\(\int \frac {(A+B x^2) (d+e x^2)^q}{a+b x^2+c x^4} \, dx\) [257]

Optimal result

Integrand size = 31, antiderivative size = 218 \[ \int \frac {\left (A+B x^2\right ) \left (d+e x^2\right )^q}{a+b x^2+c x^4} \, dx=\frac {\left (B-\frac {b B-2 A c}{\sqrt {b^2-4 a c}}\right ) x \left (d+e x^2\right )^q \left (1+\frac {e x^2}{d}\right )^{-q} \operatorname {AppellF1}\left (\frac {1}{2},-q,1,\frac {3}{2},-\frac {e x^2}{d},-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}\right )}{b-\sqrt {b^2-4 a c}}+\frac {\left (B+\frac {b B-2 A c}{\sqrt {b^2-4 a c}}\right ) x \left (d+e x^2\right )^q \left (1+\frac {e x^2}{d}\right )^{-q} \operatorname {AppellF1}\left (\frac {1}{2},-q,1,\frac {3}{2},-\frac {e x^2}{d},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{b+\sqrt {b^2-4 a c}} \] Output:

(B-(-2*A*c+B*b)/(-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)/(b-(-4*a*c+b^2)^(1/2))/((1+e*x^2/d 
)^q)+(B+(-2*A*c+B*b)/(-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)/(b+(-4*a*c+b^2)^(1/2))/((1+e* 
x^2/d)^q)
 

Mathematica [F]

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

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

Output:

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

Rubi [A] (verified)

Time = 0.51 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, 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.

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

\(\Big \downarrow \) 2256

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

((B - (b*B - 2*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)])/((b - Sqrt[b^2 - 
 4*a*c])*(1 + (e*x^2)/d)^q) + ((B + (b*B - 2*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)])/((b + Sqrt[b^2 - 4*a*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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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