x*x^m*exp(-a))/((c*x^n)^b*(8*m - 8*b*n + 8)) - (x*x^m*exp(-3*a))/((c*x^ 
n)^(3*b)*(8*m - 24*b*n + 8)) + (x*x^m*exp(3*a)*(c*x^n)^(3*b))/(8*m + 24*b* 
n + 8) - (3*x*x^m*exp(a)*(c*x^n)^b)/(8*m + 8*b*n + 8)
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 784, normalized size of antiderivative = 3.86 \[ \int x^m \sinh ^3\left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {x^{m} x \left (-1+3 x^{2 b n} e^{2 a} c^{2 b}-3 m -x^{6 b n} e^{6 a} c^{6 b} b^{2} n^{2}-3 x^{4 b n} e^{4 a} c^{4 b}-3 m^{2}+9 x^{2 b n} e^{2 a} c^{2 b} m^{2}+9 x^{2 b n} e^{2 a} c^{2 b} m +3 b^{3} n^{3}-27 x^{2 b n} e^{2 a} c^{2 b} b^{3} n^{3}+3 x^{2 b n} e^{2 a} c^{2 b} b n +b^{2} n^{2}+3 x^{4 b n} e^{4 a} c^{4 b} b n -27 x^{2 b n} e^{2 a} c^{2 b} b^{2} n^{2}-x^{6 b n} e^{6 a} c^{6 b} b^{2} m \,n^{2}-3 x^{6 b n} e^{6 a} c^{6 b} b \,m^{2} n -6 x^{6 b n} e^{6 a} c^{6 b} b m n +27 x^{4 b n} e^{4 a} c^{4 b} b^{2} m \,n^{2}+3 x^{4 b n} e^{4 a} c^{4 b} b \,m^{2} n -27 x^{2 b n} e^{2 a} c^{2 b} b^{2} m \,n^{2}+3 x^{2 b n} e^{2 a} c^{2 b} b \,m^{2} n +6 x^{2 b n} e^{2 a} c^{2 b} b m n +6 x^{4 b n} e^{4 a} c^{4 b} b m n -9 x^{4 b n} e^{4 a} c^{4 b} m^{2}-9 x^{4 b n} e^{4 a} c^{4 b} m -6 b m n +b^{2} m \,n^{2}+x^{6 b n} e^{6 a} c^{6 b}+x^{6 b n} e^{6 a} c^{6 b} m^{3}-3 b n -3 b \,m^{2} n +3 x^{6 b n} e^{6 a} c^{6 b} m^{2}+3 x^{6 b n} e^{6 a} c^{6 b} m -3 x^{4 b n} e^{4 a} c^{4 b} m^{3}+3 x^{2 b n} e^{2 a} c^{2 b} m^{3}+3 x^{6 b n} e^{6 a} c^{6 b} b^{3} n^{3}-3 x^{6 b n} e^{6 a} c^{6 b} b n -27 x^{4 b n} e^{4 a} c^{4 b} b^{3} n^{3}+27 x^{4 b n} e^{4 a} c^{4 b} b^{2} n^{2}-m^{3}\right )}{8 x^{3 b n} e^{3 a} c^{3 b} \left (9 b^{4} n^{4}-10 b^{2} m^{2} n^{2}-20 b^{2} m \,n^{2}-10 b^{2} n^{2}+m^{4}+4 m^{3}+6 m^{2}+4 m +1\right )} \] Input:

int(x^m*sinh(a+b*log(c*x^n))^3,x)
 

Output:

(x**m*x*(3*x**(6*b*n)*e**(6*a)*c**(6*b)*b**3*n**3 - x**(6*b*n)*e**(6*a)*c* 
*(6*b)*b**2*m*n**2 - x**(6*b*n)*e**(6*a)*c**(6*b)*b**2*n**2 - 3*x**(6*b*n) 
*e**(6*a)*c**(6*b)*b*m**2*n - 6*x**(6*b*n)*e**(6*a)*c**(6*b)*b*m*n - 3*x** 
(6*b*n)*e**(6*a)*c**(6*b)*b*n + x**(6*b*n)*e**(6*a)*c**(6*b)*m**3 + 3*x**( 
6*b*n)*e**(6*a)*c**(6*b)*m**2 + 3*x**(6*b*n)*e**(6*a)*c**(6*b)*m + x**(6*b 
*n)*e**(6*a)*c**(6*b) - 27*x**(4*b*n)*e**(4*a)*c**(4*b)*b**3*n**3 + 27*x** 
(4*b*n)*e**(4*a)*c**(4*b)*b**2*m*n**2 + 27*x**(4*b*n)*e**(4*a)*c**(4*b)*b* 
*2*n**2 + 3*x**(4*b*n)*e**(4*a)*c**(4*b)*b*m**2*n + 6*x**(4*b*n)*e**(4*a)* 
c**(4*b)*b*m*n + 3*x**(4*b*n)*e**(4*a)*c**(4*b)*b*n - 3*x**(4*b*n)*e**(4*a 
)*c**(4*b)*m**3 - 9*x**(4*b*n)*e**(4*a)*c**(4*b)*m**2 - 9*x**(4*b*n)*e**(4 
*a)*c**(4*b)*m - 3*x**(4*b*n)*e**(4*a)*c**(4*b) - 27*x**(2*b*n)*e**(2*a)*c 
**(2*b)*b**3*n**3 - 27*x**(2*b*n)*e**(2*a)*c**(2*b)*b**2*m*n**2 - 27*x**(2 
*b*n)*e**(2*a)*c**(2*b)*b**2*n**2 + 3*x**(2*b*n)*e**(2*a)*c**(2*b)*b*m**2* 
n + 6*x**(2*b*n)*e**(2*a)*c**(2*b)*b*m*n + 3*x**(2*b*n)*e**(2*a)*c**(2*b)* 
b*n + 3*x**(2*b*n)*e**(2*a)*c**(2*b)*m**3 + 9*x**(2*b*n)*e**(2*a)*c**(2*b) 
*m**2 + 9*x**(2*b*n)*e**(2*a)*c**(2*b)*m + 3*x**(2*b*n)*e**(2*a)*c**(2*b) 
+ 3*b**3*n**3 + b**2*m*n**2 + b**2*n**2 - 3*b*m**2*n - 6*b*m*n - 3*b*n - m 
**3 - 3*m**2 - 3*m - 1))/(8*x**(3*b*n)*e**(3*a)*c**(3*b)*(9*b**4*n**4 - 10 
*b**2*m**2*n**2 - 20*b**2*m*n**2 - 10*b**2*n**2 + m**4 + 4*m**3 + 6*m**2 + 
 4*m + 1))