Integrand size = 31, antiderivative size = 431 \[ \int (e x)^m \left (a+b x^n\right )^3 \left (A+B x^n\right ) \left (c+d x^n\right )^3 \, dx=\frac {a^3 A c^3 (e x)^{1+m}}{e (1+m)}+\frac {a^2 c^2 (a B c+3 A (b c+a d)) x^n (e x)^{1+m}}{e (1+m+n)}+\frac {3 a c \left (a B c (b c+a d)+A \left (b^2 c^2+3 a b c d+a^2 d^2\right )\right ) x^{2 n} (e x)^{1+m}}{e (1+m+2 n)}+\frac {\left (3 a B c \left (b^2 c^2+3 a b c d+a^2 d^2\right )+A \left (b^3 c^3+9 a b^2 c^2 d+9 a^2 b c d^2+a^3 d^3\right )\right ) x^{3 n} (e x)^{1+m}}{e (1+m+3 n)}+\frac {\left (a^3 B d^3+9 a b^2 c d (B c+A d)+3 a^2 b d^2 (3 B c+A d)+b^3 c^2 (B c+3 A d)\right ) x^{4 n} (e x)^{1+m}}{e (1+m+4 n)}+\frac {3 b d \left (a^2 B d^2+b^2 c (B c+A d)+a b d (3 B c+A d)\right ) x^{5 n} (e x)^{1+m}}{e (1+m+5 n)}+\frac {b^2 d^2 (3 b B c+A b d+3 a B d) x^{6 n} (e x)^{1+m}}{e (1+m+6 n)}+\frac {b^3 B d^3 x^{7 n} (e x)^{1+m}}{e (1+m+7 n)} \] Output:
a^3*A*c^3*(e*x)^(1+m)/e/(1+m)+a^2*c^2*(a*B*c+3*A*(a*d+b*c))*x^n*(e*x)^(1+m )/e/(1+m+n)+3*a*c*(a*B*c*(a*d+b*c)+A*(a^2*d^2+3*a*b*c*d+b^2*c^2))*x^(2*n)* (e*x)^(1+m)/e/(1+m+2*n)+(3*a*B*c*(a^2*d^2+3*a*b*c*d+b^2*c^2)+A*(a^3*d^3+9* a^2*b*c*d^2+9*a*b^2*c^2*d+b^3*c^3))*x^(3*n)*(e*x)^(1+m)/e/(1+m+3*n)+(a^3*B *d^3+9*a*b^2*c*d*(A*d+B*c)+3*a^2*b*d^2*(A*d+3*B*c)+b^3*c^2*(3*A*d+B*c))*x^ (4*n)*(e*x)^(1+m)/e/(1+m+4*n)+3*b*d*(a^2*B*d^2+b^2*c*(A*d+B*c)+a*b*d*(A*d+ 3*B*c))*x^(5*n)*(e*x)^(1+m)/e/(1+m+5*n)+b^2*d^2*(A*b*d+3*B*a*d+3*B*b*c)*x^ (6*n)*(e*x)^(1+m)/e/(1+m+6*n)+b^3*B*d^3*x^(7*n)*(e*x)^(1+m)/e/(1+m+7*n)
Time = 1.37 (sec) , antiderivative size = 358, normalized size of antiderivative = 0.83 \[ \int (e x)^m \left (a+b x^n\right )^3 \left (A+B x^n\right ) \left (c+d x^n\right )^3 \, dx=x (e x)^m \left (\frac {a^3 A c^3}{1+m}+\frac {a^2 c^2 (a B c+3 A (b c+a d)) x^n}{1+m+n}+\frac {3 a c \left (a B c (b c+a d)+A \left (b^2 c^2+3 a b c d+a^2 d^2\right )\right ) x^{2 n}}{1+m+2 n}+\frac {\left (3 a B c \left (b^2 c^2+3 a b c d+a^2 d^2\right )+A \left (b^3 c^3+9 a b^2 c^2 d+9 a^2 b c d^2+a^3 d^3\right )\right ) x^{3 n}}{1+m+3 n}+\frac {\left (a^3 B d^3+9 a b^2 c d (B c+A d)+3 a^2 b d^2 (3 B c+A d)+b^3 c^2 (B c+3 A d)\right ) x^{4 n}}{1+m+4 n}+\frac {3 b d \left (a^2 B d^2+b^2 c (B c+A d)+a b d (3 B c+A d)\right ) x^{5 n}}{1+m+5 n}+\frac {b^2 d^2 (3 b B c+A b d+3 a B d) x^{6 n}}{1+m+6 n}+\frac {b^3 B d^3 x^{7 n}}{1+m+7 n}\right ) \] Input:
Integrate[(e*x)^m*(a + b*x^n)^3*(A + B*x^n)*(c + d*x^n)^3,x]
Output:
x*(e*x)^m*((a^3*A*c^3)/(1 + m) + (a^2*c^2*(a*B*c + 3*A*(b*c + a*d))*x^n)/( 1 + m + n) + (3*a*c*(a*B*c*(b*c + a*d) + A*(b^2*c^2 + 3*a*b*c*d + a^2*d^2) )*x^(2*n))/(1 + m + 2*n) + ((3*a*B*c*(b^2*c^2 + 3*a*b*c*d + a^2*d^2) + A*( b^3*c^3 + 9*a*b^2*c^2*d + 9*a^2*b*c*d^2 + a^3*d^3))*x^(3*n))/(1 + m + 3*n) + ((a^3*B*d^3 + 9*a*b^2*c*d*(B*c + A*d) + 3*a^2*b*d^2*(3*B*c + A*d) + b^3 *c^2*(B*c + 3*A*d))*x^(4*n))/(1 + m + 4*n) + (3*b*d*(a^2*B*d^2 + b^2*c*(B* c + A*d) + a*b*d*(3*B*c + A*d))*x^(5*n))/(1 + m + 5*n) + (b^2*d^2*(3*b*B*c + A*b*d + 3*a*B*d)*x^(6*n))/(1 + m + 6*n) + (b^3*B*d^3*x^(7*n))/(1 + m + 7*n))
Time = 1.33 (sec) , antiderivative size = 410, normalized size of antiderivative = 0.95, 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.
\(\displaystyle \int (e x)^m \left (a+b x^n\right )^3 \left (A+B x^n\right ) \left (c+d x^n\right )^3 \, dx\) |
\(\Big \downarrow \) 1040 |
\(\displaystyle \int \left (a^3 A c^3 (e x)^m+3 a c x^{2 n} (e x)^m \left (A \left (a^2 d^2+3 a b c d+b^2 c^2\right )+a B c (a d+b c)\right )+3 b d x^{5 n} (e x)^m \left (a^2 B d^2+a b d (A d+3 B c)+b^2 c (A d+B c)\right )+a^2 c^2 x^n (e x)^m (3 A (a d+b c)+a B c)+x^{4 n} (e x)^m \left (a^3 B d^3+3 a^2 b d^2 (A d+3 B c)+9 a b^2 c d (A d+B c)+b^3 c^2 (3 A d+B c)\right )+x^{3 n} (e x)^m \left (3 a B c \left (a^2 d^2+3 a b c d+b^2 c^2\right )+A \left (a^3 d^3+9 a^2 b c d^2+9 a b^2 c^2 d+b^3 c^3\right )\right )+b^2 d^2 x^{6 n} (e x)^m (3 a B d+A b d+3 b B c)+b^3 B d^3 x^{7 n} (e x)^m\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {a^3 A c^3 (e x)^{m+1}}{e (m+1)}+\frac {3 a c x^{2 n+1} (e x)^m \left (A \left (a^2 d^2+3 a b c d+b^2 c^2\right )+a B c (a d+b c)\right )}{m+2 n+1}+\frac {3 b d x^{5 n+1} (e x)^m \left (a^2 B d^2+a b d (A d+3 B c)+b^2 c (A d+B c)\right )}{m+5 n+1}+\frac {a^2 c^2 x^{n+1} (e x)^m (3 A (a d+b c)+a B c)}{m+n+1}+\frac {x^{4 n+1} (e x)^m \left (a^3 B d^3+3 a^2 b d^2 (A d+3 B c)+9 a b^2 c d (A d+B c)+b^3 c^2 (3 A d+B c)\right )}{m+4 n+1}+\frac {x^{3 n+1} (e x)^m \left (3 a B c \left (a^2 d^2+3 a b c d+b^2 c^2\right )+A \left (a^3 d^3+9 a^2 b c d^2+9 a b^2 c^2 d+b^3 c^3\right )\right )}{m+3 n+1}+\frac {b^2 d^2 x^{6 n+1} (e x)^m (3 a B d+A b d+3 b B c)}{m+6 n+1}+\frac {b^3 B d^3 x^{7 n+1} (e x)^m}{m+7 n+1}\) |
Input:
Int[(e*x)^m*(a + b*x^n)^3*(A + B*x^n)*(c + d*x^n)^3,x]
Output:
(a^2*c^2*(a*B*c + 3*A*(b*c + a*d))*x^(1 + n)*(e*x)^m)/(1 + m + n) + (3*a*c *(a*B*c*(b*c + a*d) + A*(b^2*c^2 + 3*a*b*c*d + a^2*d^2))*x^(1 + 2*n)*(e*x) ^m)/(1 + m + 2*n) + ((3*a*B*c*(b^2*c^2 + 3*a*b*c*d + a^2*d^2) + A*(b^3*c^3 + 9*a*b^2*c^2*d + 9*a^2*b*c*d^2 + a^3*d^3))*x^(1 + 3*n)*(e*x)^m)/(1 + m + 3*n) + ((a^3*B*d^3 + 9*a*b^2*c*d*(B*c + A*d) + 3*a^2*b*d^2*(3*B*c + A*d) + b^3*c^2*(B*c + 3*A*d))*x^(1 + 4*n)*(e*x)^m)/(1 + m + 4*n) + (3*b*d*(a^2* B*d^2 + b^2*c*(B*c + A*d) + a*b*d*(3*B*c + A*d))*x^(1 + 5*n)*(e*x)^m)/(1 + m + 5*n) + (b^2*d^2*(3*b*B*c + A*b*d + 3*a*B*d)*x^(1 + 6*n)*(e*x)^m)/(1 + m + 6*n) + (b^3*B*d^3*x^(1 + 7*n)*(e*x)^m)/(1 + m + 7*n) + (a^3*A*c^3*(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 = 5.24 (sec) , antiderivative size = 20904, normalized size of antiderivative = 48.50
method | result | size |
risch | \(\text {Expression too large to display}\) | \(20904\) |
parallelrisch | \(\text {Expression too large to display}\) | \(27583\) |
orering | \(\text {Expression too large to display}\) | \(43732\) |
Input:
int((e*x)^m*(a+b*x^n)^3*(A+B*x^n)*(c+d*x^n)^3,x,method=_RETURNVERBOSE)
Output:
result too large to display
Leaf count of result is larger than twice the leaf count of optimal. 11628 vs. \(2 (431) = 862\).
Time = 0.47 (sec) , antiderivative size = 11628, normalized size of antiderivative = 26.98 \[ \int (e x)^m \left (a+b x^n\right )^3 \left (A+B x^n\right ) \left (c+d x^n\right )^3 \, dx=\text {Too large to display} \] Input:
integrate((e*x)^m*(a+b*x^n)^3*(A+B*x^n)*(c+d*x^n)^3,x, algorithm="fricas")
Output:
Too large to include
Leaf count of result is larger than twice the leaf count of optimal. 365145 vs. \(2 (422) = 844\).
Time = 49.52 (sec) , antiderivative size = 365145, normalized size of antiderivative = 847.20 \[ \int (e x)^m \left (a+b x^n\right )^3 \left (A+B x^n\right ) \left (c+d x^n\right )^3 \, dx=\text {Too large to display} \] Input:
integrate((e*x)**m*(a+b*x**n)**3*(A+B*x**n)*(c+d*x**n)**3,x)
Output:
Piecewise(((A + B)*(a + b)**3*(c + d)**3*log(x)/e, Eq(m, -1) & Eq(n, 0)), ((A*a**3*c**3*log(x) + 3*A*a**3*c**2*d*x**n/n + 3*A*a**3*c*d**2*x**(2*n)/( 2*n) + A*a**3*d**3*x**(3*n)/(3*n) + 3*A*a**2*b*c**3*x**n/n + 9*A*a**2*b*c* *2*d*x**(2*n)/(2*n) + 3*A*a**2*b*c*d**2*x**(3*n)/n + 3*A*a**2*b*d**3*x**(4 *n)/(4*n) + 3*A*a*b**2*c**3*x**(2*n)/(2*n) + 3*A*a*b**2*c**2*d*x**(3*n)/n + 9*A*a*b**2*c*d**2*x**(4*n)/(4*n) + 3*A*a*b**2*d**3*x**(5*n)/(5*n) + A*b* *3*c**3*x**(3*n)/(3*n) + 3*A*b**3*c**2*d*x**(4*n)/(4*n) + 3*A*b**3*c*d**2* x**(5*n)/(5*n) + A*b**3*d**3*x**(6*n)/(6*n) + B*a**3*c**3*x**n/n + 3*B*a** 3*c**2*d*x**(2*n)/(2*n) + B*a**3*c*d**2*x**(3*n)/n + B*a**3*d**3*x**(4*n)/ (4*n) + 3*B*a**2*b*c**3*x**(2*n)/(2*n) + 3*B*a**2*b*c**2*d*x**(3*n)/n + 9* B*a**2*b*c*d**2*x**(4*n)/(4*n) + 3*B*a**2*b*d**3*x**(5*n)/(5*n) + B*a*b**2 *c**3*x**(3*n)/n + 9*B*a*b**2*c**2*d*x**(4*n)/(4*n) + 9*B*a*b**2*c*d**2*x* *(5*n)/(5*n) + B*a*b**2*d**3*x**(6*n)/(2*n) + B*b**3*c**3*x**(4*n)/(4*n) + 3*B*b**3*c**2*d*x**(5*n)/(5*n) + B*b**3*c*d**2*x**(6*n)/(2*n) + B*b**3*d* *3*x**(7*n)/(7*n))/e, Eq(m, -1)), (A*a**3*c**3*Piecewise((0**(-7*n - 1)*x, Eq(e, 0)), (Piecewise((-1/(7*n*(e*x)**(7*n)), Ne(n, 0)), (log(e*x), True) )/e, True)) + 3*A*a**3*c**2*d*Piecewise((-x*x**n*(e*x)**(-7*n - 1)/(6*n), Ne(n, 0)), (x*x**n*(e*x)**(-7*n - 1)*log(x), True)) + 3*A*a**3*c*d**2*Piec ewise((-x*x**(2*n)*(e*x)**(-7*n - 1)/(5*n), Ne(n, 0)), (x*x**(2*n)*(e*x)** (-7*n - 1)*log(x), True)) + A*a**3*d**3*Piecewise((-x*x**(3*n)*(e*x)**(...
Leaf count of result is larger than twice the leaf count of optimal. 1032 vs. \(2 (431) = 862\).
Time = 0.14 (sec) , antiderivative size = 1032, normalized size of antiderivative = 2.39 \[ \int (e x)^m \left (a+b x^n\right )^3 \left (A+B x^n\right ) \left (c+d x^n\right )^3 \, dx=\text {Too large to display} \] Input:
integrate((e*x)^m*(a+b*x^n)^3*(A+B*x^n)*(c+d*x^n)^3,x, algorithm="maxima")
Output:
B*b^3*d^3*e^m*x*e^(m*log(x) + 7*n*log(x))/(m + 7*n + 1) + 3*B*b^3*c*d^2*e^ m*x*e^(m*log(x) + 6*n*log(x))/(m + 6*n + 1) + 3*B*a*b^2*d^3*e^m*x*e^(m*log (x) + 6*n*log(x))/(m + 6*n + 1) + A*b^3*d^3*e^m*x*e^(m*log(x) + 6*n*log(x) )/(m + 6*n + 1) + 3*B*b^3*c^2*d*e^m*x*e^(m*log(x) + 5*n*log(x))/(m + 5*n + 1) + 9*B*a*b^2*c*d^2*e^m*x*e^(m*log(x) + 5*n*log(x))/(m + 5*n + 1) + 3*A* b^3*c*d^2*e^m*x*e^(m*log(x) + 5*n*log(x))/(m + 5*n + 1) + 3*B*a^2*b*d^3*e^ m*x*e^(m*log(x) + 5*n*log(x))/(m + 5*n + 1) + 3*A*a*b^2*d^3*e^m*x*e^(m*log (x) + 5*n*log(x))/(m + 5*n + 1) + B*b^3*c^3*e^m*x*e^(m*log(x) + 4*n*log(x) )/(m + 4*n + 1) + 9*B*a*b^2*c^2*d*e^m*x*e^(m*log(x) + 4*n*log(x))/(m + 4*n + 1) + 3*A*b^3*c^2*d*e^m*x*e^(m*log(x) + 4*n*log(x))/(m + 4*n + 1) + 9*B* a^2*b*c*d^2*e^m*x*e^(m*log(x) + 4*n*log(x))/(m + 4*n + 1) + 9*A*a*b^2*c*d^ 2*e^m*x*e^(m*log(x) + 4*n*log(x))/(m + 4*n + 1) + B*a^3*d^3*e^m*x*e^(m*log (x) + 4*n*log(x))/(m + 4*n + 1) + 3*A*a^2*b*d^3*e^m*x*e^(m*log(x) + 4*n*lo g(x))/(m + 4*n + 1) + 3*B*a*b^2*c^3*e^m*x*e^(m*log(x) + 3*n*log(x))/(m + 3 *n + 1) + A*b^3*c^3*e^m*x*e^(m*log(x) + 3*n*log(x))/(m + 3*n + 1) + 9*B*a^ 2*b*c^2*d*e^m*x*e^(m*log(x) + 3*n*log(x))/(m + 3*n + 1) + 9*A*a*b^2*c^2*d* e^m*x*e^(m*log(x) + 3*n*log(x))/(m + 3*n + 1) + 3*B*a^3*c*d^2*e^m*x*e^(m*l og(x) + 3*n*log(x))/(m + 3*n + 1) + 9*A*a^2*b*c*d^2*e^m*x*e^(m*log(x) + 3* n*log(x))/(m + 3*n + 1) + A*a^3*d^3*e^m*x*e^(m*log(x) + 3*n*log(x))/(m + 3 *n + 1) + 3*B*a^2*b*c^3*e^m*x*e^(m*log(x) + 2*n*log(x))/(m + 2*n + 1) +...
Leaf count of result is larger than twice the leaf count of optimal. 143220 vs. \(2 (431) = 862\).
Time = 1.79 (sec) , antiderivative size = 143220, normalized size of antiderivative = 332.30 \[ \int (e x)^m \left (a+b x^n\right )^3 \left (A+B x^n\right ) \left (c+d x^n\right )^3 \, dx=\text {Too large to display} \] Input:
integrate((e*x)^m*(a+b*x^n)^3*(A+B*x^n)*(c+d*x^n)^3,x, algorithm="giac")
Output:
(B*b^3*d^3*m^7*x*x^(7*n)*e^(m*log(e) + m*log(x)) + 21*B*b^3*d^3*m^6*n*x*x^ (7*n)*e^(m*log(e) + m*log(x)) + 175*B*b^3*d^3*m^5*n^2*x*x^(7*n)*e^(m*log(e ) + m*log(x)) + 735*B*b^3*d^3*m^4*n^3*x*x^(7*n)*e^(m*log(e) + m*log(x)) + 1624*B*b^3*d^3*m^3*n^4*x*x^(7*n)*e^(m*log(e) + m*log(x)) + 1764*B*b^3*d^3* m^2*n^5*x*x^(7*n)*e^(m*log(e) + m*log(x)) + 720*B*b^3*d^3*m*n^6*x*x^(7*n)* e^(m*log(e) + m*log(x)) + 3*B*b^3*c*d^2*m^7*x*x^(6*n)*e^(m*log(e) + m*log( x)) + 3*B*a*b^2*d^3*m^7*x*x^(6*n)*e^(m*log(e) + m*log(x)) + A*b^3*d^3*m^7* x*x^(6*n)*e^(m*log(e) + m*log(x)) + B*b^3*d^3*m^7*x*x^(6*n)*e^(m*log(e) + m*log(x)) + 66*B*b^3*c*d^2*m^6*n*x*x^(6*n)*e^(m*log(e) + m*log(x)) + 66*B* a*b^2*d^3*m^6*n*x*x^(6*n)*e^(m*log(e) + m*log(x)) + 22*A*b^3*d^3*m^6*n*x*x ^(6*n)*e^(m*log(e) + m*log(x)) + 21*B*b^3*d^3*m^6*n*x*x^(6*n)*e^(m*log(e) + m*log(x)) + 570*B*b^3*c*d^2*m^5*n^2*x*x^(6*n)*e^(m*log(e) + m*log(x)) + 570*B*a*b^2*d^3*m^5*n^2*x*x^(6*n)*e^(m*log(e) + m*log(x)) + 190*A*b^3*d^3* m^5*n^2*x*x^(6*n)*e^(m*log(e) + m*log(x)) + 175*B*b^3*d^3*m^5*n^2*x*x^(6*n )*e^(m*log(e) + m*log(x)) + 2460*B*b^3*c*d^2*m^4*n^3*x*x^(6*n)*e^(m*log(e) + m*log(x)) + 2460*B*a*b^2*d^3*m^4*n^3*x*x^(6*n)*e^(m*log(e) + m*log(x)) + 820*A*b^3*d^3*m^4*n^3*x*x^(6*n)*e^(m*log(e) + m*log(x)) + 735*B*b^3*d^3* m^4*n^3*x*x^(6*n)*e^(m*log(e) + m*log(x)) + 5547*B*b^3*c*d^2*m^3*n^4*x*x^( 6*n)*e^(m*log(e) + m*log(x)) + 5547*B*a*b^2*d^3*m^3*n^4*x*x^(6*n)*e^(m*log (e) + m*log(x)) + 1849*A*b^3*d^3*m^3*n^4*x*x^(6*n)*e^(m*log(e) + m*log(...
Time = 7.39 (sec) , antiderivative size = 2949, normalized size of antiderivative = 6.84 \[ \int (e x)^m \left (a+b x^n\right )^3 \left (A+B x^n\right ) \left (c+d x^n\right )^3 \, dx=\text {Too large to display} \] Input:
int((e*x)^m*(A + B*x^n)*(a + b*x^n)^3*(c + d*x^n)^3,x)
Output:
(x*x^(3*n)*(e*x)^m*(A*a^3*d^3 + A*b^3*c^3 + 3*B*a*b^2*c^3 + 3*B*a^3*c*d^2 + 9*A*a*b^2*c^2*d + 9*A*a^2*b*c*d^2 + 9*B*a^2*b*c^2*d)*(6*m + 25*n + 125*m *n + 988*m*n^2 + 250*m^2*n + 3657*m*n^3 + 250*m^3*n + 6224*m*n^4 + 125*m^4 *n + 3796*m*n^5 + 25*m^5*n + 15*m^2 + 20*m^3 + 15*m^4 + 6*m^5 + m^6 + 247* n^2 + 1219*n^3 + 3112*n^4 + 3796*n^5 + 1680*n^6 + 1482*m^2*n^2 + 3657*m^2* n^3 + 988*m^3*n^2 + 3112*m^2*n^4 + 1219*m^3*n^3 + 247*m^4*n^2 + 1))/(7*m + 28*n + 168*m*n + 1610*m*n^2 + 420*m^2*n + 7840*m*n^3 + 560*m^3*n + 20307* m*n^4 + 420*m^4*n + 26264*m*n^5 + 168*m^5*n + 13068*m*n^6 + 28*m^6*n + 21* m^2 + 35*m^3 + 35*m^4 + 21*m^5 + 7*m^6 + m^7 + 322*n^2 + 1960*n^3 + 6769*n ^4 + 13132*n^5 + 13068*n^6 + 5040*n^7 + 3220*m^2*n^2 + 11760*m^2*n^3 + 322 0*m^3*n^2 + 20307*m^2*n^4 + 7840*m^3*n^3 + 1610*m^4*n^2 + 13132*m^2*n^5 + 6769*m^3*n^4 + 1960*m^4*n^3 + 322*m^5*n^2 + 1) + (x*x^(4*n)*(e*x)^m*(B*a^3 *d^3 + B*b^3*c^3 + 3*A*a^2*b*d^3 + 3*A*b^3*c^2*d + 9*A*a*b^2*c*d^2 + 9*B*a *b^2*c^2*d + 9*B*a^2*b*c*d^2)*(6*m + 24*n + 120*m*n + 904*m*n^2 + 240*m^2* n + 3168*m*n^3 + 240*m^3*n + 5090*m*n^4 + 120*m^4*n + 2952*m*n^5 + 24*m^5* n + 15*m^2 + 20*m^3 + 15*m^4 + 6*m^5 + m^6 + 226*n^2 + 1056*n^3 + 2545*n^4 + 2952*n^5 + 1260*n^6 + 1356*m^2*n^2 + 3168*m^2*n^3 + 904*m^3*n^2 + 2545* m^2*n^4 + 1056*m^3*n^3 + 226*m^4*n^2 + 1))/(7*m + 28*n + 168*m*n + 1610*m* n^2 + 420*m^2*n + 7840*m*n^3 + 560*m^3*n + 20307*m*n^4 + 420*m^4*n + 26264 *m*n^5 + 168*m^5*n + 13068*m*n^6 + 28*m^6*n + 21*m^2 + 35*m^3 + 35*m^4 ...
Time = 0.30 (sec) , antiderivative size = 12856, normalized size of antiderivative = 29.83 \[ \int (e x)^m \left (a+b x^n\right )^3 \left (A+B x^n\right ) \left (c+d x^n\right )^3 \, dx =\text {Too large to display} \] Input:
int((e*x)^m*(a+b*x^n)^3*(A+B*x^n)*(c+d*x^n)^3,x)
Output:
(x**m*e**m*x*(x**(7*n)*b**4*d**3*m**7 + 21*x**(7*n)*b**4*d**3*m**6*n + 7*x **(7*n)*b**4*d**3*m**6 + 175*x**(7*n)*b**4*d**3*m**5*n**2 + 126*x**(7*n)*b **4*d**3*m**5*n + 21*x**(7*n)*b**4*d**3*m**5 + 735*x**(7*n)*b**4*d**3*m**4 *n**3 + 875*x**(7*n)*b**4*d**3*m**4*n**2 + 315*x**(7*n)*b**4*d**3*m**4*n + 35*x**(7*n)*b**4*d**3*m**4 + 1624*x**(7*n)*b**4*d**3*m**3*n**4 + 2940*x** (7*n)*b**4*d**3*m**3*n**3 + 1750*x**(7*n)*b**4*d**3*m**3*n**2 + 420*x**(7* n)*b**4*d**3*m**3*n + 35*x**(7*n)*b**4*d**3*m**3 + 1764*x**(7*n)*b**4*d**3 *m**2*n**5 + 4872*x**(7*n)*b**4*d**3*m**2*n**4 + 4410*x**(7*n)*b**4*d**3*m **2*n**3 + 1750*x**(7*n)*b**4*d**3*m**2*n**2 + 315*x**(7*n)*b**4*d**3*m**2 *n + 21*x**(7*n)*b**4*d**3*m**2 + 720*x**(7*n)*b**4*d**3*m*n**6 + 3528*x** (7*n)*b**4*d**3*m*n**5 + 4872*x**(7*n)*b**4*d**3*m*n**4 + 2940*x**(7*n)*b* *4*d**3*m*n**3 + 875*x**(7*n)*b**4*d**3*m*n**2 + 126*x**(7*n)*b**4*d**3*m* n + 7*x**(7*n)*b**4*d**3*m + 720*x**(7*n)*b**4*d**3*n**6 + 1764*x**(7*n)*b **4*d**3*n**5 + 1624*x**(7*n)*b**4*d**3*n**4 + 735*x**(7*n)*b**4*d**3*n**3 + 175*x**(7*n)*b**4*d**3*n**2 + 21*x**(7*n)*b**4*d**3*n + x**(7*n)*b**4*d **3 + 4*x**(6*n)*a*b**3*d**3*m**7 + 88*x**(6*n)*a*b**3*d**3*m**6*n + 28*x* *(6*n)*a*b**3*d**3*m**6 + 760*x**(6*n)*a*b**3*d**3*m**5*n**2 + 528*x**(6*n )*a*b**3*d**3*m**5*n + 84*x**(6*n)*a*b**3*d**3*m**5 + 3280*x**(6*n)*a*b**3 *d**3*m**4*n**3 + 3800*x**(6*n)*a*b**3*d**3*m**4*n**2 + 1320*x**(6*n)*a*b* *3*d**3*m**4*n + 140*x**(6*n)*a*b**3*d**3*m**4 + 7396*x**(6*n)*a*b**3*d...