\(\int \frac {A+B x^n+C x^{2 n}}{(a+b x^n+c x^{2 n})^2} \, dx\) [18]

Optimal result

Integrand size = 31, antiderivative size = 382 \[ \int \frac {A+B x^n+C x^{2 n}}{\left (a+b x^n+c x^{2 n}\right )^2} \, dx=\frac {x \left (A \left (b^2-2 a c\right )-a (b B-2 a C)+(A b c-2 a B c+a b C) x^n\right )}{a \left (b^2-4 a c\right ) n \left (a+b x^n+c x^{2 n}\right )}-\frac {\left ((A b c-2 a B c+a b C) (1-n)-\frac {A c \left (4 a c (1-2 n)-b^2 (1-n)\right )-a \left (4 a c C-b^2 C (1-n)-2 b B c n\right )}{\sqrt {b^2-4 a c}}\right ) x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{a \left (b^2-4 a c\right ) \left (b-\sqrt {b^2-4 a c}\right ) n}-\frac {\left ((A b c-2 a B c+a b C) (1-n)+\frac {A c \left (4 a c (1-2 n)-b^2 (1-n)\right )-a \left (4 a c C-2 b B c n-b^2 (C-C n)\right )}{\sqrt {b^2-4 a c}}\right ) x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{a \left (b^2-4 a c\right ) \left (b+\sqrt {b^2-4 a c}\right ) n} \] Output:

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))-((A*b*c-2*B*a*c+C*a*b)*(1-n)-(A*c*(4*a*c*(1-2*n)-b^2 
*(1-n))-a*(4*a*c*C-b^2*C*(1-n)-2*b*B*c*n))/(-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/(-4*a*c+b^2)/(b-(-4*a 
*c+b^2)^(1/2))/n-((A*b*c-2*B*a*c+C*a*b)*(1-n)+(A*c*(4*a*c*(1-2*n)-b^2*(1-n 
))-a*(4*a*c*C-2*b*B*c*n-b^2*(-C*n+C)))/(-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/(-4*a*c+b^2)/(b+(-4*a*c+b 
^2)^(1/2))/n
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(4120\) vs. \(2(382)=764\).

Time = 7.33 (sec) , antiderivative size = 4120, normalized size of antiderivative = 10.79 \[ \int \frac {A+B x^n+C x^{2 n}}{\left (a+b x^n+c x^{2 n}\right )^2} \, dx=\text {Result too large to show} \] Input:

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

Output:

((-(A*b^2) + a*b*B + 2*a*A*c - 2*a^2*C + A*b^2*n - 4*a*A*c*n)*x)/(a^2*(-b^ 
2 + 4*a*c)*n) + ((A*b^2 - a*b*B - 2*a*A*c + 2*a^2*C - A*b^2*n + 4*a*A*c*n) 
*x)/(a^2*(-b^2 + 4*a*c)*n) - (x*(A*b^2 - a*b*B - 2*a*A*c + 2*a^2*C + A*b*c 
*x^n - 2*a*B*c*x^n + a*b*C*x^n))/(a*(-b^2 + 4*a*c)*n*(a + b*x^n + c*x^(2*n 
))) - (A*b*c*x^(1 + n)*(x^n)^(n^(-1) - (1 + n)/n)*(-(Hypergeometric2F1[-n^ 
(-1), -n^(-1), (-1 + n)/n, -1/2*(-b - Sqrt[b^2 - 4*a*c])/(c*(-1/2*(-b - Sq 
rt[b^2 - 4*a*c])/c + x^n))]/(Sqrt[b^2 - 4*a*c]*(x^n/(-1/2*(-b - Sqrt[b^2 - 
 4*a*c])/c + x^n))^n^(-1))) + Hypergeometric2F1[-n^(-1), -n^(-1), (-1 + n) 
/n, -1/2*(-b + Sqrt[b^2 - 4*a*c])/(c*(-1/2*(-b + Sqrt[b^2 - 4*a*c])/c + x^ 
n))]/(Sqrt[b^2 - 4*a*c]*(x^n/(-1/2*(-b + Sqrt[b^2 - 4*a*c])/c + x^n))^n^(- 
1))))/(a*(-b^2 + 4*a*c)) + (2*B*c*x^(1 + n)*(x^n)^(n^(-1) - (1 + n)/n)*(-( 
Hypergeometric2F1[-n^(-1), -n^(-1), (-1 + n)/n, -1/2*(-b - Sqrt[b^2 - 4*a* 
c])/(c*(-1/2*(-b - Sqrt[b^2 - 4*a*c])/c + x^n))]/(Sqrt[b^2 - 4*a*c]*(x^n/( 
-1/2*(-b - Sqrt[b^2 - 4*a*c])/c + x^n))^n^(-1))) + Hypergeometric2F1[-n^(- 
1), -n^(-1), (-1 + n)/n, -1/2*(-b + Sqrt[b^2 - 4*a*c])/(c*(-1/2*(-b + Sqrt 
[b^2 - 4*a*c])/c + x^n))]/(Sqrt[b^2 - 4*a*c]*(x^n/(-1/2*(-b + Sqrt[b^2 - 4 
*a*c])/c + x^n))^n^(-1))))/(-b^2 + 4*a*c) - (b*C*x^(1 + n)*(x^n)^(n^(-1) - 
 (1 + n)/n)*(-(Hypergeometric2F1[-n^(-1), -n^(-1), (-1 + n)/n, -1/2*(-b - 
Sqrt[b^2 - 4*a*c])/(c*(-1/2*(-b - Sqrt[b^2 - 4*a*c])/c + x^n))]/(Sqrt[b^2 
- 4*a*c]*(x^n/(-1/2*(-b - Sqrt[b^2 - 4*a*c])/c + x^n))^n^(-1))) + Hyper...
 

Rubi [A] (verified)

Time = 0.88 (sec) , antiderivative size = 368, normalized size of antiderivative = 0.96, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {2327, 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.

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

\(\Big \downarrow \) 2327

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

\(\Big \downarrow \) 1752

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

\(\Big \downarrow \) 778

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

Input:

Int[(A + B*x^n + C*x^(2*n))/(a + b*x^n + c*x^(2*n))^2,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))/(a 
*(b^2 - 4*a*c)*n*(a + b*x^n + c*x^(2*n))) + (-((((A*b*c - 2*a*B*c + a*b*C) 
*(1 - n) - (A*c*(4*a*c*(1 - 2*n) - b^2*(1 - n)) - a*(4*a*c*C - 2*b*B*c*n - 
 b^2*(C - C*n)))/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*c - 2*a*B*c + a*b*C)*(1 - n) + (A*c*(4*a*c*(1 - 2*n) - b^2*(1 - n)) - a* 
(4*a*c*C - 2*b*B*c*n - b^2*(C - C*n)))/Sqrt[b^2 - 4*a*c])*x*Hypergeometric 
2F1[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)
 

Defintions of rubi rules used

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 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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

((C*a*b - 2*B*a*c + A*b*c)*x*x^n + (2*C*a^2 - B*a*b + (b^2 - 2*a*c)*A)*x)/ 
(a^2*b^2*n - 4*a^3*c*n + (a*b^2*c*n - 4*a^2*c^2*n)*x^(2*n) + (a*b^3*n - 4* 
a^2*b*c*n)*x^n) - integrate((2*C*a^2 - B*a*b + (2*a*c*(2*n - 1) - b^2*(n - 
 1))*A - (C*a*b*(n - 1) - 2*B*a*c*(n - 1) + A*b*c*(n - 1))*x^n)/(a^2*b^2*n 
 - 4*a^3*c*n + (a*b^2*c*n - 4*a^2*c^2*n)*x^(2*n) + (a*b^3*n - 4*a^2*b*c*n) 
*x^n), x)
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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