\(\int (e x)^m (a+b x^n)^2 (A+B x^n) (c+d x^n) \, dx\) [20]

Optimal result

Integrand size = 29, antiderivative size = 172 \[ \int (e x)^m \left (a+b x^n\right )^2 \left (A+B x^n\right ) \left (c+d x^n\right ) \, dx=\frac {a^2 A c (e x)^{1+m}}{e (1+m)}+\frac {a (2 A b c+a B c+a A d) x^n (e x)^{1+m}}{e (1+m+n)}+\frac {(a B (2 b c+a d)+A b (b c+2 a d)) x^{2 n} (e x)^{1+m}}{e (1+m+2 n)}+\frac {b (b B c+A b d+2 a B d) x^{3 n} (e x)^{1+m}}{e (1+m+3 n)}+\frac {b^2 B d x^{4 n} (e x)^{1+m}}{e (1+m+4 n)} \] Output:

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

Mathematica [A] (verified)

Time = 0.31 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.75 \[ \int (e x)^m \left (a+b x^n\right )^2 \left (A+B x^n\right ) \left (c+d x^n\right ) \, dx=x (e x)^m \left (\frac {a^2 A c}{1+m}+\frac {a (2 A b c+a B c+a A d) x^n}{1+m+n}+\frac {(a B (2 b c+a d)+A b (b c+2 a d)) x^{2 n}}{1+m+2 n}+\frac {b (b B c+A b d+2 a B d) x^{3 n}}{1+m+3 n}+\frac {b^2 B d x^{4 n}}{1+m+4 n}\right ) \] Input:

Integrate[(e*x)^m*(a + b*x^n)^2*(A + B*x^n)*(c + d*x^n),x]
 

Output:

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))
 

Rubi [A] (verified)

Time = 0.60 (sec) , antiderivative size = 160, 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.

\(\displaystyle \int (e x)^m \left (a+b x^n\right )^2 \left (A+B x^n\right ) \left (c+d x^n\right ) \, dx\)

\(\Big \downarrow \) 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

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

Input:

Int[(e*x)^m*(a + b*x^n)^2*(A + B*x^n)*(c + d*x^n),x]
 

Output:

(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))
 

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 = 0.68 (sec) , antiderivative size = 2377, normalized size of antiderivative = 13.82

method	result	size
risch	\(\text {Expression too large to display}\)	\(2377\)
parallelrisch	\(\text {Expression too large to display}\)	\(3344\)
orering	\(\text {Expression too large to display}\)	\(4757\)

Input:

int((e*x)^m*(a+b*x^n)^2*(A+B*x^n)*(c+d*x^n),x,method=_RETURNVERBOSE)
 

Output:

x*(48*A*a*b*d*m*n*(x^n)^2+8*B*a^2*d*m^3*n*(x^n)^2+19*B*a^2*d*m^2*n^2*(x^n) 
^2+12*B*a^2*d*m*n^3*(x^n)^2+2*B*a*b*c*m^4*(x^n)^2+8*B*a*b*d*m^3*(x^n)^3+16 
*B*a*b*d*n^3*(x^n)^3+8*B*b^2*c*m*n^3*(x^n)^3+18*B*b^2*d*m^2*n*(x^n)^4+22*B 
*b^2*d*m*n^2*(x^n)^4+8*B*a^2*d*(x^n)^2*n+A*a^2*d*x^n+B*a^2*c*x^n+19*B*a^2* 
d*n^2*(x^n)^2+4*B*b^2*c*(x^n)^3*m+7*B*b^2*c*(x^n)^3*n+12*A*b^2*c*m*n^3*(x^ 
n)^2+21*A*b^2*d*m^2*n*(x^n)^3+28*A*b^2*d*m*n^2*(x^n)^3+b^2*B*d*(x^n)^4+A*b 
^2*d*(x^n)^3+B*b^2*c*(x^n)^3+A*b^2*c*(x^n)^2+B*a^2*d*(x^n)^2+24*A*a^2*c*n^ 
4+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+4* 
a^2*A*c*m+10*a^2*A*c*n+8*A*a*b*d*m^3*(x^n)^2+24*A*a*b*d*n^3*(x^n)^2+24*A*b 
^2*c*m^2*n*(x^n)^2+38*A*b^2*c*m*n^2*(x^n)^2+21*A*b^2*d*m*n*(x^n)^3+9*B*a^2 
*c*m^3*n*x^n+a^2*A*c+26*B*a^2*c*m^2*n^2*x^n+24*B*a^2*c*m*n^3*x^n+24*B*a^2* 
d*m^2*n*(x^n)^2+38*B*a^2*d*m*n^2*(x^n)^2+8*B*a*b*c*m^3*(x^n)^2+24*B*a*b*c* 
n^3*(x^n)^2+12*B*a*b*d*m^2*(x^n)^3+11*B*b^2*d*m^2*n^2*(x^n)^4+6*B*b^2*d*m* 
n^3*(x^n)^4+7*A*b^2*d*m^3*n*(x^n)^3+14*A*b^2*d*m^2*n^2*(x^n)^3+8*A*b^2*d*m 
*n^3*(x^n)^3+28*B*a*b*d*n^2*(x^n)^3+14*B*a*b*d*m^3*n*(x^n)^3+28*B*a*b*d*m^ 
2*n^2*(x^n)^3+14*B*a*b*d*(x^n)^3*n+27*A*a^2*d*m*n*x^n+12*A*a*b*c*m^2*x^n+5 
2*A*a*b*c*n^2*x^n+8*A*a*b*d*(x^n)^2*m+16*A*a*b*d*(x^n)^2*n+27*B*a^2*c*m*n* 
x^n+16*B*a*b*d*m*n^3*(x^n)^3+16*A*a*b*d*m^3*n*(x^n)^2+38*A*a*b*d*m^2*n^2*( 
x^n)^2+24*A*a*b*d*m*n^3*(x^n)^2+16*B*a*b*c*m^3*n*(x^n)^2+38*B*a*b*c*m^2*n^ 
2*(x^n)^2+8*A*a*b*c*m^3*x^n+48*A*a*b*c*n^3*x^n+12*A*a*b*d*m^2*(x^n)^2+3...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1524 vs. \(2 (172) = 344\).

Time = 0.17 (sec) , antiderivative size = 1524, normalized size of antiderivative = 8.86 \[ \int (e x)^m \left (a+b x^n\right )^2 \left (A+B x^n\right ) \left (c+d x^n\right ) \, dx=\text {Too large to display} \] Input:

integrate((e*x)^m*(a+b*x^n)^2*(A+B*x^n)*(c+d*x^n),x, algorithm="fricas")
 

Output:

((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*...
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 25315 vs. \(2 (163) = 326\).

Time = 9.84 (sec) , antiderivative size = 25315, normalized size of antiderivative = 147.18 \[ \int (e x)^m \left (a+b x^n\right )^2 \left (A+B x^n\right ) \left (c+d x^n\right ) \, dx=\text {Too large to display} \] Input:

integrate((e*x)**m*(a+b*x**n)**2*(A+B*x**n)*(c+d*x**n),x)
 

Output:

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...
 

Maxima [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 332, normalized size of antiderivative = 1.93 \[ \int (e x)^m \left (a+b x^n\right )^2 \left (A+B x^n\right ) \left (c+d x^n\right ) \, dx=\frac {B b^{2} d e^{m} x e^{\left (m \log \left (x\right ) + 4 \, n \log \left (x\right )\right )}}{m + 4 \, n + 1} + \frac {B b^{2} c e^{m} x e^{\left (m \log \left (x\right ) + 3 \, n \log \left (x\right )\right )}}{m + 3 \, n + 1} + \frac {2 \, B a b d e^{m} x e^{\left (m \log \left (x\right ) + 3 \, n \log \left (x\right )\right )}}{m + 3 \, n + 1} + \frac {A b^{2} d e^{m} x e^{\left (m \log \left (x\right ) + 3 \, n \log \left (x\right )\right )}}{m + 3 \, n + 1} + \frac {2 \, B a b c e^{m} x e^{\left (m \log \left (x\right ) + 2 \, n \log \left (x\right )\right )}}{m + 2 \, n + 1} + \frac {A b^{2} c e^{m} x e^{\left (m \log \left (x\right ) + 2 \, n \log \left (x\right )\right )}}{m + 2 \, n + 1} + \frac {B a^{2} d e^{m} x e^{\left (m \log \left (x\right ) + 2 \, n \log \left (x\right )\right )}}{m + 2 \, n + 1} + \frac {2 \, A a b d e^{m} x e^{\left (m \log \left (x\right ) + 2 \, n \log \left (x\right )\right )}}{m + 2 \, n + 1} + \frac {B a^{2} c e^{m} x e^{\left (m \log \left (x\right ) + n \log \left (x\right )\right )}}{m + n + 1} + \frac {2 \, A a b c e^{m} x e^{\left (m \log \left (x\right ) + n \log \left (x\right )\right )}}{m + n + 1} + \frac {A a^{2} d e^{m} x e^{\left (m \log \left (x\right ) + n \log \left (x\right )\right )}}{m + n + 1} + \frac {\left (e x\right )^{m + 1} A a^{2} c}{e {\left (m + 1\right )}} \] Input:

integrate((e*x)^m*(a+b*x^n)^2*(A+B*x^n)*(c+d*x^n),x, algorithm="maxima")
 

Output:

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))
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 11834 vs. \(2 (172) = 344\).

Time = 0.21 (sec) , antiderivative size = 11834, normalized size of antiderivative = 68.80 \[ \int (e x)^m \left (a+b x^n\right )^2 \left (A+B x^n\right ) \left (c+d x^n\right ) \, dx=\text {Too large to display} \] Input:

integrate((e*x)^m*(a+b*x^n)^2*(A+B*x^n)*(c+d*x^n),x, algorithm="giac")
 

Output:

(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(...
 

Mupad [B] (verification not implemented)

Time = 5.16 (sec) , antiderivative size = 588, normalized size of antiderivative = 3.42 \[ \int (e x)^m \left (a+b x^n\right )^2 \left (A+B x^n\right ) \left (c+d x^n\right ) \, dx=\frac {x\,x^{2\,n}\,{\left (e\,x\right )}^m\,\left (A\,b^2\,c+B\,a^2\,d+2\,A\,a\,b\,d+2\,B\,a\,b\,c\right )\,\left (m^3+8\,m^2\,n+3\,m^2+19\,m\,n^2+16\,m\,n+3\,m+12\,n^3+19\,n^2+8\,n+1\right )}{m^4+10\,m^3\,n+4\,m^3+35\,m^2\,n^2+30\,m^2\,n+6\,m^2+50\,m\,n^3+70\,m\,n^2+30\,m\,n+4\,m+24\,n^4+50\,n^3+35\,n^2+10\,n+1}+\frac {A\,a^2\,c\,x\,{\left (e\,x\right )}^m}{m+1}+\frac {a\,x\,x^n\,{\left (e\,x\right )}^m\,\left (A\,a\,d+2\,A\,b\,c+B\,a\,c\right )\,\left (m^3+9\,m^2\,n+3\,m^2+26\,m\,n^2+18\,m\,n+3\,m+24\,n^3+26\,n^2+9\,n+1\right )}{m^4+10\,m^3\,n+4\,m^3+35\,m^2\,n^2+30\,m^2\,n+6\,m^2+50\,m\,n^3+70\,m\,n^2+30\,m\,n+4\,m+24\,n^4+50\,n^3+35\,n^2+10\,n+1}+\frac {b\,x\,x^{3\,n}\,{\left (e\,x\right )}^m\,\left (A\,b\,d+2\,B\,a\,d+B\,b\,c\right )\,\left (m^3+7\,m^2\,n+3\,m^2+14\,m\,n^2+14\,m\,n+3\,m+8\,n^3+14\,n^2+7\,n+1\right )}{m^4+10\,m^3\,n+4\,m^3+35\,m^2\,n^2+30\,m^2\,n+6\,m^2+50\,m\,n^3+70\,m\,n^2+30\,m\,n+4\,m+24\,n^4+50\,n^3+35\,n^2+10\,n+1}+\frac {B\,b^2\,d\,x\,x^{4\,n}\,{\left (e\,x\right )}^m\,\left (m^3+6\,m^2\,n+3\,m^2+11\,m\,n^2+12\,m\,n+3\,m+6\,n^3+11\,n^2+6\,n+1\right )}{m^4+10\,m^3\,n+4\,m^3+35\,m^2\,n^2+30\,m^2\,n+6\,m^2+50\,m\,n^3+70\,m\,n^2+30\,m\,n+4\,m+24\,n^4+50\,n^3+35\,n^2+10\,n+1} \] Input:

int((e*x)^m*(A + B*x^n)*(a + b*x^n)^2*(c + d*x^n),x)
 

Output:

(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)
 

Reduce [B] (verification not implemented)

Time = 0.32 (sec) , antiderivative size = 1607, normalized size of antiderivative = 9.34 \[ \int (e x)^m \left (a+b x^n\right )^2 \left (A+B x^n\right ) \left (c+d x^n\right ) \, dx =\text {Too large to display} \] Input:

int((e*x)^m*(a+b*x^n)^2*(A+B*x^n)*(c+d*x^n),x)
 

Output:

(x**m*e**m*x*(x**(4*n)*b**3*d*m**4 + 6*x**(4*n)*b**3*d*m**3*n + 4*x**(4*n) 
*b**3*d*m**3 + 11*x**(4*n)*b**3*d*m**2*n**2 + 18*x**(4*n)*b**3*d*m**2*n + 
6*x**(4*n)*b**3*d*m**2 + 6*x**(4*n)*b**3*d*m*n**3 + 22*x**(4*n)*b**3*d*m*n 
**2 + 18*x**(4*n)*b**3*d*m*n + 4*x**(4*n)*b**3*d*m + 6*x**(4*n)*b**3*d*n** 
3 + 11*x**(4*n)*b**3*d*n**2 + 6*x**(4*n)*b**3*d*n + x**(4*n)*b**3*d + 3*x* 
*(3*n)*a*b**2*d*m**4 + 21*x**(3*n)*a*b**2*d*m**3*n + 12*x**(3*n)*a*b**2*d* 
m**3 + 42*x**(3*n)*a*b**2*d*m**2*n**2 + 63*x**(3*n)*a*b**2*d*m**2*n + 18*x 
**(3*n)*a*b**2*d*m**2 + 24*x**(3*n)*a*b**2*d*m*n**3 + 84*x**(3*n)*a*b**2*d 
*m*n**2 + 63*x**(3*n)*a*b**2*d*m*n + 12*x**(3*n)*a*b**2*d*m + 24*x**(3*n)* 
a*b**2*d*n**3 + 42*x**(3*n)*a*b**2*d*n**2 + 21*x**(3*n)*a*b**2*d*n + 3*x** 
(3*n)*a*b**2*d + x**(3*n)*b**3*c*m**4 + 7*x**(3*n)*b**3*c*m**3*n + 4*x**(3 
*n)*b**3*c*m**3 + 14*x**(3*n)*b**3*c*m**2*n**2 + 21*x**(3*n)*b**3*c*m**2*n 
 + 6*x**(3*n)*b**3*c*m**2 + 8*x**(3*n)*b**3*c*m*n**3 + 28*x**(3*n)*b**3*c* 
m*n**2 + 21*x**(3*n)*b**3*c*m*n + 4*x**(3*n)*b**3*c*m + 8*x**(3*n)*b**3*c* 
n**3 + 14*x**(3*n)*b**3*c*n**2 + 7*x**(3*n)*b**3*c*n + x**(3*n)*b**3*c + 3 
*x**(2*n)*a**2*b*d*m**4 + 24*x**(2*n)*a**2*b*d*m**3*n + 12*x**(2*n)*a**2*b 
*d*m**3 + 57*x**(2*n)*a**2*b*d*m**2*n**2 + 72*x**(2*n)*a**2*b*d*m**2*n + 1 
8*x**(2*n)*a**2*b*d*m**2 + 36*x**(2*n)*a**2*b*d*m*n**3 + 114*x**(2*n)*a**2 
*b*d*m*n**2 + 72*x**(2*n)*a**2*b*d*m*n + 12*x**(2*n)*a**2*b*d*m + 36*x**(2 
*n)*a**2*b*d*n**3 + 57*x**(2*n)*a**2*b*d*n**2 + 24*x**(2*n)*a**2*b*d*n ...