\(\int \frac {(d+e x^n)^3}{a+c x^{2 n}} \, dx\) [26]

Optimal result

Integrand size = 21, antiderivative size = 141 \[ \int \frac {\left (d+e x^n\right )^3}{a+c x^{2 n}} \, dx=\frac {3 d e^2 x}{c}+\frac {e^3 x^{1+n}}{c (1+n)}+\frac {d \left (c d^2-3 a e^2\right ) x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2 n},\frac {1}{2} \left (2+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )}{a c}+\frac {e \left (3 c d^2-a e^2\right ) x^{1+n} \operatorname {Hypergeometric2F1}\left (1,\frac {1+n}{2 n},\frac {1}{2} \left (3+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )}{a c (1+n)} \] Output:

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

Mathematica [A] (verified)

Time = 0.74 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.90 \[ \int \frac {\left (d+e x^n\right )^3}{a+c x^{2 n}} \, dx=\frac {x \left (d \left (c d^2-3 a e^2\right ) (1+n) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2 n},\frac {1}{2} \left (2+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )+e \left (a e \left (3 d (1+n)+e x^n\right )+\left (3 c d^2-a e^2\right ) x^n \operatorname {Hypergeometric2F1}\left (1,\frac {1+n}{2 n},\frac {1}{2} \left (3+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )\right )\right )}{a c (1+n)} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 141, 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.

\(\displaystyle \int \frac {\left (d+e x^n\right )^3}{a+c x^{2 n}} \, dx\)

\(\Big \downarrow \) 1755

\(\displaystyle \int \left (\frac {x^n \left (3 c d^2 e-a e^3\right )-3 a d e^2+c d^3}{c \left (a+c x^{2 n}\right )}+\frac {3 d e^2}{c}+\frac {e^3 x^n}{c}\right )dx\)

\(\Big \downarrow \) 2009

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

Input:

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

Output:

(3*d*e^2*x)/c + (e^3*x^(1 + n))/(c*(1 + n)) + (d*(c*d^2 - 3*a*e^2)*x*Hyper 
geometric2F1[1, 1/(2*n), (2 + n^(-1))/2, -((c*x^(2*n))/a)])/(a*c) + (e*(3* 
c*d^2 - a*e^2)*x^(1 + n)*Hypergeometric2F1[1, (1 + n)/(2*n), (3 + n^(-1))/ 
2, -((c*x^(2*n))/a)])/(a*c*(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]

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

Input:

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

Output:

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

Fricas [F]

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

integrate((d+e*x^n)^3/(a+c*x^(2*n)),x, algorithm="fricas")
 

Output:

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 4.49 (sec) , antiderivative size = 466, normalized size of antiderivative = 3.30 \[ \int \frac {\left (d+e x^n\right )^3}{a+c x^{2 n}} \, dx=\frac {a^{\frac {1}{2 n}} a^{-1 - \frac {1}{2 n}} d^{3} x \Phi \left (\frac {c x^{2 n} e^{i \pi }}{a}, 1, \frac {1}{2 n}\right ) \Gamma \left (\frac {1}{2 n}\right )}{4 n^{2} \Gamma \left (1 + \frac {1}{2 n}\right )} + \frac {3 a^{- \frac {5}{2} - \frac {1}{2 n}} a^{\frac {3}{2} + \frac {1}{2 n}} e^{3} x^{3 n + 1} \Phi \left (\frac {c x^{2 n} e^{i \pi }}{a}, 1, \frac {3}{2} + \frac {1}{2 n}\right ) \Gamma \left (\frac {3}{2} + \frac {1}{2 n}\right )}{4 n \Gamma \left (\frac {5}{2} + \frac {1}{2 n}\right )} + \frac {a^{- \frac {5}{2} - \frac {1}{2 n}} a^{\frac {3}{2} + \frac {1}{2 n}} e^{3} x^{3 n + 1} \Phi \left (\frac {c x^{2 n} e^{i \pi }}{a}, 1, \frac {3}{2} + \frac {1}{2 n}\right ) \Gamma \left (\frac {3}{2} + \frac {1}{2 n}\right )}{4 n^{2} \Gamma \left (\frac {5}{2} + \frac {1}{2 n}\right )} + \frac {3 a^{- \frac {3}{2} - \frac {1}{2 n}} a^{\frac {1}{2} + \frac {1}{2 n}} d^{2} e x^{n + 1} \Phi \left (\frac {c x^{2 n} e^{i \pi }}{a}, 1, \frac {1}{2} + \frac {1}{2 n}\right ) \Gamma \left (\frac {1}{2} + \frac {1}{2 n}\right )}{4 n \Gamma \left (\frac {3}{2} + \frac {1}{2 n}\right )} + \frac {3 a^{- \frac {3}{2} - \frac {1}{2 n}} a^{\frac {1}{2} + \frac {1}{2 n}} d^{2} e x^{n + 1} \Phi \left (\frac {c x^{2 n} e^{i \pi }}{a}, 1, \frac {1}{2} + \frac {1}{2 n}\right ) \Gamma \left (\frac {1}{2} + \frac {1}{2 n}\right )}{4 n^{2} \Gamma \left (\frac {3}{2} + \frac {1}{2 n}\right )} - \frac {3 a^{- \frac {1}{2 n}} a^{1 + \frac {1}{2 n}} c^{\frac {1}{2 n}} c^{-1 - \frac {1}{2 n}} d e^{2} x \Phi \left (\frac {a x^{- 2 n} e^{i \pi }}{c}, 1, \frac {e^{i \pi }}{2 n}\right ) \Gamma \left (\frac {1}{2 n}\right )}{4 a n^{2} \Gamma \left (1 + \frac {1}{2 n}\right )} \] Input:

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

Output:

a**(1/(2*n))*a**(-1 - 1/(2*n))*d**3*x*lerchphi(c*x**(2*n)*exp_polar(I*pi)/ 
a, 1, 1/(2*n))*gamma(1/(2*n))/(4*n**2*gamma(1 + 1/(2*n))) + 3*a**(-5/2 - 1 
/(2*n))*a**(3/2 + 1/(2*n))*e**3*x**(3*n + 1)*lerchphi(c*x**(2*n)*exp_polar 
(I*pi)/a, 1, 3/2 + 1/(2*n))*gamma(3/2 + 1/(2*n))/(4*n*gamma(5/2 + 1/(2*n)) 
) + a**(-5/2 - 1/(2*n))*a**(3/2 + 1/(2*n))*e**3*x**(3*n + 1)*lerchphi(c*x* 
*(2*n)*exp_polar(I*pi)/a, 1, 3/2 + 1/(2*n))*gamma(3/2 + 1/(2*n))/(4*n**2*g 
amma(5/2 + 1/(2*n))) + 3*a**(-3/2 - 1/(2*n))*a**(1/2 + 1/(2*n))*d**2*e*x** 
(n + 1)*lerchphi(c*x**(2*n)*exp_polar(I*pi)/a, 1, 1/2 + 1/(2*n))*gamma(1/2 
 + 1/(2*n))/(4*n*gamma(3/2 + 1/(2*n))) + 3*a**(-3/2 - 1/(2*n))*a**(1/2 + 1 
/(2*n))*d**2*e*x**(n + 1)*lerchphi(c*x**(2*n)*exp_polar(I*pi)/a, 1, 1/2 + 
1/(2*n))*gamma(1/2 + 1/(2*n))/(4*n**2*gamma(3/2 + 1/(2*n))) - 3*a**(1 + 1/ 
(2*n))*c**(1/(2*n))*c**(-1 - 1/(2*n))*d*e**2*x*lerchphi(a*exp_polar(I*pi)/ 
(c*x**(2*n)), 1, exp_polar(I*pi)/(2*n))*gamma(1/(2*n))/(4*a*a**(1/(2*n))*n 
**2*gamma(1 + 1/(2*n)))
 

Maxima [F]

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

integrate((d+e*x^n)^3/(a+c*x^(2*n)),x, algorithm="maxima")
 

Output:

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

Giac [F]

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

integrate((d+e*x^n)^3/(a+c*x^(2*n)),x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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