\(\int \frac {1}{(d+e x^n)^2 (a+b x^n+c x^{2 n})} \, dx\) [166]

Optimal result

Integrand size = 26, antiderivative size = 387 \[ \int \frac {1}{\left (d+e x^n\right )^2 \left (a+b x^n+c x^{2 n}\right )} \, dx=\frac {e^2 x}{d \left (c d^2-b d e+a e^2\right ) n \left (d+e x^n\right )}-\frac {c \left (2 c^2 d^2+b \left (b+\sqrt {b^2-4 a c}\right ) e^2-2 c e \left (b d+\sqrt {b^2-4 a c} d+a e\right )\right ) x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c-b \sqrt {b^2-4 a c}\right ) \left (c d^2-b d e+a e^2\right )^2}-\frac {c \left (2 c^2 d^2+b \left (b-\sqrt {b^2-4 a c}\right ) e^2-2 c e \left (b d-\sqrt {b^2-4 a c} d+a e\right )\right ) x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c+b \sqrt {b^2-4 a c}\right ) \left (c d^2-b d e+a e^2\right )^2}-\frac {e^2 \left (c d^2 (1-3 n)+e (a e (1-n)-b (d-2 d n))\right ) x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {e x^n}{d}\right )}{d^2 \left (c d^2-b d e+a e^2\right )^2 n} \] Output:

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

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 327, normalized size of antiderivative = 0.84 \[ \int \frac {1}{\left (d+e x^n\right )^2 \left (a+b x^n+c x^{2 n}\right )} \, dx=\frac {x \left (\frac {c \left (2 c^2 d^2+b \left (b+\sqrt {b^2-4 a c}\right ) e^2-2 c e \left (b d+\sqrt {b^2-4 a c} d+a e\right )\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},\frac {2 c x^n}{-b+\sqrt {b^2-4 a c}}\right )}{-b^2+4 a c+b \sqrt {b^2-4 a c}}+\frac {c \left (-2 c^2 d^2+b \left (-b+\sqrt {b^2-4 a c}\right ) e^2+2 c e \left (b d-\sqrt {b^2-4 a c} d+a e\right )\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{b^2-4 a c+b \sqrt {b^2-4 a c}}+\frac {e^2 (2 c d-b e) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {e x^n}{d}\right )}{d}+\frac {e^2 \left (c d^2+e (-b d+a e)\right ) \operatorname {Hypergeometric2F1}\left (2,\frac {1}{n},1+\frac {1}{n},-\frac {e x^n}{d}\right )}{d^2}\right )}{\left (c d^2+e (-b d+a e)\right )^2} \] Input:

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

Output:

(x*((c*(2*c^2*d^2 + b*(b + Sqrt[b^2 - 4*a*c])*e^2 - 2*c*e*(b*d + Sqrt[b^2 
- 4*a*c]*d + a*e))*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*(-2*c^2*d 
^2 + b*(-b + Sqrt[b^2 - 4*a*c])*e^2 + 2*c*e*(b*d - Sqrt[b^2 - 4*a*c]*d + a 
*e))*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]) + (e^2*(2*c*d - b*e)*Hypergeo 
metric2F1[1, n^(-1), 1 + n^(-1), -((e*x^n)/d)])/d + (e^2*(c*d^2 + e*(-(b*d 
) + a*e))*Hypergeometric2F1[2, n^(-1), 1 + n^(-1), -((e*x^n)/d)])/d^2))/(c 
*d^2 + e*(-(b*d) + a*e))^2
 

Rubi [A] (verified)

Time = 0.79 (sec) , antiderivative size = 368, normalized size of antiderivative = 0.95, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {1754, 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.

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

\(\Big \downarrow \) 1754

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

-((c*(2*c^2*d^2 + b*(b + Sqrt[b^2 - 4*a*c])*e^2 - 2*c*e*(b*d + Sqrt[b^2 - 
4*a*c]*d + a*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*c^2*d^2 + b*(b - Sqrt[b^2 - 4*a*c])*e^2 - 2*c*e*(b*d 
 - Sqrt[b^2 - 4*a*c]*d + a*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) + (e^2*(2*c*d - b*e)*x*Hypergeometric2F1[1, n^( 
-1), 1 + n^(-1), -((e*x^n)/d)])/(d*(c*d^2 - b*d*e + a*e^2)^2) + (e^2*x*Hyp 
ergeometric2F1[2, n^(-1), 1 + n^(-1), -((e*x^n)/d)])/(d^2*(c*d^2 - b*d*e + 
 a*e^2))
 

Defintions of rubi rules used

rule 1754

Int[((d_) + (e_.)*(x_)^(n_))^(q_)/((a_) + (b_.)*(x_)^(n_) + (c_.)*(x_)^(n2_ 
)), x_Symbol] :> Int[ExpandIntegrand[(d + e*x^n)^q/(a + b*x^n + c*x^(2*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] && IntegerQ[q]
 

rule 2009

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

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

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

Output:

Exception raised: HeuristicGCDFailed >> no luck
 

Maxima [F]

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

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

Output:

e^2*x/(c*d^4*n - b*d^3*e*n + a*d^2*e^2*n + (c*d^3*e*n - b*d^2*e^2*n + a*d* 
e^3*n)*x^n) + (c*d^2*e^2*(3*n - 1) - b*d*e^3*(2*n - 1) + a*e^4*(n - 1))*in 
tegrate(1/(c^2*d^6*n - 2*b*c*d^5*e*n + b^2*d^4*e^2*n + a^2*d^2*e^4*n + 2*( 
c*d^4*e^2*n - b*d^3*e^3*n)*a + (c^2*d^5*e*n - 2*b*c*d^4*e^2*n + b^2*d^3*e^ 
3*n + a^2*d*e^5*n + 2*(c*d^3*e^3*n - b*d^2*e^4*n)*a)*x^n), x) + integrate( 
(c^2*d^2 - 2*b*c*d*e + b^2*e^2 - a*c*e^2 - (2*c^2*d*e - b*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]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

int(1/(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)