\(\int \frac {(e x)^m (A+B x^n)}{(a+b x^n) (c+d x^n)^3} \, dx\) [57]

Optimal result

Integrand size = 31, antiderivative size = 366 \[ \int \frac {(e x)^m \left (A+B x^n\right )}{\left (a+b x^n\right ) \left (c+d x^n\right )^3} \, dx=\frac {(B c-A d) (e x)^{1+m}}{2 c (b c-a d) e n \left (c+d x^n\right )^2}+\frac {(b c (A d (1+m-4 n)-B c (1+m-2 n))+a d (B c (1+m)-A d (1+m-2 n))) (e x)^{1+m}}{2 c^2 (b c-a d)^2 e n^2 \left (c+d x^n\right )}+\frac {b^2 (A b-a B) (e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{n},\frac {1+m+n}{n},-\frac {b x^n}{a}\right )}{a (b c-a d)^3 e (1+m)}-\frac {\left (b^2 c^2 (A d (1+m-3 n)-B c (1+m-n)) (1+m-2 n)-a^2 d^2 (B c (1+m)-A d (1+m-2 n)) (1+m-n)+2 a b c d \left (B c (1+m) (1+m-2 n)-A d \left (1+m^2+m (2-4 n)-4 n+3 n^2\right )\right )\right ) (e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{n},\frac {1+m+n}{n},-\frac {d x^n}{c}\right )}{2 c^3 (b c-a d)^3 e (1+m) n^2} \] Output:

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

Mathematica [A] (verified)

Time = 0.41 (sec) , antiderivative size = 201, normalized size of antiderivative = 0.55 \[ \int \frac {(e x)^m \left (A+B x^n\right )}{\left (a+b x^n\right ) \left (c+d x^n\right )^3} \, dx=\frac {x (e x)^m \left (\frac {b^2 (A b-a B) \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{n},\frac {1+m+n}{n},-\frac {b x^n}{a}\right )}{a}-\frac {b (A b-a B) d \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{n},\frac {1+m+n}{n},-\frac {d x^n}{c}\right )}{c}-\frac {(A b-a B) d (b c-a d) \operatorname {Hypergeometric2F1}\left (2,\frac {1+m}{n},\frac {1+m+n}{n},-\frac {d x^n}{c}\right )}{c^2}+\frac {(b c-a d)^2 (B c-A d) \operatorname {Hypergeometric2F1}\left (3,\frac {1+m}{n},\frac {1+m+n}{n},-\frac {d x^n}{c}\right )}{c^3}\right )}{(b c-a d)^3 (1+m)} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 1.93 (sec) , antiderivative size = 404, normalized size of antiderivative = 1.10, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {1065, 25, 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.

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

\(\Big \downarrow \) 1065

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 1065

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

\(\Big \downarrow \) 1067

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-2)]

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

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

Output:

Exception raised: HeuristicGCDFailed >> no luck
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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