\(\int \frac {1}{x^2 (d+e x^n) (a+b x^n+c x^{2 n})} \, dx\) [142]

Optimal result

Integrand size = 29, antiderivative size = 269 \[ \int \frac {1}{x^2 \left (d+e x^n\right ) \left (a+b x^n+c x^{2 n}\right )} \, dx=\frac {c \left (2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e\right ) \operatorname {Hypergeometric2F1}\left (1,-\frac {1}{n},-\frac {1-n}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c-b \sqrt {b^2-4 a c}\right ) \left (c d^2-b d e+a e^2\right ) x}+\frac {c \left (e+\frac {2 c d-b e}{\sqrt {b^2-4 a c}}\right ) \operatorname {Hypergeometric2F1}\left (1,-\frac {1}{n},-\frac {1-n}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{\left (b+\sqrt {b^2-4 a c}\right ) \left (c d^2-b d e+a e^2\right ) x}-\frac {e^2 \operatorname {Hypergeometric2F1}\left (1,-\frac {1}{n},-\frac {1-n}{n},-\frac {e x^n}{d}\right )}{d \left (c d^2-b d e+a e^2\right ) x} \] Output:

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

Mathematica [A] (warning: unable to verify)

Time = 3.30 (sec) , antiderivative size = 462, normalized size of antiderivative = 1.72 \[ \int \frac {1}{x^2 \left (d+e x^n\right ) \left (a+b x^n+c x^{2 n}\right )} \, dx=\frac {-\frac {e^2 \operatorname {Hypergeometric2F1}\left (1,-\frac {1}{n},\frac {-1+n}{n},-\frac {e x^n}{d}\right )}{d}+\frac {2^{\frac {1}{n}} \left (-2 (c d-b e) x^{-n} \left (\frac {c x^n}{b-\sqrt {b^2-4 a c}+2 c x^n}\right )^{1+\frac {1}{n}} \operatorname {Hypergeometric2F1}\left (1+\frac {1}{n},1+\frac {1}{n},2+\frac {1}{n},\frac {b-\sqrt {b^2-4 a c}}{b-\sqrt {b^2-4 a c}+2 c x^n}\right )+2 (c d-b e) x^{-n} \left (\frac {c x^n}{b+\sqrt {b^2-4 a c}+2 c x^n}\right )^{1+\frac {1}{n}} \operatorname {Hypergeometric2F1}\left (1+\frac {1}{n},1+\frac {1}{n},2+\frac {1}{n},\frac {b+\sqrt {b^2-4 a c}}{b+\sqrt {b^2-4 a c}+2 c x^n}\right )+c e (1+n) \left (\left (\frac {c x^n}{b-\sqrt {b^2-4 a c}+2 c x^n}\right )^{\frac {1}{n}} \operatorname {Hypergeometric2F1}\left (\frac {1}{n},\frac {1}{n},1+\frac {1}{n},\frac {b-\sqrt {b^2-4 a c}}{b-\sqrt {b^2-4 a c}+2 c x^n}\right )-\left (\frac {c x^n}{b+\sqrt {b^2-4 a c}+2 c x^n}\right )^{\frac {1}{n}} \operatorname {Hypergeometric2F1}\left (\frac {1}{n},\frac {1}{n},1+\frac {1}{n},\frac {b+\sqrt {b^2-4 a c}}{b+\sqrt {b^2-4 a c}+2 c x^n}\right )\right )\right )}{\sqrt {b^2-4 a c} (1+n)}}{\left (c d^2+e (-b d+a e)\right ) x} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 327, normalized size of antiderivative = 1.22, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, 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.

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

\(\Big \downarrow \) 1880

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

\(\Big \downarrow \) 1010

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

\(\Big \downarrow \) 888

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

Input:

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

Output:

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

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-2)]

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

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

Output:

Exception raised: HeuristicGCDFailed >> no luck
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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