\(\int \frac {(d+e x^2)^q (A+B x^2+C x^4)}{(a+c x^4)^2} \, dx\) [130]

Optimal result

Integrand size = 31, antiderivative size = 512 \[ \int \frac {\left (d+e x^2\right )^q \left (A+B x^2+C x^4\right )}{\left (a+c x^4\right )^2} \, dx=\frac {x \left (d+e x^2\right )^{1+q} \left (A c d-a C d+a B e+(B c d-A c e+a C e) x^2\right )}{4 a \left (c d^2+a e^2\right ) \left (a+c x^4\right )}+\frac {\left (A c \left (3 c d^2+a e^2 (3-2 q)\right )-\sqrt {-a} \sqrt {c} \left (B \left (c d^2+a e^2 (1-2 q)\right )-2 (A c-a C) d e q\right )+a \left (a C e^2 (1+2 q)+c d (C d+2 B e q)\right )\right ) x \left (d+e x^2\right )^q \left (1+\frac {e x^2}{d}\right )^{-q} \operatorname {AppellF1}\left (\frac {1}{2},-q,1,\frac {3}{2},-\frac {e x^2}{d},-\frac {\sqrt {c} x^2}{\sqrt {-a}}\right )}{8 a^2 c \left (c d^2+a e^2\right )}+\frac {\left (A c \left (3 c d^2+a e^2 (3-2 q)\right )+\sqrt {-a} \sqrt {c} \left (B \left (c d^2+a e^2 (1-2 q)\right )-2 (A c-a C) d e q\right )+a \left (a C e^2 (1+2 q)+c d (C d+2 B e q)\right )\right ) x \left (d+e x^2\right )^q \left (1+\frac {e x^2}{d}\right )^{-q} \operatorname {AppellF1}\left (\frac {1}{2},-q,1,\frac {3}{2},-\frac {e x^2}{d},\frac {\sqrt {c} x^2}{\sqrt {-a}}\right )}{8 a^2 c \left (c d^2+a e^2\right )}-\frac {e (B c d-A c e+a C e) (1+2 q) x \left (d+e x^2\right )^q \left (1+\frac {e x^2}{d}\right )^{-q} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-q,\frac {3}{2},-\frac {e x^2}{d}\right )}{4 a c \left (c d^2+a e^2\right )} \] Output:

1/4*x*(e*x^2+d)^(1+q)*(A*c*d-C*a*d+B*a*e+(-A*c*e+B*c*d+C*a*e)*x^2)/a/(a*e^ 
2+c*d^2)/(c*x^4+a)+1/8*(A*c*(3*c*d^2+a*e^2*(3-2*q))-(-a)^(1/2)*c^(1/2)*(B* 
(c*d^2+a*e^2*(1-2*q))-2*(A*c-C*a)*d*e*q)+a*(a*C*e^2*(1+2*q)+c*d*(2*B*e*q+C 
*d)))*x*(e*x^2+d)^q*AppellF1(1/2,1,-q,3/2,-c^(1/2)*x^2/(-a)^(1/2),-e*x^2/d 
)/a^2/c/(a*e^2+c*d^2)/((1+e*x^2/d)^q)+1/8*(A*c*(3*c*d^2+a*e^2*(3-2*q))+(-a 
)^(1/2)*c^(1/2)*(B*(c*d^2+a*e^2*(1-2*q))-2*(A*c-C*a)*d*e*q)+a*(a*C*e^2*(1+ 
2*q)+c*d*(2*B*e*q+C*d)))*x*(e*x^2+d)^q*AppellF1(1/2,1,-q,3/2,c^(1/2)*x^2/( 
-a)^(1/2),-e*x^2/d)/a^2/c/(a*e^2+c*d^2)/((1+e*x^2/d)^q)-1/4*e*(-A*c*e+B*c* 
d+C*a*e)*(1+2*q)*x*(e*x^2+d)^q*hypergeom([1/2, -q],[3/2],-e*x^2/d)/a/c/(a* 
e^2+c*d^2)/((1+e*x^2/d)^q)
 

Mathematica [F]

\[ \int \frac {\left (d+e x^2\right )^q \left (A+B x^2+C x^4\right )}{\left (a+c x^4\right )^2} \, dx=\int \frac {\left (d+e x^2\right )^q \left (A+B x^2+C x^4\right )}{\left (a+c x^4\right )^2} \, dx \] Input:

Integrate[((d + e*x^2)^q*(A + B*x^2 + C*x^4))/(a + c*x^4)^2,x]
 

Output:

Integrate[((d + e*x^2)^q*(A + B*x^2 + C*x^4))/(a + c*x^4)^2, x]
 

Rubi [F]

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.

\(\displaystyle \int \frac {\left (A+B x^2+C x^4\right ) \left (d+e x^2\right )^q}{\left (a+c x^4\right )^2} \, dx\)

\(\Big \downarrow \) 2257

\(\displaystyle \int \left (\frac {\left (d+e x^2\right )^q \left (-a C+A c+B c x^2\right )}{c \left (a+c x^4\right )^2}+\frac {C \left (d+e x^2\right )^q}{c \left (a+c x^4\right )}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {(A c-a C) \int \frac {\left (e x^2+d\right )^q}{\left (c x^4+a\right )^2}dx}{c}+B \int \frac {x^2 \left (e x^2+d\right )^q}{\left (c x^4+a\right )^2}dx+\frac {C x \left (d+e x^2\right )^q \left (\frac {e x^2}{d}+1\right )^{-q} \operatorname {AppellF1}\left (\frac {1}{2},1,-q,\frac {3}{2},-\frac {\sqrt {c} x^2}{\sqrt {-a}},-\frac {e x^2}{d}\right )}{2 a c}+\frac {C x \left (d+e x^2\right )^q \left (\frac {e x^2}{d}+1\right )^{-q} \operatorname {AppellF1}\left (\frac {1}{2},1,-q,\frac {3}{2},\frac {\sqrt {c} x^2}{\sqrt {-a}},-\frac {e x^2}{d}\right )}{2 a c}\)

Input:

Int[((d + e*x^2)^q*(A + B*x^2 + C*x^4))/(a + c*x^4)^2,x]
 

Output:

$Aborted
 

Defintions of rubi rules used

rule 2009

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

rule 2257

Int[(Px_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol 
] :> Int[ExpandIntegrand[Px*(d + e*x^2)^q*(a + c*x^4)^p, x], x] /; FreeQ[{a 
, c, d, e, q}, x] && PolyQ[Px, x] && IntegerQ[p]
 

Maple [F]

\[\int \frac {\left (e \,x^{2}+d \right )^{q} \left (C \,x^{4}+B \,x^{2}+A \right )}{\left (c \,x^{4}+a \right )^{2}}d x\]

Input:

int((e*x^2+d)^q*(C*x^4+B*x^2+A)/(c*x^4+a)^2,x)
 

Output:

int((e*x^2+d)^q*(C*x^4+B*x^2+A)/(c*x^4+a)^2,x)
 

Fricas [F]

\[ \int \frac {\left (d+e x^2\right )^q \left (A+B x^2+C x^4\right )}{\left (a+c x^4\right )^2} \, dx=\int { \frac {{\left (C x^{4} + B x^{2} + A\right )} {\left (e x^{2} + d\right )}^{q}}{{\left (c x^{4} + a\right )}^{2}} \,d x } \] Input:

integrate((e*x^2+d)^q*(C*x^4+B*x^2+A)/(c*x^4+a)^2,x, algorithm="fricas")
 

Output:

integral((C*x^4 + B*x^2 + A)*(e*x^2 + d)^q/(c^2*x^8 + 2*a*c*x^4 + a^2), x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (d+e x^2\right )^q \left (A+B x^2+C x^4\right )}{\left (a+c x^4\right )^2} \, dx=\text {Timed out} \] Input:

integrate((e*x**2+d)**q*(C*x**4+B*x**2+A)/(c*x**4+a)**2,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {\left (d+e x^2\right )^q \left (A+B x^2+C x^4\right )}{\left (a+c x^4\right )^2} \, dx=\int { \frac {{\left (C x^{4} + B x^{2} + A\right )} {\left (e x^{2} + d\right )}^{q}}{{\left (c x^{4} + a\right )}^{2}} \,d x } \] Input:

integrate((e*x^2+d)^q*(C*x^4+B*x^2+A)/(c*x^4+a)^2,x, algorithm="maxima")
 

Output:

integrate((C*x^4 + B*x^2 + A)*(e*x^2 + d)^q/(c*x^4 + a)^2, x)
 

Giac [F]

\[ \int \frac {\left (d+e x^2\right )^q \left (A+B x^2+C x^4\right )}{\left (a+c x^4\right )^2} \, dx=\int { \frac {{\left (C x^{4} + B x^{2} + A\right )} {\left (e x^{2} + d\right )}^{q}}{{\left (c x^{4} + a\right )}^{2}} \,d x } \] Input:

integrate((e*x^2+d)^q*(C*x^4+B*x^2+A)/(c*x^4+a)^2,x, algorithm="giac")
 

Output:

integrate((C*x^4 + B*x^2 + A)*(e*x^2 + d)^q/(c*x^4 + a)^2, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (d+e x^2\right )^q \left (A+B x^2+C x^4\right )}{\left (a+c x^4\right )^2} \, dx=\int \frac {{\left (e\,x^2+d\right )}^q\,\left (C\,x^4+B\,x^2+A\right )}{{\left (c\,x^4+a\right )}^2} \,d x \] Input:

int(((d + e*x^2)^q*(A + B*x^2 + C*x^4))/(a + c*x^4)^2,x)
 

Output:

int(((d + e*x^2)^q*(A + B*x^2 + C*x^4))/(a + c*x^4)^2, x)
 

Reduce [F]

\[ \int \frac {\left (d+e x^2\right )^q \left (A+B x^2+C x^4\right )}{\left (a+c x^4\right )^2} \, dx =\text {Too large to display} \] Input:

int((e*x^2+d)^q*(C*x^4+B*x^2+A)/(c*x^4+a)^2,x)
 

Output:

( - (d + e*x**2)**q*b*e*x - (d + e*x**2)**q*c*d*x + int((d + e*x**2)**q/(a 
**2*d + a**2*e*x**2 + 2*a*c*d*x**4 + 2*a*c*e*x**6 + c**2*d*x**8 + c**2*e*x 
**10),x)*a**2*b*d*e + 4*int((d + e*x**2)**q/(a**2*d + a**2*e*x**2 + 2*a*c* 
d*x**4 + 2*a*c*e*x**6 + c**2*d*x**8 + c**2*e*x**10),x)*a**2*c*d**2 + int(( 
d + e*x**2)**q/(a**2*d + a**2*e*x**2 + 2*a*c*d*x**4 + 2*a*c*e*x**6 + c**2* 
d*x**8 + c**2*e*x**10),x)*a*b*c*d*e*x**4 + 4*int((d + e*x**2)**q/(a**2*d + 
 a**2*e*x**2 + 2*a*c*d*x**4 + 2*a*c*e*x**6 + c**2*d*x**8 + c**2*e*x**10),x 
)*a*c**2*d**2*x**4 + 2*int(((d + e*x**2)**q*x**6)/(a**2*d + a**2*e*x**2 + 
2*a*c*d*x**4 + 2*a*c*e*x**6 + c**2*d*x**8 + c**2*e*x**10),x)*a*b*c*e**2*q 
- 3*int(((d + e*x**2)**q*x**6)/(a**2*d + a**2*e*x**2 + 2*a*c*d*x**4 + 2*a* 
c*e*x**6 + c**2*d*x**8 + c**2*e*x**10),x)*a*b*c*e**2 + 2*int(((d + e*x**2) 
**q*x**6)/(a**2*d + a**2*e*x**2 + 2*a*c*d*x**4 + 2*a*c*e*x**6 + c**2*d*x** 
8 + c**2*e*x**10),x)*a*c**2*d*e*q + 2*int(((d + e*x**2)**q*x**6)/(a**2*d + 
 a**2*e*x**2 + 2*a*c*d*x**4 + 2*a*c*e*x**6 + c**2*d*x**8 + c**2*e*x**10),x 
)*b*c**2*e**2*q*x**4 - 3*int(((d + e*x**2)**q*x**6)/(a**2*d + a**2*e*x**2 
+ 2*a*c*d*x**4 + 2*a*c*e*x**6 + c**2*d*x**8 + c**2*e*x**10),x)*b*c**2*e**2 
*x**4 + 2*int(((d + e*x**2)**q*x**6)/(a**2*d + a**2*e*x**2 + 2*a*c*d*x**4 
+ 2*a*c*e*x**6 + c**2*d*x**8 + c**2*e*x**10),x)*c**3*d*e*q*x**4 + 2*int((( 
d + e*x**2)**q*x**2)/(a**2*d + a**2*e*x**2 + 2*a*c*d*x**4 + 2*a*c*e*x**6 + 
 c**2*d*x**8 + c**2*e*x**10),x)*a**2*b*e**2*q + int(((d + e*x**2)**q*x*...