3.1.9 \(\int \frac {d+e x+f x^2+g x^3}{(a+b x^n+c x^{2 n})^2} \, dx\) [9]

3.1.9.1 Optimal result

Integrand size = 32, antiderivative size = 1654 \[ \int \frac {d+e x+f x^2+g x^3}{\left (a+b x^n+c x^{2 n}\right )^2} \, dx =\text {Too large to display} \]

output

d*x*(b^2-2*a*c+b*c*x^n)/a/(-4*a*c+b^2)/n/(a+b*x^n+c*x^(2*n))+e*x^2*(b^2-2* 
a*c+b*c*x^n)/a/(-4*a*c+b^2)/n/(a+b*x^n+c*x^(2*n))+f*x^3*(b^2-2*a*c+b*c*x^n 
)/a/(-4*a*c+b^2)/n/(a+b*x^n+c*x^(2*n))+g*x^4*(b^2-2*a*c+b*c*x^n)/a/(-4*a*c 
+b^2)/n/(a+b*x^n+c*x^(2*n))-2*b*c^2*e*(2-n)*x^(2+n)*hypergeom([1, (2+n)/n] 
,[2+2/n],-2*c*x^n/(b-(-4*a*c+b^2)^(1/2)))/a/(-4*a*c+b^2)^(3/2)/n/(2+n)/(b- 
(-4*a*c+b^2)^(1/2))-2*b*c^2*f*(3-n)*x^(3+n)*hypergeom([1, (3+n)/n],[2+3/n] 
,-2*c*x^n/(b-(-4*a*c+b^2)^(1/2)))/a/(-4*a*c+b^2)^(3/2)/n/(3+n)/(b-(-4*a*c+ 
b^2)^(1/2))-2*b*c^2*g*(4-n)*x^(4+n)*hypergeom([1, (4+n)/n],[2+4/n],-2*c*x^ 
n/(b-(-4*a*c+b^2)^(1/2)))/a/(-4*a*c+b^2)^(3/2)/n/(4+n)/(b-(-4*a*c+b^2)^(1/ 
2))+2*b*c^2*e*(2-n)*x^(2+n)*hypergeom([1, (2+n)/n],[2+2/n],-2*c*x^n/(b+(-4 
*a*c+b^2)^(1/2)))/a/(-4*a*c+b^2)^(3/2)/n/(2+n)/(b+(-4*a*c+b^2)^(1/2))+2*b* 
c^2*f*(3-n)*x^(3+n)*hypergeom([1, (3+n)/n],[2+3/n],-2*c*x^n/(b+(-4*a*c+b^2 
)^(1/2)))/a/(-4*a*c+b^2)^(3/2)/n/(3+n)/(b+(-4*a*c+b^2)^(1/2))+2*b*c^2*g*(4 
-n)*x^(4+n)*hypergeom([1, (4+n)/n],[2+4/n],-2*c*x^n/(b+(-4*a*c+b^2)^(1/2)) 
)/a/(-4*a*c+b^2)^(3/2)/n/(4+n)/(b+(-4*a*c+b^2)^(1/2))-c*e*(4*a*c*(1-n)-b^2 
*(2-n))*x^2*hypergeom([1, 2/n],[(2+n)/n],-2*c*x^n/(b-(-4*a*c+b^2)^(1/2)))/ 
a/(-4*a*c+b^2)/n/(b^2-4*a*c-b*(-4*a*c+b^2)^(1/2))-2/3*c*f*(2*a*c*(3-2*n)-b 
^2*(3-n))*x^3*hypergeom([1, 3/n],[(3+n)/n],-2*c*x^n/(b-(-4*a*c+b^2)^(1/2)) 
)/a/(-4*a*c+b^2)/n/(b^2-4*a*c-b*(-4*a*c+b^2)^(1/2))-1/2*c*g*(4*a*c*(2-n)-b 
^2*(4-n))*x^4*hypergeom([1, 4/n],[(4+n)/n],-2*c*x^n/(b-(-4*a*c+b^2)^(1/...
 

3.1.9.2 Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(8737\) vs. \(2(1654)=3308\).

Time = 6.73 (sec) , antiderivative size = 8737, normalized size of antiderivative = 5.28 \[ \int \frac {d+e x+f x^2+g x^3}{\left (a+b x^n+c x^{2 n}\right )^2} \, dx=\text {Result too large to show} \]

input

Integrate[(d + e*x + f*x^2 + g*x^3)/(a + b*x^n + c*x^(2*n))^2,x]
 

output

Result too large to show
 

3.1.9.3 Rubi [A] (verified)

Time = 2.60 (sec) , antiderivative size = 1654, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {2328, 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.

\(\displaystyle \int \frac {d+e x+f x^2+g x^3}{\left (a+b x^n+c x^{2 n}\right )^2} \, dx\)

\(\Big \downarrow \) 2328

\(\displaystyle \int \left (\frac {d}{\left (a+b x^n+c x^{2 n}\right )^2}+\frac {e x}{\left (a+b x^n+c x^{2 n}\right )^2}+\frac {f x^2}{\left (a+b x^n+c x^{2 n}\right )^2}+\frac {g x^3}{\left (a+b x^n+c x^{2 n}\right )^2}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {2 b c^2 e (2-n) \operatorname {Hypergeometric2F1}\left (1,\frac {n+2}{n},2 \left (1+\frac {1}{n}\right ),-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right ) x^{n+2}}{a \left (b^2-4 a c\right )^{3/2} \left (b-\sqrt {b^2-4 a c}\right ) n (n+2)}+\frac {2 b c^2 e (2-n) \operatorname {Hypergeometric2F1}\left (1,\frac {n+2}{n},2 \left (1+\frac {1}{n}\right ),-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right ) x^{n+2}}{a \left (b^2-4 a c\right )^{3/2} \left (b+\sqrt {b^2-4 a c}\right ) n (n+2)}-\frac {2 b c^2 f (3-n) \operatorname {Hypergeometric2F1}\left (1,\frac {n+3}{n},2+\frac {3}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right ) x^{n+3}}{a \left (b^2-4 a c\right )^{3/2} \left (b-\sqrt {b^2-4 a c}\right ) n (n+3)}+\frac {2 b c^2 f (3-n) \operatorname {Hypergeometric2F1}\left (1,\frac {n+3}{n},2+\frac {3}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right ) x^{n+3}}{a \left (b^2-4 a c\right )^{3/2} \left (b+\sqrt {b^2-4 a c}\right ) n (n+3)}-\frac {2 b c^2 g (4-n) \operatorname {Hypergeometric2F1}\left (1,\frac {n+4}{n},2 \left (1+\frac {2}{n}\right ),-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right ) x^{n+4}}{a \left (b^2-4 a c\right )^{3/2} \left (b-\sqrt {b^2-4 a c}\right ) n (n+4)}+\frac {2 b c^2 g (4-n) \operatorname {Hypergeometric2F1}\left (1,\frac {n+4}{n},2 \left (1+\frac {2}{n}\right ),-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right ) x^{n+4}}{a \left (b^2-4 a c\right )^{3/2} \left (b+\sqrt {b^2-4 a c}\right ) n (n+4)}-\frac {c g \left (4 a c (2-n)-b^2 (4-n)\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {4}{n},\frac {n+4}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right ) x^4}{2 a \left (b^2-4 a c\right ) \left (b^2-\sqrt {b^2-4 a c} b-4 a c\right ) n}-\frac {c g \left (4 a c (2-n)-b^2 (4-n)\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {4}{n},\frac {n+4}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right ) x^4}{2 a \left (b^2-4 a c\right ) \left (b^2+\sqrt {b^2-4 a c} b-4 a c\right ) n}+\frac {g \left (b c x^n+b^2-2 a c\right ) x^4}{a \left (b^2-4 a c\right ) n \left (b x^n+c x^{2 n}+a\right )}-\frac {2 c f \left (2 a c (3-2 n)-b^2 (3-n)\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {3}{n},\frac {n+3}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right ) x^3}{3 a \left (b^2-4 a c\right ) \left (b^2-\sqrt {b^2-4 a c} b-4 a c\right ) n}-\frac {2 c f \left (2 a c (3-2 n)-b^2 (3-n)\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {3}{n},\frac {n+3}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right ) x^3}{3 a \left (b^2-4 a c\right ) \left (b^2+\sqrt {b^2-4 a c} b-4 a c\right ) n}+\frac {f \left (b c x^n+b^2-2 a c\right ) x^3}{a \left (b^2-4 a c\right ) n \left (b x^n+c x^{2 n}+a\right )}-\frac {c e \left (4 a c (1-n)-b^2 (2-n)\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {2}{n},\frac {n+2}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right ) x^2}{a \left (b^2-4 a c\right ) \left (b^2-\sqrt {b^2-4 a c} b-4 a c\right ) n}-\frac {c e \left (4 a c (1-n)-b^2 (2-n)\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {2}{n},\frac {n+2}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right ) x^2}{a \left (b^2-4 a c\right ) \left (b^2+\sqrt {b^2-4 a c} b-4 a c\right ) n}+\frac {e \left (b c x^n+b^2-2 a c\right ) x^2}{a \left (b^2-4 a c\right ) n \left (b x^n+c x^{2 n}+a\right )}-\frac {c d \left (-\left ((1-n) b^2\right )-\sqrt {b^2-4 a c} (1-n) b+4 a c (1-2 n)\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right ) x}{a \left (b^2-4 a c\right ) \left (b^2-\sqrt {b^2-4 a c} b-4 a c\right ) n}-\frac {c d \left (-\left ((1-n) b^2\right )+\sqrt {b^2-4 a c} (1-n) b+4 a c (1-2 n)\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right ) x}{a \left (b^2-4 a c\right ) \left (b^2+\sqrt {b^2-4 a c} b-4 a c\right ) n}+\frac {d \left (b c x^n+b^2-2 a c\right ) x}{a \left (b^2-4 a c\right ) n \left (b x^n+c x^{2 n}+a\right )}\)

input

Int[(d + e*x + f*x^2 + g*x^3)/(a + b*x^n + c*x^(2*n))^2,x]
 

output

(d*x*(b^2 - 2*a*c + b*c*x^n))/(a*(b^2 - 4*a*c)*n*(a + b*x^n + c*x^(2*n))) 
+ (e*x^2*(b^2 - 2*a*c + b*c*x^n))/(a*(b^2 - 4*a*c)*n*(a + b*x^n + c*x^(2*n 
))) + (f*x^3*(b^2 - 2*a*c + b*c*x^n))/(a*(b^2 - 4*a*c)*n*(a + b*x^n + c*x^ 
(2*n))) + (g*x^4*(b^2 - 2*a*c + b*c*x^n))/(a*(b^2 - 4*a*c)*n*(a + b*x^n + 
c*x^(2*n))) - (c*d*(4*a*c*(1 - 2*n) - b^2*(1 - n) - b*Sqrt[b^2 - 4*a*c]*(1 
 - n))*x*Hypergeometric2F1[1, n^(-1), 1 + n^(-1), (-2*c*x^n)/(b - Sqrt[b^2 
 - 4*a*c])])/(a*(b^2 - 4*a*c)*(b^2 - 4*a*c - b*Sqrt[b^2 - 4*a*c])*n) - (c* 
d*(4*a*c*(1 - 2*n) - b^2*(1 - n) + b*Sqrt[b^2 - 4*a*c]*(1 - n))*x*Hypergeo 
metric2F1[1, n^(-1), 1 + n^(-1), (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c])])/(a*( 
b^2 - 4*a*c)*(b^2 - 4*a*c + b*Sqrt[b^2 - 4*a*c])*n) - (c*e*(4*a*c*(1 - n) 
- b^2*(2 - n))*x^2*Hypergeometric2F1[1, 2/n, (2 + n)/n, (-2*c*x^n)/(b - Sq 
rt[b^2 - 4*a*c])])/(a*(b^2 - 4*a*c)*(b^2 - 4*a*c - b*Sqrt[b^2 - 4*a*c])*n) 
 - (c*e*(4*a*c*(1 - n) - b^2*(2 - n))*x^2*Hypergeometric2F1[1, 2/n, (2 + n 
)/n, (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c])])/(a*(b^2 - 4*a*c)*(b^2 - 4*a*c + 
b*Sqrt[b^2 - 4*a*c])*n) - (2*c*f*(2*a*c*(3 - 2*n) - b^2*(3 - n))*x^3*Hyper 
geometric2F1[1, 3/n, (3 + n)/n, (-2*c*x^n)/(b - Sqrt[b^2 - 4*a*c])])/(3*a* 
(b^2 - 4*a*c)*(b^2 - 4*a*c - b*Sqrt[b^2 - 4*a*c])*n) - (2*c*f*(2*a*c*(3 - 
2*n) - b^2*(3 - n))*x^3*Hypergeometric2F1[1, 3/n, (3 + n)/n, (-2*c*x^n)/(b 
 + Sqrt[b^2 - 4*a*c])])/(3*a*(b^2 - 4*a*c)*(b^2 - 4*a*c + b*Sqrt[b^2 - 4*a 
*c])*n) - (c*g*(4*a*c*(2 - n) - b^2*(4 - n))*x^4*Hypergeometric2F1[1, 4...
 

3.1.9.3.1 Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2328

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.))^(p_), x_Symbol] :> 
Int[ExpandIntegrand[Pq*(a + b*x^n + c*x^(2*n))^p, x], x] /; FreeQ[{a, b, c, 
 n}, x] && EqQ[n2, 2*n] && PolyQ[Pq, x] && ILtQ[p, -1]
 

3.1.9.4 Maple [F]

\[\int \frac {g \,x^{3}+f \,x^{2}+e x +d}{\left (a +b \,x^{n}+c \,x^{2 n}\right )^{2}}d x\]

input

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

output

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

3.1.9.5 Fricas [F]

\[ \int \frac {d+e x+f x^2+g x^3}{\left (a+b x^n+c x^{2 n}\right )^2} \, dx=\int { \frac {g x^{3} + f x^{2} + e x + d}{{\left (c x^{2 \, n} + b x^{n} + a\right )}^{2}} \,d x } \]

input

integrate((g*x^3+f*x^2+e*x+d)/(a+b*x^n+c*x^(2*n))^2,x, algorithm="fricas")
 

output

integral((g*x^3 + f*x^2 + e*x + d)/(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)
 

3.1.9.6 Sympy [F(-1)]

Timed out. \[ \int \frac {d+e x+f x^2+g x^3}{\left (a+b x^n+c x^{2 n}\right )^2} \, dx=\text {Timed out} \]

input

integrate((g*x**3+f*x**2+e*x+d)/(a+b*x**n+c*x**(2*n))**2,x)
 

output

Timed out
 

3.1.9.7 Maxima [F]

\[ \int \frac {d+e x+f x^2+g x^3}{\left (a+b x^n+c x^{2 n}\right )^2} \, dx=\int { \frac {g x^{3} + f x^{2} + e x + d}{{\left (c x^{2 \, n} + b x^{n} + a\right )}^{2}} \,d x } \]

input

integrate((g*x^3+f*x^2+e*x+d)/(a+b*x^n+c*x^(2*n))^2,x, algorithm="maxima")
 

output

((b^2*g - 2*a*c*g)*x^4 + (b^2*f - 2*a*c*f)*x^3 + (b^2*e - 2*a*c*e)*x^2 + ( 
b*c*g*x^4 + b*c*f*x^3 + b*c*e*x^2 + b*c*d*x)*x^n + (b^2*d - 2*a*c*d)*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*a*c*d*(2*n - 1) - b^2*d*(n - 1) + (4*a*c*g*(n 
 - 2) - b^2*g*(n - 4))*x^3 + (2*a*c*f*(2*n - 3) - b^2*f*(n - 3))*x^2 - (b* 
c*g*(n - 4)*x^3 + b*c*f*(n - 3)*x^2 + b*c*e*(n - 2)*x + b*c*d*(n - 1))*x^n 
 + (4*a*c*e*(n - 1) - b^2*e*(n - 2))*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), x)
 

3.1.9.8 Giac [F]

\[ \int \frac {d+e x+f x^2+g x^3}{\left (a+b x^n+c x^{2 n}\right )^2} \, dx=\int { \frac {g x^{3} + f x^{2} + e x + d}{{\left (c x^{2 \, n} + b x^{n} + a\right )}^{2}} \,d x } \]

input

integrate((g*x^3+f*x^2+e*x+d)/(a+b*x^n+c*x^(2*n))^2,x, algorithm="giac")
 

output

integrate((g*x^3 + f*x^2 + e*x + d)/(c*x^(2*n) + b*x^n + a)^2, x)
 

3.1.9.9 Mupad [F(-1)]

Timed out. \[ \int \frac {d+e x+f x^2+g x^3}{\left (a+b x^n+c x^{2 n}\right )^2} \, dx=\int \frac {g\,x^3+f\,x^2+e\,x+d}{{\left (a+b\,x^n+c\,x^{2\,n}\right )}^2} \,d x \]

input

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

output

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