Integrand size = 31, antiderivative size = 252 \[ \int (e x)^m \left (a+b x^n\right )^2 \left (A+B x^n\right ) \left (c+d x^n\right )^2 \, dx=\frac {a^2 A c^2 (e x)^{1+m}}{e (1+m)}+\frac {a c (a B c+2 A (b c+a d)) x^n (e x)^{1+m}}{e (1+m+n)}+\frac {\left (2 a B c (b c+a d)+A \left (b^2 c^2+4 a b c d+a^2 d^2\right )\right ) x^{2 n} (e x)^{1+m}}{e (1+m+2 n)}+\frac {\left (a^2 B d^2+2 a b d (2 B c+A d)+b^2 c (B c+2 A d)\right ) x^{3 n} (e x)^{1+m}}{e (1+m+3 n)}+\frac {b d (2 b B c+A b d+2 a B d) x^{4 n} (e x)^{1+m}}{e (1+m+4 n)}+\frac {b^2 B d^2 x^{5 n} (e x)^{1+m}}{e (1+m+5 n)} \] Output:
a^2*A*c^2*(e*x)^(1+m)/e/(1+m)+a*c*(a*B*c+2*A*(a*d+b*c))*x^n*(e*x)^(1+m)/e/ (1+m+n)+(2*a*B*c*(a*d+b*c)+A*(a^2*d^2+4*a*b*c*d+b^2*c^2))*x^(2*n)*(e*x)^(1 +m)/e/(1+m+2*n)+(a^2*B*d^2+2*a*b*d*(A*d+2*B*c)+b^2*c*(2*A*d+B*c))*x^(3*n)* (e*x)^(1+m)/e/(1+m+3*n)+b*d*(A*b*d+2*B*a*d+2*B*b*c)*x^(4*n)*(e*x)^(1+m)/e/ (1+m+4*n)+b^2*B*d^2*x^(5*n)*(e*x)^(1+m)/e/(1+m+5*n)
Time = 0.65 (sec) , antiderivative size = 199, normalized size of antiderivative = 0.79 \[ \int (e x)^m \left (a+b x^n\right )^2 \left (A+B x^n\right ) \left (c+d x^n\right )^2 \, dx=x (e x)^m \left (\frac {a^2 A c^2}{1+m}+\frac {a c (a B c+2 A (b c+a d)) x^n}{1+m+n}+\frac {\left (2 a B c (b c+a d)+A \left (b^2 c^2+4 a b c d+a^2 d^2\right )\right ) x^{2 n}}{1+m+2 n}+\frac {\left (a^2 B d^2+2 a b d (2 B c+A d)+b^2 c (B c+2 A d)\right ) x^{3 n}}{1+m+3 n}+\frac {b d (2 b B c+A b d+2 a B d) x^{4 n}}{1+m+4 n}+\frac {b^2 B d^2 x^{5 n}}{1+m+5 n}\right ) \] Input:
Integrate[(e*x)^m*(a + b*x^n)^2*(A + B*x^n)*(c + d*x^n)^2,x]
Output:
x*(e*x)^m*((a^2*A*c^2)/(1 + m) + (a*c*(a*B*c + 2*A*(b*c + a*d))*x^n)/(1 + m + n) + ((2*a*B*c*(b*c + a*d) + A*(b^2*c^2 + 4*a*b*c*d + a^2*d^2))*x^(2*n ))/(1 + m + 2*n) + ((a^2*B*d^2 + 2*a*b*d*(2*B*c + A*d) + b^2*c*(B*c + 2*A* d))*x^(3*n))/(1 + m + 3*n) + (b*d*(2*b*B*c + A*b*d + 2*a*B*d)*x^(4*n))/(1 + m + 4*n) + (b^2*B*d^2*x^(5*n))/(1 + m + 5*n))
Time = 0.80 (sec) , antiderivative size = 237, normalized size of antiderivative = 0.94, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, 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.
\(\displaystyle \int (e x)^m \left (a+b x^n\right )^2 \left (A+B x^n\right ) \left (c+d x^n\right )^2 \, dx\) |
\(\Big \downarrow \) 1040 |
\(\displaystyle \int \left (x^{2 n} (e x)^m \left (A \left (a^2 d^2+4 a b c d+b^2 c^2\right )+2 a B c (a d+b c)\right )+x^{3 n} (e x)^m \left (a^2 B d^2+2 a b d (A d+2 B c)+b^2 c (2 A d+B c)\right )+a^2 A c^2 (e x)^m+b d x^{4 n} (e x)^m (2 a B d+A b d+2 b B c)+a c x^n (e x)^m (2 A (a d+b c)+a B c)+b^2 B d^2 x^{5 n} (e x)^m\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {x^{2 n+1} (e x)^m \left (A \left (a^2 d^2+4 a b c d+b^2 c^2\right )+2 a B c (a d+b c)\right )}{m+2 n+1}+\frac {x^{3 n+1} (e x)^m \left (a^2 B d^2+2 a b d (A d+2 B c)+b^2 c (2 A d+B c)\right )}{m+3 n+1}+\frac {a^2 A c^2 (e x)^{m+1}}{e (m+1)}+\frac {a c x^{n+1} (e x)^m (2 A (a d+b c)+a B c)}{m+n+1}+\frac {b d x^{4 n+1} (e x)^m (2 a B d+A b d+2 b B c)}{m+4 n+1}+\frac {b^2 B d^2 x^{5 n+1} (e x)^m}{m+5 n+1}\) |
Input:
Int[(e*x)^m*(a + b*x^n)^2*(A + B*x^n)*(c + d*x^n)^2,x]
Output:
(a*c*(a*B*c + 2*A*(b*c + a*d))*x^(1 + n)*(e*x)^m)/(1 + m + n) + ((2*a*B*c* (b*c + a*d) + A*(b^2*c^2 + 4*a*b*c*d + a^2*d^2))*x^(1 + 2*n)*(e*x)^m)/(1 + m + 2*n) + ((a^2*B*d^2 + 2*a*b*d*(2*B*c + A*d) + b^2*c*(B*c + 2*A*d))*x^( 1 + 3*n)*(e*x)^m)/(1 + m + 3*n) + (b*d*(2*b*B*c + A*b*d + 2*a*B*d)*x^(1 + 4*n)*(e*x)^m)/(1 + m + 4*n) + (b^2*B*d^2*x^(1 + 5*n)*(e*x)^m)/(1 + m + 5*n ) + (a^2*A*c^2*(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 = 1.41 (sec) , antiderivative size = 5875, normalized size of antiderivative = 23.31
method | result | size |
risch | \(\text {Expression too large to display}\) | \(5875\) |
parallelrisch | \(\text {Expression too large to display}\) | \(7994\) |
orering | \(\text {Expression too large to display}\) | \(11761\) |
Input:
int((e*x)^m*(a+b*x^n)^2*(A+B*x^n)*(c+d*x^n)^2,x,method=_RETURNVERBOSE)
Output:
result too large to display
Leaf count of result is larger than twice the leaf count of optimal. 3515 vs. \(2 (252) = 504\).
Time = 0.18 (sec) , antiderivative size = 3515, normalized size of antiderivative = 13.95 \[ \int (e x)^m \left (a+b x^n\right )^2 \left (A+B x^n\right ) \left (c+d x^n\right )^2 \, dx=\text {Too large to display} \] Input:
integrate((e*x)^m*(a+b*x^n)^2*(A+B*x^n)*(c+d*x^n)^2,x, algorithm="fricas")
Output:
Too large to include
Leaf count of result is larger than twice the leaf count of optimal. 72500 vs. \(2 (241) = 482\).
Time = 18.77 (sec) , antiderivative size = 72500, normalized size of antiderivative = 287.70 \[ \int (e x)^m \left (a+b x^n\right )^2 \left (A+B x^n\right ) \left (c+d x^n\right )^2 \, dx=\text {Too large to display} \] Input:
integrate((e*x)**m*(a+b*x**n)**2*(A+B*x**n)*(c+d*x**n)**2,x)
Output:
Piecewise(((A + B)*(a + b)**2*(c + d)**2*log(x)/e, Eq(m, -1) & Eq(n, 0)), ((A*a**2*c**2*log(x) + 2*A*a**2*c*d*x**n/n + A*a**2*d**2*x**(2*n)/(2*n) + 2*A*a*b*c**2*x**n/n + 2*A*a*b*c*d*x**(2*n)/n + 2*A*a*b*d**2*x**(3*n)/(3*n) + A*b**2*c**2*x**(2*n)/(2*n) + 2*A*b**2*c*d*x**(3*n)/(3*n) + A*b**2*d**2* x**(4*n)/(4*n) + B*a**2*c**2*x**n/n + B*a**2*c*d*x**(2*n)/n + B*a**2*d**2* x**(3*n)/(3*n) + B*a*b*c**2*x**(2*n)/n + 4*B*a*b*c*d*x**(3*n)/(3*n) + B*a* b*d**2*x**(4*n)/(2*n) + B*b**2*c**2*x**(3*n)/(3*n) + B*b**2*c*d*x**(4*n)/( 2*n) + B*b**2*d**2*x**(5*n)/(5*n))/e, Eq(m, -1)), (A*a**2*c**2*Piecewise(( 0**(-5*n - 1)*x, Eq(e, 0)), (Piecewise((-1/(5*n*(e*x)**(5*n)), Ne(n, 0)), (log(e*x), True))/e, True)) + 2*A*a**2*c*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)) + A*a**2* d**2*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)) + 2*A*a*b*c**2*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)) + 4*A*a*b*c*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)) + 2*A*a*b*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)*lo g(x), True)) + A*b**2*c**2*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)) + 2*A*b**2*c*d*Pi ecewise((-x*x**(3*n)*(e*x)**(-5*n - 1)/(2*n), Ne(n, 0)), (x*x**(3*n)*(e...
Leaf count of result is larger than twice the leaf count of optimal. 540 vs. \(2 (252) = 504\).
Time = 0.09 (sec) , antiderivative size = 540, normalized size of antiderivative = 2.14 \[ \int (e x)^m \left (a+b x^n\right )^2 \left (A+B x^n\right ) \left (c+d x^n\right )^2 \, dx=\frac {B b^{2} d^{2} e^{m} x e^{\left (m \log \left (x\right ) + 5 \, n \log \left (x\right )\right )}}{m + 5 \, n + 1} + \frac {2 \, B b^{2} c d e^{m} x e^{\left (m \log \left (x\right ) + 4 \, n \log \left (x\right )\right )}}{m + 4 \, n + 1} + \frac {2 \, B a 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 {A b^{2} 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 b^{2} c^{2} e^{m} x e^{\left (m \log \left (x\right ) + 3 \, n \log \left (x\right )\right )}}{m + 3 \, n + 1} + \frac {4 \, B a 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 {2 \, A b^{2} 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^{2} d^{2} e^{m} x e^{\left (m \log \left (x\right ) + 3 \, n \log \left (x\right )\right )}}{m + 3 \, n + 1} + \frac {2 \, A 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 {2 \, B a 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 {A b^{2} 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^{2} c d e^{m} x e^{\left (m \log \left (x\right ) + 2 \, n \log \left (x\right )\right )}}{m + 2 \, n + 1} + \frac {4 \, A 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^{2} 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^{2} 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 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^{2} 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^{2} c^{2}}{e {\left (m + 1\right )}} \] Input:
integrate((e*x)^m*(a+b*x^n)^2*(A+B*x^n)*(c+d*x^n)^2,x, algorithm="maxima")
Output:
B*b^2*d^2*e^m*x*e^(m*log(x) + 5*n*log(x))/(m + 5*n + 1) + 2*B*b^2*c*d*e^m* x*e^(m*log(x) + 4*n*log(x))/(m + 4*n + 1) + 2*B*a*b*d^2*e^m*x*e^(m*log(x) + 4*n*log(x))/(m + 4*n + 1) + A*b^2*d^2*e^m*x*e^(m*log(x) + 4*n*log(x))/(m + 4*n + 1) + B*b^2*c^2*e^m*x*e^(m*log(x) + 3*n*log(x))/(m + 3*n + 1) + 4* B*a*b*c*d*e^m*x*e^(m*log(x) + 3*n*log(x))/(m + 3*n + 1) + 2*A*b^2*c*d*e^m* x*e^(m*log(x) + 3*n*log(x))/(m + 3*n + 1) + B*a^2*d^2*e^m*x*e^(m*log(x) + 3*n*log(x))/(m + 3*n + 1) + 2*A*a*b*d^2*e^m*x*e^(m*log(x) + 3*n*log(x))/(m + 3*n + 1) + 2*B*a*b*c^2*e^m*x*e^(m*log(x) + 2*n*log(x))/(m + 2*n + 1) + A*b^2*c^2*e^m*x*e^(m*log(x) + 2*n*log(x))/(m + 2*n + 1) + 2*B*a^2*c*d*e^m* x*e^(m*log(x) + 2*n*log(x))/(m + 2*n + 1) + 4*A*a*b*c*d*e^m*x*e^(m*log(x) + 2*n*log(x))/(m + 2*n + 1) + A*a^2*d^2*e^m*x*e^(m*log(x) + 2*n*log(x))/(m + 2*n + 1) + B*a^2*c^2*e^m*x*e^(m*log(x) + n*log(x))/(m + n + 1) + 2*A*a* b*c^2*e^m*x*e^(m*log(x) + n*log(x))/(m + n + 1) + 2*A*a^2*c*d*e^m*x*e^(m*l og(x) + n*log(x))/(m + n + 1) + (e*x)^(m + 1)*A*a^2*c^2/(e*(m + 1))
Leaf count of result is larger than twice the leaf count of optimal. 32523 vs. \(2 (252) = 504\).
Time = 0.35 (sec) , antiderivative size = 32523, normalized size of antiderivative = 129.06 \[ \int (e x)^m \left (a+b x^n\right )^2 \left (A+B x^n\right ) \left (c+d x^n\right )^2 \, dx=\text {Too large to display} \] Input:
integrate((e*x)^m*(a+b*x^n)^2*(A+B*x^n)*(c+d*x^n)^2,x, algorithm="giac")
Output:
(B*b^2*d^2*m^5*x*x^(5*n)*e^(m*log(e) + m*log(x)) + 10*B*b^2*d^2*m^4*n*x*x^ (5*n)*e^(m*log(e) + m*log(x)) + 35*B*b^2*d^2*m^3*n^2*x*x^(5*n)*e^(m*log(e) + m*log(x)) + 50*B*b^2*d^2*m^2*n^3*x*x^(5*n)*e^(m*log(e) + m*log(x)) + 24 *B*b^2*d^2*m*n^4*x*x^(5*n)*e^(m*log(e) + m*log(x)) + 2*B*b^2*c*d*m^5*x*x^( 4*n)*e^(m*log(e) + m*log(x)) + 2*B*a*b*d^2*m^5*x*x^(4*n)*e^(m*log(e) + m*l og(x)) + A*b^2*d^2*m^5*x*x^(4*n)*e^(m*log(e) + m*log(x)) + B*b^2*d^2*m^5*x *x^(4*n)*e^(m*log(e) + m*log(x)) + 22*B*b^2*c*d*m^4*n*x*x^(4*n)*e^(m*log(e ) + m*log(x)) + 22*B*a*b*d^2*m^4*n*x*x^(4*n)*e^(m*log(e) + m*log(x)) + 11* A*b^2*d^2*m^4*n*x*x^(4*n)*e^(m*log(e) + m*log(x)) + 10*B*b^2*d^2*m^4*n*x*x ^(4*n)*e^(m*log(e) + m*log(x)) + 82*B*b^2*c*d*m^3*n^2*x*x^(4*n)*e^(m*log(e ) + m*log(x)) + 82*B*a*b*d^2*m^3*n^2*x*x^(4*n)*e^(m*log(e) + m*log(x)) + 4 1*A*b^2*d^2*m^3*n^2*x*x^(4*n)*e^(m*log(e) + m*log(x)) + 35*B*b^2*d^2*m^3*n ^2*x*x^(4*n)*e^(m*log(e) + m*log(x)) + 122*B*b^2*c*d*m^2*n^3*x*x^(4*n)*e^( m*log(e) + m*log(x)) + 122*B*a*b*d^2*m^2*n^3*x*x^(4*n)*e^(m*log(e) + m*log (x)) + 61*A*b^2*d^2*m^2*n^3*x*x^(4*n)*e^(m*log(e) + m*log(x)) + 50*B*b^2*d ^2*m^2*n^3*x*x^(4*n)*e^(m*log(e) + m*log(x)) + 60*B*b^2*c*d*m*n^4*x*x^(4*n )*e^(m*log(e) + m*log(x)) + 60*B*a*b*d^2*m*n^4*x*x^(4*n)*e^(m*log(e) + m*l og(x)) + 30*A*b^2*d^2*m*n^4*x*x^(4*n)*e^(m*log(e) + m*log(x)) + 24*B*b^2*d ^2*m*n^4*x*x^(4*n)*e^(m*log(e) + m*log(x)) + B*b^2*c^2*m^5*x*x^(3*n)*e^(m* log(e) + m*log(x)) + 4*B*a*b*c*d*m^5*x*x^(3*n)*e^(m*log(e) + m*log(x)) ...
Time = 5.81 (sec) , antiderivative size = 1119, normalized size of antiderivative = 4.44 \[ \int (e x)^m \left (a+b x^n\right )^2 \left (A+B x^n\right ) \left (c+d x^n\right )^2 \, dx =\text {Too large to display} \] Input:
int((e*x)^m*(A + B*x^n)*(a + b*x^n)^2*(c + d*x^n)^2,x)
Output:
(x*x^(2*n)*(e*x)^m*(A*a^2*d^2 + A*b^2*c^2 + 2*B*a*b*c^2 + 2*B*a^2*c*d + 4* A*a*b*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) + (x*x^(3*n)*(e* x)^m*(B*a^2*d^2 + B*b^2*c^2 + 2*A*a*b*d^2 + 2*A*b^2*c*d + 4*B*a*b*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) + (A*a^2*c^2*x*(e*x)^m)/(m + 1) + (b*d*x*x^(4*n)*(e*x)^m*(A*b*d + 2*B*a*d + 2*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) + (B*b^2*d^2*x*x^(5*n)*(e*x)^m*(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))/(5*m + 15*n + 60*m*n + 255*m*n^2 ...
Time = 0.27 (sec) , antiderivative size = 3949, normalized size of antiderivative = 15.67 \[ \int (e x)^m \left (a+b x^n\right )^2 \left (A+B x^n\right ) \left (c+d x^n\right )^2 \, dx =\text {Too large to display} \] Input:
int((e*x)^m*(a+b*x^n)^2*(A+B*x^n)*(c+d*x^n)^2,x)
Output:
(x**m*e**m*x*(x**(5*n)*b**3*d**2*m**5 + 10*x**(5*n)*b**3*d**2*m**4*n + 5*x **(5*n)*b**3*d**2*m**4 + 35*x**(5*n)*b**3*d**2*m**3*n**2 + 40*x**(5*n)*b** 3*d**2*m**3*n + 10*x**(5*n)*b**3*d**2*m**3 + 50*x**(5*n)*b**3*d**2*m**2*n* *3 + 105*x**(5*n)*b**3*d**2*m**2*n**2 + 60*x**(5*n)*b**3*d**2*m**2*n + 10* x**(5*n)*b**3*d**2*m**2 + 24*x**(5*n)*b**3*d**2*m*n**4 + 100*x**(5*n)*b**3 *d**2*m*n**3 + 105*x**(5*n)*b**3*d**2*m*n**2 + 40*x**(5*n)*b**3*d**2*m*n + 5*x**(5*n)*b**3*d**2*m + 24*x**(5*n)*b**3*d**2*n**4 + 50*x**(5*n)*b**3*d* *2*n**3 + 35*x**(5*n)*b**3*d**2*n**2 + 10*x**(5*n)*b**3*d**2*n + x**(5*n)* b**3*d**2 + 3*x**(4*n)*a*b**2*d**2*m**5 + 33*x**(4*n)*a*b**2*d**2*m**4*n + 15*x**(4*n)*a*b**2*d**2*m**4 + 123*x**(4*n)*a*b**2*d**2*m**3*n**2 + 132*x **(4*n)*a*b**2*d**2*m**3*n + 30*x**(4*n)*a*b**2*d**2*m**3 + 183*x**(4*n)*a *b**2*d**2*m**2*n**3 + 369*x**(4*n)*a*b**2*d**2*m**2*n**2 + 198*x**(4*n)*a *b**2*d**2*m**2*n + 30*x**(4*n)*a*b**2*d**2*m**2 + 90*x**(4*n)*a*b**2*d**2 *m*n**4 + 366*x**(4*n)*a*b**2*d**2*m*n**3 + 369*x**(4*n)*a*b**2*d**2*m*n** 2 + 132*x**(4*n)*a*b**2*d**2*m*n + 15*x**(4*n)*a*b**2*d**2*m + 90*x**(4*n) *a*b**2*d**2*n**4 + 183*x**(4*n)*a*b**2*d**2*n**3 + 123*x**(4*n)*a*b**2*d* *2*n**2 + 33*x**(4*n)*a*b**2*d**2*n + 3*x**(4*n)*a*b**2*d**2 + 2*x**(4*n)* b**3*c*d*m**5 + 22*x**(4*n)*b**3*c*d*m**4*n + 10*x**(4*n)*b**3*c*d*m**4 + 82*x**(4*n)*b**3*c*d*m**3*n**2 + 88*x**(4*n)*b**3*c*d*m**3*n + 20*x**(4*n) *b**3*c*d*m**3 + 122*x**(4*n)*b**3*c*d*m**2*n**3 + 246*x**(4*n)*b**3*c*...