Integrand size = 29, antiderivative size = 160 \[ \int (e x)^m \left (a+b x^n\right )^2 \left (A+B x^n\right ) \left (c+d x^n\right ) \, dx=\frac {a (2 A b c+a B c+a A d) x^{1+n} (e x)^m}{1+m+n}+\frac {(a B (2 b c+a d)+A b (b c+2 a d)) x^{1+2 n} (e x)^m}{1+m+2 n}+\frac {b (b B c+A b d+2 a B d) x^{1+3 n} (e x)^m}{1+m+3 n}+\frac {b^2 B d x^{1+4 n} (e x)^m}{1+m+4 n}+\frac {a^2 A c (e x)^{1+m}}{e (1+m)} \]
a*(A*a*d+2*A*b*c+B*a*c)*x^(1+n)*(e*x)^m/(1+m+n)+(a*B*(a*d+2*b*c)+A*b*(2*a* d+b*c))*x^(1+2*n)*(e*x)^m/(1+m+2*n)+b*(A*b*d+2*B*a*d+B*b*c)*x^(1+3*n)*(e*x )^m/(1+m+3*n)+b^2*B*d*x^(1+4*n)*(e*x)^m/(1+m+4*n)+a^2*A*c*(e*x)^(1+m)/e/(1 +m)
Time = 0.39 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.81 \[ \int (e x)^m \left (a+b x^n\right )^2 \left (A+B x^n\right ) \left (c+d x^n\right ) \, dx=x (e x)^m \left (\frac {a^2 A c}{1+m}+\frac {a (2 A b c+a B c+a A d) x^n}{1+m+n}+\frac {(a B (2 b c+a d)+A b (b c+2 a d)) x^{2 n}}{1+m+2 n}+\frac {b (b B c+A b d+2 a B d) x^{3 n}}{1+m+3 n}+\frac {b^2 B d x^{4 n}}{1+m+4 n}\right ) \]
x*(e*x)^m*((a^2*A*c)/(1 + m) + (a*(2*A*b*c + a*B*c + a*A*d)*x^n)/(1 + m + n) + ((a*B*(2*b*c + a*d) + A*b*(b*c + 2*a*d))*x^(2*n))/(1 + m + 2*n) + (b* (b*B*c + A*b*d + 2*a*B*d)*x^(3*n))/(1 + m + 3*n) + (b^2*B*d*x^(4*n))/(1 + m + 4*n))
Time = 0.34 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {1040, 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.
\(\displaystyle \int (e x)^m \left (a+b x^n\right )^2 \left (A+B x^n\right ) \left (c+d x^n\right ) \, dx\) |
\(\Big \downarrow \) 1040 |
\(\displaystyle \int \left (a^2 A c (e x)^m+x^{2 n} (e x)^m (A b (2 a d+b c)+a B (a d+2 b c))+b x^{3 n} (e x)^m (2 a B d+A b d+b B c)+a x^n (e x)^m (a A d+a B c+2 A b c)+b^2 B d x^{4 n} (e x)^m\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {a^2 A c (e x)^{m+1}}{e (m+1)}+\frac {a x^{n+1} (e x)^m (a A d+a B c+2 A b c)}{m+n+1}+\frac {x^{2 n+1} (e x)^m (A b (2 a d+b c)+a B (a d+2 b c))}{m+2 n+1}+\frac {b x^{3 n+1} (e x)^m (2 a B d+A b d+b B c)}{m+3 n+1}+\frac {b^2 B d x^{4 n+1} (e x)^m}{m+4 n+1}\) |
(a*(2*A*b*c + a*B*c + a*A*d)*x^(1 + n)*(e*x)^m)/(1 + m + n) + ((a*B*(2*b*c + a*d) + A*b*(b*c + 2*a*d))*x^(1 + 2*n)*(e*x)^m)/(1 + m + 2*n) + (b*(b*B* c + A*b*d + 2*a*B*d)*x^(1 + 3*n)*(e*x)^m)/(1 + m + 3*n) + (b^2*B*d*x^(1 + 4*n)*(e*x)^m)/(1 + m + 4*n) + (a^2*A*c*(e*x)^(1 + m))/(e*(1 + m))
3.1.2.3.1 Defintions of rubi rules used
Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n _))^(q_.)*((e_) + (f_.)*(x_)^(n_))^(r_.), x_Symbol] :> Int[ExpandIntegrand[ (g*x)^m*(a + b*x^n)^p*(c + d*x^n)^q*(e + f*x^n)^r, x], x] /; FreeQ[{a, b, c , d, e, f, g, m, n}, x] && IGtQ[p, -2] && IGtQ[q, 0] && IGtQ[r, 0]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 2.40 (sec) , antiderivative size = 2377, normalized size of antiderivative = 14.86
method | result | size |
risch | \(\text {Expression too large to display}\) | \(2377\) |
parallelrisch | \(\text {Expression too large to display}\) | \(3344\) |
x*(8*B*a*b*c*m^3*(x^n)^2+48*A*a*b*d*m^2*n*(x^n)^2+19*A*b^2*c*m^2*n^2*(x^n) ^2+38*A*a*b*d*n^2*(x^n)^2+4*A*b^2*d*(x^n)^3*m+26*A*a^2*d*m^2*n^2*x^n+21*A* b^2*d*m*n*(x^n)^3+2*B*a*b*d*m^4*(x^n)^3+21*B*b^2*c*m^2*n*(x^n)^3+6*b^2*B*d *(x^n)^4*n+4*A*a^2*d*m^3*x^n+24*A*a^2*d*n^3*x^n+16*B*a*b*d*n^3*(x^n)^3+24* B*a*b*c*m*n^3*(x^n)^2+56*B*a*b*d*m*n^2*(x^n)^3+4*A*a^2*c*m+8*A*b^2*c*m^3*n *(x^n)^2+12*B*a*b*d*m^2*(x^n)^3+14*B*a*b*d*m^3*n*(x^n)^3+A*a^2*d*m^4*x^n+2 4*A*a^2*c*n^4+8*B*a^2*d*m^3*n*(x^n)^2+6*A*a^2*d*m^2*x^n+18*B*b^2*d*m*n*(x^ n)^4+42*B*a*b*d*m^2*n*(x^n)^3+B*a^2*d*(x^n)^2+A*a^2*c*m^4+4*A*a^2*c*m^3+50 *A*a^2*c*n^3+6*A*a^2*c*m^2+35*A*a^2*c*n^2+27*B*a^2*c*m^2*n*x^n+52*B*a^2*c* m*n^2*x^n+9*A*a^2*d*m^3*n*x^n+b^2*B*d*(x^n)^4+8*B*b^2*c*m*n^3*(x^n)^3+18*B *b^2*d*m^2*n*(x^n)^4+24*A*a*b*d*m*n^3*(x^n)^2+12*B*a*b*c*m^2*(x^n)^2+38*B* a*b*c*n^2*(x^n)^2+8*B*a*b*d*(x^n)^3*m+48*B*a*b*c*m^2*n*(x^n)^2+38*A*b^2*c* m*n^2*(x^n)^2+24*B*a^2*d*m^2*n*(x^n)^2+24*A*b^2*c*m^2*n*(x^n)^2+10*A*a^2*c *n+12*A*b^2*c*m*n^3*(x^n)^2+24*B*a^2*c*m*n^3*x^n+8*A*b^2*d*n^3*(x^n)^3+18* A*a*b*c*m^3*n*x^n+48*B*a*b*c*m*n*(x^n)^2+54*A*a*b*c*m^2*n*x^n+12*B*a^2*d*m *n^3*(x^n)^2+2*B*a*b*c*m^4*(x^n)^2+4*A*a^2*d*x^n*m+9*A*a^2*d*x^n*n+4*B*b^2 *c*(x^n)^3*m+7*B*b^2*c*(x^n)^3*n+48*A*a*b*c*n^3*x^n+22*B*b^2*d*m*n^2*(x^n) ^4+16*B*a*b*c*m^3*n*(x^n)^2+14*A*b^2*d*m^2*n^2*(x^n)^3+8*A*a*b*d*m^3*(x^n) ^2+24*B*a*b*c*n^3*(x^n)^2+16*A*a*b*d*m^3*n*(x^n)^2+7*A*b^2*d*m^3*n*(x^n)^3 +2*A*a*b*d*m^4*(x^n)^2+21*B*b^2*c*m*n*(x^n)^3+52*A*a*b*c*m^2*n^2*x^n+38...
Leaf count of result is larger than twice the leaf count of optimal. 1524 vs. \(2 (160) = 320\).
Time = 0.32 (sec) , antiderivative size = 1524, normalized size of antiderivative = 9.52 \[ \int (e x)^m \left (a+b x^n\right )^2 \left (A+B x^n\right ) \left (c+d x^n\right ) \, dx=\text {Too large to display} \]
((B*b^2*d*m^4 + 4*B*b^2*d*m^3 + 6*B*b^2*d*m^2 + 4*B*b^2*d*m + B*b^2*d + 6* (B*b^2*d*m + B*b^2*d)*n^3 + 11*(B*b^2*d*m^2 + 2*B*b^2*d*m + B*b^2*d)*n^2 + 6*(B*b^2*d*m^3 + 3*B*b^2*d*m^2 + 3*B*b^2*d*m + B*b^2*d)*n)*x*x^(4*n)*e^(m *log(e) + m*log(x)) + ((B*b^2*c + (2*B*a*b + A*b^2)*d)*m^4 + B*b^2*c + 4*( B*b^2*c + (2*B*a*b + A*b^2)*d)*m^3 + 8*(B*b^2*c + (2*B*a*b + A*b^2)*d + (B *b^2*c + (2*B*a*b + A*b^2)*d)*m)*n^3 + 6*(B*b^2*c + (2*B*a*b + A*b^2)*d)*m ^2 + 14*(B*b^2*c + (B*b^2*c + (2*B*a*b + A*b^2)*d)*m^2 + (2*B*a*b + A*b^2) *d + 2*(B*b^2*c + (2*B*a*b + A*b^2)*d)*m)*n^2 + (2*B*a*b + A*b^2)*d + 4*(B *b^2*c + (2*B*a*b + A*b^2)*d)*m + 7*(B*b^2*c + (B*b^2*c + (2*B*a*b + A*b^2 )*d)*m^3 + 3*(B*b^2*c + (2*B*a*b + A*b^2)*d)*m^2 + (2*B*a*b + A*b^2)*d + 3 *(B*b^2*c + (2*B*a*b + A*b^2)*d)*m)*n)*x*x^(3*n)*e^(m*log(e) + m*log(x)) + (((2*B*a*b + A*b^2)*c + (B*a^2 + 2*A*a*b)*d)*m^4 + 4*((2*B*a*b + A*b^2)*c + (B*a^2 + 2*A*a*b)*d)*m^3 + 12*((2*B*a*b + A*b^2)*c + (B*a^2 + 2*A*a*b)* d + ((2*B*a*b + A*b^2)*c + (B*a^2 + 2*A*a*b)*d)*m)*n^3 + 6*((2*B*a*b + A*b ^2)*c + (B*a^2 + 2*A*a*b)*d)*m^2 + 19*(((2*B*a*b + A*b^2)*c + (B*a^2 + 2*A *a*b)*d)*m^2 + (2*B*a*b + A*b^2)*c + (B*a^2 + 2*A*a*b)*d + 2*((2*B*a*b + A *b^2)*c + (B*a^2 + 2*A*a*b)*d)*m)*n^2 + (2*B*a*b + A*b^2)*c + (B*a^2 + 2*A *a*b)*d + 4*((2*B*a*b + A*b^2)*c + (B*a^2 + 2*A*a*b)*d)*m + 8*(((2*B*a*b + A*b^2)*c + (B*a^2 + 2*A*a*b)*d)*m^3 + 3*((2*B*a*b + A*b^2)*c + (B*a^2 + 2 *A*a*b)*d)*m^2 + (2*B*a*b + A*b^2)*c + (B*a^2 + 2*A*a*b)*d + 3*((2*B*a*...
Leaf count of result is larger than twice the leaf count of optimal. 25315 vs. \(2 (156) = 312\).
Time = 7.33 (sec) , antiderivative size = 25315, normalized size of antiderivative = 158.22 \[ \int (e x)^m \left (a+b x^n\right )^2 \left (A+B x^n\right ) \left (c+d x^n\right ) \, dx=\text {Too large to display} \]
Piecewise(((A + B)*(a + b)**2*(c + d)*log(x)/e, Eq(m, -1) & Eq(n, 0)), ((A *a**2*c*log(x) + A*a**2*d*x**n/n + 2*A*a*b*c*x**n/n + A*a*b*d*x**(2*n)/n + A*b**2*c*x**(2*n)/(2*n) + A*b**2*d*x**(3*n)/(3*n) + B*a**2*c*x**n/n + B*a **2*d*x**(2*n)/(2*n) + B*a*b*c*x**(2*n)/n + 2*B*a*b*d*x**(3*n)/(3*n) + B*b **2*c*x**(3*n)/(3*n) + B*b**2*d*x**(4*n)/(4*n))/e, Eq(m, -1)), (A*a**2*c*P iecewise((0**(-4*n - 1)*x, Eq(e, 0)), (Piecewise((-1/(4*n*(e*x)**(4*n)), N e(n, 0)), (log(e*x), True))/e, True)) + A*a**2*d*Piecewise((-x*x**n*(e*x)* *(-4*n - 1)/(3*n), Ne(n, 0)), (x*x**n*(e*x)**(-4*n - 1)*log(x), True)) + 2 *A*a*b*c*Piecewise((-x*x**n*(e*x)**(-4*n - 1)/(3*n), Ne(n, 0)), (x*x**n*(e *x)**(-4*n - 1)*log(x), True)) + 2*A*a*b*d*Piecewise((-x*x**(2*n)*(e*x)**( -4*n - 1)/(2*n), Ne(n, 0)), (x*x**(2*n)*(e*x)**(-4*n - 1)*log(x), True)) + A*b**2*c*Piecewise((-x*x**(2*n)*(e*x)**(-4*n - 1)/(2*n), Ne(n, 0)), (x*x* *(2*n)*(e*x)**(-4*n - 1)*log(x), True)) + A*b**2*d*Piecewise((-x*x**(3*n)* (e*x)**(-4*n - 1)/n, Ne(n, 0)), (x*x**(3*n)*(e*x)**(-4*n - 1)*log(x), True )) + B*a**2*c*Piecewise((-x*x**n*(e*x)**(-4*n - 1)/(3*n), Ne(n, 0)), (x*x* *n*(e*x)**(-4*n - 1)*log(x), True)) + B*a**2*d*Piecewise((-x*x**(2*n)*(e*x )**(-4*n - 1)/(2*n), Ne(n, 0)), (x*x**(2*n)*(e*x)**(-4*n - 1)*log(x), True )) + 2*B*a*b*c*Piecewise((-x*x**(2*n)*(e*x)**(-4*n - 1)/(2*n), Ne(n, 0)), (x*x**(2*n)*(e*x)**(-4*n - 1)*log(x), True)) + 2*B*a*b*d*Piecewise((-x*x** (3*n)*(e*x)**(-4*n - 1)/n, Ne(n, 0)), (x*x**(3*n)*(e*x)**(-4*n - 1)*log...
Leaf count of result is larger than twice the leaf count of optimal. 332 vs. \(2 (160) = 320\).
Time = 0.24 (sec) , antiderivative size = 332, normalized size of antiderivative = 2.08 \[ \int (e x)^m \left (a+b x^n\right )^2 \left (A+B x^n\right ) \left (c+d x^n\right ) \, dx=\frac {B b^{2} d e^{m} x e^{\left (m \log \left (x\right ) + 4 \, n \log \left (x\right )\right )}}{m + 4 \, n + 1} + \frac {B b^{2} c e^{m} x e^{\left (m \log \left (x\right ) + 3 \, n \log \left (x\right )\right )}}{m + 3 \, n + 1} + \frac {2 \, B a b d e^{m} x e^{\left (m \log \left (x\right ) + 3 \, n \log \left (x\right )\right )}}{m + 3 \, n + 1} + \frac {A b^{2} d e^{m} x e^{\left (m \log \left (x\right ) + 3 \, n \log \left (x\right )\right )}}{m + 3 \, n + 1} + \frac {2 \, B a b c e^{m} x e^{\left (m \log \left (x\right ) + 2 \, n \log \left (x\right )\right )}}{m + 2 \, n + 1} + \frac {A b^{2} c e^{m} x e^{\left (m \log \left (x\right ) + 2 \, n \log \left (x\right )\right )}}{m + 2 \, n + 1} + \frac {B a^{2} d e^{m} x e^{\left (m \log \left (x\right ) + 2 \, n \log \left (x\right )\right )}}{m + 2 \, n + 1} + \frac {2 \, A a b d e^{m} x e^{\left (m \log \left (x\right ) + 2 \, n \log \left (x\right )\right )}}{m + 2 \, n + 1} + \frac {B a^{2} c e^{m} x e^{\left (m \log \left (x\right ) + n \log \left (x\right )\right )}}{m + n + 1} + \frac {2 \, A a b c e^{m} x e^{\left (m \log \left (x\right ) + n \log \left (x\right )\right )}}{m + n + 1} + \frac {A a^{2} d e^{m} x e^{\left (m \log \left (x\right ) + n \log \left (x\right )\right )}}{m + n + 1} + \frac {\left (e x\right )^{m + 1} A a^{2} c}{e {\left (m + 1\right )}} \]
B*b^2*d*e^m*x*e^(m*log(x) + 4*n*log(x))/(m + 4*n + 1) + B*b^2*c*e^m*x*e^(m *log(x) + 3*n*log(x))/(m + 3*n + 1) + 2*B*a*b*d*e^m*x*e^(m*log(x) + 3*n*lo g(x))/(m + 3*n + 1) + A*b^2*d*e^m*x*e^(m*log(x) + 3*n*log(x))/(m + 3*n + 1 ) + 2*B*a*b*c*e^m*x*e^(m*log(x) + 2*n*log(x))/(m + 2*n + 1) + A*b^2*c*e^m* x*e^(m*log(x) + 2*n*log(x))/(m + 2*n + 1) + B*a^2*d*e^m*x*e^(m*log(x) + 2* n*log(x))/(m + 2*n + 1) + 2*A*a*b*d*e^m*x*e^(m*log(x) + 2*n*log(x))/(m + 2 *n + 1) + B*a^2*c*e^m*x*e^(m*log(x) + n*log(x))/(m + n + 1) + 2*A*a*b*c*e^ m*x*e^(m*log(x) + n*log(x))/(m + n + 1) + A*a^2*d*e^m*x*e^(m*log(x) + n*lo g(x))/(m + n + 1) + (e*x)^(m + 1)*A*a^2*c/(e*(m + 1))
Leaf count of result is larger than twice the leaf count of optimal. 11834 vs. \(2 (160) = 320\).
Time = 0.38 (sec) , antiderivative size = 11834, normalized size of antiderivative = 73.96 \[ \int (e x)^m \left (a+b x^n\right )^2 \left (A+B x^n\right ) \left (c+d x^n\right ) \, dx=\text {Too large to display} \]
(B*b^2*d*m^4*x*x^(4*n)*e^(m*log(e) + m*log(x)) + 6*B*b^2*d*m^3*n*x*x^(4*n) *e^(m*log(e) + m*log(x)) + 11*B*b^2*d*m^2*n^2*x*x^(4*n)*e^(m*log(e) + m*lo g(x)) + 6*B*b^2*d*m*n^3*x*x^(4*n)*e^(m*log(e) + m*log(x)) + B*b^2*c*m^4*x* x^(3*n)*e^(m*log(e) + m*log(x)) + 2*B*a*b*d*m^4*x*x^(3*n)*e^(m*log(e) + m* log(x)) + A*b^2*d*m^4*x*x^(3*n)*e^(m*log(e) + m*log(x)) + B*b^2*d*m^4*x*x^ (3*n)*e^(m*log(e) + m*log(x)) + 7*B*b^2*c*m^3*n*x*x^(3*n)*e^(m*log(e) + m* log(x)) + 14*B*a*b*d*m^3*n*x*x^(3*n)*e^(m*log(e) + m*log(x)) + 7*A*b^2*d*m ^3*n*x*x^(3*n)*e^(m*log(e) + m*log(x)) + 6*B*b^2*d*m^3*n*x*x^(3*n)*e^(m*lo g(e) + m*log(x)) + 14*B*b^2*c*m^2*n^2*x*x^(3*n)*e^(m*log(e) + m*log(x)) + 28*B*a*b*d*m^2*n^2*x*x^(3*n)*e^(m*log(e) + m*log(x)) + 14*A*b^2*d*m^2*n^2* x*x^(3*n)*e^(m*log(e) + m*log(x)) + 11*B*b^2*d*m^2*n^2*x*x^(3*n)*e^(m*log( e) + m*log(x)) + 8*B*b^2*c*m*n^3*x*x^(3*n)*e^(m*log(e) + m*log(x)) + 16*B* a*b*d*m*n^3*x*x^(3*n)*e^(m*log(e) + m*log(x)) + 8*A*b^2*d*m*n^3*x*x^(3*n)* e^(m*log(e) + m*log(x)) + 6*B*b^2*d*m*n^3*x*x^(3*n)*e^(m*log(e) + m*log(x) ) + 2*B*a*b*c*m^4*x*x^(2*n)*e^(m*log(e) + m*log(x)) + A*b^2*c*m^4*x*x^(2*n )*e^(m*log(e) + m*log(x)) + B*b^2*c*m^4*x*x^(2*n)*e^(m*log(e) + m*log(x)) + B*a^2*d*m^4*x*x^(2*n)*e^(m*log(e) + m*log(x)) + 2*A*a*b*d*m^4*x*x^(2*n)* e^(m*log(e) + m*log(x)) + 2*B*a*b*d*m^4*x*x^(2*n)*e^(m*log(e) + m*log(x)) + A*b^2*d*m^4*x*x^(2*n)*e^(m*log(e) + m*log(x)) + B*b^2*d*m^4*x*x^(2*n)*e^ (m*log(e) + m*log(x)) + 16*B*a*b*c*m^3*n*x*x^(2*n)*e^(m*log(e) + m*log(...
Time = 9.50 (sec) , antiderivative size = 588, normalized size of antiderivative = 3.68 \[ \int (e x)^m \left (a+b x^n\right )^2 \left (A+B x^n\right ) \left (c+d x^n\right ) \, dx=\frac {x\,x^{2\,n}\,{\left (e\,x\right )}^m\,\left (A\,b^2\,c+B\,a^2\,d+2\,A\,a\,b\,d+2\,B\,a\,b\,c\right )\,\left (m^3+8\,m^2\,n+3\,m^2+19\,m\,n^2+16\,m\,n+3\,m+12\,n^3+19\,n^2+8\,n+1\right )}{m^4+10\,m^3\,n+4\,m^3+35\,m^2\,n^2+30\,m^2\,n+6\,m^2+50\,m\,n^3+70\,m\,n^2+30\,m\,n+4\,m+24\,n^4+50\,n^3+35\,n^2+10\,n+1}+\frac {A\,a^2\,c\,x\,{\left (e\,x\right )}^m}{m+1}+\frac {a\,x\,x^n\,{\left (e\,x\right )}^m\,\left (A\,a\,d+2\,A\,b\,c+B\,a\,c\right )\,\left (m^3+9\,m^2\,n+3\,m^2+26\,m\,n^2+18\,m\,n+3\,m+24\,n^3+26\,n^2+9\,n+1\right )}{m^4+10\,m^3\,n+4\,m^3+35\,m^2\,n^2+30\,m^2\,n+6\,m^2+50\,m\,n^3+70\,m\,n^2+30\,m\,n+4\,m+24\,n^4+50\,n^3+35\,n^2+10\,n+1}+\frac {b\,x\,x^{3\,n}\,{\left (e\,x\right )}^m\,\left (A\,b\,d+2\,B\,a\,d+B\,b\,c\right )\,\left (m^3+7\,m^2\,n+3\,m^2+14\,m\,n^2+14\,m\,n+3\,m+8\,n^3+14\,n^2+7\,n+1\right )}{m^4+10\,m^3\,n+4\,m^3+35\,m^2\,n^2+30\,m^2\,n+6\,m^2+50\,m\,n^3+70\,m\,n^2+30\,m\,n+4\,m+24\,n^4+50\,n^3+35\,n^2+10\,n+1}+\frac {B\,b^2\,d\,x\,x^{4\,n}\,{\left (e\,x\right )}^m\,\left (m^3+6\,m^2\,n+3\,m^2+11\,m\,n^2+12\,m\,n+3\,m+6\,n^3+11\,n^2+6\,n+1\right )}{m^4+10\,m^3\,n+4\,m^3+35\,m^2\,n^2+30\,m^2\,n+6\,m^2+50\,m\,n^3+70\,m\,n^2+30\,m\,n+4\,m+24\,n^4+50\,n^3+35\,n^2+10\,n+1} \]
(x*x^(2*n)*(e*x)^m*(A*b^2*c + B*a^2*d + 2*A*a*b*d + 2*B*a*b*c)*(3*m + 8*n + 16*m*n + 19*m*n^2 + 8*m^2*n + 3*m^2 + m^3 + 19*n^2 + 12*n^3 + 1))/(4*m + 10*n + 30*m*n + 70*m*n^2 + 30*m^2*n + 50*m*n^3 + 10*m^3*n + 6*m^2 + 4*m^3 + m^4 + 35*n^2 + 50*n^3 + 24*n^4 + 35*m^2*n^2 + 1) + (A*a^2*c*x*(e*x)^m)/ (m + 1) + (a*x*x^n*(e*x)^m*(A*a*d + 2*A*b*c + B*a*c)*(3*m + 9*n + 18*m*n + 26*m*n^2 + 9*m^2*n + 3*m^2 + m^3 + 26*n^2 + 24*n^3 + 1))/(4*m + 10*n + 30 *m*n + 70*m*n^2 + 30*m^2*n + 50*m*n^3 + 10*m^3*n + 6*m^2 + 4*m^3 + m^4 + 3 5*n^2 + 50*n^3 + 24*n^4 + 35*m^2*n^2 + 1) + (b*x*x^(3*n)*(e*x)^m*(A*b*d + 2*B*a*d + B*b*c)*(3*m + 7*n + 14*m*n + 14*m*n^2 + 7*m^2*n + 3*m^2 + m^3 + 14*n^2 + 8*n^3 + 1))/(4*m + 10*n + 30*m*n + 70*m*n^2 + 30*m^2*n + 50*m*n^3 + 10*m^3*n + 6*m^2 + 4*m^3 + m^4 + 35*n^2 + 50*n^3 + 24*n^4 + 35*m^2*n^2 + 1) + (B*b^2*d*x*x^(4*n)*(e*x)^m*(3*m + 6*n + 12*m*n + 11*m*n^2 + 6*m^2*n + 3*m^2 + m^3 + 11*n^2 + 6*n^3 + 1))/(4*m + 10*n + 30*m*n + 70*m*n^2 + 30 *m^2*n + 50*m*n^3 + 10*m^3*n + 6*m^2 + 4*m^3 + m^4 + 35*n^2 + 50*n^3 + 24* n^4 + 35*m^2*n^2 + 1)