\(\int \frac {1}{(d+e x^n) (a+c x^{2 n})} \, dx\) [28]

Optimal result

Integrand size = 21, antiderivative size = 152 \[ \int \frac {1}{\left (d+e x^n\right ) \left (a+c x^{2 n}\right )} \, dx=\frac {c d x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2 n},\frac {1}{2} \left (2+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )}{a \left (c d^2+a e^2\right )}+\frac {e^2 x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {e x^n}{d}\right )}{d \left (c d^2+a e^2\right )}-\frac {c e x^{1+n} \operatorname {Hypergeometric2F1}\left (1,\frac {1+n}{2 n},\frac {1}{2} \left (3+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )}{a \left (c d^2+a e^2\right ) (1+n)} \] Output:

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

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.86 \[ \int \frac {1}{\left (d+e x^n\right ) \left (a+c x^{2 n}\right )} \, dx=\frac {x \left (c d^2 (1+n) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2 n},\frac {1}{2} \left (2+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )+e \left (a e (1+n) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {e x^n}{d}\right )-c d x^n \operatorname {Hypergeometric2F1}\left (1,\frac {1+n}{2 n},\frac {1}{2} \left (3+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )\right )\right )}{a d \left (c d^2+a e^2\right ) (1+n)} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.31 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {1755, 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.

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

\(\Big \downarrow \) 1755

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 1755

Int[((d_) + (e_.)*(x_)^(n_))^(q_)/((a_) + (c_.)*(x_)^(n2_)), x_Symbol] :> I 
nt[ExpandIntegrand[(d + e*x^n)^q/(a + c*x^(2*n)), x], x] /; FreeQ[{a, c, d, 
 e, n}, x] && EqQ[n2, 2*n] && NeQ[c*d^2 + a*e^2, 0] && IntegerQ[q]
 

rule 2009

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

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-2)]

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

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

Output:

Exception raised: HeuristicGCDFailed >> no luck
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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