\(\int \frac {d+e x^n}{(a+b x^n+c x^{2 n})^2} \, dx\) [71]

Optimal result

Integrand size = 24, antiderivative size = 362 \[ \int \frac {d+e x^n}{\left (a+b x^n+c x^{2 n}\right )^2} \, dx=\frac {x \left (b^2 d-2 a c d-a b e+c (b d-2 a e) x^n\right )}{a \left (b^2-4 a c\right ) n \left (a+b x^n+c x^{2 n}\right )}-\frac {c \left (2 a \left (2 c d (1-2 n)+\sqrt {b^2-4 a c} e (1-n)\right )-b^2 (d-d n)-b \left (\sqrt {b^2-4 a c} d (1-n)-2 a e n\right )\right ) x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{a \left (b^2-4 a c\right ) \left (b^2-4 a c-b \sqrt {b^2-4 a c}\right ) n}-\frac {c \left (2 a \left (c d (2-4 n)-\sqrt {b^2-4 a c} e (1-n)\right )-b^2 d (1-n)+b \left (\sqrt {b^2-4 a c} d (1-n)+2 a e n\right )\right ) x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{a \left (b^2-4 a c\right ) \left (b^2-4 a c+b \sqrt {b^2-4 a c}\right ) n} \] Output:

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

Mathematica [A] (warning: unable to verify)

Time = 6.63 (sec) , antiderivative size = 603, normalized size of antiderivative = 1.67 \[ \int \frac {d+e x^n}{\left (a+b x^n+c x^{2 n}\right )^2} \, dx=\frac {c x \left (\frac {4 \left (b^2-4 a c\right ) \left (b^2 d (-1+n) x^n \left (b+c x^n\right )-2 a^2 c \left (2 d n+e x^n\right )+a \left (-2 c^2 d (-1+2 n) x^{2 n}+b c x^n \left (3 d-4 d n+e x^n\right )+b^2 \left (d n+e x^n\right )\right )\right )}{\left (b^2-4 a c-b \sqrt {b^2-4 a c}\right ) \left (b^2-4 a c+b \sqrt {b^2-4 a c}\right ) \left (a+x^n \left (b+c x^n\right )\right )}+\frac {2^{-1/n} \left (4 a c \left (\sqrt {b^2-4 a c} d (1-2 n)+2 a e (-1+n)\right )+b^3 d (-1+n)+b^2 \left (\sqrt {b^2-4 a c} d-2 a e\right ) (-1+n)+2 a b \left (-2 c d (-1+n)+\sqrt {b^2-4 a c} e n\right )\right ) \left (\frac {c x^n}{b-\sqrt {b^2-4 a c}+2 c x^n}\right )^{-1/n} \operatorname {Hypergeometric2F1}\left (-\frac {1}{n},-\frac {1}{n},\frac {-1+n}{n},\frac {b-\sqrt {b^2-4 a c}}{b-\sqrt {b^2-4 a c}+2 c x^n}\right )}{\sqrt {b^2-4 a c} \left (-b^2+4 a c+b \sqrt {b^2-4 a c}\right )}+\frac {2^{-1/n} \left (b \sqrt {b^2-4 a c} d (-1+n)-2 a \sqrt {b^2-4 a c} e (-1+n)-2 a b e n+4 a c d (-1+2 n)+b^2 (d-d n)\right ) \left (\frac {c x^n}{b+\sqrt {b^2-4 a c}+2 c x^n}\right )^{-1/n} \operatorname {Hypergeometric2F1}\left (-\frac {1}{n},-\frac {1}{n},\frac {-1+n}{n},\frac {b+\sqrt {b^2-4 a c}}{b+\sqrt {b^2-4 a c}+2 c x^n}\right )}{\sqrt {b^2-4 a c} \left (b+\sqrt {b^2-4 a c}\right )}\right )}{a \left (-b^2+4 a c\right ) n} \] Input:

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

Output:

(c*x*((4*(b^2 - 4*a*c)*(b^2*d*(-1 + n)*x^n*(b + c*x^n) - 2*a^2*c*(2*d*n + 
e*x^n) + a*(-2*c^2*d*(-1 + 2*n)*x^(2*n) + b*c*x^n*(3*d - 4*d*n + e*x^n) + 
b^2*(d*n + e*x^n))))/((b^2 - 4*a*c - b*Sqrt[b^2 - 4*a*c])*(b^2 - 4*a*c + b 
*Sqrt[b^2 - 4*a*c])*(a + x^n*(b + c*x^n))) + ((4*a*c*(Sqrt[b^2 - 4*a*c]*d* 
(1 - 2*n) + 2*a*e*(-1 + n)) + b^3*d*(-1 + n) + b^2*(Sqrt[b^2 - 4*a*c]*d - 
2*a*e)*(-1 + n) + 2*a*b*(-2*c*d*(-1 + n) + Sqrt[b^2 - 4*a*c]*e*n))*Hyperge 
ometric2F1[-n^(-1), -n^(-1), (-1 + n)/n, (b - Sqrt[b^2 - 4*a*c])/(b - Sqrt 
[b^2 - 4*a*c] + 2*c*x^n)])/(2^n^(-1)*Sqrt[b^2 - 4*a*c]*(-b^2 + 4*a*c + b*S 
qrt[b^2 - 4*a*c])*((c*x^n)/(b - Sqrt[b^2 - 4*a*c] + 2*c*x^n))^n^(-1)) + (( 
b*Sqrt[b^2 - 4*a*c]*d*(-1 + n) - 2*a*Sqrt[b^2 - 4*a*c]*e*(-1 + n) - 2*a*b* 
e*n + 4*a*c*d*(-1 + 2*n) + b^2*(d - d*n))*Hypergeometric2F1[-n^(-1), -n^(- 
1), (-1 + n)/n, (b + Sqrt[b^2 - 4*a*c])/(b + Sqrt[b^2 - 4*a*c] + 2*c*x^n)] 
)/(2^n^(-1)*Sqrt[b^2 - 4*a*c]*(b + Sqrt[b^2 - 4*a*c])*((c*x^n)/(b + Sqrt[b 
^2 - 4*a*c] + 2*c*x^n))^n^(-1))))/(a*(-b^2 + 4*a*c)*n)
 

Rubi [A] (verified)

Time = 1.19 (sec) , antiderivative size = 312, normalized size of antiderivative = 0.86, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {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.

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

\(\Big \downarrow \) 1760

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 1752

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

\(\Big \downarrow \) 778

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

Input:

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

Output:

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

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral((e*x^n + 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)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

((b*c*d - 2*a*c*e)*x*x^n + (b^2*d - (2*c*d + b*e)*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((b^2*d*(n - 1) - (2*c*d*(2*n - 1) - b*e)*a + (b*c*d*(n - 1) - 2* 
a*c*e*(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]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

int(x**n/(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)*e + 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)*d