\(\int \frac {(e x)^m (A+B x^n)}{(a+b x^n)^2 (c+d x^n)^3} \, dx\) [58]

Optimal result

Integrand size = 31, antiderivative size = 482 \[ \int \frac {(e x)^m \left (A+B x^n\right )}{\left (a+b x^n\right )^2 \left (c+d x^n\right )^3} \, dx=\frac {d (2 A b c-3 a B c+a A d) (e x)^{1+m}}{2 a c (b c-a d)^2 e n \left (c+d x^n\right )^2}+\frac {(A b-a B) (e x)^{1+m}}{a (b c-a d) e n \left (a+b x^n\right ) \left (c+d x^n\right )^2}-\frac {d \left (a^2 d (B c (1+m)-A d (1+m-2 n))-a b c (B c-A d) (1+m-6 n)-2 A b^2 c^2 n\right ) (e x)^{1+m}}{2 a c^2 (b c-a d)^3 e n^2 \left (c+d x^n\right )}+\frac {b^2 (a B (b c (1+m)-a d (1+m-3 n))+A b (a d (1+m-4 n)-b c (1+m-n))) (e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{n},\frac {1+m+n}{n},-\frac {b x^n}{a}\right )}{a^2 (b c-a d)^4 e (1+m) n}+\frac {d \left (b^2 c^2 (A d (1+m-4 n)-B c (1+m-2 n)) (1+m-3 n)-a^2 d^2 (B c (1+m)-A d (1+m-2 n)) (1+m-n)+2 a b c d \left (B c (1+m) (1+m-3 n)-A d \left (1+m^2+m (2-5 n)-5 n+4 n^2\right )\right )\right ) (e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{n},\frac {1+m+n}{n},-\frac {d x^n}{c}\right )}{2 c^3 (b c-a d)^4 e (1+m) n^2} \] Output:

1/2*d*(A*a*d+2*A*b*c-3*B*a*c)*(e*x)^(1+m)/a/c/(-a*d+b*c)^2/e/n/(c+d*x^n)^2 
+(A*b-B*a)*(e*x)^(1+m)/a/(-a*d+b*c)/e/n/(a+b*x^n)/(c+d*x^n)^2-1/2*d*(a^2*d 
*(B*c*(1+m)-A*d*(1+m-2*n))-a*b*c*(-A*d+B*c)*(1+m-6*n)-2*A*b^2*c^2*n)*(e*x) 
^(1+m)/a/c^2/(-a*d+b*c)^3/e/n^2/(c+d*x^n)+b^2*(a*B*(b*c*(1+m)-a*d*(1+m-3*n 
))+A*b*(a*d*(1+m-4*n)-b*c*(1+m-n)))*(e*x)^(1+m)*hypergeom([1, (1+m)/n],[(1 
+m+n)/n],-b*x^n/a)/a^2/(-a*d+b*c)^4/e/(1+m)/n+1/2*d*(b^2*c^2*(A*d*(1+m-4*n 
)-B*c*(1+m-2*n))*(1+m-3*n)-a^2*d^2*(B*c*(1+m)-A*d*(1+m-2*n))*(1+m-n)+2*a*b 
*c*d*(B*c*(1+m)*(1+m-3*n)-A*d*(1+m^2+m*(2-5*n)-5*n+4*n^2)))*(e*x)^(1+m)*hy 
pergeom([1, (1+m)/n],[(1+m+n)/n],-d*x^n/c)/c^3/(-a*d+b*c)^4/e/(1+m)/n^2
 

Mathematica [A] (verified)

Time = 0.77 (sec) , antiderivative size = 271, normalized size of antiderivative = 0.56 \[ \int \frac {(e x)^m \left (A+B x^n\right )}{\left (a+b x^n\right )^2 \left (c+d x^n\right )^3} \, dx=\frac {x (e x)^m \left (\frac {b^2 (b B c-3 A b d+2 a B d) \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{n},\frac {1+m+n}{n},-\frac {b x^n}{a}\right )}{a}-\frac {b d (b B c-3 A b d+2 a B d) \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{n},\frac {1+m+n}{n},-\frac {d x^n}{c}\right )}{c}+\frac {b^2 (-A b+a B) (-b c+a d) \operatorname {Hypergeometric2F1}\left (2,\frac {1+m}{n},\frac {1+m+n}{n},-\frac {b x^n}{a}\right )}{a^2}-\frac {d (b c-a d) (b B c-2 A b d+a B d) \operatorname {Hypergeometric2F1}\left (2,\frac {1+m}{n},\frac {1+m+n}{n},-\frac {d x^n}{c}\right )}{c^2}+\frac {d (b c-a d)^2 (-B c+A d) \operatorname {Hypergeometric2F1}\left (3,\frac {1+m}{n},\frac {1+m+n}{n},-\frac {d x^n}{c}\right )}{c^3}\right )}{(b c-a d)^4 (1+m)} \] Input:

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

Output:

(x*(e*x)^m*((b^2*(b*B*c - 3*A*b*d + 2*a*B*d)*Hypergeometric2F1[1, (1 + m)/ 
n, (1 + m + n)/n, -((b*x^n)/a)])/a - (b*d*(b*B*c - 3*A*b*d + 2*a*B*d)*Hype 
rgeometric2F1[1, (1 + m)/n, (1 + m + n)/n, -((d*x^n)/c)])/c + (b^2*(-(A*b) 
 + a*B)*(-(b*c) + a*d)*Hypergeometric2F1[2, (1 + m)/n, (1 + m + n)/n, -((b 
*x^n)/a)])/a^2 - (d*(b*c - a*d)*(b*B*c - 2*A*b*d + a*B*d)*Hypergeometric2F 
1[2, (1 + m)/n, (1 + m + n)/n, -((d*x^n)/c)])/c^2 + (d*(b*c - a*d)^2*(-(B* 
c) + A*d)*Hypergeometric2F1[3, (1 + m)/n, (1 + m + n)/n, -((d*x^n)/c)])/c^ 
3))/((b*c - a*d)^4*(1 + m))
 

Rubi [A] (verified)

Time = 2.74 (sec) , antiderivative size = 525, normalized size of antiderivative = 1.09, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {1065, 25, 1065, 1065, 1067, 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.

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

\(\Big \downarrow \) 1065

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 1065

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

\(\Big \downarrow \) 1065

\(\displaystyle \frac {\frac {\frac {\int \frac {(e x)^m \left (b d (m-n+1) n \left (d (B c (m+1)-A d (m-2 n+1)) a^2-b c (B c-A d) (m-6 n+1) a-2 A b^2 c^2 n\right ) x^n+n \left (a d (m+1) \left (d (B c (m+1)-A d (m-2 n+1)) a^2-b c (B c-A d) (m-6 n+1) a-2 A b^2 c^2 n\right )+(b c-a d) n \left (a B c (2 b c+a d) (m+1)-A \left (2 b^2 (m-n+1) c^2+4 a b d n c+a^2 d^2 (m-2 n+1)\right )\right )\right )\right )}{\left (b x^n+a\right ) \left (d x^n+c\right )}dx}{c n (b c-a d)}-\frac {d (e x)^{m+1} \left (a^2 d (B c (m+1)-A d (m-2 n+1))-a b c (m-6 n+1) (B c-A d)-2 A b^2 c^2 n\right )}{c e (b c-a d) \left (c+d x^n\right )}}{2 c n (b c-a d)}+\frac {d (e x)^{m+1} (a A d-3 a B c+2 A b c)}{2 c e (b c-a d) \left (c+d x^n\right )^2}}{a n (b c-a d)}+\frac {(e x)^{m+1} (A b-a B)}{a e n (b c-a d) \left (a+b x^n\right ) \left (c+d x^n\right )^2}\)

\(\Big \downarrow \) 1067

\(\displaystyle \frac {\frac {\frac {\int \left (\frac {2 b^2 c^2 (a B (b c (m+1)-a d (m-3 n+1))+A b (a d (m-4 n+1)-b c (m-n+1))) n^2 (e x)^m}{(b c-a d) \left (b x^n+a\right )}+\frac {a d n \left (b^2 (A d (m-4 n+1)-B c (m-2 n+1)) (m-3 n+1) c^2+2 a b d \left (B c (m+1) (m-3 n+1)-A d \left (m^2+(2-5 n) m+4 n^2-5 n+1\right )\right ) c-a^2 d^2 (B c (m+1)-A d (m-2 n+1)) (m-n+1)\right ) (e x)^m}{(b c-a d) \left (d x^n+c\right )}\right )dx}{c n (b c-a d)}-\frac {d (e x)^{m+1} \left (a^2 d (B c (m+1)-A d (m-2 n+1))-a b c (m-6 n+1) (B c-A d)-2 A b^2 c^2 n\right )}{c e (b c-a d) \left (c+d x^n\right )}}{2 c n (b c-a d)}+\frac {d (e x)^{m+1} (a A d-3 a B c+2 A b c)}{2 c e (b c-a d) \left (c+d x^n\right )^2}}{a n (b c-a d)}+\frac {(e x)^{m+1} (A b-a B)}{a e n (b c-a d) \left (a+b x^n\right ) \left (c+d x^n\right )^2}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\frac {\frac {\frac {a d n (e x)^{m+1} \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{n},\frac {m+n+1}{n},-\frac {d x^n}{c}\right ) \left (-a^2 d^2 (m-n+1) (B c (m+1)-A d (m-2 n+1))+2 a b c d \left (B c (m+1) (m-3 n+1)-A d \left (m^2+m (2-5 n)+4 n^2-5 n+1\right )\right )+b^2 c^2 (m-3 n+1) (A d (m-4 n+1)-B c (m-2 n+1))\right )}{c e (m+1) (b c-a d)}+\frac {2 b^2 c^2 n^2 (e x)^{m+1} \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{n},\frac {m+n+1}{n},-\frac {b x^n}{a}\right ) (A b (a d (m-4 n+1)-b c (m-n+1))+a B (b c (m+1)-a d (m-3 n+1)))}{a e (m+1) (b c-a d)}}{c n (b c-a d)}-\frac {d (e x)^{m+1} \left (a^2 d (B c (m+1)-A d (m-2 n+1))-a b c (m-6 n+1) (B c-A d)-2 A b^2 c^2 n\right )}{c e (b c-a d) \left (c+d x^n\right )}}{2 c n (b c-a d)}+\frac {d (e x)^{m+1} (a A d-3 a B c+2 A b c)}{2 c e (b c-a d) \left (c+d x^n\right )^2}}{a n (b c-a d)}+\frac {(e x)^{m+1} (A b-a B)}{a e n (b c-a d) \left (a+b x^n\right ) \left (c+d x^n\right )^2}\)

Input:

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

Output:

((A*b - a*B)*(e*x)^(1 + m))/(a*(b*c - a*d)*e*n*(a + b*x^n)*(c + d*x^n)^2) 
+ ((d*(2*A*b*c - 3*a*B*c + a*A*d)*(e*x)^(1 + m))/(2*c*(b*c - a*d)*e*(c + d 
*x^n)^2) + (-((d*(a^2*d*(B*c*(1 + m) - A*d*(1 + m - 2*n)) - a*b*c*(B*c - A 
*d)*(1 + m - 6*n) - 2*A*b^2*c^2*n)*(e*x)^(1 + m))/(c*(b*c - a*d)*e*(c + d* 
x^n))) + ((2*b^2*c^2*(a*B*(b*c*(1 + m) - a*d*(1 + m - 3*n)) + A*b*(a*d*(1 
+ m - 4*n) - b*c*(1 + m - n)))*n^2*(e*x)^(1 + m)*Hypergeometric2F1[1, (1 + 
 m)/n, (1 + m + n)/n, -((b*x^n)/a)])/(a*(b*c - a*d)*e*(1 + m)) + (a*d*n*(b 
^2*c^2*(A*d*(1 + m - 4*n) - B*c*(1 + m - 2*n))*(1 + m - 3*n) - a^2*d^2*(B* 
c*(1 + m) - A*d*(1 + m - 2*n))*(1 + m - n) + 2*a*b*c*d*(B*c*(1 + m)*(1 + m 
 - 3*n) - A*d*(1 + m^2 + m*(2 - 5*n) - 5*n + 4*n^2)))*(e*x)^(1 + m)*Hyperg 
eometric2F1[1, (1 + m)/n, (1 + m + n)/n, -((d*x^n)/c)])/(c*(b*c - a*d)*e*( 
1 + m)))/(c*(b*c - a*d)*n))/(2*c*(b*c - a*d)*n))/(a*(b*c - a*d)*n)
 

Defintions of rubi rules used

rule 25

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

rule 1065

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 + 1)/(a*g*n*(b*c - a*d)*(p + 1))) 
, x] + Simp[1/(a*n*(b*c - a*d)*(p + 1))   Int[(g*x)^m*(a + b*x^n)^(p + 1)*( 
c + d*x^n)^q*Simp[c*(b*e - a*f)*(m + 1) + e*n*(b*c - a*d)*(p + 1) + d*(b*e 
- a*f)*(m + n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, 
g, m, n, q}, x] && LtQ[p, -1]
 

rule 1067

Int[(((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((e_) + (f_.)*(x_)^(n 
_)))/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Int[ExpandIntegrand[(g*x)^m*(a 
+ b*x^n)^p*((e + f*x^n)/(c + d*x^n)), x], x] /; FreeQ[{a, b, c, d, e, f, g, 
 m, n, p}, x]
 

rule 2009

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

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral((B*x^n + A)*(e*x)^m/(b^2*d^3*x^(5*n) + a^2*c^3 + (3*b^2*c*d^2 + 2 
*a*b*d^3)*x^(4*n) + (3*b^2*c^2*d + 6*a*b*c*d^2 + a^2*d^3)*x^(3*n) + (b^2*c 
^3 + 6*a*b*c^2*d + 3*a^2*c*d^2)*x^(2*n) + (2*a*b*c^3 + 3*a^2*c^2*d)*x^n), 
x)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

(((m^2 - m*(7*n - 2) + 12*n^2 - 7*n + 1)*b^2*c^2*d^2*e^m - 2*(m^2 - m*(5*n 
 - 2) + 4*n^2 - 5*n + 1)*a*b*c*d^3*e^m + (m^2 - m*(3*n - 2) + 2*n^2 - 3*n 
+ 1)*a^2*d^4*e^m)*A - ((m^2 - m*(5*n - 2) + 6*n^2 - 5*n + 1)*b^2*c^3*d*e^m 
 - 2*(m^2 - m*(3*n - 2) - 3*n + 1)*a*b*c^2*d^2*e^m + (m^2 - m*(n - 2) - n 
+ 1)*a^2*c*d^3*e^m)*B)*integrate(1/2*x^m/(b^4*c^7*n^2 - 4*a*b^3*c^6*d*n^2 
+ 6*a^2*b^2*c^5*d^2*n^2 - 4*a^3*b*c^4*d^3*n^2 + a^4*c^3*d^4*n^2 + (b^4*c^6 
*d*n^2 - 4*a*b^3*c^5*d^2*n^2 + 6*a^2*b^2*c^4*d^3*n^2 - 4*a^3*b*c^3*d^4*n^2 
 + a^4*c^2*d^5*n^2)*x^n), x) - ((b^4*c*e^m*(m - n + 1) - a*b^3*d*e^m*(m - 
4*n + 1))*A + (a^2*b^2*d*e^m*(m - 3*n + 1) - a*b^3*c*e^m*(m + 1))*B)*integ 
rate(x^m/(a^2*b^4*c^4*n - 4*a^3*b^3*c^3*d*n + 6*a^4*b^2*c^2*d^2*n - 4*a^5* 
b*c*d^3*n + a^6*d^4*n + (a*b^5*c^4*n - 4*a^2*b^4*c^3*d*n + 6*a^3*b^3*c^2*d 
^2*n - 4*a^4*b^2*c*d^3*n + a^5*b*d^4*n)*x^n), x) + 1/2*(((a^3*c*d^3*e^m*(m 
 - 3*n + 1) - a^2*b*c^2*d^2*e^m*(m - 7*n + 1) + 2*b^3*c^4*e^m*n)*A - (a^3* 
c^2*d^2*e^m*(m - n + 1) - a^2*b*c^3*d*e^m*(m - 5*n + 1) + 2*a*b^2*c^4*e^m* 
n)*B)*x*x^m + ((a^2*b*d^4*e^m*(m - 2*n + 1) - a*b^2*c*d^3*e^m*(m - 6*n + 1 
) + 2*b^3*c^2*d^2*e^m*n)*A + (a*b^2*c^2*d^2*e^m*(m - 6*n + 1) - a^2*b*c*d^ 
3*e^m*(m + 1))*B)*x*e^(m*log(x) + 2*n*log(x)) + ((a^3*d^4*e^m*(m - 2*n + 1 
) - a*b^2*c^2*d^2*e^m*(m - 7*n + 1) + 4*b^3*c^3*d*e^m*n + 3*a^2*b*c*d^3*e^ 
m*n)*A + (a*b^2*c^3*d*e^m*(m - 9*n + 1) - a^3*c*d^3*e^m*(m + 1) - 3*a^2*b* 
c^2*d^2*e^m*n)*B)*x*e^(m*log(x) + n*log(x)))/(a^2*b^3*c^7*n^2 - 3*a^3*b...
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {(e x)^m \left (A+B x^n\right )}{\left (a+b x^n\right )^2 \left (c+d x^n\right )^3} \, dx=e^{m} \left (\int \frac {x^{m}}{x^{4 n} b \,d^{3}+x^{3 n} a \,d^{3}+3 x^{3 n} b c \,d^{2}+3 x^{2 n} a c \,d^{2}+3 x^{2 n} b \,c^{2} d +3 x^{n} a \,c^{2} d +x^{n} b \,c^{3}+a \,c^{3}}d x \right ) \] Input:

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

Output:

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