\(\int \frac {1}{(d+e x^n) (a+b x^n+c x^{2 n})^2} \, dx\) [72]

Optimal result

Integrand size = 26, antiderivative size = 889 \[ \int \frac {1}{\left (d+e x^n\right ) \left (a+b x^n+c x^{2 n}\right )^2} \, dx=\frac {x \left (c \left (b^2-2 a c\right ) d-b \left (b^2-3 a c\right ) e+c \left (b c d-b^2 e+2 a c e\right ) x^n\right )}{a \left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right ) n \left (a+b x^n+c x^{2 n}\right )}+\frac {c \left (b c \left (a e^2 \left (2 a e (2-5 n)-\sqrt {b^2-4 a c} d (1-n)\right )+c d^2 \left (2 a e (4-7 n)+\sqrt {b^2-4 a c} d (1-n)\right )\right )-2 a c \left (c d e \left (2 a e (1-4 n)-\sqrt {b^2-4 a c} d (1-n)\right )-a \sqrt {b^2-4 a c} e^3 (1-3 n)+2 c^2 d^3 (1-2 n)\right )-b^3 e \left (e \left (a e (1-2 n)-\sqrt {b^2-4 a c} d (1-n)\right )+2 c d^2 (1-n)\right )-b^2 \left (a \sqrt {b^2-4 a c} e^3 (1-2 n)-c^2 d^3 (1-n)+c d e \left (2 \sqrt {b^2-4 a c} d+3 a e\right ) (1-n)\right )+b^4 d e^2 (1-n)\right ) x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{a \left (b^2-4 a c\right ) \left (b^2-4 a c-b \sqrt {b^2-4 a c}\right ) \left (c d^2-b d e+a e^2\right )^2 n}+\frac {c \left (b c \left (c d^2 \left (2 a e (4-7 n)-\sqrt {b^2-4 a c} d (1-n)\right )+a e^2 \left (2 a e (2-5 n)+\sqrt {b^2-4 a c} d (1-n)\right )\right )-2 a c \left (c d e \left (2 a e (1-4 n)+\sqrt {b^2-4 a c} d (1-n)\right )+a \sqrt {b^2-4 a c} e^3 (1-3 n)+2 c^2 d^3 (1-2 n)\right )-b^3 e \left (e \left (a e (1-2 n)+\sqrt {b^2-4 a c} d (1-n)\right )+2 c d^2 (1-n)\right )+b^2 \left (a \sqrt {b^2-4 a c} e^3 (1-2 n)+c^2 d^3 (1-n)+c d e \left (2 \sqrt {b^2-4 a c} d-3 a e\right ) (1-n)\right )+b^4 d e^2 (1-n)\right ) x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{a \left (b^2-4 a c\right ) \left (b^2-4 a c+b \sqrt {b^2-4 a c}\right ) \left (c d^2-b d e+a e^2\right )^2 n}+\frac {e^4 x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {e x^n}{d}\right )}{d \left (c d^2-b d e+a e^2\right )^2} \] Output:

x*(c*(-2*a*c+b^2)*d-b*(-3*a*c+b^2)*e+c*(2*a*c*e-b^2*e+b*c*d)*x^n)/a/(-4*a* 
c+b^2)/(a*e^2-b*d*e+c*d^2)/n/(a+b*x^n+c*x^(2*n))+c*(b*c*(a*e^2*(2*a*e*(2-5 
*n)-(-4*a*c+b^2)^(1/2)*d*(1-n))+c*d^2*(2*a*e*(4-7*n)+(-4*a*c+b^2)^(1/2)*d* 
(1-n)))-2*a*c*(c*d*e*(2*a*e*(1-4*n)-(-4*a*c+b^2)^(1/2)*d*(1-n))-a*(-4*a*c+ 
b^2)^(1/2)*e^3*(1-3*n)+2*c^2*d^3*(1-2*n))-b^3*e*(e*(a*e*(1-2*n)-(-4*a*c+b^ 
2)^(1/2)*d*(1-n))+2*c*d^2*(1-n))-b^2*(a*(-4*a*c+b^2)^(1/2)*e^3*(1-2*n)-c^2 
*d^3*(1-n)+c*d*e*(2*(-4*a*c+b^2)^(1/2)*d+3*a*e)*(1-n))+b^4*d*e^2*(1-n))*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^2-4*a*c-b*(-4*a*c+b^2)^(1/2))/(a*e^2-b*d*e+c*d^2)^2/n+c*(b*c*(c*d^2*(2 
*a*e*(4-7*n)-(-4*a*c+b^2)^(1/2)*d*(1-n))+a*e^2*(2*a*e*(2-5*n)+(-4*a*c+b^2) 
^(1/2)*d*(1-n)))-2*a*c*(c*d*e*(2*a*e*(1-4*n)+(-4*a*c+b^2)^(1/2)*d*(1-n))+a 
*(-4*a*c+b^2)^(1/2)*e^3*(1-3*n)+2*c^2*d^3*(1-2*n))-b^3*e*(e*(a*e*(1-2*n)+( 
-4*a*c+b^2)^(1/2)*d*(1-n))+2*c*d^2*(1-n))+b^2*(a*(-4*a*c+b^2)^(1/2)*e^3*(1 
-2*n)+c^2*d^3*(1-n)+c*d*e*(2*(-4*a*c+b^2)^(1/2)*d-3*a*e)*(1-n))+b^4*d*e^2* 
(1-n))*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)-4*a*c+b^2)/(a*e^2-b*d*e+c*d^2)^2/n+e^4*x*h 
ypergeom([1, 1/n],[1+1/n],-e*x^n/d)/d/(a*e^2-b*d*e+c*d^2)^2
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(11767\) vs. \(2(889)=1778\).

Time = 7.46 (sec) , antiderivative size = 11767, normalized size of antiderivative = 13.24 \[ \int \frac {1}{\left (d+e x^n\right ) \left (a+b x^n+c x^{2 n}\right )^2} \, dx=\text {Result too large to show} \] Input:

Integrate[1/((d + e*x^n)*(a + b*x^n + c*x^(2*n))^2),x]
 

Output:

Result too large to show
 

Rubi [A] (verified)

Time = 1.67 (sec) , antiderivative size = 726, normalized size of antiderivative = 0.82, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {1766, 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.

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

\(\Big \downarrow \) 1766

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

\(\Big \downarrow \) 2009

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

Input:

Int[1/((d + e*x^n)*(a + b*x^n + c*x^(2*n))^2),x]
 

Output:

(x*(b^2*c*d - 2*a*c^2*d - b^3*e + 3*a*b*c*e + c*(b*c*d - b^2*e + 2*a*c*e)* 
x^n))/(a*(b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^2)*n*(a + b*x^n + c*x^(2*n))) 
- (c*e^2*(2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)*x*Hypergeometric2F1[1, n^(-1) 
, 1 + n^(-1), (-2*c*x^n)/(b - Sqrt[b^2 - 4*a*c])])/((b^2 - 4*a*c - b*Sqrt[ 
b^2 - 4*a*c])*(c*d^2 - b*d*e + a*e^2)^2) - (c*((2*a*b*c*e*(2 - 3*n) - 4*a* 
c^2*d*(1 - 2*n) + b^2*c*d*(1 - n) - b^3*e*(1 - n))/Sqrt[b^2 - 4*a*c] + (b* 
c*d - b^2*e + 2*a*c*e)*(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 - Sqrt[b^2 - 4*a 
*c])*(c*d^2 - b*d*e + a*e^2)*n) - (c*e^2*(2*c*d - (b - Sqrt[b^2 - 4*a*c])* 
e)*x*Hypergeometric2F1[1, n^(-1), 1 + n^(-1), (-2*c*x^n)/(b + Sqrt[b^2 - 4 
*a*c])])/((b^2 - 4*a*c + b*Sqrt[b^2 - 4*a*c])*(c*d^2 - b*d*e + a*e^2)^2) + 
 (c*(b*c*(2*a*e*(2 - 3*n) - Sqrt[b^2 - 4*a*c]*d*(1 - n)) - 2*a*c*(2*c*d*(1 
 - 2*n) + Sqrt[b^2 - 4*a*c]*e*(1 - n)) - b^3*e*(1 - n) + b^2*(c*d + Sqrt[b 
^2 - 4*a*c]*e)*(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])*(c*d^2 - b*d*e + a*e^2)*n) + (e^4*x*Hypergeometric2F1[1, n^(-1), 
1 + n^(-1), -((e*x^n)/d)])/(d*(c*d^2 - b*d*e + a*e^2)^2)
 

Defintions of rubi rules used

rule 1766

Int[((d_) + (e_.)*(x_)^(n_))^(q_)*((a_) + (b_.)*(x_)^(n_) + (c_.)*(x_)^(n2_ 
))^(p_), x_Symbol] :> Int[ExpandIntegrand[(d + e*x^n)^q*(a + b*x^n + c*x^(2 
*n))^p, x], x] /; FreeQ[{a, b, c, d, e, n, p, q}, x] && EqQ[n2, 2*n] && NeQ 
[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && ((IntegersQ[p, q] && 
!IntegerQ[n]) || IGtQ[p, 0] || (IGtQ[q, 0] &&  !IntegerQ[n]))
 

rule 2009

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

Maple [F]

\[\int \frac {1}{\left (d +e \,x^{n}\right ) \left (a +b \,x^{n}+c \,x^{2 n}\right )^{2}}d x\]

Input:

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

Output:

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

Fricas [F]

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

integrate(1/(d+e*x^n)/(a+b*x^n+c*x^(2*n))^2,x, algorithm="fricas")
 

Output:

integral(1/(b^2*e*x^(3*n) + a^2*d + (c^2*e*x^n + c^2*d)*x^(4*n) + 2*(b*c*e 
*x^(2*n) + a*c*d + (b*c*d + a*c*e)*x^n)*x^(2*n) + (b^2*d + 2*a*b*e)*x^(2*n 
) + (2*a*b*d + a^2*e)*x^n), x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{\left (d+e x^n\right ) \left (a+b x^n+c x^{2 n}\right )^2} \, dx=\text {Timed out} \] Input:

integrate(1/(d+e*x**n)/(a+b*x**n+c*x**(2*n))**2,x)
 

Output:

Timed out
 

Maxima [F]

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

integrate(1/(d+e*x^n)/(a+b*x^n+c*x^(2*n))^2,x, algorithm="maxima")
 

Output:

e^4*integrate(1/(c^2*d^5 - 2*b*c*d^4*e + b^2*d^3*e^2 + a^2*d*e^4 + 2*(c*d^ 
3*e^2 - b*d^2*e^3)*a + (c^2*d^4*e - 2*b*c*d^3*e^2 + b^2*d^2*e^3 + a^2*e^5 
+ 2*(c*d^2*e^3 - b*d*e^4)*a)*x^n), x) - ((b*c^2*d - b^2*c*e + 2*a*c^2*e)*x 
*x^n + (b^2*c*d - b^3*e - (2*c^2*d - 3*b*c*e)*a)*x)/(4*a^4*c*e^2*n + (4*c^ 
2*d^2*n - 4*b*c*d*e*n - b^2*e^2*n)*a^3 - (b^2*c*d^2*n - b^3*d*e*n)*a^2 + ( 
4*a^3*c^2*e^2*n + (4*c^3*d^2*n - 4*b*c^2*d*e*n - b^2*c*e^2*n)*a^2 - (b^2*c 
^2*d^2*n - b^3*c*d*e*n)*a)*x^(2*n) + (4*a^3*b*c*e^2*n + (4*b*c^2*d^2*n - 4 
*b^2*c*d*e*n - b^3*e^2*n)*a^2 - (b^3*c*d^2*n - b^4*d*e*n)*a)*x^n) - integr 
ate((b^2*c^2*d^3*(n - 1) - 2*b^3*c*d^2*e*(n - 1) + b^4*d*e^2*(n - 1) + (b* 
c*e^3*(8*n - 3) - 2*c^2*d*e^2*(4*n - 1))*a^2 + (b*c^2*d^2*e*(8*n - 5) - 2* 
c^3*d^3*(2*n - 1) - b^3*e^3*(2*n - 1) - 2*b^2*c*d*e^2*(n - 1))*a + (2*a^2* 
c^2*e^3*(3*n - 1) + b*c^3*d^3*(n - 1) - 2*b^2*c^2*d^2*e*(n - 1) + b^3*c*d* 
e^2*(n - 1) - (b^2*c*e^3*(2*n - 1) - 2*c^3*d^2*e*(n - 1) + b*c^2*d*e^2*(n 
- 1))*a)*x^n)/(4*a^5*c*e^4*n + (8*c^2*d^2*e^2*n - 8*b*c*d*e^3*n - b^2*e^4* 
n)*a^4 + 2*(2*c^3*d^4*n - 4*b*c^2*d^3*e*n + b^2*c*d^2*e^2*n + b^3*d*e^3*n) 
*a^3 - (b^2*c^2*d^4*n - 2*b^3*c*d^3*e*n + b^4*d^2*e^2*n)*a^2 + (4*a^4*c^2* 
e^4*n + (8*c^3*d^2*e^2*n - 8*b*c^2*d*e^3*n - b^2*c*e^4*n)*a^3 + 2*(2*c^4*d 
^4*n - 4*b*c^3*d^3*e*n + b^2*c^2*d^2*e^2*n + b^3*c*d*e^3*n)*a^2 - (b^2*c^3 
*d^4*n - 2*b^3*c^2*d^3*e*n + b^4*c*d^2*e^2*n)*a)*x^(2*n) + (4*a^4*b*c*e^4* 
n + (8*b*c^2*d^2*e^2*n - 8*b^2*c*d*e^3*n - b^3*e^4*n)*a^3 + 2*(2*b*c^3*...
 

Giac [F]

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

integrate(1/(d+e*x^n)/(a+b*x^n+c*x^(2*n))^2,x, algorithm="giac")
 

Output:

integrate(1/((c*x^(2*n) + b*x^n + a)^2*(e*x^n + d)), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\left (d+e x^n\right ) \left (a+b x^n+c x^{2 n}\right )^2} \, dx=\int \frac {1}{\left (d+e\,x^n\right )\,{\left (a+b\,x^n+c\,x^{2\,n}\right )}^2} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {1}{\left (d+e x^n\right ) \left (a+b x^n+c x^{2 n}\right )^2} \, dx=\int \frac {1}{x^{5 n} c^{2} e +2 x^{4 n} b c e +x^{4 n} c^{2} d +2 x^{3 n} a c e +x^{3 n} b^{2} e +2 x^{3 n} b c d +2 x^{2 n} a b e +2 x^{2 n} a c d +x^{2 n} b^{2} d +x^{n} a^{2} e +2 x^{n} a b d +a^{2} d}d x \] Input:

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

Output:

int(1/(x**(5*n)*c**2*e + 2*x**(4*n)*b*c*e + x**(4*n)*c**2*d + 2*x**(3*n)*a 
*c*e + x**(3*n)*b**2*e + 2*x**(3*n)*b*c*d + 2*x**(2*n)*a*b*e + 2*x**(2*n)* 
a*c*d + x**(2*n)*b**2*d + x**n*a**2*e + 2*x**n*a*b*d + a**2*d),x)