\(\int \frac {A+B x^n+C x^{2 n}}{(a+b x^n+c x^{2 n})^3} \, dx\) [19]

Optimal result

Integrand size = 31, antiderivative size = 802 \[ \int \frac {A+B x^n+C x^{2 n}}{\left (a+b x^n+c x^{2 n}\right )^3} \, dx=\frac {x \left (A \left (b^2-2 a c\right )-a (b B-2 a C)+(A b c-2 a B c+a b C) x^n\right )}{2 a \left (b^2-4 a c\right ) n \left (a+b x^n+c x^{2 n}\right )^2}+\frac {x \left (\left (b^2-2 a c\right ) \left (a (b B-2 a C)+A \left (2 a c (1-4 n)-b^2 (1-2 n)\right )\right )+a b (A b c-2 a B c+a b C) (1-3 n)+c \left (A b \left (2 a c (2-7 n)-b^2 (1-2 n)\right )+a \left (b^2 B-4 a B c (1-3 n)-6 a b C n\right )\right ) x^n\right )}{2 a^2 \left (b^2-4 a c\right )^2 n^2 \left (a+b x^n+c x^{2 n}\right )}-\frac {c \left (A \left (b^4 \left (1-3 n+2 n^2\right )+b^3 \sqrt {b^2-4 a c} \left (1-3 n+2 n^2\right )-6 a b^2 c \left (1-4 n+3 n^2\right )-2 a b c \sqrt {b^2-4 a c} \left (2-9 n+7 n^2\right )+8 a^2 c^2 \left (1-6 n+8 n^2\right )\right )-a \left (b^3 B (1-n)-2 a b \left (3 \sqrt {b^2-4 a c} C (1-n) n+2 B c \left (1-n-3 n^2\right )\right )+4 a c \left (2 a C (1-2 n)-B \sqrt {b^2-4 a c} \left (1-4 n+3 n^2\right )\right )+b^2 \left (B \sqrt {b^2-4 a c} (1-n)-2 a C \left (1-2 n+3 n^2\right )\right )\right )\right ) x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{2 a^2 \left (b^2-4 a c\right )^2 \left (b^2-4 a c-b \sqrt {b^2-4 a c}\right ) n^2}-\frac {c \left ((1-n) \left (A b \left (2 a c (2-7 n)-b^2 (1-2 n)\right )+a \left (b^2 B-4 a B c (1-3 n)-6 a b C n\right )\right )-\frac {a \left (8 a^2 c C (1-2 n)+b^3 B (1-n)-4 a b B c \left (1-n-3 n^2\right )-2 a b^2 C \left (1-2 n+3 n^2\right )\right )-A \left (b^4 \left (1-3 n+2 n^2\right )-6 a b^2 c \left (1-4 n+3 n^2\right )+8 a^2 c^2 \left (1-6 n+8 n^2\right )\right )}{\sqrt {b^2-4 a c}}\right ) x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{2 a^2 \left (b^2-4 a c\right )^2 \left (b+\sqrt {b^2-4 a c}\right ) n^2} \] Output:

1/2*x*(A*(-2*a*c+b^2)-a*(B*b-2*C*a)+(A*b*c-2*B*a*c+C*a*b)*x^n)/a/(-4*a*c+b 
^2)/n/(a+b*x^n+c*x^(2*n))^2+1/2*x*((-2*a*c+b^2)*(a*(B*b-2*C*a)+A*(2*a*c*(1 
-4*n)-b^2*(1-2*n)))+a*b*(A*b*c-2*B*a*c+C*a*b)*(1-3*n)+c*(A*b*(2*a*c*(2-7*n 
)-b^2*(1-2*n))+a*(b^2*B-4*a*B*c*(1-3*n)-6*a*b*C*n))*x^n)/a^2/(-4*a*c+b^2)^ 
2/n^2/(a+b*x^n+c*x^(2*n))-1/2*c*(A*(b^4*(2*n^2-3*n+1)+b^3*(-4*a*c+b^2)^(1/ 
2)*(2*n^2-3*n+1)-6*a*b^2*c*(3*n^2-4*n+1)-2*a*b*c*(-4*a*c+b^2)^(1/2)*(7*n^2 
-9*n+2)+8*a^2*c^2*(8*n^2-6*n+1))-a*(b^3*B*(1-n)-2*a*b*(3*(-4*a*c+b^2)^(1/2 
)*C*(1-n)*n+2*B*c*(-3*n^2-n+1))+4*a*c*(2*a*C*(1-2*n)-B*(-4*a*c+b^2)^(1/2)* 
(3*n^2-4*n+1))+b^2*(B*(-4*a*c+b^2)^(1/2)*(1-n)-2*a*C*(3*n^2-2*n+1))))*x*hy 
pergeom([1, 1/n],[1+1/n],-2*c*x^n/(b-(-4*a*c+b^2)^(1/2)))/a^2/(-4*a*c+b^2) 
^2/(b^2-4*a*c-b*(-4*a*c+b^2)^(1/2))/n^2-1/2*c*((1-n)*(A*b*(2*a*c*(2-7*n)-b 
^2*(1-2*n))+a*(b^2*B-4*a*B*c*(1-3*n)-6*a*b*C*n))-(a*(8*a^2*c*C*(1-2*n)+b^3 
*B*(1-n)-4*a*b*B*c*(-3*n^2-n+1)-2*a*b^2*C*(3*n^2-2*n+1))-A*(b^4*(2*n^2-3*n 
+1)-6*a*b^2*c*(3*n^2-4*n+1)+8*a^2*c^2*(8*n^2-6*n+1)))/(-4*a*c+b^2)^(1/2))* 
x*hypergeom([1, 1/n],[1+1/n],-2*c*x^n/(b+(-4*a*c+b^2)^(1/2)))/a^2/(-4*a*c+ 
b^2)^2/(b+(-4*a*c+b^2)^(1/2))/n^2
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(10759\) vs. \(2(802)=1604\).

Time = 8.40 (sec) , antiderivative size = 10759, normalized size of antiderivative = 13.42 \[ \int \frac {A+B x^n+C x^{2 n}}{\left (a+b x^n+c x^{2 n}\right )^3} \, dx=\text {Result too large to show} \] Input:

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

Output:

Result too large to show
 

Rubi [A] (verified)

Time = 1.58 (sec) , antiderivative size = 752, normalized size of antiderivative = 0.94, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {2327, 1760, 25, 1752, 778}

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.

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

\(\Big \downarrow \) 2327

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

\(\Big \downarrow \) 1760

\(\displaystyle \frac {-\frac {\int -\frac {-c (1-n) \left (A \left (2 a b c (2-7 n)-b^3 (1-2 n)\right )+a \left (B b^2-6 a C n b-4 a B c (1-3 n)\right )\right ) x^n+a \left (-B (1-n) b^3+a C (n+1) b^2+2 a B c (2-5 n) b-4 a^2 c C (1-2 n)\right )+A \left (\left (2 n^2-3 n+1\right ) b^4-a c \left (16 n^2-21 n+5\right ) b^2+4 a^2 c^2 \left (8 n^2-6 n+1\right )\right )}{b x^n+c x^{2 n}+a}dx}{a n \left (b^2-4 a c\right )}-\frac {x \left (A \left (4 a^2 c^2 (1-4 n)-5 a b^2 c (1-3 n)+b^4 (1-2 n)\right )-a \left (4 a^2 c C-a b^2 C (3 n+1)-2 a b B c (2-3 n)+b^3 B\right )-c x^n \left (A \left (2 a b c (2-7 n)-b^3 (1-2 n)\right )+a \left (-6 a b C n-4 a B c (1-3 n)+b^2 B\right )\right )\right )}{a n \left (b^2-4 a c\right ) \left (a+b x^n+c x^{2 n}\right )}}{2 a n \left (b^2-4 a c\right )}+\frac {x \left (A \left (b^2-2 a c\right )+x^n (a b C-2 a B c+A b c)-a (b B-2 a C)\right )}{2 a n \left (b^2-4 a c\right ) \left (a+b x^n+c x^{2 n}\right )^2}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {\int \frac {-c (1-n) \left (A \left (2 a b c (2-7 n)-b^3 (1-2 n)\right )+a \left (B b^2-6 a C n b-4 a B c (1-3 n)\right )\right ) x^n+a \left (-B (1-n) b^3+a C (n+1) b^2+2 a B c (2-5 n) b-4 a^2 c C (1-2 n)\right )+A \left (\left (2 n^2-3 n+1\right ) b^4-a c \left (16 n^2-21 n+5\right ) b^2+4 a^2 c^2 \left (8 n^2-6 n+1\right )\right )}{b x^n+c x^{2 n}+a}dx}{a n \left (b^2-4 a c\right )}-\frac {x \left (A \left (4 a^2 c^2 (1-4 n)-5 a b^2 c (1-3 n)+b^4 (1-2 n)\right )-a \left (4 a^2 c C-a b^2 C (3 n+1)-2 a b B c (2-3 n)+b^3 B\right )-c x^n \left (A \left (2 a b c (2-7 n)-b^3 (1-2 n)\right )+a \left (-6 a b C n-4 a B c (1-3 n)+b^2 B\right )\right )\right )}{a n \left (b^2-4 a c\right ) \left (a+b x^n+c x^{2 n}\right )}}{2 a n \left (b^2-4 a c\right )}+\frac {x \left (A \left (b^2-2 a c\right )+x^n (a b C-2 a B c+A b c)-a (b B-2 a C)\right )}{2 a n \left (b^2-4 a c\right ) \left (a+b x^n+c x^{2 n}\right )^2}\)

\(\Big \downarrow \) 1752

\(\displaystyle \frac {\frac {-\frac {1}{2} c \left (\frac {a \left (8 a^2 c C (1-2 n)-2 a b^2 C \left (3 n^2-2 n+1\right )-4 a b B c \left (-3 n^2-n+1\right )+b^3 B (1-n)\right )-A \left (8 a^2 c^2 \left (8 n^2-6 n+1\right )-6 a b^2 c \left (3 n^2-4 n+1\right )+b^4 \left (2 n^2-3 n+1\right )\right )}{\sqrt {b^2-4 a c}}+(1-n) \left (A \left (2 a b c (2-7 n)-b^3 (1-2 n)\right )+a \left (-6 a b C n-4 a B c (1-3 n)+b^2 B\right )\right )\right ) \int \frac {1}{c x^n+\frac {1}{2} \left (b-\sqrt {b^2-4 a c}\right )}dx-\frac {1}{2} c \left ((1-n) \left (A \left (2 a b c (2-7 n)-b^3 (1-2 n)\right )+a \left (-6 a b C n-4 a B c (1-3 n)+b^2 B\right )\right )-\frac {a \left (8 a^2 c C (1-2 n)-2 a b^2 C \left (3 n^2-2 n+1\right )-4 a b B c \left (-3 n^2-n+1\right )+b^3 B (1-n)\right )-A \left (8 a^2 c^2 \left (8 n^2-6 n+1\right )-6 a b^2 c \left (3 n^2-4 n+1\right )+b^4 \left (2 n^2-3 n+1\right )\right )}{\sqrt {b^2-4 a c}}\right ) \int \frac {1}{c x^n+\frac {1}{2} \left (b+\sqrt {b^2-4 a c}\right )}dx}{a n \left (b^2-4 a c\right )}-\frac {x \left (A \left (4 a^2 c^2 (1-4 n)-5 a b^2 c (1-3 n)+b^4 (1-2 n)\right )-a \left (4 a^2 c C-a b^2 C (3 n+1)-2 a b B c (2-3 n)+b^3 B\right )-c x^n \left (A \left (2 a b c (2-7 n)-b^3 (1-2 n)\right )+a \left (-6 a b C n-4 a B c (1-3 n)+b^2 B\right )\right )\right )}{a n \left (b^2-4 a c\right ) \left (a+b x^n+c x^{2 n}\right )}}{2 a n \left (b^2-4 a c\right )}+\frac {x \left (A \left (b^2-2 a c\right )+x^n (a b C-2 a B c+A b c)-a (b B-2 a C)\right )}{2 a n \left (b^2-4 a c\right ) \left (a+b x^n+c x^{2 n}\right )^2}\)

\(\Big \downarrow \) 778

\(\displaystyle \frac {\frac {-\frac {c x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right ) \left (\frac {a \left (8 a^2 c C (1-2 n)-2 a b^2 C \left (3 n^2-2 n+1\right )-4 a b B c \left (-3 n^2-n+1\right )+b^3 B (1-n)\right )-A \left (8 a^2 c^2 \left (8 n^2-6 n+1\right )-6 a b^2 c \left (3 n^2-4 n+1\right )+b^4 \left (2 n^2-3 n+1\right )\right )}{\sqrt {b^2-4 a c}}+(1-n) \left (A \left (2 a b c (2-7 n)-b^3 (1-2 n)\right )+a \left (-6 a b C n-4 a B c (1-3 n)+b^2 B\right )\right )\right )}{b-\sqrt {b^2-4 a c}}-\frac {c x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right ) \left ((1-n) \left (A \left (2 a b c (2-7 n)-b^3 (1-2 n)\right )+a \left (-6 a b C n-4 a B c (1-3 n)+b^2 B\right )\right )-\frac {a \left (8 a^2 c C (1-2 n)-2 a b^2 C \left (3 n^2-2 n+1\right )-4 a b B c \left (-3 n^2-n+1\right )+b^3 B (1-n)\right )-A \left (8 a^2 c^2 \left (8 n^2-6 n+1\right )-6 a b^2 c \left (3 n^2-4 n+1\right )+b^4 \left (2 n^2-3 n+1\right )\right )}{\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c}+b}}{a n \left (b^2-4 a c\right )}-\frac {x \left (A \left (4 a^2 c^2 (1-4 n)-5 a b^2 c (1-3 n)+b^4 (1-2 n)\right )-a \left (4 a^2 c C-a b^2 C (3 n+1)-2 a b B c (2-3 n)+b^3 B\right )-c x^n \left (A \left (2 a b c (2-7 n)-b^3 (1-2 n)\right )+a \left (-6 a b C n-4 a B c (1-3 n)+b^2 B\right )\right )\right )}{a n \left (b^2-4 a c\right ) \left (a+b x^n+c x^{2 n}\right )}}{2 a n \left (b^2-4 a c\right )}+\frac {x \left (A \left (b^2-2 a c\right )+x^n (a b C-2 a B c+A b c)-a (b B-2 a C)\right )}{2 a n \left (b^2-4 a c\right ) \left (a+b x^n+c x^{2 n}\right )^2}\)

Input:

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

Output:

(x*(A*(b^2 - 2*a*c) - a*(b*B - 2*a*C) + (A*b*c - 2*a*B*c + a*b*C)*x^n))/(2 
*a*(b^2 - 4*a*c)*n*(a + b*x^n + c*x^(2*n))^2) + (-((x*(A*(4*a^2*c^2*(1 - 4 
*n) - 5*a*b^2*c*(1 - 3*n) + b^4*(1 - 2*n)) - a*(b^3*B + 4*a^2*c*C - 2*a*b* 
B*c*(2 - 3*n) - a*b^2*C*(1 + 3*n)) - c*(A*(2*a*b*c*(2 - 7*n) - b^3*(1 - 2* 
n)) + a*(b^2*B - 4*a*B*c*(1 - 3*n) - 6*a*b*C*n))*x^n))/(a*(b^2 - 4*a*c)*n* 
(a + b*x^n + c*x^(2*n)))) + (-((c*((1 - n)*(A*(2*a*b*c*(2 - 7*n) - b^3*(1 
- 2*n)) + a*(b^2*B - 4*a*B*c*(1 - 3*n) - 6*a*b*C*n)) + (a*(8*a^2*c*C*(1 - 
2*n) + b^3*B*(1 - n) - 4*a*b*B*c*(1 - n - 3*n^2) - 2*a*b^2*C*(1 - 2*n + 3* 
n^2)) - A*(b^4*(1 - 3*n + 2*n^2) - 6*a*b^2*c*(1 - 4*n + 3*n^2) + 8*a^2*c^2 
*(1 - 6*n + 8*n^2)))/Sqrt[b^2 - 4*a*c])*x*Hypergeometric2F1[1, n^(-1), 1 + 
 n^(-1), (-2*c*x^n)/(b - Sqrt[b^2 - 4*a*c])])/(b - Sqrt[b^2 - 4*a*c])) - ( 
c*((1 - n)*(A*(2*a*b*c*(2 - 7*n) - b^3*(1 - 2*n)) + a*(b^2*B - 4*a*B*c*(1 
- 3*n) - 6*a*b*C*n)) - (a*(8*a^2*c*C*(1 - 2*n) + b^3*B*(1 - n) - 4*a*b*B*c 
*(1 - n - 3*n^2) - 2*a*b^2*C*(1 - 2*n + 3*n^2)) - A*(b^4*(1 - 3*n + 2*n^2) 
 - 6*a*b^2*c*(1 - 4*n + 3*n^2) + 8*a^2*c^2*(1 - 6*n + 8*n^2)))/Sqrt[b^2 - 
4*a*c])*x*Hypergeometric2F1[1, n^(-1), 1 + n^(-1), (-2*c*x^n)/(b + Sqrt[b^ 
2 - 4*a*c])])/(b + Sqrt[b^2 - 4*a*c]))/(a*(b^2 - 4*a*c)*n))/(2*a*(b^2 - 4* 
a*c)*n)
 

Defintions of rubi rules used

rule 25

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

rule 778

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*x*Hypergeometric2F 
1[-p, 1/n, 1/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p 
, 0] &&  !IntegerQ[1/n] &&  !ILtQ[Simplify[1/n + p], 0] && (IntegerQ[p] || 
GtQ[a, 0])
 

rule 1752

Int[((d_) + (e_.)*(x_)^(n_))/((a_) + (b_.)*(x_)^(n_) + (c_.)*(x_)^(n2_)), x 
_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[(e/2 + (2*c*d - b*e)/(2*q)) 
   Int[1/(b/2 - q/2 + c*x^n), x], x] + Simp[(e/2 - (2*c*d - b*e)/(2*q))   I 
nt[1/(b/2 + q/2 + c*x^n), x], x]] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[n2 
, 2*n] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && (PosQ[b^2 
 - 4*a*c] ||  !IGtQ[n/2, 0])
 

rule 1760

Int[((d_) + (e_.)*(x_)^(n_))*((a_) + (b_.)*(x_)^(n_) + (c_.)*(x_)^(n2_))^(p 
_), x_Symbol] :> Simp[(-x)*(d*b^2 - a*b*e - 2*a*c*d + (b*d - 2*a*e)*c*x^n)* 
((a + b*x^n + c*x^(2*n))^(p + 1)/(a*n*(p + 1)*(b^2 - 4*a*c))), x] + Simp[1/ 
(a*n*(p + 1)*(b^2 - 4*a*c))   Int[Simp[(n*p + n + 1)*d*b^2 - a*b*e - 2*a*c* 
d*(2*n*p + 2*n + 1) + (2*n*p + 3*n + 1)*(d*b - 2*a*e)*c*x^n, x]*(a + b*x^n 
+ c*x^(2*n))^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[n2, 2*n 
] && NeQ[b^2 - 4*a*c, 0] && ILtQ[p, -1]
 

rule 2327

Int[(P2_)*((a_) + (b_.)*(x_)^(n_) + (c_.)*(x_)^(n2_))^(p_), x_Symbol] :> Wi 
th[{d = Coeff[P2, x^n, 0], e = Coeff[P2, x^n, 1], f = Coeff[P2, x^n, 2]}, S 
imp[(-x)*(b^2*d - 2*a*(c*d - a*f) - a*b*e + (b*(c*d + a*f) - 2*a*c*e)*x^n)* 
((a + b*x^n + c*x^(2*n))^(p + 1)/(a*n*(p + 1)*(b^2 - 4*a*c))), x] - Simp[1/ 
(a*n*(p + 1)*(b^2 - 4*a*c))   Int[(a + b*x^n + c*x^(2*n))^(p + 1)*Simp[a*b* 
e - b^2*d*(n + n*p + 1) - 2*a*(a*f - c*d*(2*n*(p + 1) + 1)) - (b*(c*d + a*f 
)*(n*(2*p + 3) + 1) - 2*a*c*e*(n*(2*p + 3) + 1))*x^n, x], x], x]] /; FreeQ[ 
{a, b, c, n}, x] && EqQ[n2, 2*n] && PolyQ[P2, x^n, 2] && NeQ[b^2 - 4*a*c, 0 
] && ILtQ[p, -1]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral((C*x^(2*n) + B*x^n + A)/(c^3*x^(6*n) + b^3*x^(3*n) + 3*a*b^2*x^(2 
*n) + 3*a^2*b*x^n + a^3 + 3*(b*c^2*x^n + a*c^2)*x^(4*n) + 3*(b^2*c*x^(2*n) 
 + 2*a*b*c*x^n + a^2*c)*x^(2*n)), x)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

-1/2*((6*C*a^2*b*c^2*n + (2*a*b*c^3*(7*n - 2) - b^3*c^2*(2*n - 1))*A - (4* 
a^2*c^3*(3*n - 1) + a*b^2*c^2)*B)*x*x^(3*n) + ((a*b^2*c^2*(29*n - 9) - 4*a 
^2*c^3*(4*n - 1) - 2*b^4*c*(2*n - 1))*A - 2*(a^2*b*c^2*(9*n - 4) + a*b^3*c 
)*B + (a^2*b^2*c*(9*n + 1) - 4*a^3*c^2)*C)*x*x^(2*n) + ((4*a*b^3*c*(3*n - 
1) - b^5*(2*n - 1) + 2*a^2*b*c^2*n)*A - (4*a^3*c^2*(5*n - 1) + a^2*b^2*c*( 
4*n - 3) + a*b^4)*B + (2*a^3*b*c*(5*n - 2) + a^2*b^3*(2*n + 1))*C)*x*x^n + 
 ((a^2*b^2*c*(21*n - 5) - 4*a^3*c^2*(6*n - 1) - a*b^4*(3*n - 1))*A - (2*a^ 
3*b*c*(5*n - 2) - a^2*b^3*(n - 1))*B + (4*a^4*c*(2*n - 1) + a^3*b^2*(n + 1 
))*C)*x)/(a^4*b^4*n^2 - 8*a^5*b^2*c*n^2 + 16*a^6*c^2*n^2 + (a^2*b^4*c^2*n^ 
2 - 8*a^3*b^2*c^3*n^2 + 16*a^4*c^4*n^2)*x^(4*n) + 2*(a^2*b^5*c*n^2 - 8*a^3 
*b^3*c^2*n^2 + 16*a^4*b*c^3*n^2)*x^(3*n) + (a^2*b^6*n^2 - 6*a^3*b^4*c*n^2 
+ 32*a^5*c^3*n^2)*x^(2*n) + 2*(a^3*b^5*n^2 - 8*a^4*b^3*c*n^2 + 16*a^5*b*c^ 
2*n^2)*x^n) - integrate(-1/2*(((2*n^2 - 3*n + 1)*b^4 - (16*n^2 - 21*n + 5) 
*a*b^2*c + 4*(8*n^2 - 6*n + 1)*a^2*c^2)*A - (2*a^2*b*c*(5*n - 2) - a*b^3*( 
n - 1))*B + (4*a^3*c*(2*n - 1) + a^2*b^2*(n + 1))*C - (6*(n^2 - n)*C*a^2*b 
*c - ((2*n^2 - 3*n + 1)*b^3*c - 2*(7*n^2 - 9*n + 2)*a*b*c^2)*A - (4*(3*n^2 
 - 4*n + 1)*a^2*c^2 + a*b^2*c*(n - 1))*B)*x^n)/(a^3*b^4*n^2 - 8*a^4*b^2*c* 
n^2 + 16*a^5*c^2*n^2 + (a^2*b^4*c*n^2 - 8*a^3*b^2*c^2*n^2 + 16*a^4*c^3*n^2 
)*x^(2*n) + (a^2*b^5*n^2 - 8*a^3*b^3*c*n^2 + 16*a^4*b*c^2*n^2)*x^n), x)
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {A+B x^n+C x^{2 n}}{\left (a+b x^n+c x^{2 n}\right )^3} \, dx=\int \frac {1}{x^{4 n} c^{2}+2 x^{3 n} b c +2 x^{2 n} a c +x^{2 n} b^{2}+2 x^{n} a b +a^{2}}d x \] Input:

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

Output:

int(1/(x**(4*n)*c**2 + 2*x**(3*n)*b*c + 2*x**(2*n)*a*c + x**(2*n)*b**2 + 2 
*x**n*a*b + a**2),x)