Integrand size = 29, antiderivative size = 225 \[ \int (e x)^m \left (a+b x^n\right )^3 \left (A+B x^n\right ) \left (c+d x^n\right ) \, dx=\frac {a^3 A c (e x)^{1+m}}{e (1+m)}+\frac {a^2 (3 A b c+a B c+a A d) x^n (e x)^{1+m}}{e (1+m+n)}+\frac {a (3 A b (b c+a d)+a B (3 b c+a d)) x^{2 n} (e x)^{1+m}}{e (1+m+2 n)}+\frac {b (3 a B (b c+a d)+A b (b c+3 a d)) x^{3 n} (e x)^{1+m}}{e (1+m+3 n)}+\frac {b^2 (b B c+A b d+3 a B d) x^{4 n} (e x)^{1+m}}{e (1+m+4 n)}+\frac {b^3 B d x^{5 n} (e x)^{1+m}}{e (1+m+5 n)} \] Output:
a^3*A*c*(e*x)^(1+m)/e/(1+m)+a^2*(A*a*d+3*A*b*c+B*a*c)*x^n*(e*x)^(1+m)/e/(1 +m+n)+a*(3*A*b*(a*d+b*c)+a*B*(a*d+3*b*c))*x^(2*n)*(e*x)^(1+m)/e/(1+m+2*n)+ b*(3*a*B*(a*d+b*c)+A*b*(3*a*d+b*c))*x^(3*n)*(e*x)^(1+m)/e/(1+m+3*n)+b^2*(A *b*d+3*B*a*d+B*b*c)*x^(4*n)*(e*x)^(1+m)/e/(1+m+4*n)+b^3*B*d*x^(5*n)*(e*x)^ (1+m)/e/(1+m+5*n)
Time = 0.79 (sec) , antiderivative size = 172, normalized size of antiderivative = 0.76 \[ \int (e x)^m \left (a+b x^n\right )^3 \left (A+B x^n\right ) \left (c+d x^n\right ) \, dx=x (e x)^m \left (\frac {a^3 A c}{1+m}+\frac {a^2 (3 A b c+a B c+a A d) x^n}{1+m+n}+\frac {a (3 A b (b c+a d)+a B (3 b c+a d)) x^{2 n}}{1+m+2 n}+\frac {b (3 a B (b c+a d)+A b (b c+3 a d)) x^{3 n}}{1+m+3 n}+\frac {b^2 (b B c+A b d+3 a B d) x^{4 n}}{1+m+4 n}+\frac {b^3 B d x^{5 n}}{1+m+5 n}\right ) \] Input:
Integrate[(e*x)^m*(a + b*x^n)^3*(A + B*x^n)*(c + d*x^n),x]
Output:
x*(e*x)^m*((a^3*A*c)/(1 + m) + (a^2*(3*A*b*c + a*B*c + a*A*d)*x^n)/(1 + m + n) + (a*(3*A*b*(b*c + a*d) + a*B*(3*b*c + a*d))*x^(2*n))/(1 + m + 2*n) + (b*(3*a*B*(b*c + a*d) + A*b*(b*c + 3*a*d))*x^(3*n))/(1 + m + 3*n) + (b^2* (b*B*c + A*b*d + 3*a*B*d)*x^(4*n))/(1 + m + 4*n) + (b^3*B*d*x^(5*n))/(1 + m + 5*n))
Time = 0.74 (sec) , antiderivative size = 210, normalized size of antiderivative = 0.93, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, 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.
\(\displaystyle \int (e x)^m \left (a+b x^n\right )^3 \left (A+B x^n\right ) \left (c+d x^n\right ) \, dx\) |
\(\Big \downarrow \) 1040 |
\(\displaystyle \int \left (a^3 A c (e x)^m+a^2 x^n (e x)^m (a A d+a B c+3 A b c)+b^2 x^{4 n} (e x)^m (3 a B d+A b d+b B c)+a x^{2 n} (e x)^m (3 A b (a d+b c)+a B (a d+3 b c))+b x^{3 n} (e x)^m (A b (3 a d+b c)+3 a B (a d+b c))+b^3 B d x^{5 n} (e x)^m\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {a^3 A c (e x)^{m+1}}{e (m+1)}+\frac {a^2 x^{n+1} (e x)^m (a A d+a B c+3 A b c)}{m+n+1}+\frac {b^2 x^{4 n+1} (e x)^m (3 a B d+A b d+b B c)}{m+4 n+1}+\frac {a x^{2 n+1} (e x)^m (3 A b (a d+b c)+a B (a d+3 b c))}{m+2 n+1}+\frac {b x^{3 n+1} (e x)^m (A b (3 a d+b c)+3 a B (a d+b c))}{m+3 n+1}+\frac {b^3 B d x^{5 n+1} (e x)^m}{m+5 n+1}\) |
Input:
Int[(e*x)^m*(a + b*x^n)^3*(A + B*x^n)*(c + d*x^n),x]
Output:
(a^2*(3*A*b*c + a*B*c + a*A*d)*x^(1 + n)*(e*x)^m)/(1 + m + n) + (a*(3*A*b* (b*c + a*d) + a*B*(3*b*c + a*d))*x^(1 + 2*n)*(e*x)^m)/(1 + m + 2*n) + (b*( 3*a*B*(b*c + a*d) + A*b*(b*c + 3*a*d))*x^(1 + 3*n)*(e*x)^m)/(1 + m + 3*n) + (b^2*(b*B*c + A*b*d + 3*a*B*d)*x^(1 + 4*n)*(e*x)^m)/(1 + m + 4*n) + (b^3 *B*d*x^(1 + 5*n)*(e*x)^m)/(1 + m + 5*n) + (a^3*A*c*(e*x)^(1 + m))/(e*(1 + m))
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 = 1.37 (sec) , antiderivative size = 4939, normalized size of antiderivative = 21.95
method | result | size |
risch | \(\text {Expression too large to display}\) | \(4939\) |
parallelrisch | \(\text {Expression too large to display}\) | \(6818\) |
orering | \(\text {Expression too large to display}\) | \(10171\) |
Input:
int((e*x)^m*(a+b*x^n)^3*(A+B*x^n)*(c+d*x^n),x,method=_RETURNVERBOSE)
Output:
x*(36*A*a*b^2*d*m^4*n*(x^n)^3+147*A*a*b^2*d*m^3*n^2*(x^n)^3+234*A*a*b^2*d* m^2*n^3*(x^n)^3+120*A*a*b^2*d*m*n^4*(x^n)^3+120*B*a*b^2*c*m*n^4*(x^n)^3+13 2*B*a*b^2*d*m^3*n*(x^n)^4+144*B*a*b^2*c*m^3*n*(x^n)^3+441*B*a*b^2*c*m^2*n^ 2*(x^n)^3+168*A*a^2*b*c*m*n*x^n+30*B*a*b^2*d*m^2*(x^n)^4+144*A*a*b^2*d*m*n *(x^n)^3+234*B*a^2*b*c*m^2*n*(x^n)^2+308*B*a^3*c*m*n^3*x^n+78*B*a^3*d*m^2* n*(x^n)^2+123*B*b^3*c*m^2*n^2*(x^n)^4+122*B*b^3*c*m*n^3*(x^n)^4+60*B*b^3*d *m^2*n*(x^n)^5+105*B*b^3*d*m*n^2*(x^n)^5+3*A*a^2*b*d*m^5*(x^n)^2+3*A*a*b^2 *c*m^5*(x^n)^2+15*A*a*b^2*d*m^4*(x^n)^3+120*A*a*b^2*d*n^4*(x^n)^3+123*B*b^ 3*c*m*n^2*(x^n)^4+183*B*a*b^2*d*m^2*n^3*(x^n)^4+90*B*a*b^2*d*m*n^4*(x^n)^4 +441*A*a*b^2*d*m^2*n^2*(x^n)^3+468*A*a*b^2*d*m*n^3*(x^n)^3+39*B*a^2*b*c*m^ 4*n*(x^n)^2+321*A*a^2*b*d*m^2*n^3*(x^n)^2+30*A*a*b^2*d*m^2*(x^n)^3+147*A*a *b^2*d*n^2*(x^n)^3+48*A*b^3*c*m*n*(x^n)^3+56*B*a^3*c*m^3*n*x^n+213*B*a^3*c *m^2*n^2*x^n+24*B*b^3*d*n^4*(x^n)^5+A*b^3*c*m^5*(x^n)^3+5*A*b^3*d*m^4*(x^n )^4+156*B*a^2*b*c*m^3*n*(x^n)^2+531*B*a^2*b*c*m*n^2*(x^n)^2+441*A*a*b^2*d* m*n^2*(x^n)^3+180*B*a^2*b*c*n^4*(x^n)^2+30*B*a^2*b*d*m^3*(x^n)^3+234*B*a^2 *b*d*n^3*(x^n)^3+40*B*b^3*d*m^3*n*(x^n)^5+105*B*b^3*d*m^2*n^2*(x^n)^5+100* B*b^3*d*m*n^3*(x^n)^5+3*A*a*b^2*d*m^5*(x^n)^3+12*A*b^3*c*m^4*n*(x^n)^3+49* A*b^3*c*m^3*n^2*(x^n)^3+(x^n)^5*b^3*B*d+(x^n)^4*A*b^3*d+39*A*a*b^2*c*m^4*n *(x^n)^2+147*B*a^2*b*d*m^3*n^2*(x^n)^3+234*B*a^2*b*d*m^2*n^3*(x^n)^3+10*B* b^3*d*m^4*n*(x^n)^5+35*B*b^3*d*m^3*n^2*(x^n)^5+50*B*b^3*d*m^2*n^3*(x^n)...
Leaf count of result is larger than twice the leaf count of optimal. 3073 vs. \(2 (225) = 450\).
Time = 0.20 (sec) , antiderivative size = 3073, normalized size of antiderivative = 13.66 \[ \int (e x)^m \left (a+b x^n\right )^3 \left (A+B x^n\right ) \left (c+d x^n\right ) \, dx=\text {Too large to display} \] Input:
integrate((e*x)^m*(a+b*x^n)^3*(A+B*x^n)*(c+d*x^n),x, algorithm="fricas")
Output:
Too large to include
Leaf count of result is larger than twice the leaf count of optimal. 64068 vs. \(2 (214) = 428\).
Time = 17.32 (sec) , antiderivative size = 64068, normalized size of antiderivative = 284.75 \[ \int (e x)^m \left (a+b x^n\right )^3 \left (A+B x^n\right ) \left (c+d x^n\right ) \, dx=\text {Too large to display} \] Input:
integrate((e*x)**m*(a+b*x**n)**3*(A+B*x**n)*(c+d*x**n),x)
Output:
Piecewise(((A + B)*(a + b)**3*(c + d)*log(x)/e, Eq(m, -1) & Eq(n, 0)), ((A *a**3*c*log(x) + A*a**3*d*x**n/n + 3*A*a**2*b*c*x**n/n + 3*A*a**2*b*d*x**( 2*n)/(2*n) + 3*A*a*b**2*c*x**(2*n)/(2*n) + A*a*b**2*d*x**(3*n)/n + A*b**3* c*x**(3*n)/(3*n) + A*b**3*d*x**(4*n)/(4*n) + B*a**3*c*x**n/n + B*a**3*d*x* *(2*n)/(2*n) + 3*B*a**2*b*c*x**(2*n)/(2*n) + B*a**2*b*d*x**(3*n)/n + B*a*b **2*c*x**(3*n)/n + 3*B*a*b**2*d*x**(4*n)/(4*n) + B*b**3*c*x**(4*n)/(4*n) + B*b**3*d*x**(5*n)/(5*n))/e, Eq(m, -1)), (A*a**3*c*Piecewise((0**(-5*n - 1 )*x, Eq(e, 0)), (Piecewise((-1/(5*n*(e*x)**(5*n)), Ne(n, 0)), (log(e*x), T rue))/e, True)) + A*a**3*d*Piecewise((-x*x**n*(e*x)**(-5*n - 1)/(4*n), Ne( n, 0)), (x*x**n*(e*x)**(-5*n - 1)*log(x), True)) + 3*A*a**2*b*c*Piecewise( (-x*x**n*(e*x)**(-5*n - 1)/(4*n), Ne(n, 0)), (x*x**n*(e*x)**(-5*n - 1)*log (x), True)) + 3*A*a**2*b*d*Piecewise((-x*x**(2*n)*(e*x)**(-5*n - 1)/(3*n), Ne(n, 0)), (x*x**(2*n)*(e*x)**(-5*n - 1)*log(x), True)) + 3*A*a*b**2*c*Pi ecewise((-x*x**(2*n)*(e*x)**(-5*n - 1)/(3*n), Ne(n, 0)), (x*x**(2*n)*(e*x) **(-5*n - 1)*log(x), True)) + 3*A*a*b**2*d*Piecewise((-x*x**(3*n)*(e*x)**( -5*n - 1)/(2*n), Ne(n, 0)), (x*x**(3*n)*(e*x)**(-5*n - 1)*log(x), True)) + A*b**3*c*Piecewise((-x*x**(3*n)*(e*x)**(-5*n - 1)/(2*n), Ne(n, 0)), (x*x* *(3*n)*(e*x)**(-5*n - 1)*log(x), True)) + A*b**3*d*Piecewise((-x*x**(4*n)* (e*x)**(-5*n - 1)/n, Ne(n, 0)), (x*x**(4*n)*(e*x)**(-5*n - 1)*log(x), True )) + B*a**3*c*Piecewise((-x*x**n*(e*x)**(-5*n - 1)/(4*n), Ne(n, 0)), (x...
Leaf count of result is larger than twice the leaf count of optimal. 464 vs. \(2 (225) = 450\).
Time = 0.08 (sec) , antiderivative size = 464, normalized size of antiderivative = 2.06 \[ \int (e x)^m \left (a+b x^n\right )^3 \left (A+B x^n\right ) \left (c+d x^n\right ) \, dx=\frac {B b^{3} d e^{m} x e^{\left (m \log \left (x\right ) + 5 \, n \log \left (x\right )\right )}}{m + 5 \, n + 1} + \frac {B b^{3} c e^{m} x e^{\left (m \log \left (x\right ) + 4 \, n \log \left (x\right )\right )}}{m + 4 \, n + 1} + \frac {3 \, B a b^{2} d e^{m} x e^{\left (m \log \left (x\right ) + 4 \, n \log \left (x\right )\right )}}{m + 4 \, n + 1} + \frac {A b^{3} d e^{m} x e^{\left (m \log \left (x\right ) + 4 \, n \log \left (x\right )\right )}}{m + 4 \, n + 1} + \frac {3 \, B a b^{2} c e^{m} x e^{\left (m \log \left (x\right ) + 3 \, n \log \left (x\right )\right )}}{m + 3 \, n + 1} + \frac {A b^{3} c e^{m} x e^{\left (m \log \left (x\right ) + 3 \, n \log \left (x\right )\right )}}{m + 3 \, n + 1} + \frac {3 \, B a^{2} b d e^{m} x e^{\left (m \log \left (x\right ) + 3 \, n \log \left (x\right )\right )}}{m + 3 \, n + 1} + \frac {3 \, A a b^{2} d e^{m} x e^{\left (m \log \left (x\right ) + 3 \, n \log \left (x\right )\right )}}{m + 3 \, n + 1} + \frac {3 \, B a^{2} b c e^{m} x e^{\left (m \log \left (x\right ) + 2 \, n \log \left (x\right )\right )}}{m + 2 \, n + 1} + \frac {3 \, A a b^{2} c e^{m} x e^{\left (m \log \left (x\right ) + 2 \, n \log \left (x\right )\right )}}{m + 2 \, n + 1} + \frac {B a^{3} d e^{m} x e^{\left (m \log \left (x\right ) + 2 \, n \log \left (x\right )\right )}}{m + 2 \, n + 1} + \frac {3 \, A a^{2} b d e^{m} x e^{\left (m \log \left (x\right ) + 2 \, n \log \left (x\right )\right )}}{m + 2 \, n + 1} + \frac {B a^{3} c e^{m} x e^{\left (m \log \left (x\right ) + n \log \left (x\right )\right )}}{m + n + 1} + \frac {3 \, A a^{2} b c e^{m} x e^{\left (m \log \left (x\right ) + n \log \left (x\right )\right )}}{m + n + 1} + \frac {A a^{3} d e^{m} x e^{\left (m \log \left (x\right ) + n \log \left (x\right )\right )}}{m + n + 1} + \frac {\left (e x\right )^{m + 1} A a^{3} c}{e {\left (m + 1\right )}} \] Input:
integrate((e*x)^m*(a+b*x^n)^3*(A+B*x^n)*(c+d*x^n),x, algorithm="maxima")
Output:
B*b^3*d*e^m*x*e^(m*log(x) + 5*n*log(x))/(m + 5*n + 1) + B*b^3*c*e^m*x*e^(m *log(x) + 4*n*log(x))/(m + 4*n + 1) + 3*B*a*b^2*d*e^m*x*e^(m*log(x) + 4*n* log(x))/(m + 4*n + 1) + A*b^3*d*e^m*x*e^(m*log(x) + 4*n*log(x))/(m + 4*n + 1) + 3*B*a*b^2*c*e^m*x*e^(m*log(x) + 3*n*log(x))/(m + 3*n + 1) + A*b^3*c* e^m*x*e^(m*log(x) + 3*n*log(x))/(m + 3*n + 1) + 3*B*a^2*b*d*e^m*x*e^(m*log (x) + 3*n*log(x))/(m + 3*n + 1) + 3*A*a*b^2*d*e^m*x*e^(m*log(x) + 3*n*log( x))/(m + 3*n + 1) + 3*B*a^2*b*c*e^m*x*e^(m*log(x) + 2*n*log(x))/(m + 2*n + 1) + 3*A*a*b^2*c*e^m*x*e^(m*log(x) + 2*n*log(x))/(m + 2*n + 1) + B*a^3*d* e^m*x*e^(m*log(x) + 2*n*log(x))/(m + 2*n + 1) + 3*A*a^2*b*d*e^m*x*e^(m*log (x) + 2*n*log(x))/(m + 2*n + 1) + B*a^3*c*e^m*x*e^(m*log(x) + n*log(x))/(m + n + 1) + 3*A*a^2*b*c*e^m*x*e^(m*log(x) + n*log(x))/(m + n + 1) + A*a^3* d*e^m*x*e^(m*log(x) + n*log(x))/(m + n + 1) + (e*x)^(m + 1)*A*a^3*c/(e*(m + 1))
Leaf count of result is larger than twice the leaf count of optimal. 27992 vs. \(2 (225) = 450\).
Time = 0.32 (sec) , antiderivative size = 27992, normalized size of antiderivative = 124.41 \[ \int (e x)^m \left (a+b x^n\right )^3 \left (A+B x^n\right ) \left (c+d x^n\right ) \, dx=\text {Too large to display} \] Input:
integrate((e*x)^m*(a+b*x^n)^3*(A+B*x^n)*(c+d*x^n),x, algorithm="giac")
Output:
(B*b^3*d*m^5*x*x^(5*n)*e^(m*log(e) + m*log(x)) + 10*B*b^3*d*m^4*n*x*x^(5*n )*e^(m*log(e) + m*log(x)) + 35*B*b^3*d*m^3*n^2*x*x^(5*n)*e^(m*log(e) + m*l og(x)) + 50*B*b^3*d*m^2*n^3*x*x^(5*n)*e^(m*log(e) + m*log(x)) + 24*B*b^3*d *m*n^4*x*x^(5*n)*e^(m*log(e) + m*log(x)) + B*b^3*c*m^5*x*x^(4*n)*e^(m*log( e) + m*log(x)) + 3*B*a*b^2*d*m^5*x*x^(4*n)*e^(m*log(e) + m*log(x)) + A*b^3 *d*m^5*x*x^(4*n)*e^(m*log(e) + m*log(x)) + B*b^3*d*m^5*x*x^(4*n)*e^(m*log( e) + m*log(x)) + 11*B*b^3*c*m^4*n*x*x^(4*n)*e^(m*log(e) + m*log(x)) + 33*B *a*b^2*d*m^4*n*x*x^(4*n)*e^(m*log(e) + m*log(x)) + 11*A*b^3*d*m^4*n*x*x^(4 *n)*e^(m*log(e) + m*log(x)) + 10*B*b^3*d*m^4*n*x*x^(4*n)*e^(m*log(e) + m*l og(x)) + 41*B*b^3*c*m^3*n^2*x*x^(4*n)*e^(m*log(e) + m*log(x)) + 123*B*a*b^ 2*d*m^3*n^2*x*x^(4*n)*e^(m*log(e) + m*log(x)) + 41*A*b^3*d*m^3*n^2*x*x^(4* n)*e^(m*log(e) + m*log(x)) + 35*B*b^3*d*m^3*n^2*x*x^(4*n)*e^(m*log(e) + m* log(x)) + 61*B*b^3*c*m^2*n^3*x*x^(4*n)*e^(m*log(e) + m*log(x)) + 183*B*a*b ^2*d*m^2*n^3*x*x^(4*n)*e^(m*log(e) + m*log(x)) + 61*A*b^3*d*m^2*n^3*x*x^(4 *n)*e^(m*log(e) + m*log(x)) + 50*B*b^3*d*m^2*n^3*x*x^(4*n)*e^(m*log(e) + m *log(x)) + 30*B*b^3*c*m*n^4*x*x^(4*n)*e^(m*log(e) + m*log(x)) + 90*B*a*b^2 *d*m*n^4*x*x^(4*n)*e^(m*log(e) + m*log(x)) + 30*A*b^3*d*m*n^4*x*x^(4*n)*e^ (m*log(e) + m*log(x)) + 24*B*b^3*d*m*n^4*x*x^(4*n)*e^(m*log(e) + m*log(x)) + 3*B*a*b^2*c*m^5*x*x^(3*n)*e^(m*log(e) + m*log(x)) + A*b^3*c*m^5*x*x^(3* n)*e^(m*log(e) + m*log(x)) + B*b^3*c*m^5*x*x^(3*n)*e^(m*log(e) + m*log(...
Time = 5.55 (sec) , antiderivative size = 1089, normalized size of antiderivative = 4.84 \[ \int (e x)^m \left (a+b x^n\right )^3 \left (A+B x^n\right ) \left (c+d x^n\right ) \, dx =\text {Too large to display} \] Input:
int((e*x)^m*(A + B*x^n)*(a + b*x^n)^3*(c + d*x^n),x)
Output:
(A*a^3*c*x*(e*x)^m)/(m + 1) + (b^2*x*x^(4*n)*(e*x)^m*(A*b*d + 3*B*a*d + B* b*c)*(4*m + 11*n + 33*m*n + 82*m*n^2 + 33*m^2*n + 61*m*n^3 + 11*m^3*n + 6* m^2 + 4*m^3 + m^4 + 41*n^2 + 61*n^3 + 30*n^4 + 41*m^2*n^2 + 1))/(5*m + 15* n + 60*m*n + 255*m*n^2 + 90*m^2*n + 450*m*n^3 + 60*m^3*n + 274*m*n^4 + 15* m^4*n + 10*m^2 + 10*m^3 + 5*m^4 + m^5 + 85*n^2 + 225*n^3 + 274*n^4 + 120*n ^5 + 255*m^2*n^2 + 225*m^2*n^3 + 85*m^3*n^2 + 1) + (a*x*x^(2*n)*(e*x)^m*(3 *A*b^2*c + B*a^2*d + 3*A*a*b*d + 3*B*a*b*c)*(4*m + 13*n + 39*m*n + 118*m*n ^2 + 39*m^2*n + 107*m*n^3 + 13*m^3*n + 6*m^2 + 4*m^3 + m^4 + 59*n^2 + 107* n^3 + 60*n^4 + 59*m^2*n^2 + 1))/(5*m + 15*n + 60*m*n + 255*m*n^2 + 90*m^2* n + 450*m*n^3 + 60*m^3*n + 274*m*n^4 + 15*m^4*n + 10*m^2 + 10*m^3 + 5*m^4 + m^5 + 85*n^2 + 225*n^3 + 274*n^4 + 120*n^5 + 255*m^2*n^2 + 225*m^2*n^3 + 85*m^3*n^2 + 1) + (b*x*x^(3*n)*(e*x)^m*(A*b^2*c + 3*B*a^2*d + 3*A*a*b*d + 3*B*a*b*c)*(4*m + 12*n + 36*m*n + 98*m*n^2 + 36*m^2*n + 78*m*n^3 + 12*m^3 *n + 6*m^2 + 4*m^3 + m^4 + 49*n^2 + 78*n^3 + 40*n^4 + 49*m^2*n^2 + 1))/(5* m + 15*n + 60*m*n + 255*m*n^2 + 90*m^2*n + 450*m*n^3 + 60*m^3*n + 274*m*n^ 4 + 15*m^4*n + 10*m^2 + 10*m^3 + 5*m^4 + m^5 + 85*n^2 + 225*n^3 + 274*n^4 + 120*n^5 + 255*m^2*n^2 + 225*m^2*n^3 + 85*m^3*n^2 + 1) + (a^2*x*x^n*(e*x) ^m*(A*a*d + 3*A*b*c + B*a*c)*(4*m + 14*n + 42*m*n + 142*m*n^2 + 42*m^2*n + 154*m*n^3 + 14*m^3*n + 6*m^2 + 4*m^3 + m^4 + 71*n^2 + 154*n^3 + 120*n^4 + 71*m^2*n^2 + 1))/(5*m + 15*n + 60*m*n + 255*m*n^2 + 90*m^2*n + 450*m*n...
Time = 0.26 (sec) , antiderivative size = 3053, normalized size of antiderivative = 13.57 \[ \int (e x)^m \left (a+b x^n\right )^3 \left (A+B x^n\right ) \left (c+d x^n\right ) \, dx =\text {Too large to display} \] Input:
int((e*x)^m*(a+b*x^n)^3*(A+B*x^n)*(c+d*x^n),x)
Output:
(x**m*e**m*x*(x**(5*n)*b**4*d*m**5 + 10*x**(5*n)*b**4*d*m**4*n + 5*x**(5*n )*b**4*d*m**4 + 35*x**(5*n)*b**4*d*m**3*n**2 + 40*x**(5*n)*b**4*d*m**3*n + 10*x**(5*n)*b**4*d*m**3 + 50*x**(5*n)*b**4*d*m**2*n**3 + 105*x**(5*n)*b** 4*d*m**2*n**2 + 60*x**(5*n)*b**4*d*m**2*n + 10*x**(5*n)*b**4*d*m**2 + 24*x **(5*n)*b**4*d*m*n**4 + 100*x**(5*n)*b**4*d*m*n**3 + 105*x**(5*n)*b**4*d*m *n**2 + 40*x**(5*n)*b**4*d*m*n + 5*x**(5*n)*b**4*d*m + 24*x**(5*n)*b**4*d* n**4 + 50*x**(5*n)*b**4*d*n**3 + 35*x**(5*n)*b**4*d*n**2 + 10*x**(5*n)*b** 4*d*n + x**(5*n)*b**4*d + 4*x**(4*n)*a*b**3*d*m**5 + 44*x**(4*n)*a*b**3*d* m**4*n + 20*x**(4*n)*a*b**3*d*m**4 + 164*x**(4*n)*a*b**3*d*m**3*n**2 + 176 *x**(4*n)*a*b**3*d*m**3*n + 40*x**(4*n)*a*b**3*d*m**3 + 244*x**(4*n)*a*b** 3*d*m**2*n**3 + 492*x**(4*n)*a*b**3*d*m**2*n**2 + 264*x**(4*n)*a*b**3*d*m* *2*n + 40*x**(4*n)*a*b**3*d*m**2 + 120*x**(4*n)*a*b**3*d*m*n**4 + 488*x**( 4*n)*a*b**3*d*m*n**3 + 492*x**(4*n)*a*b**3*d*m*n**2 + 176*x**(4*n)*a*b**3* d*m*n + 20*x**(4*n)*a*b**3*d*m + 120*x**(4*n)*a*b**3*d*n**4 + 244*x**(4*n) *a*b**3*d*n**3 + 164*x**(4*n)*a*b**3*d*n**2 + 44*x**(4*n)*a*b**3*d*n + 4*x **(4*n)*a*b**3*d + x**(4*n)*b**4*c*m**5 + 11*x**(4*n)*b**4*c*m**4*n + 5*x* *(4*n)*b**4*c*m**4 + 41*x**(4*n)*b**4*c*m**3*n**2 + 44*x**(4*n)*b**4*c*m** 3*n + 10*x**(4*n)*b**4*c*m**3 + 61*x**(4*n)*b**4*c*m**2*n**3 + 123*x**(4*n )*b**4*c*m**2*n**2 + 66*x**(4*n)*b**4*c*m**2*n + 10*x**(4*n)*b**4*c*m**2 + 30*x**(4*n)*b**4*c*m*n**4 + 122*x**(4*n)*b**4*c*m*n**3 + 123*x**(4*n)*...