\(\int \frac {(d+e x^2)^q (A+B x^2+C x^4)}{a+b x^2+c x^4} \, dx\) [261]

Optimal result

Integrand size = 36, antiderivative size = 312 \[ \int \frac {\left (d+e x^2\right )^q \left (A+B x^2+C x^4\right )}{a+b x^2+c x^4} \, dx=\frac {\left (B c-b C-\frac {b B c-b^2 C-2 c (A c-a C)}{\sqrt {b^2-4 a c}}\right ) x \left (d+e x^2\right )^q \left (1+\frac {e x^2}{d}\right )^{-q} \operatorname {AppellF1}\left (\frac {1}{2},-q,1,\frac {3}{2},-\frac {e x^2}{d},-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}\right )}{c \left (b-\sqrt {b^2-4 a c}\right )}+\frac {\left (B c-b C+\frac {b B c-b^2 C-2 c (A c-a C)}{\sqrt {b^2-4 a c}}\right ) x \left (d+e x^2\right )^q \left (1+\frac {e x^2}{d}\right )^{-q} \operatorname {AppellF1}\left (\frac {1}{2},-q,1,\frac {3}{2},-\frac {e x^2}{d},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{c \left (b+\sqrt {b^2-4 a c}\right )}+\frac {C x \left (d+e x^2\right )^q \left (1+\frac {e x^2}{d}\right )^{-q} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-q,\frac {3}{2},-\frac {e x^2}{d}\right )}{c} \] Output:

(B*c-C*b-(B*b*c-b^2*C-2*c*(A*c-C*a))/(-4*a*c+b^2)^(1/2))*x*(e*x^2+d)^q*App 
ellF1(1/2,1,-q,3/2,-2*c*x^2/(b-(-4*a*c+b^2)^(1/2)),-e*x^2/d)/c/(b-(-4*a*c+ 
b^2)^(1/2))/((1+e*x^2/d)^q)+(B*c-C*b+(B*b*c-b^2*C-2*c*(A*c-C*a))/(-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)/c/(b+(-4*a*c+b^2)^(1/2))/((1+e*x^2/d)^q)+C*x*(e*x^2+d)^q*hy 
pergeom([1/2, -q],[3/2],-e*x^2/d)/c/((1+e*x^2/d)^q)
 

Mathematica [F]

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

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

Output:

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

Rubi [A] (warning: unable to verify)

Time = 0.83 (sec) , antiderivative size = 312, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {2256, 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.

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

\(\Big \downarrow \) 2256

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {x \left (d+e x^2\right )^q \left (\frac {e x^2}{d}+1\right )^{-q} \left (-\frac {-2 c (A c-a C)+b^2 (-C)+b B c}{\sqrt {b^2-4 a c}}-b C+B c\right ) \operatorname {AppellF1}\left (\frac {1}{2},1,-q,\frac {3}{2},-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}},-\frac {e x^2}{d}\right )}{c \left (b-\sqrt {b^2-4 a c}\right )}+\frac {x \left (d+e x^2\right )^q \left (\frac {e x^2}{d}+1\right )^{-q} \left (\frac {-2 c (A c-a C)+b^2 (-C)+b B c}{\sqrt {b^2-4 a c}}-b C+B c\right ) \operatorname {AppellF1}\left (\frac {1}{2},1,-q,\frac {3}{2},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}},-\frac {e x^2}{d}\right )}{c \left (\sqrt {b^2-4 a c}+b\right )}+\frac {C x \left (d+e x^2\right )^q \left (\frac {e x^2}{d}+1\right )^{-q} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-q,\frac {3}{2},-\frac {e x^2}{d}\right )}{c}\)

Input:

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

Output:

((B*c - b*C - (b*B*c - b^2*C - 2*c*(A*c - a*C))/Sqrt[b^2 - 4*a*c])*x*(d + 
e*x^2)^q*AppellF1[1/2, 1, -q, 3/2, (-2*c*x^2)/(b - Sqrt[b^2 - 4*a*c]), -(( 
e*x^2)/d)])/(c*(b - Sqrt[b^2 - 4*a*c])*(1 + (e*x^2)/d)^q) + ((B*c - b*C + 
(b*B*c - b^2*C - 2*c*(A*c - a*C))/Sqrt[b^2 - 4*a*c])*x*(d + e*x^2)^q*Appel 
lF1[1/2, 1, -q, 3/2, (-2*c*x^2)/(b + Sqrt[b^2 - 4*a*c]), -((e*x^2)/d)])/(c 
*(b + Sqrt[b^2 - 4*a*c])*(1 + (e*x^2)/d)^q) + (C*x*(d + e*x^2)^q*Hypergeom 
etric2F1[1/2, -q, 3/2, -((e*x^2)/d)])/(c*(1 + (e*x^2)/d)^q)
 

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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\left (d+e x^2\right )^q \left (A+B x^2+C x^4\right )}{a+b x^2+c x^4} \, dx=\frac {\left (e \,x^{2}+d \right )^{q} x +4 \left (\int \frac {\left (e \,x^{2}+d \right )^{q}}{2 e q \,x^{2}+e \,x^{2}+2 d q +d}d x \right ) d \,q^{2}+2 \left (\int \frac {\left (e \,x^{2}+d \right )^{q}}{2 e q \,x^{2}+e \,x^{2}+2 d q +d}d x \right ) d q}{2 q +1} \] Input:

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

Output:

((d + e*x**2)**q*x + 4*int((d + e*x**2)**q/(2*d*q + d + 2*e*q*x**2 + e*x** 
2),x)*d*q**2 + 2*int((d + e*x**2)**q/(2*d*q + d + 2*e*q*x**2 + e*x**2),x)* 
d*q)/(2*q + 1)