\(\int \frac {x^n}{(d+e x^n)^2 (a+b x^n+c x^{2 n})} \, dx\) [165]

Optimal result

Integrand size = 29, antiderivative size = 354 \[ \int \frac {x^n}{\left (d+e x^n\right )^2 \left (a+b x^n+c x^{2 n}\right )} \, dx=-\frac {e x}{\left (c d^2-b d e+a e^2\right ) n \left (d+e x^n\right )}+\frac {c \left (a \sqrt {b^2-4 a c} e^2-c d \left (\sqrt {b^2-4 a c} d+4 a e\right )+b \left (c d^2+a e^2\right )\right ) x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c-b \sqrt {b^2-4 a c}\right ) \left (c d^2-b d e+a e^2\right )^2}+\frac {c \left (c d^2-a e^2+\frac {b c d^2-4 a c d e+a b e^2}{\sqrt {b^2-4 a c}}\right ) x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{\left (b+\sqrt {b^2-4 a c}\right ) \left (c d^2-b d e+a e^2\right )^2}+\frac {e \left (e (a e-b d (1-n))+c d^2 (1-2 n)\right ) x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {e x^n}{d}\right )}{d \left (c d^2-b d e+a e^2\right )^2 n} \] Output:

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

Mathematica [A] (verified)

Time = 1.21 (sec) , antiderivative size = 326, normalized size of antiderivative = 0.92 \[ \int \frac {x^n}{\left (d+e x^n\right )^2 \left (a+b x^n+c x^{2 n}\right )} \, dx=\frac {x^{1+n} \left (-\frac {c \left (2 c d e-b e^2-\frac {2 c^2 d^2+b^2 e^2-2 c e (b d+a e)}{\sqrt {b^2-4 a c}}\right ) \operatorname {Hypergeometric2F1}\left (1,1+\frac {1}{n},2+\frac {1}{n},\frac {2 c x^n}{-b+\sqrt {b^2-4 a c}}\right )}{b-\sqrt {b^2-4 a c}}-\frac {c \left (2 c d e-b e^2+\frac {2 c^2 d^2+b^2 e^2-2 c e (b d+a e)}{\sqrt {b^2-4 a c}}\right ) \operatorname {Hypergeometric2F1}\left (1,1+\frac {1}{n},2+\frac {1}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{b+\sqrt {b^2-4 a c}}+\frac {e^2 (2 c d-b e) \operatorname {Hypergeometric2F1}\left (1,1+\frac {1}{n},2+\frac {1}{n},-\frac {e x^n}{d}\right )}{d}+\frac {e^2 \left (c d^2+e (-b d+a e)\right ) \operatorname {Hypergeometric2F1}\left (2,1+\frac {1}{n},2+\frac {1}{n},-\frac {e x^n}{d}\right )}{d^2}\right )}{\left (c d^2+e (-b d+a e)\right )^2 (1+n)} \] Input:

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

Output:

(x^(1 + n)*(-((c*(2*c*d*e - b*e^2 - (2*c^2*d^2 + b^2*e^2 - 2*c*e*(b*d + a* 
e))/Sqrt[b^2 - 4*a*c])*Hypergeometric2F1[1, 1 + n^(-1), 2 + n^(-1), (2*c*x 
^n)/(-b + Sqrt[b^2 - 4*a*c])])/(b - Sqrt[b^2 - 4*a*c])) - (c*(2*c*d*e - b* 
e^2 + (2*c^2*d^2 + b^2*e^2 - 2*c*e*(b*d + a*e))/Sqrt[b^2 - 4*a*c])*Hyperge 
ometric2F1[1, 1 + n^(-1), 2 + n^(-1), (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c])]) 
/(b + Sqrt[b^2 - 4*a*c]) + (e^2*(2*c*d - b*e)*Hypergeometric2F1[1, 1 + n^( 
-1), 2 + n^(-1), -((e*x^n)/d)])/d + (e^2*(c*d^2 + e*(-(b*d) + a*e))*Hyperg 
eometric2F1[2, 1 + n^(-1), 2 + n^(-1), -((e*x^n)/d)])/d^2))/((c*d^2 + e*(- 
(b*d) + a*e))^2*(1 + n))
 

Rubi [A] (verified)

Time = 1.17 (sec) , antiderivative size = 559, normalized size of antiderivative = 1.58, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {1880, 1006, 1067, 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.

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

\(\Big \downarrow \) 1880

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

\(\Big \downarrow \) 1006

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

\(\Big \downarrow \) 1067

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

(2*c*(-((e*x^(1 + n))/(d*(2*c*d - b*e + Sqrt[b^2 - 4*a*c]*e)*n*(d + e*x^n) 
)) + ((4*c^2*d*n*x^(1 + n)*Hypergeometric2F1[1, 1 + n^(-1), 2 + 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 + n)) - (e*((b - Sqrt[b^2 - 4*a*c])*e - 2*c*d*(1 
- n))*x^(1 + n)*Hypergeometric2F1[1, 1 + n^(-1), 2 + n^(-1), -((e*x^n)/d)] 
)/(d*(2*c*d - (b - Sqrt[b^2 - 4*a*c])*e)*(1 + n)))/(d*(2*c*d - b*e + Sqrt[ 
b^2 - 4*a*c]*e)*n)))/Sqrt[b^2 - 4*a*c] - (2*c*(-((e*x^(1 + n))/(d*(2*c*d - 
 (b + Sqrt[b^2 - 4*a*c])*e)*n*(d + e*x^n))) + ((4*c^2*d*n*x^(1 + n)*Hyperg 
eometric2F1[1, 1 + n^(-1), 2 + 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 + n)) - 
(e*((b + Sqrt[b^2 - 4*a*c])*e - 2*c*d*(1 - n))*x^(1 + n)*Hypergeometric2F1 
[1, 1 + n^(-1), 2 + n^(-1), -((e*x^n)/d)])/(d*(2*c*d - (b + Sqrt[b^2 - 4*a 
*c])*e)*(1 + n)))/(d*(2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)*n)))/Sqrt[b^2 - 4 
*a*c]
 

Defintions of rubi rules used

rule 1006

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_ 
))^(q_), x_Symbol] :> Simp[(-b)*(e*x)^(m + 1)*(a + b*x^n)^(p + 1)*((c + d*x 
^n)^(q + 1)/(a*e*n*(b*c - a*d)*(p + 1))), x] + Simp[1/(a*n*(b*c - a*d)*(p + 
 1))   Int[(e*x)^m*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*b*(m + 1) + n*( 
b*c - a*d)*(p + 1) + d*b*(m + n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{ 
a, b, c, d, e, m, n, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && IntBinomi 
alQ[a, b, c, d, e, m, n, p, q, x]
 

rule 1067

Int[(((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((e_) + (f_.)*(x_)^(n 
_)))/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Int[ExpandIntegrand[(g*x)^m*(a 
+ b*x^n)^p*((e + f*x^n)/(c + d*x^n)), x], x] /; FreeQ[{a, b, c, d, e, f, g, 
 m, n, p}, x]
 

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]
 

rule 2009

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

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-2)]

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

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

Output:

Exception raised: HeuristicGCDFailed >> no luck
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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