\(\int \frac {(e x)^m (A+B x^n)}{(a+b x^n)^3 (c+d x^n)} \, dx\) [46]

Optimal result

Integrand size = 31, antiderivative size = 407 \[ \int \frac {(e x)^m \left (A+B x^n\right )}{\left (a+b x^n\right )^3 \left (c+d x^n\right )} \, dx=\frac {(A b-a B) (e x)^{1+m}}{2 a (b c-a d) e n \left (a+b x^n\right )^2}+\frac {(A b (a d (1+m-4 n)-b c (1+m-2 n))+a B (b c (1+m)-a d (1+m-2 n))) (e x)^{1+m}}{2 a^2 (b c-a d)^2 e n^2 \left (a+b x^n\right )}+\frac {\left (a B \left (2 a b c d (1+m) (1+m-2 n)-b^2 c^2 (1+m) (1+m-n)-a^2 d^2 \left (1+m^2+m (2-3 n)-3 n+2 n^2\right )\right )+A b \left (b^2 c^2 \left (1+m^2+m (2-3 n)-3 n+2 n^2\right )-2 a b c d \left (1+m^2+m (2-4 n)-4 n+3 n^2\right )+a^2 d^2 \left (1+m^2+m (2-5 n)-5 n+6 n^2\right )\right )\right ) (e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{n},\frac {1+m+n}{n},-\frac {b x^n}{a}\right )}{2 a^3 (b c-a d)^3 e (1+m) n^2}+\frac {d^2 (B c-A d) (e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{n},\frac {1+m+n}{n},-\frac {d x^n}{c}\right )}{c (b c-a d)^3 e (1+m)} \] Output:

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

Mathematica [A] (verified)

Time = 0.41 (sec) , antiderivative size = 199, normalized size of antiderivative = 0.49 \[ \int \frac {(e x)^m \left (A+B x^n\right )}{\left (a+b x^n\right )^3 \left (c+d x^n\right )} \, dx=\frac {x (e x)^m \left (\frac {b d (-B c+A d) \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{n},\frac {1+m+n}{n},-\frac {b x^n}{a}\right )}{a}+\frac {d^2 (B c-A d) \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{n},\frac {1+m+n}{n},-\frac {d x^n}{c}\right )}{c}+\frac {b (b c-a d) (B c-A d) \operatorname {Hypergeometric2F1}\left (2,\frac {1+m}{n},\frac {1+m+n}{n},-\frac {b x^n}{a}\right )}{a^2}+\frac {(A b-a B) (b c-a d)^2 \operatorname {Hypergeometric2F1}\left (3,\frac {1+m}{n},\frac {1+m+n}{n},-\frac {b x^n}{a}\right )}{a^3}\right )}{(b c-a d)^3 (1+m)} \] Input:

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

Output:

(x*(e*x)^m*((b*d*(-(B*c) + A*d)*Hypergeometric2F1[1, (1 + m)/n, (1 + m + n 
)/n, -((b*x^n)/a)])/a + (d^2*(B*c - A*d)*Hypergeometric2F1[1, (1 + m)/n, ( 
1 + m + n)/n, -((d*x^n)/c)])/c + (b*(b*c - a*d)*(B*c - A*d)*Hypergeometric 
2F1[2, (1 + m)/n, (1 + m + n)/n, -((b*x^n)/a)])/a^2 + ((A*b - a*B)*(b*c - 
a*d)^2*Hypergeometric2F1[3, (1 + m)/n, (1 + m + n)/n, -((b*x^n)/a)])/a^3)) 
/((b*c - a*d)^3*(1 + m))
 

Rubi [A] (verified)

Time = 2.13 (sec) , antiderivative size = 446, 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.

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

\(\Big \downarrow \) 1065

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 1065

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

\(\Big \downarrow \) 1067

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-2)]

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

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

Output:

Exception raised: HeuristicGCDFailed >> no luck
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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