\(\int \frac {(A+B x^2) (d+e x^2)^q}{a+c x^4} \, dx\) [125]

Optimal result

Integrand size = 26, antiderivative size = 169 \[ \int \frac {\left (A+B x^2\right ) \left (d+e x^2\right )^q}{a+c x^4} \, dx=\frac {\left (A-\frac {\sqrt {-a} B}{\sqrt {c}}\right ) x \left (d+e x^2\right )^q \left (1+\frac {e x^2}{d}\right )^{-q} \operatorname {AppellF1}\left (\frac {1}{2},-q,1,\frac {3}{2},-\frac {e x^2}{d},-\frac {\sqrt {c} x^2}{\sqrt {-a}}\right )}{2 a}+\frac {\left (A+\frac {\sqrt {-a} B}{\sqrt {c}}\right ) x \left (d+e x^2\right )^q \left (1+\frac {e x^2}{d}\right )^{-q} \operatorname {AppellF1}\left (\frac {1}{2},-q,1,\frac {3}{2},-\frac {e x^2}{d},\frac {\sqrt {c} x^2}{\sqrt {-a}}\right )}{2 a} \] Output:

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

Mathematica [F]

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

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

Output:

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

Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {2257, 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.

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

\(\Big \downarrow \) 2257

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

((A - (Sqrt[-a]*B)/Sqrt[c])*x*(d + e*x^2)^q*AppellF1[1/2, 1, -q, 3/2, -((S 
qrt[c]*x^2)/Sqrt[-a]), -((e*x^2)/d)])/(2*a*(1 + (e*x^2)/d)^q) + ((A + (Sqr 
t[-a]*B)/Sqrt[c])*x*(d + e*x^2)^q*AppellF1[1/2, 1, -q, 3/2, (Sqrt[c]*x^2)/ 
Sqrt[-a], -((e*x^2)/d)])/(2*a*(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 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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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