\(\int (e x)^m (a+b x^n) (A+B x^n) (c+d x^n)^2 \, dx\) [28]

Optimal result

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

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

Mathematica [A] (verified)

Time = 0.26 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.75 \[ \int (e x)^m \left (a+b x^n\right ) \left (A+B x^n\right ) \left (c+d x^n\right )^2 \, dx=x (e x)^m \left (\frac {a A c^2}{1+m}+\frac {c (A b c+a B c+2 a A d) x^n}{1+m+n}+\frac {(a d (2 B c+A d)+b c (B c+2 A d)) x^{2 n}}{1+m+2 n}+\frac {d (2 b B c+A b d+a B d) x^{3 n}}{1+m+3 n}+\frac {b B d^2 x^{4 n}}{1+m+4 n}\right ) \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.57 (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.

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

\(\Big \downarrow \) 1040

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

(c*(A*b*c + a*B*c + 2*a*A*d)*x^(1 + n)*(e*x)^m)/(1 + m + n) + ((a*d*(2*B*c 
 + A*d) + b*c*(B*c + 2*A*d))*x^(1 + 2*n)*(e*x)^m)/(1 + m + 2*n) + (d*(2*b* 
B*c + A*b*d + a*B*d)*x^(1 + 3*n)*(e*x)^m)/(1 + m + 3*n) + (b*B*d^2*x^(1 + 
4*n)*(e*x)^m)/(1 + m + 4*n) + (a*A*c^2*(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 = 0.71 (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)*(A+B*x^n)*(c+d*x^n)^2,x,method=_RETURNVERBOSE)
 

Output:

x*(7*A*b*d^2*(x^n)^3*n+4*B*a*c^2*m^3*x^n+24*B*a*c^2*n^3*x^n+4*B*a*d^2*(x^n 
)^3*m+7*B*a*d^2*(x^n)^3*n+38*B*a*c*d*n^2*(x^n)^2+24*B*b*c^2*m*n*(x^n)^2+8* 
A*a*d^2*m^3*n*(x^n)^2+6*B*b*d^2*m*n^3*(x^n)^4+7*A*b*d^2*m^3*n*(x^n)^3+6*B* 
a*c^2*m^2*x^n+26*B*a*c^2*n^2*x^n+4*B*b*c^2*(x^n)^2*m+8*B*b*c^2*(x^n)^2*n+4 
*A*b*c^2*x^n*m+9*A*b*c^2*x^n*n+6*A*b*c^2*m^2*x^n+50*A*a*c^2*n^3+6*A*a*c^2* 
m^2+35*A*a*c^2*n^2+10*A*a*c^2*m^3*n+38*B*a*c*d*m^2*n^2*(x^n)^2+24*B*a*c*d* 
m*n^3*(x^n)^2+42*B*b*c*d*m^2*n*(x^n)^3+14*B*a*d^2*m^2*n^2*(x^n)^3+8*B*a*d^ 
2*m*n^3*(x^n)^3+2*B*b*c*d*m^4*(x^n)^3+24*A*a*c^2*n^4+A*a*c^2*m^4+4*A*a*c^2 
*m^3+4*A*a*c^2*m+10*A*a*c^2*n+21*A*b*d^2*m*n*(x^n)^3+9*B*a*c^2*m^3*n*x^n+2 
6*B*a*c^2*m^2*n^2*x^n+B*a*d^2*m^4*(x^n)^3+4*B*b*d^2*m^3*(x^n)^4+8*A*a*d^2* 
(x^n)^2*n+8*A*a*c*d*m^3*x^n+48*A*a*c*d*n^3*x^n+24*A*a*d^2*m*n*(x^n)^2+27*A 
*b*c^2*m^2*n*x^n+52*A*b*c^2*m*n^2*x^n+24*B*a*c^2*m*n^3*x^n+38*A*b*c*d*m^2* 
n^2*(x^n)^2+24*A*b*c*d*m*n^3*(x^n)^2+16*B*a*c*d*m^3*n*(x^n)^2+24*A*a*d^2*m 
^2*n*(x^n)^2+8*A*a*c*d*x^n*m+18*A*a*c*d*x^n*n+8*B*a*c*d*m^3*(x^n)^2+2*B*a* 
c*d*m^4*(x^n)^2+21*B*a*d^2*m^2*n*(x^n)^3+28*B*a*d^2*m*n^2*(x^n)^3+8*B*b*c^ 
2*m^3*n*(x^n)^2+6*B*b*d^2*m^3*n*(x^n)^4+11*B*b*d^2*m^2*n^2*(x^n)^4+56*B*b* 
c*d*m*n^2*(x^n)^3+18*A*a*c*d*m^3*n*x^n+52*A*a*c*d*m^2*n^2*x^n+48*A*a*c*d*m 
*n^3*x^n+48*A*b*c*d*m^2*n*(x^n)^2+76*A*b*c*d*m*n^2*(x^n)^2+48*B*a*c*d*m^2* 
n*(x^n)^2+24*B*a*c*d*n^3*(x^n)^2+14*B*a*d^2*n^2*(x^n)^3+4*B*b*c^2*m^3*(x^n 
)^2+12*B*b*c^2*n^3*(x^n)^2+4*m*b*B*d^2*(x^n)^4+6*b*B*d^2*(x^n)^4*n+6*A*...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1426 vs. \(2 (172) = 344\).

Time = 0.14 (sec) , antiderivative size = 1426, normalized size of antiderivative = 8.29 \[ \int (e x)^m \left (a+b x^n\right ) \left (A+B x^n\right ) \left (c+d x^n\right )^2 \, dx=\text {Too large to display} \] Input:

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

Output:

((B*b*d^2*m^4 + 4*B*b*d^2*m^3 + 6*B*b*d^2*m^2 + 4*B*b*d^2*m + B*b*d^2 + 6* 
(B*b*d^2*m + B*b*d^2)*n^3 + 11*(B*b*d^2*m^2 + 2*B*b*d^2*m + B*b*d^2)*n^2 + 
 6*(B*b*d^2*m^3 + 3*B*b*d^2*m^2 + 3*B*b*d^2*m + B*b*d^2)*n)*x*x^(4*n)*e^(m 
*log(e) + m*log(x)) + ((2*B*b*c*d + (B*a + A*b)*d^2)*m^4 + 2*B*b*c*d + 4*( 
2*B*b*c*d + (B*a + A*b)*d^2)*m^3 + 8*(2*B*b*c*d + (B*a + A*b)*d^2 + (2*B*b 
*c*d + (B*a + A*b)*d^2)*m)*n^3 + (B*a + A*b)*d^2 + 6*(2*B*b*c*d + (B*a + A 
*b)*d^2)*m^2 + 14*(2*B*b*c*d + (B*a + A*b)*d^2 + (2*B*b*c*d + (B*a + A*b)* 
d^2)*m^2 + 2*(2*B*b*c*d + (B*a + A*b)*d^2)*m)*n^2 + 4*(2*B*b*c*d + (B*a + 
A*b)*d^2)*m + 7*(2*B*b*c*d + (2*B*b*c*d + (B*a + A*b)*d^2)*m^3 + (B*a + A* 
b)*d^2 + 3*(2*B*b*c*d + (B*a + A*b)*d^2)*m^2 + 3*(2*B*b*c*d + (B*a + A*b)* 
d^2)*m)*n)*x*x^(3*n)*e^(m*log(e) + m*log(x)) + ((B*b*c^2 + A*a*d^2 + 2*(B* 
a + A*b)*c*d)*m^4 + B*b*c^2 + A*a*d^2 + 4*(B*b*c^2 + A*a*d^2 + 2*(B*a + A* 
b)*c*d)*m^3 + 12*(B*b*c^2 + A*a*d^2 + 2*(B*a + A*b)*c*d + (B*b*c^2 + A*a*d 
^2 + 2*(B*a + A*b)*c*d)*m)*n^3 + 2*(B*a + A*b)*c*d + 6*(B*b*c^2 + A*a*d^2 
+ 2*(B*a + A*b)*c*d)*m^2 + 19*(B*b*c^2 + A*a*d^2 + 2*(B*a + A*b)*c*d + (B* 
b*c^2 + A*a*d^2 + 2*(B*a + A*b)*c*d)*m^2 + 2*(B*b*c^2 + A*a*d^2 + 2*(B*a + 
 A*b)*c*d)*m)*n^2 + 4*(B*b*c^2 + A*a*d^2 + 2*(B*a + A*b)*c*d)*m + 8*(B*b*c 
^2 + A*a*d^2 + (B*b*c^2 + A*a*d^2 + 2*(B*a + A*b)*c*d)*m^3 + 2*(B*a + A*b) 
*c*d + 3*(B*b*c^2 + A*a*d^2 + 2*(B*a + A*b)*c*d)*m^2 + 3*(B*b*c^2 + A*a*d^ 
2 + 2*(B*a + A*b)*c*d)*m)*n)*x*x^(2*n)*e^(m*log(e) + m*log(x)) + ((2*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 \[ \int (e x)^m \left (a+b x^n\right ) \left (A+B x^n\right ) \left (c+d x^n\right )^2 \, dx=\text {Too large to display} \] Input:

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

Output:

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

Maxima [A] (verification not implemented)

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

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

Output:

B*b*d^2*e^m*x*e^(m*log(x) + 4*n*log(x))/(m + 4*n + 1) + 2*B*b*c*d*e^m*x*e^ 
(m*log(x) + 3*n*log(x))/(m + 3*n + 1) + B*a*d^2*e^m*x*e^(m*log(x) + 3*n*lo 
g(x))/(m + 3*n + 1) + A*b*d^2*e^m*x*e^(m*log(x) + 3*n*log(x))/(m + 3*n + 1 
) + B*b*c^2*e^m*x*e^(m*log(x) + 2*n*log(x))/(m + 2*n + 1) + 2*B*a*c*d*e^m* 
x*e^(m*log(x) + 2*n*log(x))/(m + 2*n + 1) + 2*A*b*c*d*e^m*x*e^(m*log(x) + 
2*n*log(x))/(m + 2*n + 1) + A*a*d^2*e^m*x*e^(m*log(x) + 2*n*log(x))/(m + 2 
*n + 1) + B*a*c^2*e^m*x*e^(m*log(x) + n*log(x))/(m + n + 1) + A*b*c^2*e^m* 
x*e^(m*log(x) + n*log(x))/(m + n + 1) + 2*A*a*c*d*e^m*x*e^(m*log(x) + n*lo 
g(x))/(m + n + 1) + (e*x)^(m + 1)*A*a*c^2/(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 = 0.22 (sec) , antiderivative size = 11834, normalized size of antiderivative = 68.80 \[ \int (e x)^m \left (a+b x^n\right ) \left (A+B x^n\right ) \left (c+d x^n\right )^2 \, dx=\text {Too large to display} \] Input:

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

Output:

(B*b*d^2*m^4*x*x^(4*n)*e^(m*log(e) + m*log(x)) + 6*B*b*d^2*m^3*n*x*x^(4*n) 
*e^(m*log(e) + m*log(x)) + 11*B*b*d^2*m^2*n^2*x*x^(4*n)*e^(m*log(e) + m*lo 
g(x)) + 6*B*b*d^2*m*n^3*x*x^(4*n)*e^(m*log(e) + m*log(x)) + 2*B*b*c*d*m^4* 
x*x^(3*n)*e^(m*log(e) + m*log(x)) + B*a*d^2*m^4*x*x^(3*n)*e^(m*log(e) + m* 
log(x)) + A*b*d^2*m^4*x*x^(3*n)*e^(m*log(e) + m*log(x)) + B*b*d^2*m^4*x*x^ 
(3*n)*e^(m*log(e) + m*log(x)) + 14*B*b*c*d*m^3*n*x*x^(3*n)*e^(m*log(e) + m 
*log(x)) + 7*B*a*d^2*m^3*n*x*x^(3*n)*e^(m*log(e) + m*log(x)) + 7*A*b*d^2*m 
^3*n*x*x^(3*n)*e^(m*log(e) + m*log(x)) + 6*B*b*d^2*m^3*n*x*x^(3*n)*e^(m*lo 
g(e) + m*log(x)) + 28*B*b*c*d*m^2*n^2*x*x^(3*n)*e^(m*log(e) + m*log(x)) + 
14*B*a*d^2*m^2*n^2*x*x^(3*n)*e^(m*log(e) + m*log(x)) + 14*A*b*d^2*m^2*n^2* 
x*x^(3*n)*e^(m*log(e) + m*log(x)) + 11*B*b*d^2*m^2*n^2*x*x^(3*n)*e^(m*log( 
e) + m*log(x)) + 16*B*b*c*d*m*n^3*x*x^(3*n)*e^(m*log(e) + m*log(x)) + 8*B* 
a*d^2*m*n^3*x*x^(3*n)*e^(m*log(e) + m*log(x)) + 8*A*b*d^2*m*n^3*x*x^(3*n)* 
e^(m*log(e) + m*log(x)) + 6*B*b*d^2*m*n^3*x*x^(3*n)*e^(m*log(e) + m*log(x) 
) + B*b*c^2*m^4*x*x^(2*n)*e^(m*log(e) + m*log(x)) + 2*B*a*c*d*m^4*x*x^(2*n 
)*e^(m*log(e) + m*log(x)) + 2*A*b*c*d*m^4*x*x^(2*n)*e^(m*log(e) + m*log(x) 
) + 2*B*b*c*d*m^4*x*x^(2*n)*e^(m*log(e) + m*log(x)) + A*a*d^2*m^4*x*x^(2*n 
)*e^(m*log(e) + m*log(x)) + B*a*d^2*m^4*x*x^(2*n)*e^(m*log(e) + m*log(x)) 
+ A*b*d^2*m^4*x*x^(2*n)*e^(m*log(e) + m*log(x)) + B*b*d^2*m^4*x*x^(2*n)*e^ 
(m*log(e) + m*log(x)) + 8*B*b*c^2*m^3*n*x*x^(2*n)*e^(m*log(e) + m*log(x...
 

Mupad [B] (verification not implemented)

Time = 5.32 (sec) , antiderivative size = 588, normalized size of antiderivative = 3.42 \[ \int (e x)^m \left (a+b x^n\right ) \left (A+B x^n\right ) \left (c+d x^n\right )^2 \, dx=\frac {x\,x^{2\,n}\,{\left (e\,x\right )}^m\,\left (A\,a\,d^2+B\,b\,c^2+2\,A\,b\,c\,d+2\,B\,a\,c\,d\right )\,\left (m^3+8\,m^2\,n+3\,m^2+19\,m\,n^2+16\,m\,n+3\,m+12\,n^3+19\,n^2+8\,n+1\right )}{m^4+10\,m^3\,n+4\,m^3+35\,m^2\,n^2+30\,m^2\,n+6\,m^2+50\,m\,n^3+70\,m\,n^2+30\,m\,n+4\,m+24\,n^4+50\,n^3+35\,n^2+10\,n+1}+\frac {A\,a\,c^2\,x\,{\left (e\,x\right )}^m}{m+1}+\frac {c\,x\,x^n\,{\left (e\,x\right )}^m\,\left (2\,A\,a\,d+A\,b\,c+B\,a\,c\right )\,\left (m^3+9\,m^2\,n+3\,m^2+26\,m\,n^2+18\,m\,n+3\,m+24\,n^3+26\,n^2+9\,n+1\right )}{m^4+10\,m^3\,n+4\,m^3+35\,m^2\,n^2+30\,m^2\,n+6\,m^2+50\,m\,n^3+70\,m\,n^2+30\,m\,n+4\,m+24\,n^4+50\,n^3+35\,n^2+10\,n+1}+\frac {d\,x\,x^{3\,n}\,{\left (e\,x\right )}^m\,\left (A\,b\,d+B\,a\,d+2\,B\,b\,c\right )\,\left (m^3+7\,m^2\,n+3\,m^2+14\,m\,n^2+14\,m\,n+3\,m+8\,n^3+14\,n^2+7\,n+1\right )}{m^4+10\,m^3\,n+4\,m^3+35\,m^2\,n^2+30\,m^2\,n+6\,m^2+50\,m\,n^3+70\,m\,n^2+30\,m\,n+4\,m+24\,n^4+50\,n^3+35\,n^2+10\,n+1}+\frac {B\,b\,d^2\,x\,x^{4\,n}\,{\left (e\,x\right )}^m\,\left (m^3+6\,m^2\,n+3\,m^2+11\,m\,n^2+12\,m\,n+3\,m+6\,n^3+11\,n^2+6\,n+1\right )}{m^4+10\,m^3\,n+4\,m^3+35\,m^2\,n^2+30\,m^2\,n+6\,m^2+50\,m\,n^3+70\,m\,n^2+30\,m\,n+4\,m+24\,n^4+50\,n^3+35\,n^2+10\,n+1} \] Input:

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

Output:

(x*x^(2*n)*(e*x)^m*(A*a*d^2 + B*b*c^2 + 2*A*b*c*d + 2*B*a*c*d)*(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*c^2*x*(e*x)^m)/ 
(m + 1) + (c*x*x^n*(e*x)^m*(2*A*a*d + 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) + (d*x*x^(3*n)*(e*x)^m*(A*b*d + 
B*a*d + 2*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*d^2*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 = 0.25 (sec) , antiderivative size = 1916, normalized size of antiderivative = 11.14 \[ \int (e x)^m \left (a+b x^n\right ) \left (A+B x^n\right ) \left (c+d x^n\right )^2 \, dx =\text {Too large to display} \] Input:

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

Output:

(x**m*e**m*x*(x**(4*n)*b**2*d**2*m**4 + 6*x**(4*n)*b**2*d**2*m**3*n + 4*x* 
*(4*n)*b**2*d**2*m**3 + 11*x**(4*n)*b**2*d**2*m**2*n**2 + 18*x**(4*n)*b**2 
*d**2*m**2*n + 6*x**(4*n)*b**2*d**2*m**2 + 6*x**(4*n)*b**2*d**2*m*n**3 + 2 
2*x**(4*n)*b**2*d**2*m*n**2 + 18*x**(4*n)*b**2*d**2*m*n + 4*x**(4*n)*b**2* 
d**2*m + 6*x**(4*n)*b**2*d**2*n**3 + 11*x**(4*n)*b**2*d**2*n**2 + 6*x**(4* 
n)*b**2*d**2*n + x**(4*n)*b**2*d**2 + 2*x**(3*n)*a*b*d**2*m**4 + 14*x**(3* 
n)*a*b*d**2*m**3*n + 8*x**(3*n)*a*b*d**2*m**3 + 28*x**(3*n)*a*b*d**2*m**2* 
n**2 + 42*x**(3*n)*a*b*d**2*m**2*n + 12*x**(3*n)*a*b*d**2*m**2 + 16*x**(3* 
n)*a*b*d**2*m*n**3 + 56*x**(3*n)*a*b*d**2*m*n**2 + 42*x**(3*n)*a*b*d**2*m* 
n + 8*x**(3*n)*a*b*d**2*m + 16*x**(3*n)*a*b*d**2*n**3 + 28*x**(3*n)*a*b*d* 
*2*n**2 + 14*x**(3*n)*a*b*d**2*n + 2*x**(3*n)*a*b*d**2 + 2*x**(3*n)*b**2*c 
*d*m**4 + 14*x**(3*n)*b**2*c*d*m**3*n + 8*x**(3*n)*b**2*c*d*m**3 + 28*x**( 
3*n)*b**2*c*d*m**2*n**2 + 42*x**(3*n)*b**2*c*d*m**2*n + 12*x**(3*n)*b**2*c 
*d*m**2 + 16*x**(3*n)*b**2*c*d*m*n**3 + 56*x**(3*n)*b**2*c*d*m*n**2 + 42*x 
**(3*n)*b**2*c*d*m*n + 8*x**(3*n)*b**2*c*d*m + 16*x**(3*n)*b**2*c*d*n**3 + 
 28*x**(3*n)*b**2*c*d*n**2 + 14*x**(3*n)*b**2*c*d*n + 2*x**(3*n)*b**2*c*d 
+ x**(2*n)*a**2*d**2*m**4 + 8*x**(2*n)*a**2*d**2*m**3*n + 4*x**(2*n)*a**2* 
d**2*m**3 + 19*x**(2*n)*a**2*d**2*m**2*n**2 + 24*x**(2*n)*a**2*d**2*m**2*n 
 + 6*x**(2*n)*a**2*d**2*m**2 + 12*x**(2*n)*a**2*d**2*m*n**3 + 38*x**(2*n)* 
a**2*d**2*m*n**2 + 24*x**(2*n)*a**2*d**2*m*n + 4*x**(2*n)*a**2*d**2*m +...