\(\int \frac {x^{2 n}}{(d+e x^n) (a+b x^n+c x^{2 n})} \, dx\) [157]

Optimal result

Integrand size = 31, antiderivative size = 284 \[ \int \frac {x^{2 n}}{\left (d+e x^n\right ) \left (a+b x^n+c x^{2 n}\right )} \, dx=-\frac {c \left (e-\frac {2 c d-b e}{\sqrt {b^2-4 a c}}\right ) x^{1+2 n} \operatorname {Hypergeometric2F1}\left (1,2+\frac {1}{n},3+\frac {1}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{\left (b-\sqrt {b^2-4 a c}\right ) \left (c d^2-b d e+a e^2\right ) (1+2 n)}-\frac {c \left (e+\frac {2 c d-b e}{\sqrt {b^2-4 a c}}\right ) x^{1+2 n} \operatorname {Hypergeometric2F1}\left (1,2+\frac {1}{n},3+\frac {1}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{\left (b+\sqrt {b^2-4 a c}\right ) \left (c d^2-b d e+a e^2\right ) (1+2 n)}+\frac {e^2 x^{1+2 n} \operatorname {Hypergeometric2F1}\left (1,2+\frac {1}{n},3+\frac {1}{n},-\frac {e x^n}{d}\right )}{d \left (c d^2-b d e+a e^2\right ) (1+2 n)} \] Output:

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

Mathematica [A] (verified)

Time = 0.95 (sec) , antiderivative size = 219, normalized size of antiderivative = 0.77 \[ \int \frac {x^{2 n}}{\left (d+e x^n\right ) \left (a+b x^n+c x^{2 n}\right )} \, dx=\frac {x^{1+2 n} \left (-\frac {c \left (e+\frac {-2 c d+b e}{\sqrt {b^2-4 a c}}\right ) \operatorname {Hypergeometric2F1}\left (1,2+\frac {1}{n},3+\frac {1}{n},\frac {2 c x^n}{-b+\sqrt {b^2-4 a c}}\right )}{b-\sqrt {b^2-4 a c}}-\frac {c \left (e+\frac {2 c d-b e}{\sqrt {b^2-4 a c}}\right ) \operatorname {Hypergeometric2F1}\left (1,2+\frac {1}{n},3+\frac {1}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{b+\sqrt {b^2-4 a c}}+\frac {e^2 \operatorname {Hypergeometric2F1}\left (1,2+\frac {1}{n},3+\frac {1}{n},-\frac {e x^n}{d}\right )}{d}\right )}{\left (c d^2+e (-b d+a e)\right ) (1+2 n)} \] Input:

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

Output:

(x^(1 + 2*n)*(-((c*(e + (-2*c*d + b*e)/Sqrt[b^2 - 4*a*c])*Hypergeometric2F 
1[1, 2 + n^(-1), 3 + n^(-1), (2*c*x^n)/(-b + Sqrt[b^2 - 4*a*c])])/(b - Sqr 
t[b^2 - 4*a*c])) - (c*(e + (2*c*d - b*e)/Sqrt[b^2 - 4*a*c])*Hypergeometric 
2F1[1, 2 + n^(-1), 3 + n^(-1), (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c])])/(b + S 
qrt[b^2 - 4*a*c]) + (e^2*Hypergeometric2F1[1, 2 + n^(-1), 3 + n^(-1), -((e 
*x^n)/d)])/d))/((c*d^2 + e*(-(b*d) + a*e))*(1 + 2*n))
 

Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 353, normalized size of antiderivative = 1.24, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {1880, 1010, 888}

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.

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

\(\Big \downarrow \) 1880

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

\(\Big \downarrow \) 1010

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

\(\Big \downarrow \) 888

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 888

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p 
*((c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/n, (m + 1)/n + 1 
, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] && (ILt 
Q[p, 0] || GtQ[a, 0])
 

rule 1010

Int[((e_.)*(x_))^(m_.)/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), 
 x_Symbol] :> Simp[b/(b*c - a*d)   Int[(e*x)^m/(a + b*x^n), x], x] - Simp[d 
/(b*c - a*d)   Int[(e*x)^m/(c + d*x^n), x], x] /; FreeQ[{a, b, c, d, e, n, 
m}, x] && NeQ[b*c - a*d, 0]
 

rule 1880

Int[(((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(n_))^(q_))/((a_) + (c_.)*(x_)^( 
n2_.) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{r = Rt[b^2 - 4*a*c, 2]}, Simp[ 
2*(c/r)   Int[(f*x)^m*((d + e*x^n)^q/(b - r + 2*c*x^n)), x], x] - Simp[2*(c 
/r)   Int[(f*x)^m*((d + e*x^n)^q/(b + r + 2*c*x^n)), x], x]] /; FreeQ[{a, b 
, c, d, e, f, m, n, q}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] &&  !Rati 
onalQ[n]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-2)]

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

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

Output:

Exception raised: HeuristicGCDFailed >> no luck
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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