\(\int \frac {(e x)^m (a+b x^n)^3 (A+B x^n)}{(c+d x^n)^2} \, dx\) [47]

Optimal result

Integrand size = 31, antiderivative size = 387 \[ \int \frac {(e x)^m \left (a+b x^n\right )^3 \left (A+B x^n\right )}{\left (c+d x^n\right )^2} \, dx=-\frac {b \left (3 a^2 d^2 (A d (1+m)-B c (1+m+n))-3 a b c d (A d (1+m+n)-B c (1+m+2 n))+b^2 c^2 (A d (1+m+2 n)-B c (1+m+3 n))\right ) (e x)^{1+m}}{c d^4 e (1+m) n}-\frac {b^2 (3 a d (A d (1+m+n)-B c (1+m+2 n))-b c (A d (1+m+2 n)-B c (1+m+3 n))) x^n (e x)^{1+m}}{c d^3 e n (1+m+n)}-\frac {b^3 \left (A-\frac {B c (1+m+3 n)}{d (1+m+2 n)}\right ) x^{2 n} (e x)^{1+m}}{c d e n}-\frac {(B c-A d) (e x)^{1+m} \left (a+b x^n\right )^3}{c d e n \left (c+d x^n\right )}+\frac {(b c-a d)^2 (a d (B c (1+m)-A d (1+m-n))+b c (A d (1+m+2 n)-B c (1+m+3 n))) (e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{n},\frac {1+m+n}{n},-\frac {d x^n}{c}\right )}{c^2 d^4 e (1+m) n} \] Output:

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

Mathematica [A] (verified)

Time = 0.97 (sec) , antiderivative size = 220, normalized size of antiderivative = 0.57 \[ \int \frac {(e x)^m \left (a+b x^n\right )^3 \left (A+B x^n\right )}{\left (c+d x^n\right )^2} \, dx=\frac {x (e x)^m \left (\frac {b \left (3 a^2 B d^2+b^2 c (3 B c-2 A d)+3 a b d (-2 B c+A d)\right )}{1+m}+\frac {b^2 d (-2 b B c+A b d+3 a B d) x^n}{1+m+n}+\frac {b^3 B d^2 x^{2 n}}{1+m+2 n}-\frac {(b c-a d)^2 (4 b B c-3 A b d-a B d) \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{n},\frac {1+m+n}{n},-\frac {d x^n}{c}\right )}{c (1+m)}+\frac {(b c-a d)^3 (B c-A d) \operatorname {Hypergeometric2F1}\left (2,\frac {1+m}{n},\frac {1+m+n}{n},-\frac {d x^n}{c}\right )}{c^2 (1+m)}\right )}{d^4} \] Input:

Integrate[((e*x)^m*(a + b*x^n)^3*(A + B*x^n))/(c + d*x^n)^2,x]
 

Output:

(x*(e*x)^m*((b*(3*a^2*B*d^2 + b^2*c*(3*B*c - 2*A*d) + 3*a*b*d*(-2*B*c + A* 
d)))/(1 + m) + (b^2*d*(-2*b*B*c + A*b*d + 3*a*B*d)*x^n)/(1 + m + n) + (b^3 
*B*d^2*x^(2*n))/(1 + m + 2*n) - ((b*c - a*d)^2*(4*b*B*c - 3*A*b*d - a*B*d) 
*Hypergeometric2F1[1, (1 + m)/n, (1 + m + n)/n, -((d*x^n)/c)])/(c*(1 + m)) 
 + ((b*c - a*d)^3*(B*c - A*d)*Hypergeometric2F1[2, (1 + m)/n, (1 + m + n)/ 
n, -((d*x^n)/c)])/(c^2*(1 + m))))/d^4
 

Rubi [A] (verified)

Time = 1.64 (sec) , antiderivative size = 368, normalized size of antiderivative = 0.95, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {1064, 25, 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.

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

\(\Big \downarrow \) 1064

\(\displaystyle -\frac {\int -\frac {(e x)^m \left (b x^n+a\right )^2 \left (a (B c (m+1)-A d (m-n+1))-b (A d (m+2 n+1)-B c (m+3 n+1)) x^n\right )}{d x^n+c}dx}{c d n}-\frac {(e x)^{m+1} \left (a+b x^n\right )^3 (B c-A d)}{c d e n \left (c+d x^n\right )}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\int \frac {(e x)^m \left (b x^n+a\right )^2 \left (a (B c (m+1)-A d (m-n+1))-b (A d (m+2 n+1)-B c (m+3 n+1)) x^n\right )}{d x^n+c}dx}{c d n}-\frac {(e x)^{m+1} \left (a+b x^n\right )^3 (B c-A d)}{c d e n \left (c+d x^n\right )}\)

\(\Big \downarrow \) 1040

\(\displaystyle \frac {\int \left (\frac {b^2 (b c (A d (m+2 n+1)-B c (m+3 n+1))-3 a d (A d (m+n+1)-B c (m+2 n+1))) x^n (e x)^m}{d^2}+\frac {b^3 (B c (m+3 n+1)-A d (m+2 n+1)) x^{2 n} (e x)^m}{d}+\frac {b \left (-b^2 (A d (m+2 n+1)-B c (m+3 n+1)) c^2+3 a b d (A d (m+n+1)-B c (m+2 n+1)) c-3 a^2 d^2 (A d (m+1)-B c (m+n+1))\right ) (e x)^m}{d^3}+\frac {(b c-a d)^2 (a d (B c (m+1)-A d (m-n+1))+b c (A d (m+2 n+1)-B c (m+3 n+1))) (e x)^m}{d^3 \left (d x^n+c\right )}\right )dx}{c d n}-\frac {(e x)^{m+1} \left (a+b x^n\right )^3 (B c-A d)}{c d e n \left (c+d x^n\right )}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {-\frac {b (e x)^{m+1} \left (3 a^2 d^2 (A d (m+1)-B c (m+n+1))-3 a b c d (A d (m+n+1)-B c (m+2 n+1))+b^2 c^2 (A d (m+2 n+1)-B c (m+3 n+1))\right )}{d^3 e (m+1)}-\frac {b^2 x^{n+1} (e x)^m (3 a d (A d (m+n+1)-B c (m+2 n+1))-b c (A d (m+2 n+1)-B c (m+3 n+1)))}{d^2 (m+n+1)}+\frac {(e x)^{m+1} (b c-a d)^2 \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{n},\frac {m+n+1}{n},-\frac {d x^n}{c}\right ) (a d (B c (m+1)-A d (m-n+1))+b c (A d (m+2 n+1)-B c (m+3 n+1)))}{c d^3 e (m+1)}-b^3 x^{2 n+1} (e x)^m \left (A-\frac {B c (m+3 n+1)}{d (m+2 n+1)}\right )}{c d n}-\frac {(e x)^{m+1} \left (a+b x^n\right )^3 (B c-A d)}{c d e n \left (c+d x^n\right )}\)

Input:

Int[((e*x)^m*(a + b*x^n)^3*(A + B*x^n))/(c + d*x^n)^2,x]
 

Output:

-(((B*c - A*d)*(e*x)^(1 + m)*(a + b*x^n)^3)/(c*d*e*n*(c + d*x^n))) + (-((b 
^2*(3*a*d*(A*d*(1 + m + n) - B*c*(1 + m + 2*n)) - b*c*(A*d*(1 + m + 2*n) - 
 B*c*(1 + m + 3*n)))*x^(1 + n)*(e*x)^m)/(d^2*(1 + m + n))) - b^3*(A - (B*c 
*(1 + m + 3*n))/(d*(1 + m + 2*n)))*x^(1 + 2*n)*(e*x)^m - (b*(3*a^2*d^2*(A* 
d*(1 + m) - B*c*(1 + m + n)) - 3*a*b*c*d*(A*d*(1 + m + n) - B*c*(1 + m + 2 
*n)) + b^2*c^2*(A*d*(1 + m + 2*n) - B*c*(1 + m + 3*n)))*(e*x)^(1 + m))/(d^ 
3*e*(1 + m)) + ((b*c - a*d)^2*(a*d*(B*c*(1 + m) - A*d*(1 + m - n)) + b*c*( 
A*d*(1 + m + 2*n) - B*c*(1 + m + 3*n)))*(e*x)^(1 + m)*Hypergeometric2F1[1, 
 (1 + m)/n, (1 + m + n)/n, -((d*x^n)/c)])/(c*d^3*e*(1 + m)))/(c*d*n)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 1064

Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_ 
))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> Simp[(-(b*e - a*f))*(g*x)^( 
m + 1)*(a + b*x^n)^(p + 1)*((c + d*x^n)^q/(a*b*g*n*(p + 1))), x] + Simp[1/( 
a*b*n*(p + 1))   Int[(g*x)^m*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q - 1)*Simp[c 
*(b*e*n*(p + 1) + (b*e - a*f)*(m + 1)) + d*(b*e*n*(p + 1) + (b*e - a*f)*(m 
+ n*q + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n}, x] && LtQ 
[p, -1] && GtQ[q, 0] &&  !(EqQ[q, 1] && SimplerQ[b*c - a*d, b*e - a*f])
 

rule 2009

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

Maple [F]

\[\int \frac {\left (e x \right )^{m} \left (a +b \,x^{n}\right )^{3} \left (A +B \,x^{n}\right )}{\left (c +d \,x^{n}\right )^{2}}d x\]

Input:

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

Output:

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

Fricas [F]

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

integrate((e*x)^m*(a+b*x^n)^3*(A+B*x^n)/(c+d*x^n)^2,x, algorithm="fricas")
 

Output:

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

Sympy [F(-2)]

Exception generated. \[ \int \frac {(e x)^m \left (a+b x^n\right )^3 \left (A+B x^n\right )}{\left (c+d x^n\right )^2} \, dx=\text {Exception raised: HeuristicGCDFailed} \] Input:

integrate((e*x)**m*(a+b*x**n)**3*(A+B*x**n)/(c+d*x**n)**2,x)
 

Output:

Exception raised: HeuristicGCDFailed >> no luck
 

Maxima [F]

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

integrate((e*x)^m*(a+b*x^n)^3*(A+B*x^n)/(c+d*x^n)^2,x, algorithm="maxima")
 

Output:

((b^3*c^3*d*e^m*(m + 2*n + 1) - 3*a*b^2*c^2*d^2*e^m*(m + n + 1) - a^3*d^4* 
e^m*(m - n + 1) + 3*a^2*b*c*d^3*e^m*(m + 1))*A - (b^3*c^4*e^m*(m + 3*n + 1 
) - 3*a*b^2*c^3*d*e^m*(m + 2*n + 1) + 3*a^2*b*c^2*d^2*e^m*(m + n + 1) - a^ 
3*c*d^3*e^m*(m + 1))*B)*integrate(x^m/(c*d^5*n*x^n + c^2*d^4*n), x) + ((m^ 
2*n + (n^2 + 2*n)*m + n^2 + n)*B*b^3*c*d^3*e^m*x*e^(m*log(x) + 3*n*log(x)) 
 - (((m^3 + m^2*(5*n + 3) + 4*n^3 + (8*n^2 + 10*n + 3)*m + 8*n^2 + 5*n + 1 
)*b^3*c^3*d*e^m - 3*(m^3 + m^2*(4*n + 3) + 2*n^3 + (5*n^2 + 8*n + 3)*m + 5 
*n^2 + 4*n + 1)*a*b^2*c^2*d^2*e^m + 3*(m^3 + 3*m^2*(n + 1) + (2*n^2 + 6*n 
+ 3)*m + 2*n^2 + 3*n + 1)*a^2*b*c*d^3*e^m - (m^3 + 3*m^2*(n + 1) + (2*n^2 
+ 6*n + 3)*m + 2*n^2 + 3*n + 1)*a^3*d^4*e^m)*A - ((m^3 + 3*m^2*(2*n + 1) + 
 6*n^3 + (11*n^2 + 12*n + 3)*m + 11*n^2 + 6*n + 1)*b^3*c^4*e^m - 3*(m^3 + 
m^2*(5*n + 3) + 4*n^3 + (8*n^2 + 10*n + 3)*m + 8*n^2 + 5*n + 1)*a*b^2*c^3* 
d*e^m + 3*(m^3 + m^2*(4*n + 3) + 2*n^3 + (5*n^2 + 8*n + 3)*m + 5*n^2 + 4*n 
 + 1)*a^2*b*c^2*d^2*e^m - (m^3 + 3*m^2*(n + 1) + (2*n^2 + 6*n + 3)*m + 2*n 
^2 + 3*n + 1)*a^3*c*d^3*e^m)*B)*x*x^m + ((m^2*n + 2*(n^2 + n)*m + 2*n^2 + 
n)*A*b^3*c*d^3*e^m - ((m^2*n + (3*n^2 + 2*n)*m + 3*n^2 + n)*b^3*c^2*d^2*e^ 
m - 3*(m^2*n + 2*(n^2 + n)*m + 2*n^2 + n)*a*b^2*c*d^3*e^m)*B)*x*e^(m*log(x 
) + 2*n*log(x)) - (((m^2*n + 4*n^3 + 2*(2*n^2 + n)*m + 4*n^2 + n)*b^3*c^2* 
d^2*e^m - 3*(m^2*n + 2*n^3 + (3*n^2 + 2*n)*m + 3*n^2 + n)*a*b^2*c*d^3*e^m) 
*A - ((m^2*n + 6*n^3 + (5*n^2 + 2*n)*m + 5*n^2 + n)*b^3*c^3*d*e^m - 3*(...
 

Giac [F]

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

integrate((e*x)^m*(a+b*x^n)^3*(A+B*x^n)/(c+d*x^n)^2,x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {(e x)^m \left (a+b x^n\right )^3 \left (A+B x^n\right )}{\left (c+d x^n\right )^2} \, dx=\text {too large to display} \] Input:

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

Output:

(e**m*(x**(m + 3*n)*b**4*c**2*d*m**2*x + x**(m + 3*n)*b**4*c**2*d*m*n*x + 
2*x**(m + 3*n)*b**4*c**2*d*m*x + x**(m + 3*n)*b**4*c**2*d*n*x + x**(m + 3* 
n)*b**4*c**2*d*x + 4*x**(m + 2*n)*a*b**3*c**2*d*m**2*x + 8*x**(m + 2*n)*a* 
b**3*c**2*d*m*n*x + 8*x**(m + 2*n)*a*b**3*c**2*d*m*x + 8*x**(m + 2*n)*a*b* 
*3*c**2*d*n*x + 4*x**(m + 2*n)*a*b**3*c**2*d*x - x**(m + 2*n)*b**4*c**3*m* 
*2*x - 3*x**(m + 2*n)*b**4*c**3*m*n*x - 2*x**(m + 2*n)*b**4*c**3*m*x - 3*x 
**(m + 2*n)*b**4*c**3*n*x - x**(m + 2*n)*b**4*c**3*x - x**(m + n)*a**4*d** 
3*m**2*x - x**(m + n)*a**4*d**3*m*n*x - 2*x**(m + n)*a**4*d**3*m*x + 2*x** 
(m + n)*a**4*d**3*n**2*x - x**(m + n)*a**4*d**3*n*x - x**(m + n)*a**4*d**3 
*x + 4*x**(m + n)*a**3*b*c*d**2*m**2*x + 8*x**(m + n)*a**3*b*c*d**2*m*n*x 
+ 8*x**(m + n)*a**3*b*c*d**2*m*x + 8*x**(m + n)*a**3*b*c*d**2*n*x + 4*x**( 
m + n)*a**3*b*c*d**2*x + x**m*a**4*c*d**2*m**2*x + 3*x**m*a**4*c*d**2*m*n* 
x + 2*x**m*a**4*c*d**2*m*x + 2*x**m*a**4*c*d**2*n**2*x + 3*x**m*a**4*c*d** 
2*n*x + x**m*a**4*c*d**2*x + x**n*int(x**(m + 2*n)/(x**(2*n)*d**2*m + x**( 
2*n)*d**2*n + x**(2*n)*d**2 + 2*x**n*c*d*m + 2*x**n*c*d*n + 2*x**n*c*d + c 
**2*m + c**2*n + c**2),x)*a**4*d**5*m**4 + 2*x**n*int(x**(m + 2*n)/(x**(2* 
n)*d**2*m + x**(2*n)*d**2*n + x**(2*n)*d**2 + 2*x**n*c*d*m + 2*x**n*c*d*n 
+ 2*x**n*c*d + c**2*m + c**2*n + c**2),x)*a**4*d**5*m**3*n + 4*x**n*int(x* 
*(m + 2*n)/(x**(2*n)*d**2*m + x**(2*n)*d**2*n + x**(2*n)*d**2 + 2*x**n*c*d 
*m + 2*x**n*c*d*n + 2*x**n*c*d + c**2*m + c**2*n + c**2),x)*a**4*d**5*m...