\(\int (e x)^m (a+b x^n) (A+B x^n) (c+d x^n)^3 \, dx\) [35]

Optimal result

Integrand size = 29, antiderivative size = 225 \[ \int (e x)^m \left (a+b x^n\right ) \left (A+B x^n\right ) \left (c+d x^n\right )^3 \, dx=\frac {a A c^3 (e x)^{1+m}}{e (1+m)}+\frac {c^2 (A b c+a B c+3 a A d) x^n (e x)^{1+m}}{e (1+m+n)}+\frac {c (3 a d (B c+A d)+b c (B c+3 A d)) x^{2 n} (e x)^{1+m}}{e (1+m+2 n)}+\frac {d (3 b c (B c+A d)+a d (3 B c+A d)) x^{3 n} (e x)^{1+m}}{e (1+m+3 n)}+\frac {d^2 (3 b B c+A b d+a B d) x^{4 n} (e x)^{1+m}}{e (1+m+4 n)}+\frac {b B d^3 x^{5 n} (e x)^{1+m}}{e (1+m+5 n)} \] Output:

a*A*c^3*(e*x)^(1+m)/e/(1+m)+c^2*(3*A*a*d+A*b*c+B*a*c)*x^n*(e*x)^(1+m)/e/(1 
+m+n)+c*(3*a*d*(A*d+B*c)+b*c*(3*A*d+B*c))*x^(2*n)*(e*x)^(1+m)/e/(1+m+2*n)+ 
d*(3*b*c*(A*d+B*c)+a*d*(A*d+3*B*c))*x^(3*n)*(e*x)^(1+m)/e/(1+m+3*n)+d^2*(A 
*b*d+B*a*d+3*B*b*c)*x^(4*n)*(e*x)^(1+m)/e/(1+m+4*n)+b*B*d^3*x^(5*n)*(e*x)^ 
(1+m)/e/(1+m+5*n)
 

Mathematica [A] (verified)

Time = 0.64 (sec) , antiderivative size = 172, normalized size of antiderivative = 0.76 \[ \int (e x)^m \left (a+b x^n\right ) \left (A+B x^n\right ) \left (c+d x^n\right )^3 \, dx=x (e x)^m \left (\frac {a A c^3}{1+m}+\frac {c^2 (A b c+a B c+3 a A d) x^n}{1+m+n}+\frac {c (3 a d (B c+A d)+b c (B c+3 A d)) x^{2 n}}{1+m+2 n}+\frac {d (3 b c (B c+A d)+a d (3 B c+A d)) x^{3 n}}{1+m+3 n}+\frac {d^2 (3 b B c+A b d+a B d) x^{4 n}}{1+m+4 n}+\frac {b B d^3 x^{5 n}}{1+m+5 n}\right ) \] Input:

Integrate[(e*x)^m*(a + b*x^n)*(A + B*x^n)*(c + d*x^n)^3,x]
 

Output:

x*(e*x)^m*((a*A*c^3)/(1 + m) + (c^2*(A*b*c + a*B*c + 3*a*A*d)*x^n)/(1 + m 
+ n) + (c*(3*a*d*(B*c + A*d) + b*c*(B*c + 3*A*d))*x^(2*n))/(1 + m + 2*n) + 
 (d*(3*b*c*(B*c + A*d) + a*d*(3*B*c + A*d))*x^(3*n))/(1 + m + 3*n) + (d^2* 
(3*b*B*c + A*b*d + a*B*d)*x^(4*n))/(1 + m + 4*n) + (b*B*d^3*x^(5*n))/(1 + 
m + 5*n))
 

Rubi [A] (verified)

Time = 0.73 (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.

\(\displaystyle \int (e x)^m \left (a+b x^n\right ) \left (A+B x^n\right ) \left (c+d x^n\right )^3 \, dx\)

\(\Big \downarrow \) 1040

\(\displaystyle \int \left (c^2 x^n (e x)^m (3 a A d+a B c+A b c)+d^2 x^{4 n} (e x)^m (a B d+A b d+3 b B c)+c x^{2 n} (e x)^m (3 a d (A d+B c)+b c (3 A d+B c))+d x^{3 n} (e x)^m (a d (A d+3 B c)+3 b c (A d+B c))+a A c^3 (e x)^m+b B d^3 x^{5 n} (e x)^m\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {c^2 x^{n+1} (e x)^m (3 a A d+a B c+A b c)}{m+n+1}+\frac {d^2 x^{4 n+1} (e x)^m (a B d+A b d+3 b B c)}{m+4 n+1}+\frac {c x^{2 n+1} (e x)^m (3 a d (A d+B c)+b c (3 A d+B c))}{m+2 n+1}+\frac {d x^{3 n+1} (e x)^m (a d (A d+3 B c)+3 b c (A d+B c))}{m+3 n+1}+\frac {a A c^3 (e x)^{m+1}}{e (m+1)}+\frac {b B d^3 x^{5 n+1} (e x)^m}{m+5 n+1}\)

Input:

Int[(e*x)^m*(a + b*x^n)*(A + B*x^n)*(c + d*x^n)^3,x]
 

Output:

(c^2*(A*b*c + a*B*c + 3*a*A*d)*x^(1 + n)*(e*x)^m)/(1 + m + n) + (c*(3*a*d* 
(B*c + A*d) + b*c*(B*c + 3*A*d))*x^(1 + 2*n)*(e*x)^m)/(1 + m + 2*n) + (d*( 
3*b*c*(B*c + A*d) + a*d*(3*B*c + A*d))*x^(1 + 3*n)*(e*x)^m)/(1 + m + 3*n) 
+ (d^2*(3*b*B*c + A*b*d + a*B*d)*x^(1 + 4*n)*(e*x)^m)/(1 + m + 4*n) + (b*B 
*d^3*x^(1 + 5*n)*(e*x)^m)/(1 + m + 5*n) + (a*A*c^3*(e*x)^(1 + m))/(e*(1 + 
m))
 

Defintions of rubi rules used

rule 1040

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]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 1.34 (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)*(A+B*x^n)*(c+d*x^n)^3,x,method=_RETURNVERBOSE)
 

Output:

x*(15*A*a*c*d^2*(x^n)^2*m+39*A*a*c*d^2*(x^n)^2*n+56*A*b*c^3*m*n*x^n+15*A*b 
*c^2*d*(x^n)^2*m+39*A*b*c^2*d*(x^n)^2*n+71*B*a*c^3*n^2*x^n+156*A*b*c^2*d*m 
^3*n*(x^n)^2+531*A*b*c^2*d*m^2*n^2*(x^n)^2+642*A*b*c^2*d*m*n^3*(x^n)^2+216 
*A*b*c*d^2*m^2*n*(x^n)^3+10*B*a*c^3*m^2*x^n+5*m*b*B*d^3*(x^n)^5+11*B*a*d^3 
*m^4*n*(x^n)^4+41*B*a*d^3*m^3*n^2*(x^n)^4+61*B*a*d^3*m^2*n^3*(x^n)^4+30*B* 
a*d^3*m*n^4*(x^n)^4+3*B*b*c*d^2*m^5*(x^n)^4+40*B*b*d^3*m^3*n*(x^n)^5+105*B 
*b*d^3*m^2*n^2*(x^n)^5+100*B*b*d^3*m*n^3*(x^n)^5+321*A*a*c*d^2*m^2*n^3*(x^ 
n)^2+180*A*a*c*d^2*m*n^4*(x^n)^2+39*A*b*c^2*d*m^4*n*(x^n)^2+177*A*b*c^2*d* 
m^3*n^2*(x^n)^2+321*A*b*c^2*d*m^2*n^3*(x^n)^2+441*A*b*c*d^2*m*n^2*(x^n)^3+ 
156*B*a*c^2*d*m^3*n*(x^n)^2+531*B*a*c^2*d*m^2*n^2*(x^n)^2+642*B*a*c^2*d*m* 
n^3*(x^n)^2+216*B*a*c*d^2*m^2*n*(x^n)^3+123*B*a*d^3*m^2*n^2*(x^n)^4+122*B* 
a*d^3*m*n^3*(x^n)^4+3*B*b*c^2*d*m^5*(x^n)^3+15*B*b*c*d^2*m^4*(x^n)^4+90*B* 
b*c*d^2*n^4*(x^n)^4+60*B*b*d^3*m^2*n*(x^n)^5+105*B*b*d^3*m*n^2*(x^n)^5+3*A 
*a*c*d^2*m^5*(x^n)^2+48*A*a*d^3*m^3*n*(x^n)^3+147*A*a*d^3*m^2*n^2*(x^n)^3+ 
156*A*a*d^3*m*n^3*(x^n)^3+213*A*a*c^2*d*m^3*n^2*x^n+462*A*a*c^2*d*m^2*n^3* 
x^n+360*A*a*c^2*d*m*n^4*x^n+156*A*a*c*d^2*m^3*n*(x^n)^2+531*A*a*c*d^2*m^2* 
n^2*(x^n)^2+642*A*a*c*d^2*m*n^3*(x^n)^2+308*A*b*c^3*m*n^3*x^n+30*A*b*c^2*d 
*m^3*(x^n)^2+321*A*b*c^2*d*n^3*(x^n)^2+30*A*b*c*d^2*m^2*(x^n)^3+147*A*b*c* 
d^2*n^2*(x^n)^3+56*B*a*c^3*m^3*n*x^n+213*B*a*c^3*m^2*n^2*x^n+44*B*a*d^3*m^ 
3*n*(x^n)^4+120*A*a*c^3*n^5+A*a*c^3*m^5+5*A*a*c^3*m^4+274*A*a*c^3*n^4+1...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2833 vs. \(2 (225) = 450\).

Time = 0.17 (sec) , antiderivative size = 2833, normalized size of antiderivative = 12.59 \[ \int (e x)^m \left (a+b x^n\right ) \left (A+B x^n\right ) \left (c+d x^n\right )^3 \, dx=\text {Too large to display} \] Input:

integrate((e*x)^m*(a+b*x^n)*(A+B*x^n)*(c+d*x^n)^3,x, algorithm="fricas")
 

Output:

((B*b*d^3*m^5 + 5*B*b*d^3*m^4 + 10*B*b*d^3*m^3 + 10*B*b*d^3*m^2 + 5*B*b*d^ 
3*m + B*b*d^3 + 24*(B*b*d^3*m + B*b*d^3)*n^4 + 50*(B*b*d^3*m^2 + 2*B*b*d^3 
*m + B*b*d^3)*n^3 + 35*(B*b*d^3*m^3 + 3*B*b*d^3*m^2 + 3*B*b*d^3*m + B*b*d^ 
3)*n^2 + 10*(B*b*d^3*m^4 + 4*B*b*d^3*m^3 + 6*B*b*d^3*m^2 + 4*B*b*d^3*m + B 
*b*d^3)*n)*x*x^(5*n)*e^(m*log(e) + m*log(x)) + ((3*B*b*c*d^2 + (B*a + A*b) 
*d^3)*m^5 + 3*B*b*c*d^2 + 5*(3*B*b*c*d^2 + (B*a + A*b)*d^3)*m^4 + 30*(3*B* 
b*c*d^2 + (B*a + A*b)*d^3 + (3*B*b*c*d^2 + (B*a + A*b)*d^3)*m)*n^4 + (B*a 
+ A*b)*d^3 + 10*(3*B*b*c*d^2 + (B*a + A*b)*d^3)*m^3 + 61*(3*B*b*c*d^2 + (B 
*a + A*b)*d^3 + (3*B*b*c*d^2 + (B*a + A*b)*d^3)*m^2 + 2*(3*B*b*c*d^2 + (B* 
a + A*b)*d^3)*m)*n^3 + 10*(3*B*b*c*d^2 + (B*a + A*b)*d^3)*m^2 + 41*(3*B*b* 
c*d^2 + (B*a + A*b)*d^3 + (3*B*b*c*d^2 + (B*a + A*b)*d^3)*m^3 + 3*(3*B*b*c 
*d^2 + (B*a + A*b)*d^3)*m^2 + 3*(3*B*b*c*d^2 + (B*a + A*b)*d^3)*m)*n^2 + 5 
*(3*B*b*c*d^2 + (B*a + A*b)*d^3)*m + 11*(3*B*b*c*d^2 + (3*B*b*c*d^2 + (B*a 
 + A*b)*d^3)*m^4 + (B*a + A*b)*d^3 + 4*(3*B*b*c*d^2 + (B*a + A*b)*d^3)*m^3 
 + 6*(3*B*b*c*d^2 + (B*a + A*b)*d^3)*m^2 + 4*(3*B*b*c*d^2 + (B*a + A*b)*d^ 
3)*m)*n)*x*x^(4*n)*e^(m*log(e) + m*log(x)) + ((3*B*b*c^2*d + A*a*d^3 + 3*( 
B*a + A*b)*c*d^2)*m^5 + 3*B*b*c^2*d + A*a*d^3 + 5*(3*B*b*c^2*d + A*a*d^3 + 
 3*(B*a + A*b)*c*d^2)*m^4 + 40*(3*B*b*c^2*d + A*a*d^3 + 3*(B*a + A*b)*c*d^ 
2 + (3*B*b*c^2*d + A*a*d^3 + 3*(B*a + A*b)*c*d^2)*m)*n^4 + 3*(B*a + A*b)*c 
*d^2 + 10*(3*B*b*c^2*d + A*a*d^3 + 3*(B*a + A*b)*c*d^2)*m^3 + 78*(3*B*b...
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 64068 vs. \(2 (214) = 428\).

Time = 16.55 (sec) , antiderivative size = 64068, normalized size of antiderivative = 284.75 \[ \int (e x)^m \left (a+b x^n\right ) \left (A+B x^n\right ) \left (c+d x^n\right )^3 \, dx=\text {Too large to display} \] Input:

integrate((e*x)**m*(a+b*x**n)*(A+B*x**n)*(c+d*x**n)**3,x)
 

Output:

Piecewise(((A + B)*(a + b)*(c + d)**3*log(x)/e, Eq(m, -1) & Eq(n, 0)), ((A 
*a*c**3*log(x) + 3*A*a*c**2*d*x**n/n + 3*A*a*c*d**2*x**(2*n)/(2*n) + A*a*d 
**3*x**(3*n)/(3*n) + A*b*c**3*x**n/n + 3*A*b*c**2*d*x**(2*n)/(2*n) + A*b*c 
*d**2*x**(3*n)/n + A*b*d**3*x**(4*n)/(4*n) + B*a*c**3*x**n/n + 3*B*a*c**2* 
d*x**(2*n)/(2*n) + B*a*c*d**2*x**(3*n)/n + B*a*d**3*x**(4*n)/(4*n) + B*b*c 
**3*x**(2*n)/(2*n) + B*b*c**2*d*x**(3*n)/n + 3*B*b*c*d**2*x**(4*n)/(4*n) + 
 B*b*d**3*x**(5*n)/(5*n))/e, Eq(m, -1)), (A*a*c**3*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)) + 3*A*a*c**2*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*c*d**2*Piecew 
ise((-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)) + A*a*d**3*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*c** 
3*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*b*c**2*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 
*b*c*d**2*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*d**3*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*c**3*Piecewise((-x*x**n*(e*x)**(-5*n - 1)/(4*n), Ne(n, 0)), (x...
 

Maxima [B] (verification not implemented)

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 \[ \int (e x)^m \left (a+b x^n\right ) \left (A+B x^n\right ) \left (c+d x^n\right )^3 \, dx=\frac {B b d^{3} e^{m} x e^{\left (m \log \left (x\right ) + 5 \, n \log \left (x\right )\right )}}{m + 5 \, n + 1} + \frac {3 \, B b c d^{2} e^{m} x e^{\left (m \log \left (x\right ) + 4 \, n \log \left (x\right )\right )}}{m + 4 \, n + 1} + \frac {B a d^{3} e^{m} x e^{\left (m \log \left (x\right ) + 4 \, n \log \left (x\right )\right )}}{m + 4 \, n + 1} + \frac {A b d^{3} e^{m} x e^{\left (m \log \left (x\right ) + 4 \, n \log \left (x\right )\right )}}{m + 4 \, n + 1} + \frac {3 \, B b c^{2} d e^{m} x e^{\left (m \log \left (x\right ) + 3 \, n \log \left (x\right )\right )}}{m + 3 \, n + 1} + \frac {3 \, B a c d^{2} e^{m} x e^{\left (m \log \left (x\right ) + 3 \, n \log \left (x\right )\right )}}{m + 3 \, n + 1} + \frac {3 \, A b c d^{2} e^{m} x e^{\left (m \log \left (x\right ) + 3 \, n \log \left (x\right )\right )}}{m + 3 \, n + 1} + \frac {A a d^{3} e^{m} x e^{\left (m \log \left (x\right ) + 3 \, n \log \left (x\right )\right )}}{m + 3 \, n + 1} + \frac {B b c^{3} e^{m} x e^{\left (m \log \left (x\right ) + 2 \, n \log \left (x\right )\right )}}{m + 2 \, n + 1} + \frac {3 \, B a c^{2} d e^{m} x e^{\left (m \log \left (x\right ) + 2 \, n \log \left (x\right )\right )}}{m + 2 \, n + 1} + \frac {3 \, A b c^{2} d e^{m} x e^{\left (m \log \left (x\right ) + 2 \, n \log \left (x\right )\right )}}{m + 2 \, n + 1} + \frac {3 \, A a c d^{2} e^{m} x e^{\left (m \log \left (x\right ) + 2 \, n \log \left (x\right )\right )}}{m + 2 \, n + 1} + \frac {B a c^{3} e^{m} x e^{\left (m \log \left (x\right ) + n \log \left (x\right )\right )}}{m + n + 1} + \frac {A b c^{3} e^{m} x e^{\left (m \log \left (x\right ) + n \log \left (x\right )\right )}}{m + n + 1} + \frac {3 \, A a c^{2} d e^{m} x e^{\left (m \log \left (x\right ) + n \log \left (x\right )\right )}}{m + n + 1} + \frac {\left (e x\right )^{m + 1} A a c^{3}}{e {\left (m + 1\right )}} \] Input:

integrate((e*x)^m*(a+b*x^n)*(A+B*x^n)*(c+d*x^n)^3,x, algorithm="maxima")
 

Output:

B*b*d^3*e^m*x*e^(m*log(x) + 5*n*log(x))/(m + 5*n + 1) + 3*B*b*c*d^2*e^m*x* 
e^(m*log(x) + 4*n*log(x))/(m + 4*n + 1) + B*a*d^3*e^m*x*e^(m*log(x) + 4*n* 
log(x))/(m + 4*n + 1) + A*b*d^3*e^m*x*e^(m*log(x) + 4*n*log(x))/(m + 4*n + 
 1) + 3*B*b*c^2*d*e^m*x*e^(m*log(x) + 3*n*log(x))/(m + 3*n + 1) + 3*B*a*c* 
d^2*e^m*x*e^(m*log(x) + 3*n*log(x))/(m + 3*n + 1) + 3*A*b*c*d^2*e^m*x*e^(m 
*log(x) + 3*n*log(x))/(m + 3*n + 1) + A*a*d^3*e^m*x*e^(m*log(x) + 3*n*log( 
x))/(m + 3*n + 1) + B*b*c^3*e^m*x*e^(m*log(x) + 2*n*log(x))/(m + 2*n + 1) 
+ 3*B*a*c^2*d*e^m*x*e^(m*log(x) + 2*n*log(x))/(m + 2*n + 1) + 3*A*b*c^2*d* 
e^m*x*e^(m*log(x) + 2*n*log(x))/(m + 2*n + 1) + 3*A*a*c*d^2*e^m*x*e^(m*log 
(x) + 2*n*log(x))/(m + 2*n + 1) + B*a*c^3*e^m*x*e^(m*log(x) + n*log(x))/(m 
 + n + 1) + A*b*c^3*e^m*x*e^(m*log(x) + n*log(x))/(m + n + 1) + 3*A*a*c^2* 
d*e^m*x*e^(m*log(x) + n*log(x))/(m + n + 1) + (e*x)^(m + 1)*A*a*c^3/(e*(m 
+ 1))
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 27992 vs. \(2 (225) = 450\).

Time = 0.31 (sec) , antiderivative size = 27992, normalized size of antiderivative = 124.41 \[ \int (e x)^m \left (a+b x^n\right ) \left (A+B x^n\right ) \left (c+d x^n\right )^3 \, dx=\text {Too large to display} \] Input:

integrate((e*x)^m*(a+b*x^n)*(A+B*x^n)*(c+d*x^n)^3,x, algorithm="giac")
 

Output:

(B*b*d^3*m^5*x*x^(5*n)*e^(m*log(e) + m*log(x)) + 10*B*b*d^3*m^4*n*x*x^(5*n 
)*e^(m*log(e) + m*log(x)) + 35*B*b*d^3*m^3*n^2*x*x^(5*n)*e^(m*log(e) + m*l 
og(x)) + 50*B*b*d^3*m^2*n^3*x*x^(5*n)*e^(m*log(e) + m*log(x)) + 24*B*b*d^3 
*m*n^4*x*x^(5*n)*e^(m*log(e) + m*log(x)) + 3*B*b*c*d^2*m^5*x*x^(4*n)*e^(m* 
log(e) + m*log(x)) + B*a*d^3*m^5*x*x^(4*n)*e^(m*log(e) + m*log(x)) + A*b*d 
^3*m^5*x*x^(4*n)*e^(m*log(e) + m*log(x)) + B*b*d^3*m^5*x*x^(4*n)*e^(m*log( 
e) + m*log(x)) + 33*B*b*c*d^2*m^4*n*x*x^(4*n)*e^(m*log(e) + m*log(x)) + 11 
*B*a*d^3*m^4*n*x*x^(4*n)*e^(m*log(e) + m*log(x)) + 11*A*b*d^3*m^4*n*x*x^(4 
*n)*e^(m*log(e) + m*log(x)) + 10*B*b*d^3*m^4*n*x*x^(4*n)*e^(m*log(e) + m*l 
og(x)) + 123*B*b*c*d^2*m^3*n^2*x*x^(4*n)*e^(m*log(e) + m*log(x)) + 41*B*a* 
d^3*m^3*n^2*x*x^(4*n)*e^(m*log(e) + m*log(x)) + 41*A*b*d^3*m^3*n^2*x*x^(4* 
n)*e^(m*log(e) + m*log(x)) + 35*B*b*d^3*m^3*n^2*x*x^(4*n)*e^(m*log(e) + m* 
log(x)) + 183*B*b*c*d^2*m^2*n^3*x*x^(4*n)*e^(m*log(e) + m*log(x)) + 61*B*a 
*d^3*m^2*n^3*x*x^(4*n)*e^(m*log(e) + m*log(x)) + 61*A*b*d^3*m^2*n^3*x*x^(4 
*n)*e^(m*log(e) + m*log(x)) + 50*B*b*d^3*m^2*n^3*x*x^(4*n)*e^(m*log(e) + m 
*log(x)) + 90*B*b*c*d^2*m*n^4*x*x^(4*n)*e^(m*log(e) + m*log(x)) + 30*B*a*d 
^3*m*n^4*x*x^(4*n)*e^(m*log(e) + m*log(x)) + 30*A*b*d^3*m*n^4*x*x^(4*n)*e^ 
(m*log(e) + m*log(x)) + 24*B*b*d^3*m*n^4*x*x^(4*n)*e^(m*log(e) + m*log(x)) 
 + 3*B*b*c^2*d*m^5*x*x^(3*n)*e^(m*log(e) + m*log(x)) + 3*B*a*c*d^2*m^5*x*x 
^(3*n)*e^(m*log(e) + m*log(x)) + 3*A*b*c*d^2*m^5*x*x^(3*n)*e^(m*log(e) ...
 

Mupad [B] (verification not implemented)

Time = 5.77 (sec) , antiderivative size = 1089, normalized size of antiderivative = 4.84 \[ \int (e x)^m \left (a+b x^n\right ) \left (A+B x^n\right ) \left (c+d x^n\right )^3 \, dx =\text {Too large to display} \] Input:

int((e*x)^m*(A + B*x^n)*(a + b*x^n)*(c + d*x^n)^3,x)
 

Output:

(A*a*c^3*x*(e*x)^m)/(m + 1) + (d^2*x*x^(4*n)*(e*x)^m*(A*b*d + B*a*d + 3*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) + (c*x*x^(2*n)*(e*x)^m*(3 
*A*a*d^2 + B*b*c^2 + 3*A*b*c*d + 3*B*a*c*d)*(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) + (d*x*x^(3*n)*(e*x)^m*(A*a*d^2 + 3*B*b*c^2 + 3*A*b*c*d + 
 3*B*a*c*d)*(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) + (c^2*x*x^n*(e*x) 
^m*(3*A*a*d + 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...
 

Reduce [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 3949, normalized size of antiderivative = 17.55 \[ \int (e x)^m \left (a+b x^n\right ) \left (A+B x^n\right ) \left (c+d x^n\right )^3 \, dx =\text {Too large to display} \] Input:

int((e*x)^m*(a+b*x^n)*(A+B*x^n)*(c+d*x^n)^3,x)
 

Output:

(x**m*e**m*x*(x**(5*n)*b**2*d**3*m**5 + 10*x**(5*n)*b**2*d**3*m**4*n + 5*x 
**(5*n)*b**2*d**3*m**4 + 35*x**(5*n)*b**2*d**3*m**3*n**2 + 40*x**(5*n)*b** 
2*d**3*m**3*n + 10*x**(5*n)*b**2*d**3*m**3 + 50*x**(5*n)*b**2*d**3*m**2*n* 
*3 + 105*x**(5*n)*b**2*d**3*m**2*n**2 + 60*x**(5*n)*b**2*d**3*m**2*n + 10* 
x**(5*n)*b**2*d**3*m**2 + 24*x**(5*n)*b**2*d**3*m*n**4 + 100*x**(5*n)*b**2 
*d**3*m*n**3 + 105*x**(5*n)*b**2*d**3*m*n**2 + 40*x**(5*n)*b**2*d**3*m*n + 
 5*x**(5*n)*b**2*d**3*m + 24*x**(5*n)*b**2*d**3*n**4 + 50*x**(5*n)*b**2*d* 
*3*n**3 + 35*x**(5*n)*b**2*d**3*n**2 + 10*x**(5*n)*b**2*d**3*n + x**(5*n)* 
b**2*d**3 + 2*x**(4*n)*a*b*d**3*m**5 + 22*x**(4*n)*a*b*d**3*m**4*n + 10*x* 
*(4*n)*a*b*d**3*m**4 + 82*x**(4*n)*a*b*d**3*m**3*n**2 + 88*x**(4*n)*a*b*d* 
*3*m**3*n + 20*x**(4*n)*a*b*d**3*m**3 + 122*x**(4*n)*a*b*d**3*m**2*n**3 + 
246*x**(4*n)*a*b*d**3*m**2*n**2 + 132*x**(4*n)*a*b*d**3*m**2*n + 20*x**(4* 
n)*a*b*d**3*m**2 + 60*x**(4*n)*a*b*d**3*m*n**4 + 244*x**(4*n)*a*b*d**3*m*n 
**3 + 246*x**(4*n)*a*b*d**3*m*n**2 + 88*x**(4*n)*a*b*d**3*m*n + 10*x**(4*n 
)*a*b*d**3*m + 60*x**(4*n)*a*b*d**3*n**4 + 122*x**(4*n)*a*b*d**3*n**3 + 82 
*x**(4*n)*a*b*d**3*n**2 + 22*x**(4*n)*a*b*d**3*n + 2*x**(4*n)*a*b*d**3 + 3 
*x**(4*n)*b**2*c*d**2*m**5 + 33*x**(4*n)*b**2*c*d**2*m**4*n + 15*x**(4*n)* 
b**2*c*d**2*m**4 + 123*x**(4*n)*b**2*c*d**2*m**3*n**2 + 132*x**(4*n)*b**2* 
c*d**2*m**3*n + 30*x**(4*n)*b**2*c*d**2*m**3 + 183*x**(4*n)*b**2*c*d**2*m* 
*2*n**3 + 369*x**(4*n)*b**2*c*d**2*m**2*n**2 + 198*x**(4*n)*b**2*c*d**2...