\(\int \frac {A+B x^n+C x^{2 n}+D x^{3 n}}{(a+b x^n+c x^{2 n})^2} \, dx\) [21]

Optimal result

Integrand size = 38, antiderivative size = 486 \[ \int \frac {A+B x^n+C x^{2 n}+D x^{3 n}}{\left (a+b x^n+c x^{2 n}\right )^2} \, dx=\frac {x \left (A c \left (b^2-2 a c\right )-a (b B c-2 a c C+a b D)+\left (A b c^2-a \left (2 B c^2-b c C+b^2 D-2 a c D\right )\right ) x^n\right )}{a c \left (b^2-4 a c\right ) n \left (a+b x^n+c x^{2 n}\right )}-\frac {\left (c (A b c-2 a B c+a b C) (1-n)-a D \left (b^2-2 a c (1+n)\right )-\frac {A c^2 \left (4 a c (1-2 n)-b^2 (1-n)\right )-a \left (4 a c^2 C+b^3 D-b^2 c C (1-n)-2 b c (B c n+a D (2+n))\right )}{\sqrt {b^2-4 a c}}\right ) x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{a c \left (b^2-4 a c\right ) \left (b-\sqrt {b^2-4 a c}\right ) n}-\frac {\left (c (A b c-2 a B c+a b C) (1-n)-a D \left (b^2-2 a c (1+n)\right )+\frac {A c^2 \left (4 a c (1-2 n)-b^2 (1-n)\right )-a \left (4 a c^2 C+b^3 D-b^2 c C (1-n)-2 b c (B c n+a D (2+n))\right )}{\sqrt {b^2-4 a c}}\right ) x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{a c \left (b^2-4 a c\right ) \left (b+\sqrt {b^2-4 a c}\right ) n} \] Output:

x*(A*c*(-2*a*c+b^2)-a*(B*b*c-2*C*a*c+D*a*b)+(A*b*c^2-a*(2*B*c^2-C*b*c-2*D* 
a*c+D*b^2))*x^n)/a/c/(-4*a*c+b^2)/n/(a+b*x^n+c*x^(2*n))-(c*(A*b*c-2*B*a*c+ 
C*a*b)*(1-n)-a*D*(b^2-2*a*c*(1+n))-(A*c^2*(4*a*c*(1-2*n)-b^2*(1-n))-a*(4*a 
*c^2*C+b^3*D-b^2*c*C*(1-n)-2*b*c*(B*c*n+a*D*(2+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/c/(-4*a*c+b 
^2)/(b-(-4*a*c+b^2)^(1/2))/n-(c*(A*b*c-2*B*a*c+C*a*b)*(1-n)-a*D*(b^2-2*a*c 
*(1+n))+(A*c^2*(4*a*c*(1-2*n)-b^2*(1-n))-a*(4*a*c^2*C+b^3*D-b^2*c*C*(1-n)- 
2*b*c*(B*c*n+a*D*(2+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/c/(-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. \(5439\) vs. \(2(486)=972\).

Time = 8.98 (sec) , antiderivative size = 5439, normalized size of antiderivative = 11.19 \[ \int \frac {A+B x^n+C x^{2 n}+D x^{3 n}}{\left (a+b x^n+c x^{2 n}\right )^2} \, dx=\text {Result too large to show} \] Input:

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

Output:

Result too large to show
 

Rubi [A] (verified)

Time = 1.23 (sec) , antiderivative size = 477, normalized size of antiderivative = 0.98, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.079, Rules used = {2326, 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.

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

\(\Big \downarrow \) 2326

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

\(\Big \downarrow \) 1752

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

\(\Big \downarrow \) 778

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

Input:

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

Output:

(x*(A*c*(b^2 - 2*a*c) - a*(b*B*c - 2*a*c*C + a*b*D) + (b*c*(A*c + a*C) - a 
*b^2*D - 2*a*c*(B*c - a*D))*x^n))/(a*c*(b^2 - 4*a*c)*n*(a + b*x^n + c*x^(2 
*n))) + (((a*b^2*D - b*c*(A*c + a*C)*(1 - n) + 2*a*c*(B*c*(1 - n) - a*D*(1 
 + n)) + (A*c^2*(4*a*c*(1 - 2*n) - b^2*(1 - n)) - a*(4*a*c^2*C + b^3*D - b 
^2*c*C*(1 - n) - 2*b*c*(B*c*n + a*D*(2 + n))))/Sqrt[b^2 - 4*a*c])*x*Hyperg 
eometric2F1[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*D - b*c*(A*c + a*C)*(1 - n) + 2*a*c*(B*c*( 
1 - n) - a*D*(1 + n)) - (A*c^2*(4*a*c*(1 - 2*n) - b^2*(1 - n)) - a*(4*a*c^ 
2*C + b^3*D - b^2*c*C*(1 - n) - 2*b*c*(B*c*n + a*D*(2 + 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*c*(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 2326

Int[(P3_)*((a_) + (b_.)*(x_)^(n_) + (c_.)*(x_)^(n2_))^(p_), x_Symbol] :> Wi 
th[{d = Coeff[P3, x^n, 0], e = Coeff[P3, x^n, 1], f = Coeff[P3, x^n, 2], g 
= Coeff[P3, x^n, 3]}, Simp[(-x)*(b^2*c*d - 2*a*c*(c*d - a*f) - a*b*(c*e + a 
*g) + (b*c*(c*d + a*f) - a*b^2*g - 2*a*c*(c*e - a*g))*x^n)*((a + b*x^n + c* 
x^(2*n))^(p + 1)/(a*c*n*(p + 1)*(b^2 - 4*a*c))), x] - Simp[1/(a*c*n*(p + 1) 
*(b^2 - 4*a*c))   Int[(a + b*x^n + c*x^(2*n))^(p + 1)*Simp[a*b*(c*e + a*g) 
- b^2*c*d*(n + n*p + 1) - 2*a*c*(a*f - c*d*(2*n*(p + 1) + 1)) + (a*b^2*g*(n 
*(p + 2) + 1) - b*c*(c*d + a*f)*(n*(2*p + 3) + 1) - 2*a*c*(a*g*(n + 1) - c* 
e*(n*(2*p + 3) + 1)))*x^n, x], x], x]] /; FreeQ[{a, b, c, n}, x] && EqQ[n2, 
 2*n] && PolyQ[P3, x^n, 3] && NeQ[b^2 - 4*a*c, 0] && ILtQ[p, -1]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

( - 2*x**(2*n)*int(x**(2*n)/(x**(4*n)*c**2*n**2 - 2*x**(4*n)*c**2*n + x**( 
4*n)*c**2 + 2*x**(3*n)*b*c*n**2 - 4*x**(3*n)*b*c*n + 2*x**(3*n)*b*c + 2*x* 
*(2*n)*a*c*n**2 - 4*x**(2*n)*a*c*n + 2*x**(2*n)*a*c + x**(2*n)*b**2*n**2 - 
 2*x**(2*n)*b**2*n + x**(2*n)*b**2 + 2*x**n*a*b*n**2 - 4*x**n*a*b*n + 2*x* 
*n*a*b + a**2*n**2 - 2*a**2*n + a**2),x)*a*c**2*d*n**4 + 3*x**(2*n)*int(x* 
*(2*n)/(x**(4*n)*c**2*n**2 - 2*x**(4*n)*c**2*n + x**(4*n)*c**2 + 2*x**(3*n 
)*b*c*n**2 - 4*x**(3*n)*b*c*n + 2*x**(3*n)*b*c + 2*x**(2*n)*a*c*n**2 - 4*x 
**(2*n)*a*c*n + 2*x**(2*n)*a*c + x**(2*n)*b**2*n**2 - 2*x**(2*n)*b**2*n + 
x**(2*n)*b**2 + 2*x**n*a*b*n**2 - 4*x**n*a*b*n + 2*x**n*a*b + a**2*n**2 - 
2*a**2*n + a**2),x)*a*c**2*d*n**3 + x**(2*n)*int(x**(2*n)/(x**(4*n)*c**2*n 
**2 - 2*x**(4*n)*c**2*n + x**(4*n)*c**2 + 2*x**(3*n)*b*c*n**2 - 4*x**(3*n) 
*b*c*n + 2*x**(3*n)*b*c + 2*x**(2*n)*a*c*n**2 - 4*x**(2*n)*a*c*n + 2*x**(2 
*n)*a*c + x**(2*n)*b**2*n**2 - 2*x**(2*n)*b**2*n + x**(2*n)*b**2 + 2*x**n* 
a*b*n**2 - 4*x**n*a*b*n + 2*x**n*a*b + a**2*n**2 - 2*a**2*n + a**2),x)*a*c 
**2*d*n**2 - 3*x**(2*n)*int(x**(2*n)/(x**(4*n)*c**2*n**2 - 2*x**(4*n)*c**2 
*n + x**(4*n)*c**2 + 2*x**(3*n)*b*c*n**2 - 4*x**(3*n)*b*c*n + 2*x**(3*n)*b 
*c + 2*x**(2*n)*a*c*n**2 - 4*x**(2*n)*a*c*n + 2*x**(2*n)*a*c + x**(2*n)*b* 
*2*n**2 - 2*x**(2*n)*b**2*n + x**(2*n)*b**2 + 2*x**n*a*b*n**2 - 4*x**n*a*b 
*n + 2*x**n*a*b + a**2*n**2 - 2*a**2*n + a**2),x)*a*c**2*d*n + x**(2*n)*in 
t(x**(2*n)/(x**(4*n)*c**2*n**2 - 2*x**(4*n)*c**2*n + x**(4*n)*c**2 + 2*...