\(\int \frac {(f x)^m (d+e x^n)}{(a+c x^{2 n})^3} \, dx\) [51]

Optimal result

Integrand size = 24, antiderivative size = 242 \[ \int \frac {(f x)^m \left (d+e x^n\right )}{\left (a+c x^{2 n}\right )^3} \, dx=\frac {(f x)^{1+m} \left (d+e x^n\right )}{4 a f n \left (a+c x^{2 n}\right )^2}-\frac {(f x)^{1+m} \left (d (1+m-4 n)+e (1+m-3 n) x^n\right )}{8 a^2 f n^2 \left (a+c x^{2 n}\right )}+\frac {d (1+m-4 n) (1+m-2 n) (f x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2 n},1+\frac {1+m}{2 n},-\frac {c x^{2 n}}{a}\right )}{8 a^3 f (1+m) n^2}+\frac {e (1+m-3 n) (1+m-n) x^n (f x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m+n}{2 n},\frac {1+m+3 n}{2 n},-\frac {c x^{2 n}}{a}\right )}{8 a^3 f n^2 (1+m+n)} \] Output:

1/4*(f*x)^(1+m)*(d+e*x^n)/a/f/n/(a+c*x^(2*n))^2-1/8*(f*x)^(1+m)*(d*(1+m-4* 
n)+e*(1+m-3*n)*x^n)/a^2/f/n^2/(a+c*x^(2*n))+1/8*d*(1+m-4*n)*(1+m-2*n)*(f*x 
)^(1+m)*hypergeom([1, 1/2*(1+m)/n],[1+1/2*(1+m)/n],-c*x^(2*n)/a)/a^3/f/(1+ 
m)/n^2+1/8*e*(1+m-3*n)*(1+m-n)*x^n*(f*x)^(1+m)*hypergeom([1, 1/2*(1+m+n)/n 
],[1/2*(1+m+3*n)/n],-c*x^(2*n)/a)/a^3/f/n^2/(1+m+n)
 

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.45 \[ \int \frac {(f x)^m \left (d+e x^n\right )}{\left (a+c x^{2 n}\right )^3} \, dx=\frac {d x (f x)^m \operatorname {Hypergeometric2F1}\left (3,\frac {1+m}{2 n},1+\frac {1+m}{2 n},-\frac {c x^{2 n}}{a}\right )}{a^3 (1+m)}+\frac {e x^{1+n} (f x)^m \operatorname {Hypergeometric2F1}\left (3,\frac {1+m+n}{2 n},\frac {1+m+3 n}{2 n},-\frac {c x^{2 n}}{a}\right )}{a^3 (1+m+n)} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.60 (sec) , antiderivative size = 306, normalized size of antiderivative = 1.26, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1883, 1883, 1885, 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.

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

\(\Big \downarrow \) 1883

\(\displaystyle \frac {(f x)^{m+1} \left (d+e x^n\right )}{4 a f n \left (a+c x^{2 n}\right )^2}-\frac {\int \frac {(f x)^m \left (e (m-3 n+1) x^n+d (m-4 n+1)\right )}{\left (c x^{2 n}+a\right )^2}dx}{4 a n}\)

\(\Big \downarrow \) 1883

\(\displaystyle \frac {(f x)^{m+1} \left (d+e x^n\right )}{4 a f n \left (a+c x^{2 n}\right )^2}-\frac {\frac {(f x)^{m+1} \left (d (m-4 n+1)+e (m-3 n+1) x^n\right )}{2 a f n \left (a+c x^{2 n}\right )}-\frac {\int \frac {(f x)^m \left (e (m-3 n+1) (m-n+1) x^n+d (m-4 n+1) (m-2 n+1)\right )}{c x^{2 n}+a}dx}{2 a n}}{4 a n}\)

\(\Big \downarrow \) 1885

\(\displaystyle \frac {(f x)^{m+1} \left (d+e x^n\right )}{4 a f n \left (a+c x^{2 n}\right )^2}-\frac {\frac {(f x)^{m+1} \left (d (m-4 n+1)+e (m-3 n+1) x^n\right )}{2 a f n \left (a+c x^{2 n}\right )}-\frac {\int \left (\frac {\left (\frac {\sqrt {-a} e (m-3 n+1) (m-n+1)}{\sqrt {c}}-d (m-4 n+1) (m-2 n+1)\right ) (f x)^m}{2 \sqrt {-a} \sqrt {c} \left (x^n+\frac {\sqrt {-a}}{\sqrt {c}}\right )}-\frac {\left (d (m-4 n+1) (m-2 n+1)+\frac {\sqrt {-a} e (m-3 n+1) (m-n+1)}{\sqrt {c}}\right ) (f x)^m}{2 \sqrt {-a} \sqrt {c} \left (\frac {\sqrt {-a}}{\sqrt {c}}-x^n\right )}\right )dx}{2 a n}}{4 a n}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {(f x)^{m+1} \left (d+e x^n\right )}{4 a f n \left (a+c x^{2 n}\right )^2}-\frac {\frac {(f x)^{m+1} \left (d (m-4 n+1)+e (m-3 n+1) x^n\right )}{2 a f n \left (a+c x^{2 n}\right )}-\frac {\frac {(f x)^{m+1} \left (d (m-4 n+1) (m-2 n+1)-\frac {\sqrt {-a} e (m-3 n+1) (m-n+1)}{\sqrt {c}}\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{n},\frac {m+n+1}{n},-\frac {\sqrt {c} x^n}{\sqrt {-a}}\right )}{2 a f (m+1)}+\frac {(f x)^{m+1} \left (\frac {\sqrt {-a} e (m-3 n+1) (m-n+1)}{\sqrt {c}}+d (m-4 n+1) (m-2 n+1)\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{n},\frac {m+n+1}{n},\frac {\sqrt {c} x^n}{\sqrt {-a}}\right )}{2 a f (m+1)}}{2 a n}}{4 a n}\)

Input:

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

Output:

((f*x)^(1 + m)*(d + e*x^n))/(4*a*f*n*(a + c*x^(2*n))^2) - (((f*x)^(1 + m)* 
(d*(1 + m - 4*n) + e*(1 + m - 3*n)*x^n))/(2*a*f*n*(a + c*x^(2*n))) - (((d* 
(1 + m - 4*n)*(1 + m - 2*n) - (Sqrt[-a]*e*(1 + m - 3*n)*(1 + m - n))/Sqrt[ 
c])*(f*x)^(1 + m)*Hypergeometric2F1[1, (1 + m)/n, (1 + m + n)/n, -((Sqrt[c 
]*x^n)/Sqrt[-a])])/(2*a*f*(1 + m)) + ((d*(1 + m - 4*n)*(1 + m - 2*n) + (Sq 
rt[-a]*e*(1 + m - 3*n)*(1 + m - n))/Sqrt[c])*(f*x)^(1 + m)*Hypergeometric2 
F1[1, (1 + m)/n, (1 + m + n)/n, (Sqrt[c]*x^n)/Sqrt[-a]])/(2*a*f*(1 + m)))/ 
(2*a*n))/(4*a*n)
 

Defintions of rubi rules used

rule 1883

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(n_))*((a_) + (c_.)*(x_)^(n2_))^( 
p_), x_Symbol] :> Simp[(-(f*x)^(m + 1))*(a + c*x^(2*n))^(p + 1)*((d + e*x^n 
)/(2*a*f*n*(p + 1))), x] + Simp[1/(2*a*n*(p + 1))   Int[(f*x)^m*(a + c*x^(2 
*n))^(p + 1)*Simp[d*(m + 2*n*(p + 1) + 1) + e*(m + n*(2*p + 3) + 1)*x^n, x] 
, x], x] /; FreeQ[{a, c, d, e, f, m, n}, x] && EqQ[n2, 2*n] &&  !RationalQ[ 
n] && ILtQ[p + 1, 0]
 

rule 1885

Int[((f_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^(n2_.))^(p_.)*((d_) + (e_.)*(x_)^ 
(n_))^(q_.), x_Symbol] :> Int[ExpandIntegrand[(f*x)^m*(d + e*x^n)^q*(a + c* 
x^(2*n))^p, x], x] /; FreeQ[{a, c, d, e, f, m, n, p, q}, x] && EqQ[n2, 2*n] 
 &&  !RationalQ[n] && (IGtQ[p, 0] || IGtQ[q, 0])
 

rule 2009

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

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral((e*x^n + d)*(f*x)^m/(c^3*x^(6*n) + 3*a*c^2*x^(4*n) + 3*a^2*c*x^(2 
*n) + a^3), x)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

-1/8*(a*d*f^m*(m - 6*n + 1)*x*x^m + c*e*f^m*(m - 3*n + 1)*x*e^(m*log(x) + 
3*n*log(x)) + c*d*f^m*(m - 4*n + 1)*x*e^(m*log(x) + 2*n*log(x)) + a*e*f^m* 
(m - 5*n + 1)*x*e^(m*log(x) + n*log(x)))/(a^2*c^2*n^2*x^(4*n) + 2*a^3*c*n^ 
2*x^(2*n) + a^4*n^2) + integrate(1/8*((m^2 - 2*m*(3*n - 1) + 8*n^2 - 6*n + 
 1)*d*f^m*x^m + (m^2 - 2*m*(2*n - 1) + 3*n^2 - 4*n + 1)*e*f^m*e^(m*log(x) 
+ n*log(x)))/(a^2*c*n^2*x^(2*n) + a^3*n^2), x)
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {(f x)^m \left (d+e x^n\right )}{\left (a+c x^{2 n}\right )^3} \, dx=f^{m} \left (\left (\int \frac {x^{m +n}}{x^{6 n} c^{3}+3 x^{4 n} a \,c^{2}+3 x^{2 n} a^{2} c +a^{3}}d x \right ) e +\left (\int \frac {x^{m}}{x^{6 n} c^{3}+3 x^{4 n} a \,c^{2}+3 x^{2 n} a^{2} c +a^{3}}d x \right ) d \right ) \] Input:

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

Output:

f**m*(int(x**(m + n)/(x**(6*n)*c**3 + 3*x**(4*n)*a*c**2 + 3*x**(2*n)*a**2* 
c + a**3),x)*e + int(x**m/(x**(6*n)*c**3 + 3*x**(4*n)*a*c**2 + 3*x**(2*n)* 
a**2*c + a**3),x)*d)