\(\int \frac {(d+e x^2)^q (A+B x^2+C x^4)}{(a+b x^2+c x^4)^2} \, dx\) [262]

Optimal result

Integrand size = 36, antiderivative size = 1114 \[ \int \frac {\left (d+e x^2\right )^q \left (A+B x^2+C x^4\right )}{\left (a+b x^2+c x^4\right )^2} \, dx =\text {Too large to display} \] Output:

1/2*x*(e*x^2+d)^(1+q)*(A*(3*a*b*c*e-2*a*c^2*d-b^3*e+b^2*c*d)-a*(2*B*a*c*e- 
B*b^2*e+B*b*c*d+C*a*b*e-2*C*a*c*d)+c*(A*(2*a*c*e-b^2*e+b*c*d)-a*(-B*b*e+2* 
B*c*d+2*C*a*e-C*b*d))*x^2)/a/(-4*a*c+b^2)/(a*e^2-b*d*e+c*d^2)/(c*x^4+b*x^2 
+a)-1/2*(A*c*(b^2*d*e-b*(c*d^2+a*e^2*(1-2*q))-4*a*c*d*e*q)+a*(2*B*c*(c*d^2 
+e*(a*e*(1-2*q)-b*d*(1-q)))-C*(b*(c*d^2+a*e^2*(1-2*q))-b^2*d*e*(1-2*q)-4*a 
*c*d*e*q))+(A*c*(b^3*d*e-12*a*b*c*d*e-b^2*(c*d^2+a*e^2*(1-2*q))+4*a*c*(3*c 
*d^2+a*e^2*(3-2*q)))+a*(b^2*(c*d*(C*d+2*B*e*(2-q))+a*C*e^2*(1-2*q))-b^3*C* 
d*e*(1-2*q)-4*b*c*(B*(a*e^2+c*d^2)+a*C*d*e*(1+2*q))+4*a*c*(a*C*e^2*(1+2*q) 
+c*d*(2*B*e*q+C*d))))/(-4*a*c+b^2)^(1/2))*x*(e*x^2+d)^q*AppellF1(1/2,1,-q, 
3/2,-2*c*x^2/(b-(-4*a*c+b^2)^(1/2)),-e*x^2/d)/a/(-4*a*c+b^2)/(b-(-4*a*c+b^ 
2)^(1/2))/(a*e^2-b*d*e+c*d^2)/((1+e*x^2/d)^q)-1/2*(A*c*(b^2*d*e-b*(c*d^2+a 
*e^2*(1-2*q))-4*a*c*d*e*q)+a*(2*B*c*(c*d^2+e*(a*e*(1-2*q)-b*d*(1-q)))-C*(b 
*(c*d^2+a*e^2*(1-2*q))-b^2*d*e*(1-2*q)-4*a*c*d*e*q))-(A*c*(b^3*d*e-12*a*b* 
c*d*e-b^2*(c*d^2+a*e^2*(1-2*q))+4*a*c*(3*c*d^2+a*e^2*(3-2*q)))+a*(b^2*(c*d 
*(C*d+2*B*e*(2-q))+a*C*e^2*(1-2*q))-b^3*C*d*e*(1-2*q)-4*b*c*(B*(a*e^2+c*d^ 
2)+a*C*d*e*(1+2*q))+4*a*c*(a*C*e^2*(1+2*q)+c*d*(2*B*e*q+C*d))))/(-4*a*c+b^ 
2)^(1/2))*x*(e*x^2+d)^q*AppellF1(1/2,1,-q,3/2,-2*c*x^2/(b+(-4*a*c+b^2)^(1/ 
2)),-e*x^2/d)/a/(-4*a*c+b^2)/(b+(-4*a*c+b^2)^(1/2))/(a*e^2-b*d*e+c*d^2)/(( 
1+e*x^2/d)^q)-1/2*e*(A*(2*a*c*e-b^2*e+b*c*d)-a*(-B*b*e+2*B*c*d+2*C*a*e-C*b 
*d))*(1+2*q)*x*(e*x^2+d)^q*hypergeom([1/2, -q],[3/2],-e*x^2/d)/a/(-4*a*...
 

Mathematica [F]

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

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

Output:

Integrate[((d + e*x^2)^q*(A + B*x^2 + C*x^4))/(a + b*x^2 + 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.

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

\(\Big \downarrow \) 2256

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

\(\Big \downarrow \) 2009

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

Input:

Int[((d + e*x^2)^q*(A + B*x^2 + C*x^4))/(a + b*x^2 + 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 2256

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

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral((C*x^4 + B*x^2 + A)*(e*x^2 + d)^q/(c^2*x^8 + 2*b*c*x^6 + (b^2 + 2 
*a*c)*x^4 + 2*a*b*x^2 + a^2), x)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

int((d + e*x**2)**q/(a + b*x**2 + c*x**4),x)