\(\int \frac {(A+B x^2) (d+e x^2)^q}{(a+c x^4)^2} \, dx\) [126]

Optimal result

Integrand size = 26, antiderivative size = 449 \[ \int \frac {\left (A+B x^2\right ) \left (d+e x^2\right )^q}{\left (a+c x^4\right )^2} \, dx=\frac {x \left (d+e x^2\right )^{1+q} \left (A c d+a B e+c (B d-A e) x^2\right )}{4 a \left (c d^2+a e^2\right ) \left (a+c x^4\right )}+\frac {\left (\sqrt {c} \left (A \left (3 c d^2+a e^2 (3-2 q)\right )+2 a B d e q\right )-\sqrt {-a} \left (B \left (c d^2+a e^2 (1-2 q)\right )-2 A c d e q\right )\right ) x \left (d+e x^2\right )^q \left (1+\frac {e x^2}{d}\right )^{-q} \operatorname {AppellF1}\left (\frac {1}{2},-q,1,\frac {3}{2},-\frac {e x^2}{d},-\frac {\sqrt {c} x^2}{\sqrt {-a}}\right )}{8 a^2 \sqrt {c} \left (c d^2+a e^2\right )}+\frac {\left (\sqrt {c} \left (A \left (3 c d^2+a e^2 (3-2 q)\right )+2 a B d e q\right )+\sqrt {-a} \left (B \left (c d^2+a e^2 (1-2 q)\right )-2 A c d e q\right )\right ) x \left (d+e x^2\right )^q \left (1+\frac {e x^2}{d}\right )^{-q} \operatorname {AppellF1}\left (\frac {1}{2},-q,1,\frac {3}{2},-\frac {e x^2}{d},\frac {\sqrt {c} x^2}{\sqrt {-a}}\right )}{8 a^2 \sqrt {c} \left (c d^2+a e^2\right )}-\frac {e (B d-A e) (1+2 q) x \left (d+e x^2\right )^q \left (1+\frac {e x^2}{d}\right )^{-q} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-q,\frac {3}{2},-\frac {e x^2}{d}\right )}{4 a \left (c d^2+a e^2\right )} \] Output:

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

Mathematica [F]

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

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

Output:

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

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

\(\Big \downarrow \) 2257

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

\(\Big \downarrow \) 2009

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

Input:

Int[((A + B*x^2)*(d + e*x^2)^q)/(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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\left (A+B x^2\right ) \left (d+e x^2\right )^q}{\left (a+c x^4\right )^2} \, dx=\left (\int \frac {\left (e \,x^{2}+d \right )^{q}}{c^{2} x^{8}+2 a c \,x^{4}+a^{2}}d x \right ) a +\left (\int \frac {\left (e \,x^{2}+d \right )^{q} x^{2}}{c^{2} x^{8}+2 a c \,x^{4}+a^{2}}d x \right ) b \] Input:

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

Output:

int((d + e*x**2)**q/(a**2 + 2*a*c*x**4 + c**2*x**8),x)*a + int(((d + e*x** 
2)**q*x**2)/(a**2 + 2*a*c*x**4 + c**2*x**8),x)*b