Integrand size = 21, antiderivative size = 221 \[ \int \frac {\left (d+e x^n\right )^2}{\left (a+c x^{2 n}\right )^3} \, dx=\frac {x \left (d+e x^n\right )^2}{4 a n \left (a+c x^{2 n}\right )^2}-\frac {x \left (c d^2 (1-4 n)-a e^2 (1-2 n)+2 c d e (1-3 n) x^n\right )}{8 a^2 c n^2 \left (a+c x^{2 n}\right )}-\frac {\left (a e^2-c d^2 (1-4 n)\right ) (1-2 n) x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2 n},\frac {1}{2} \left (2+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )}{8 a^3 c n^2}+\frac {d e (1-3 n) (1-n) x^{1+n} \operatorname {Hypergeometric2F1}\left (1,\frac {1+n}{2 n},\frac {1}{2} \left (3+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )}{4 a^3 n^2 (1+n)} \] Output:
1/4*x*(d+e*x^n)^2/a/n/(a+c*x^(2*n))^2-1/8*x*(c*d^2*(1-4*n)-a*e^2*(1-2*n)+2 *c*d*e*(1-3*n)*x^n)/a^2/c/n^2/(a+c*x^(2*n))-1/8*(a*e^2-c*d^2*(1-4*n))*(1-2 *n)*x*hypergeom([1, 1/2/n],[1+1/2/n],-c*x^(2*n)/a)/a^3/c/n^2+1/4*d*e*(1-3* n)*(1-n)*x^(1+n)*hypergeom([1, 1/2*(1+n)/n],[3/2+1/2/n],-c*x^(2*n)/a)/a^3/ n^2/(1+n)
Time = 0.53 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.62 \[ \int \frac {\left (d+e x^n\right )^2}{\left (a+c x^{2 n}\right )^3} \, dx=\frac {x \left (a e^2 (1+n) \operatorname {Hypergeometric2F1}\left (2,\frac {1}{2 n},\frac {1}{2} \left (2+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )+\left (c d^2-a e^2\right ) (1+n) \operatorname {Hypergeometric2F1}\left (3,\frac {1}{2 n},\frac {1}{2} \left (2+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )+2 c d e x^n \operatorname {Hypergeometric2F1}\left (3,\frac {1+n}{2 n},\frac {1}{2} \left (3+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )\right )}{a^3 c (1+n)} \] Input:
Integrate[(d + e*x^n)^2/(a + c*x^(2*n))^3,x]
Output:
(x*(a*e^2*(1 + n)*Hypergeometric2F1[2, 1/(2*n), (2 + n^(-1))/2, -((c*x^(2* n))/a)] + (c*d^2 - a*e^2)*(1 + n)*Hypergeometric2F1[3, 1/(2*n), (2 + n^(-1 ))/2, -((c*x^(2*n))/a)] + 2*c*d*e*x^n*Hypergeometric2F1[3, (1 + n)/(2*n), (3 + n^(-1))/2, -((c*x^(2*n))/a)]))/(a^3*c*(1 + n))
Time = 0.46 (sec) , antiderivative size = 272, normalized size of antiderivative = 1.23, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {1767, 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.
| \(\displaystyle \int \frac {\left (d+e x^n\right )^2}{\left (a+c x^{2 n}\right )^3} \, dx\) |
| \(\Big \downarrow \) 1767 |
| \(\displaystyle \int \left (\frac {-a e^2+c d^2+2 c d e x^n}{c \left (a+c x^{2 n}\right )^3}+\frac {e^2}{c \left (a+c x^{2 n}\right )^2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {(1-4 n) (1-2 n) x \left (c d^2-a e^2\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2 n},\frac {1}{2} \left (2+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )}{8 a^3 c n^2}+\frac {d e (1-3 n) (1-n) x^{n+1} \operatorname {Hypergeometric2F1}\left (1,\frac {n+1}{2 n},\frac {1}{2} \left (3+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )}{4 a^3 n^2 (n+1)}-\frac {x \left ((1-4 n) \left (c d^2-a e^2\right )+2 c d e (1-3 n) x^n\right )}{8 a^2 c n^2 \left (a+c x^{2 n}\right )}+\frac {e^2 x \operatorname {Hypergeometric2F1}\left (2,\frac {1}{2 n},\frac {1}{2} \left (2+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )}{a^2 c}+\frac {x \left (-a e^2+c d^2+2 c d e x^n\right )}{4 a c n \left (a+c x^{2 n}\right )^2}\) |
Input:
Int[(d + e*x^n)^2/(a + c*x^(2*n))^3,x]
Output:
(x*(c*d^2 - a*e^2 + 2*c*d*e*x^n))/(4*a*c*n*(a + c*x^(2*n))^2) - (x*((c*d^2 - a*e^2)*(1 - 4*n) + 2*c*d*e*(1 - 3*n)*x^n))/(8*a^2*c*n^2*(a + c*x^(2*n)) ) + ((c*d^2 - a*e^2)*(1 - 4*n)*(1 - 2*n)*x*Hypergeometric2F1[1, 1/(2*n), ( 2 + n^(-1))/2, -((c*x^(2*n))/a)])/(8*a^3*c*n^2) + (d*e*(1 - 3*n)*(1 - n)*x ^(1 + n)*Hypergeometric2F1[1, (1 + n)/(2*n), (3 + n^(-1))/2, -((c*x^(2*n)) /a)])/(4*a^3*n^2*(1 + n)) + (e^2*x*Hypergeometric2F1[2, 1/(2*n), (2 + n^(- 1))/2, -((c*x^(2*n))/a)])/(a^2*c)
Int[((d_) + (e_.)*(x_)^(n_))^(q_)*((a_) + (c_.)*(x_)^(n2_))^(p_), x_Symbol] :> Int[ExpandIntegrand[(d + e*x^n)^q*(a + c*x^(2*n))^p, x], x] /; FreeQ[{a , c, d, e, n, p, q}, x] && EqQ[n2, 2*n] && NeQ[c*d^2 + a*e^2, 0] && ((Integ ersQ[p, q] && !IntegerQ[n]) || IGtQ[p, 0] || (IGtQ[q, 0] && !IntegerQ[n]) )
\[\int \frac {\left (d +e \,x^{n}\right )^{2}}{\left (a +c \,x^{2 n}\right )^{3}}d x\]
Input:
int((d+e*x^n)^2/(a+c*x^(2*n))^3,x)
Output:
int((d+e*x^n)^2/(a+c*x^(2*n))^3,x)
\[ \int \frac {\left (d+e x^n\right )^2}{\left (a+c x^{2 n}\right )^3} \, dx=\int { \frac {{\left (e x^{n} + d\right )}^{2}}{{\left (c x^{2 \, n} + a\right )}^{3}} \,d x } \] Input:
integrate((d+e*x^n)^2/(a+c*x^(2*n))^3,x, algorithm="fricas")
Output:
integral((e^2*x^(2*n) + 2*d*e*x^n + d^2)/(c^3*x^(6*n) + 3*a*c^2*x^(4*n) + 3*a^2*c*x^(2*n) + a^3), x)
Timed out. \[ \int \frac {\left (d+e x^n\right )^2}{\left (a+c x^{2 n}\right )^3} \, dx=\text {Timed out} \] Input:
integrate((d+e*x**n)**2/(a+c*x**(2*n))**3,x)
Output:
Timed out
\[ \int \frac {\left (d+e x^n\right )^2}{\left (a+c x^{2 n}\right )^3} \, dx=\int { \frac {{\left (e x^{n} + d\right )}^{2}}{{\left (c x^{2 \, n} + a\right )}^{3}} \,d x } \] Input:
integrate((d+e*x^n)^2/(a+c*x^(2*n))^3,x, algorithm="maxima")
Output:
1/8*(2*c^2*d*e*(3*n - 1)*x*x^(3*n) + 2*a*c*d*e*(5*n - 1)*x*x^n + (c^2*d^2* (4*n - 1) + a*c*e^2)*x*x^(2*n) + (a*c*d^2*(6*n - 1) - a^2*e^2*(2*n - 1))*x )/(a^2*c^3*n^2*x^(4*n) + 2*a^3*c^2*n^2*x^(2*n) + a^4*c*n^2) + integrate(1/ 8*(2*(3*n^2 - 4*n + 1)*c*d*e*x^n + (8*n^2 - 6*n + 1)*c*d^2 + a*e^2*(2*n - 1))/(a^2*c^2*n^2*x^(2*n) + a^3*c*n^2), x)
\[ \int \frac {\left (d+e x^n\right )^2}{\left (a+c x^{2 n}\right )^3} \, dx=\int { \frac {{\left (e x^{n} + d\right )}^{2}}{{\left (c x^{2 \, n} + a\right )}^{3}} \,d x } \] Input:
integrate((d+e*x^n)^2/(a+c*x^(2*n))^3,x, algorithm="giac")
Output:
integrate((e*x^n + d)^2/(c*x^(2*n) + a)^3, x)
Timed out. \[ \int \frac {\left (d+e x^n\right )^2}{\left (a+c x^{2 n}\right )^3} \, dx=\int \frac {{\left (d+e\,x^n\right )}^2}{{\left (a+c\,x^{2\,n}\right )}^3} \,d x \] Input:
int((d + e*x^n)^2/(a + c*x^(2*n))^3,x)
Output:
int((d + e*x^n)^2/(a + c*x^(2*n))^3, x)
\[ \int \frac {\left (d+e x^n\right )^2}{\left (a+c x^{2 n}\right )^3} \, dx=\frac {x^{4 n} \left (\int \frac {x^{2 n}}{x^{6 n} c^{3}+3 x^{4 n} a \,c^{2}+3 x^{2 n} a^{2} c +a^{3}}d x \right ) a \,c^{2} e^{2}+4 x^{4 n} \left (\int \frac {x^{2 n}}{x^{6 n} c^{3}+3 x^{4 n} a \,c^{2}+3 x^{2 n} a^{2} c +a^{3}}d x \right ) c^{3} d^{2} n -x^{4 n} \left (\int \frac {x^{2 n}}{x^{6 n} c^{3}+3 x^{4 n} a \,c^{2}+3 x^{2 n} a^{2} c +a^{3}}d x \right ) c^{3} d^{2}+2 x^{4 n} \left (\int \frac {x^{n}}{x^{6 n} c^{3}+3 x^{4 n} a \,c^{2}+3 x^{2 n} a^{2} c +a^{3}}d x \right ) a \,c^{2} d e +2 x^{2 n} \left (\int \frac {x^{2 n}}{x^{6 n} c^{3}+3 x^{4 n} a \,c^{2}+3 x^{2 n} a^{2} c +a^{3}}d x \right ) a^{2} c \,e^{2}+8 x^{2 n} \left (\int \frac {x^{2 n}}{x^{6 n} c^{3}+3 x^{4 n} a \,c^{2}+3 x^{2 n} a^{2} c +a^{3}}d x \right ) a \,c^{2} d^{2} n -2 x^{2 n} \left (\int \frac {x^{2 n}}{x^{6 n} c^{3}+3 x^{4 n} a \,c^{2}+3 x^{2 n} a^{2} c +a^{3}}d x \right ) a \,c^{2} d^{2}+4 x^{2 n} \left (\int \frac {x^{n}}{x^{6 n} c^{3}+3 x^{4 n} a \,c^{2}+3 x^{2 n} a^{2} c +a^{3}}d x \right ) a^{2} c d e +\left (\int \frac {x^{2 n}}{x^{6 n} c^{3}+3 x^{4 n} a \,c^{2}+3 x^{2 n} a^{2} c +a^{3}}d x \right ) a^{3} e^{2}+4 \left (\int \frac {x^{2 n}}{x^{6 n} c^{3}+3 x^{4 n} a \,c^{2}+3 x^{2 n} a^{2} c +a^{3}}d x \right ) a^{2} c \,d^{2} n -\left (\int \frac {x^{2 n}}{x^{6 n} c^{3}+3 x^{4 n} a \,c^{2}+3 x^{2 n} a^{2} c +a^{3}}d x \right ) a^{2} c \,d^{2}+2 \left (\int \frac {x^{n}}{x^{6 n} c^{3}+3 x^{4 n} a \,c^{2}+3 x^{2 n} a^{2} c +a^{3}}d x \right ) a^{3} d e +d^{2} x}{a \left (x^{4 n} c^{2}+2 x^{2 n} a c +a^{2}\right )} \] Input:
int((d+e*x^n)^2/(a+c*x^(2*n))^3,x)
Output:
(x**(4*n)*int(x**(2*n)/(x**(6*n)*c**3 + 3*x**(4*n)*a*c**2 + 3*x**(2*n)*a** 2*c + a**3),x)*a*c**2*e**2 + 4*x**(4*n)*int(x**(2*n)/(x**(6*n)*c**3 + 3*x* *(4*n)*a*c**2 + 3*x**(2*n)*a**2*c + a**3),x)*c**3*d**2*n - x**(4*n)*int(x* *(2*n)/(x**(6*n)*c**3 + 3*x**(4*n)*a*c**2 + 3*x**(2*n)*a**2*c + a**3),x)*c **3*d**2 + 2*x**(4*n)*int(x**n/(x**(6*n)*c**3 + 3*x**(4*n)*a*c**2 + 3*x**( 2*n)*a**2*c + a**3),x)*a*c**2*d*e + 2*x**(2*n)*int(x**(2*n)/(x**(6*n)*c**3 + 3*x**(4*n)*a*c**2 + 3*x**(2*n)*a**2*c + a**3),x)*a**2*c*e**2 + 8*x**(2* n)*int(x**(2*n)/(x**(6*n)*c**3 + 3*x**(4*n)*a*c**2 + 3*x**(2*n)*a**2*c + a **3),x)*a*c**2*d**2*n - 2*x**(2*n)*int(x**(2*n)/(x**(6*n)*c**3 + 3*x**(4*n )*a*c**2 + 3*x**(2*n)*a**2*c + a**3),x)*a*c**2*d**2 + 4*x**(2*n)*int(x**n/ (x**(6*n)*c**3 + 3*x**(4*n)*a*c**2 + 3*x**(2*n)*a**2*c + a**3),x)*a**2*c*d *e + int(x**(2*n)/(x**(6*n)*c**3 + 3*x**(4*n)*a*c**2 + 3*x**(2*n)*a**2*c + a**3),x)*a**3*e**2 + 4*int(x**(2*n)/(x**(6*n)*c**3 + 3*x**(4*n)*a*c**2 + 3*x**(2*n)*a**2*c + a**3),x)*a**2*c*d**2*n - int(x**(2*n)/(x**(6*n)*c**3 + 3*x**(4*n)*a*c**2 + 3*x**(2*n)*a**2*c + a**3),x)*a**2*c*d**2 + 2*int(x**n /(x**(6*n)*c**3 + 3*x**(4*n)*a*c**2 + 3*x**(2*n)*a**2*c + a**3),x)*a**3*d* e + d**2*x)/(a*(x**(4*n)*c**2 + 2*x**(2*n)*a*c + a**2))