Integrand size = 13, antiderivative size = 140 \[ \int \frac {\sinh ^7(x)}{a+b \cosh (x)} \, dx=-\frac {a \left (a^4-3 a^2 b^2+3 b^4\right ) \cosh (x)}{b^6}+\frac {\left (a^4-3 a^2 b^2+3 b^4\right ) \cosh ^2(x)}{2 b^5}-\frac {a \left (a^2-3 b^2\right ) \cosh ^3(x)}{3 b^4}+\frac {\left (a^2-3 b^2\right ) \cosh ^4(x)}{4 b^3}-\frac {a \cosh ^5(x)}{5 b^2}+\frac {\cosh ^6(x)}{6 b}+\frac {\left (a^2-b^2\right )^3 \log (a+b \cosh (x))}{b^7} \] Output:
-a*(a^4-3*a^2*b^2+3*b^4)*cosh(x)/b^6+1/2*(a^4-3*a^2*b^2+3*b^4)*cosh(x)^2/b ^5-1/3*a*(a^2-3*b^2)*cosh(x)^3/b^4+1/4*(a^2-3*b^2)*cosh(x)^4/b^3-1/5*a*cos h(x)^5/b^2+1/6*cosh(x)^6/b+(a^2-b^2)^3*ln(a+b*cosh(x))/b^7
Time = 0.16 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.03 \[ \int \frac {\sinh ^7(x)}{a+b \cosh (x)} \, dx=\frac {-120 a b \left (8 a^4-22 a^2 b^2+19 b^4\right ) \cosh (x)+15 b^2 \left (16 a^4-40 a^2 b^2+29 b^4\right ) \cosh (2 x)-20 a (2 a-3 b) b^3 (2 a+3 b) \cosh (3 x)-30 b^4 \left (-a^2+2 b^2\right ) \cosh (4 x)-12 a b^5 \cosh (5 x)+5 b^6 \cosh (6 x)+960 \left (a^2-b^2\right )^3 \log (a+b \cosh (x))}{960 b^7} \] Input:
Integrate[Sinh[x]^7/(a + b*Cosh[x]),x]
Output:
(-120*a*b*(8*a^4 - 22*a^2*b^2 + 19*b^4)*Cosh[x] + 15*b^2*(16*a^4 - 40*a^2* b^2 + 29*b^4)*Cosh[2*x] - 20*a*(2*a - 3*b)*b^3*(2*a + 3*b)*Cosh[3*x] - 30* b^4*(-a^2 + 2*b^2)*Cosh[4*x] - 12*a*b^5*Cosh[5*x] + 5*b^6*Cosh[6*x] + 960* (a^2 - b^2)^3*Log[a + b*Cosh[x]])/(960*b^7)
Time = 0.44 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {3042, 26, 3147, 476, 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.
| \(\displaystyle \int \frac {\sinh ^7(x)}{a+b \cosh (x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {i \cos \left (-\frac {\pi }{2}+i x\right )^7}{a-b \sin \left (-\frac {\pi }{2}+i x\right )}dx\) |
| \(\Big \downarrow \) 26 |
| \(\displaystyle i \int \frac {\cos \left (i x-\frac {\pi }{2}\right )^7}{a-b \sin \left (i x-\frac {\pi }{2}\right )}dx\) |
| \(\Big \downarrow \) 3147 |
| \(\displaystyle -\frac {\int \frac {\left (b^2-b^2 \cosh ^2(x)\right )^3}{a+b \cosh (x)}d(b \cosh (x))}{b^7}\) |
| \(\Big \downarrow \) 476 |
| \(\displaystyle -\frac {\int \left (\left (\frac {3 \left (b^2-a^2\right ) b^2}{a^4}+1\right ) a^5+b^4 \cosh ^4(x) a+b^2 \left (a^2-3 b^2\right ) \cosh ^2(x) a-b^5 \cosh ^5(x)-b^3 \left (a^2-3 b^2\right ) \cosh ^3(x)-b \left (a^4-3 b^2 a^2+3 b^4\right ) \cosh (x)-\frac {\left (a^2-b^2\right )^3}{a+b \cosh (x)}\right )d(b \cosh (x))}{b^7}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {-\left (a^2-b^2\right )^3 \log (a+b \cosh (x))-\frac {1}{4} b^4 \left (a^2-3 b^2\right ) \cosh ^4(x)+\frac {1}{3} a b^3 \left (a^2-3 b^2\right ) \cosh ^3(x)-\frac {1}{2} b^2 \left (a^4-3 a^2 b^2+3 b^4\right ) \cosh ^2(x)+a b \left (a^4-3 a^2 b^2+3 b^4\right ) \cosh (x)+\frac {1}{5} a b^5 \cosh ^5(x)-\frac {1}{6} b^6 \cosh ^6(x)}{b^7}\) |
Input:
Int[Sinh[x]^7/(a + b*Cosh[x]),x]
Output:
-((a*b*(a^4 - 3*a^2*b^2 + 3*b^4)*Cosh[x] - (b^2*(a^4 - 3*a^2*b^2 + 3*b^4)* Cosh[x]^2)/2 + (a*b^3*(a^2 - 3*b^2)*Cosh[x]^3)/3 - (b^4*(a^2 - 3*b^2)*Cosh [x]^4)/4 + (a*b^5*Cosh[x]^5)/5 - (b^6*Cosh[x]^6)/6 - (a^2 - b^2)^3*Log[a + b*Cosh[x]])/b^7)
Int[(Complex[0, a_])*(Fx_), x_Symbol] :> Simp[(Complex[Identity[0], a]) I nt[Fx, x], x] /; FreeQ[a, x] && EqQ[a^2, 1]
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ ExpandIntegrand[(c + d*x)^n*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, n}, x] && IGtQ[p, 0]
Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m _.), x_Symbol] :> Simp[1/(b^p*f) Subst[Int[(a + x)^m*(b^2 - x^2)^((p - 1) /2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && IntegerQ[(p - 1)/2] && NeQ[a^2 - b^2, 0]
Time = 0.09 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.06
\[\frac {\frac {\cosh \left (x \right )^{6} b^{5}}{6}-\frac {a \cosh \left (x \right )^{5} b^{4}}{5}+\frac {b \left (a^{2} b^{2}-3 b^{4}\right ) \cosh \left (x \right )^{4}}{4}-\frac {a \left (a^{2} b^{2}-3 b^{4}\right ) \cosh \left (x \right )^{3}}{3}+\frac {\left (a^{4}-3 a^{2} b^{2}+3 b^{4}\right ) \cosh \left (x \right )^{2} b}{2}-a \left (a^{4}-3 a^{2} b^{2}+3 b^{4}\right ) \cosh \left (x \right )}{b^{6}}+\frac {\left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right ) \ln \left (a +b \cosh \left (x \right )\right )}{b^{7}}\]
Input:
int(sinh(x)^7/(a+b*cosh(x)),x)
Output:
1/b^6*(1/6*cosh(x)^6*b^5-1/5*a*cosh(x)^5*b^4+1/4*b*(a^2*b^2-3*b^4)*cosh(x) ^4-1/3*a*(a^2*b^2-3*b^4)*cosh(x)^3+1/2*(a^4-3*a^2*b^2+3*b^4)*cosh(x)^2*b-a *(a^4-3*a^2*b^2+3*b^4)*cosh(x))+(a^6-3*a^4*b^2+3*a^2*b^4-b^6)/b^7*ln(a+b*c osh(x))
Leaf count of result is larger than twice the leaf count of optimal. 2134 vs. \(2 (130) = 260\).
Time = 0.11 (sec) , antiderivative size = 2134, normalized size of antiderivative = 15.24 \[ \int \frac {\sinh ^7(x)}{a+b \cosh (x)} \, dx=\text {Too large to display} \] Input:
integrate(sinh(x)^7/(a+b*cosh(x)),x, algorithm="fricas")
Output:
1/1920*(5*b^6*cosh(x)^12 + 5*b^6*sinh(x)^12 - 12*a*b^5*cosh(x)^11 + 12*(5* b^6*cosh(x) - a*b^5)*sinh(x)^11 + 30*(a^2*b^4 - 2*b^6)*cosh(x)^10 + 6*(55* b^6*cosh(x)^2 - 22*a*b^5*cosh(x) + 5*a^2*b^4 - 10*b^6)*sinh(x)^10 - 20*(4* a^3*b^3 - 9*a*b^5)*cosh(x)^9 + 20*(55*b^6*cosh(x)^3 - 33*a*b^5*cosh(x)^2 - 4*a^3*b^3 + 9*a*b^5 + 15*(a^2*b^4 - 2*b^6)*cosh(x))*sinh(x)^9 + 15*(16*a^ 4*b^2 - 40*a^2*b^4 + 29*b^6)*cosh(x)^8 + 15*(165*b^6*cosh(x)^4 - 132*a*b^5 *cosh(x)^3 + 16*a^4*b^2 - 40*a^2*b^4 + 29*b^6 + 90*(a^2*b^4 - 2*b^6)*cosh( x)^2 - 12*(4*a^3*b^3 - 9*a*b^5)*cosh(x))*sinh(x)^8 - 1920*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*x*cosh(x)^6 - 120*(8*a^5*b - 22*a^3*b^3 + 19*a*b^5)*co sh(x)^7 + 120*(33*b^6*cosh(x)^5 - 33*a*b^5*cosh(x)^4 - 8*a^5*b + 22*a^3*b^ 3 - 19*a*b^5 + 30*(a^2*b^4 - 2*b^6)*cosh(x)^3 - 6*(4*a^3*b^3 - 9*a*b^5)*co sh(x)^2 + (16*a^4*b^2 - 40*a^2*b^4 + 29*b^6)*cosh(x))*sinh(x)^7 - 12*a*b^5 *cosh(x) + 12*(385*b^6*cosh(x)^6 - 462*a*b^5*cosh(x)^5 + 525*(a^2*b^4 - 2* b^6)*cosh(x)^4 - 140*(4*a^3*b^3 - 9*a*b^5)*cosh(x)^3 + 35*(16*a^4*b^2 - 40 *a^2*b^4 + 29*b^6)*cosh(x)^2 - 160*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*x - 70*(8*a^5*b - 22*a^3*b^3 + 19*a*b^5)*cosh(x))*sinh(x)^6 + 5*b^6 - 120*(8* a^5*b - 22*a^3*b^3 + 19*a*b^5)*cosh(x)^5 + 24*(165*b^6*cosh(x)^7 - 231*a*b ^5*cosh(x)^6 - 40*a^5*b + 110*a^3*b^3 - 95*a*b^5 + 315*(a^2*b^4 - 2*b^6)*c osh(x)^5 - 105*(4*a^3*b^3 - 9*a*b^5)*cosh(x)^4 + 35*(16*a^4*b^2 - 40*a^2*b ^4 + 29*b^6)*cosh(x)^3 - 480*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*x*cosh...
Timed out. \[ \int \frac {\sinh ^7(x)}{a+b \cosh (x)} \, dx=\text {Timed out} \] Input:
integrate(sinh(x)**7/(a+b*cosh(x)),x)
Output:
Timed out
Leaf count of result is larger than twice the leaf count of optimal. 310 vs. \(2 (130) = 260\).
Time = 0.05 (sec) , antiderivative size = 310, normalized size of antiderivative = 2.21 \[ \int \frac {\sinh ^7(x)}{a+b \cosh (x)} \, dx=-\frac {{\left (12 \, a b^{4} e^{\left (-x\right )} - 5 \, b^{5} - 30 \, {\left (a^{2} b^{3} - 2 \, b^{5}\right )} e^{\left (-2 \, x\right )} + 20 \, {\left (4 \, a^{3} b^{2} - 9 \, a b^{4}\right )} e^{\left (-3 \, x\right )} - 15 \, {\left (16 \, a^{4} b - 40 \, a^{2} b^{3} + 29 \, b^{5}\right )} e^{\left (-4 \, x\right )} + 120 \, {\left (8 \, a^{5} - 22 \, a^{3} b^{2} + 19 \, a b^{4}\right )} e^{\left (-5 \, x\right )}\right )} e^{\left (6 \, x\right )}}{1920 \, b^{6}} - \frac {12 \, a b^{4} e^{\left (-5 \, x\right )} - 5 \, b^{5} e^{\left (-6 \, x\right )} + 120 \, {\left (8 \, a^{5} - 22 \, a^{3} b^{2} + 19 \, a b^{4}\right )} e^{\left (-x\right )} - 15 \, {\left (16 \, a^{4} b - 40 \, a^{2} b^{3} + 29 \, b^{5}\right )} e^{\left (-2 \, x\right )} + 20 \, {\left (4 \, a^{3} b^{2} - 9 \, a b^{4}\right )} e^{\left (-3 \, x\right )} - 30 \, {\left (a^{2} b^{3} - 2 \, b^{5}\right )} e^{\left (-4 \, x\right )}}{1920 \, b^{6}} + \frac {{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} x}{b^{7}} + \frac {{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} \log \left (2 \, a e^{\left (-x\right )} + b e^{\left (-2 \, x\right )} + b\right )}{b^{7}} \] Input:
integrate(sinh(x)^7/(a+b*cosh(x)),x, algorithm="maxima")
Output:
-1/1920*(12*a*b^4*e^(-x) - 5*b^5 - 30*(a^2*b^3 - 2*b^5)*e^(-2*x) + 20*(4*a ^3*b^2 - 9*a*b^4)*e^(-3*x) - 15*(16*a^4*b - 40*a^2*b^3 + 29*b^5)*e^(-4*x) + 120*(8*a^5 - 22*a^3*b^2 + 19*a*b^4)*e^(-5*x))*e^(6*x)/b^6 - 1/1920*(12*a *b^4*e^(-5*x) - 5*b^5*e^(-6*x) + 120*(8*a^5 - 22*a^3*b^2 + 19*a*b^4)*e^(-x ) - 15*(16*a^4*b - 40*a^2*b^3 + 29*b^5)*e^(-2*x) + 20*(4*a^3*b^2 - 9*a*b^4 )*e^(-3*x) - 30*(a^2*b^3 - 2*b^5)*e^(-4*x))/b^6 + (a^6 - 3*a^4*b^2 + 3*a^2 *b^4 - b^6)*x/b^7 + (a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*log(2*a*e^(-x) + b *e^(-2*x) + b)/b^7
Time = 0.12 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.64 \[ \int \frac {\sinh ^7(x)}{a+b \cosh (x)} \, dx=\frac {5 \, b^{5} {\left (e^{\left (-x\right )} + e^{x}\right )}^{6} - 12 \, a b^{4} {\left (e^{\left (-x\right )} + e^{x}\right )}^{5} + 30 \, a^{2} b^{3} {\left (e^{\left (-x\right )} + e^{x}\right )}^{4} - 90 \, b^{5} {\left (e^{\left (-x\right )} + e^{x}\right )}^{4} - 80 \, a^{3} b^{2} {\left (e^{\left (-x\right )} + e^{x}\right )}^{3} + 240 \, a b^{4} {\left (e^{\left (-x\right )} + e^{x}\right )}^{3} + 240 \, a^{4} b {\left (e^{\left (-x\right )} + e^{x}\right )}^{2} - 720 \, a^{2} b^{3} {\left (e^{\left (-x\right )} + e^{x}\right )}^{2} + 720 \, b^{5} {\left (e^{\left (-x\right )} + e^{x}\right )}^{2} - 960 \, a^{5} {\left (e^{\left (-x\right )} + e^{x}\right )} + 2880 \, a^{3} b^{2} {\left (e^{\left (-x\right )} + e^{x}\right )} - 2880 \, a b^{4} {\left (e^{\left (-x\right )} + e^{x}\right )}}{1920 \, b^{6}} + \frac {{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} \log \left ({\left | b {\left (e^{\left (-x\right )} + e^{x}\right )} + 2 \, a \right |}\right )}{b^{7}} \] Input:
integrate(sinh(x)^7/(a+b*cosh(x)),x, algorithm="giac")
Output:
1/1920*(5*b^5*(e^(-x) + e^x)^6 - 12*a*b^4*(e^(-x) + e^x)^5 + 30*a^2*b^3*(e ^(-x) + e^x)^4 - 90*b^5*(e^(-x) + e^x)^4 - 80*a^3*b^2*(e^(-x) + e^x)^3 + 2 40*a*b^4*(e^(-x) + e^x)^3 + 240*a^4*b*(e^(-x) + e^x)^2 - 720*a^2*b^3*(e^(- x) + e^x)^2 + 720*b^5*(e^(-x) + e^x)^2 - 960*a^5*(e^(-x) + e^x) + 2880*a^3 *b^2*(e^(-x) + e^x) - 2880*a*b^4*(e^(-x) + e^x))/b^6 + (a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*log(abs(b*(e^(-x) + e^x) + 2*a))/b^7
Time = 2.77 (sec) , antiderivative size = 289, normalized size of antiderivative = 2.06 \[ \int \frac {\sinh ^7(x)}{a+b \cosh (x)} \, dx=\frac {{\mathrm {e}}^{-6\,x}}{384\,b}+\frac {{\mathrm {e}}^{6\,x}}{384\,b}-\frac {x\,{\left (a^2-b^2\right )}^3}{b^7}-\frac {{\mathrm {e}}^{-x}\,\left (8\,a^5-22\,a^3\,b^2+19\,a\,b^4\right )}{16\,b^6}+\frac {{\mathrm {e}}^{-3\,x}\,\left (9\,a\,b^2-4\,a^3\right )}{96\,b^4}+\frac {{\mathrm {e}}^{3\,x}\,\left (9\,a\,b^2-4\,a^3\right )}{96\,b^4}+\frac {{\mathrm {e}}^{-4\,x}\,\left (a^2-2\,b^2\right )}{64\,b^3}+\frac {{\mathrm {e}}^{4\,x}\,\left (a^2-2\,b^2\right )}{64\,b^3}-\frac {a\,{\mathrm {e}}^{-5\,x}}{160\,b^2}-\frac {a\,{\mathrm {e}}^{5\,x}}{160\,b^2}+\frac {{\mathrm {e}}^{-2\,x}\,\left (16\,a^4-40\,a^2\,b^2+29\,b^4\right )}{128\,b^5}+\frac {{\mathrm {e}}^{2\,x}\,\left (16\,a^4-40\,a^2\,b^2+29\,b^4\right )}{128\,b^5}-\frac {{\mathrm {e}}^x\,\left (8\,a^5-22\,a^3\,b^2+19\,a\,b^4\right )}{16\,b^6}+\frac {\ln \left (b+2\,a\,{\mathrm {e}}^x+b\,{\mathrm {e}}^{2\,x}\right )\,\left (a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6\right )}{b^7} \] Input:
int(sinh(x)^7/(a + b*cosh(x)),x)
Output:
exp(-6*x)/(384*b) + exp(6*x)/(384*b) - (x*(a^2 - b^2)^3)/b^7 - (exp(-x)*(1 9*a*b^4 + 8*a^5 - 22*a^3*b^2))/(16*b^6) + (exp(-3*x)*(9*a*b^2 - 4*a^3))/(9 6*b^4) + (exp(3*x)*(9*a*b^2 - 4*a^3))/(96*b^4) + (exp(-4*x)*(a^2 - 2*b^2)) /(64*b^3) + (exp(4*x)*(a^2 - 2*b^2))/(64*b^3) - (a*exp(-5*x))/(160*b^2) - (a*exp(5*x))/(160*b^2) + (exp(-2*x)*(16*a^4 + 29*b^4 - 40*a^2*b^2))/(128*b ^5) + (exp(2*x)*(16*a^4 + 29*b^4 - 40*a^2*b^2))/(128*b^5) - (exp(x)*(19*a* b^4 + 8*a^5 - 22*a^3*b^2))/(16*b^6) + (log(b + 2*a*exp(x) + b*exp(2*x))*(a ^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2))/b^7
Time = 0.33 (sec) , antiderivative size = 444, normalized size of antiderivative = 3.17 \[ \int \frac {\sinh ^7(x)}{a+b \cosh (x)} \, dx=\frac {5 b^{6}-12 e^{11 x} a \,b^{5}+30 e^{10 x} a^{2} b^{4}-80 e^{9 x} a^{3} b^{3}+180 e^{9 x} a \,b^{5}+240 e^{8 x} a^{4} b^{2}-600 e^{8 x} a^{2} b^{4}-960 e^{7 x} a^{5} b +2640 e^{7 x} a^{3} b^{3}-2280 e^{7 x} a \,b^{5}+1920 e^{6 x} \mathrm {log}\left (e^{2 x} b +2 e^{x} a +b \right ) a^{6}-1920 e^{6 x} \mathrm {log}\left (e^{2 x} b +2 e^{x} a +b \right ) b^{6}-1920 e^{6 x} a^{6} x +1920 e^{6 x} b^{6} x -960 e^{5 x} a^{5} b +2640 e^{5 x} a^{3} b^{3}-2280 e^{5 x} a \,b^{5}-80 e^{3 x} a^{3} b^{3}+180 e^{3 x} a \,b^{5}+435 e^{4 x} b^{6}-60 e^{2 x} b^{6}+240 e^{4 x} a^{4} b^{2}+30 e^{2 x} a^{2} b^{4}-12 e^{x} a \,b^{5}+5 e^{12 x} b^{6}-60 e^{10 x} b^{6}+435 e^{8 x} b^{6}-600 e^{4 x} a^{2} b^{4}-5760 e^{6 x} \mathrm {log}\left (e^{2 x} b +2 e^{x} a +b \right ) a^{4} b^{2}+5760 e^{6 x} \mathrm {log}\left (e^{2 x} b +2 e^{x} a +b \right ) a^{2} b^{4}+5760 e^{6 x} a^{4} b^{2} x -5760 e^{6 x} a^{2} b^{4} x}{1920 e^{6 x} b^{7}} \] Input:
int(sinh(x)^7/(a+b*cosh(x)),x)
Output:
(5*e**(12*x)*b**6 - 12*e**(11*x)*a*b**5 + 30*e**(10*x)*a**2*b**4 - 60*e**( 10*x)*b**6 - 80*e**(9*x)*a**3*b**3 + 180*e**(9*x)*a*b**5 + 240*e**(8*x)*a* *4*b**2 - 600*e**(8*x)*a**2*b**4 + 435*e**(8*x)*b**6 - 960*e**(7*x)*a**5*b + 2640*e**(7*x)*a**3*b**3 - 2280*e**(7*x)*a*b**5 + 1920*e**(6*x)*log(e**( 2*x)*b + 2*e**x*a + b)*a**6 - 5760*e**(6*x)*log(e**(2*x)*b + 2*e**x*a + b) *a**4*b**2 + 5760*e**(6*x)*log(e**(2*x)*b + 2*e**x*a + b)*a**2*b**4 - 1920 *e**(6*x)*log(e**(2*x)*b + 2*e**x*a + b)*b**6 - 1920*e**(6*x)*a**6*x + 576 0*e**(6*x)*a**4*b**2*x - 5760*e**(6*x)*a**2*b**4*x + 1920*e**(6*x)*b**6*x - 960*e**(5*x)*a**5*b + 2640*e**(5*x)*a**3*b**3 - 2280*e**(5*x)*a*b**5 + 2 40*e**(4*x)*a**4*b**2 - 600*e**(4*x)*a**2*b**4 + 435*e**(4*x)*b**6 - 80*e* *(3*x)*a**3*b**3 + 180*e**(3*x)*a*b**5 + 30*e**(2*x)*a**2*b**4 - 60*e**(2* x)*b**6 - 12*e**x*a*b**5 + 5*b**6)/(1920*e**(6*x)*b**7)