\(\int (A+B x^2) (d+e x^2)^q (a+b x^2+c x^4) \, dx\) [255]

Optimal result

Integrand size = 29, antiderivative size = 318 \[ \int \left (A+B x^2\right ) \left (d+e x^2\right )^q \left (a+b x^2+c x^4\right ) \, dx=-\frac {\left (A e (7+2 q) (3 c d-b e (5+2 q))-B \left (15 c d^2-e (7+2 q) (3 b d-a e (5+2 q))\right )\right ) x \left (d+e x^2\right )^{1+q}}{e^3 (3+2 q) (5+2 q) (7+2 q)}+\frac {(A c e (7+2 q)-B (5 c d-b e (7+2 q))) x^3 \left (d+e x^2\right )^{1+q}}{e^2 (5+2 q) (7+2 q)}+\frac {B c x^5 \left (d+e x^2\right )^{1+q}}{e (7+2 q)}+\frac {\left (A e (7+2 q) \left (3 c d^2-e (5+2 q) (b d-a e (3+2 q))\right )-B d \left (15 c d^2-e (7+2 q) (3 b d-a e (5+2 q))\right )\right ) x \left (d+e x^2\right )^q \left (1+\frac {e x^2}{d}\right )^{-q} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-q,\frac {3}{2},-\frac {e x^2}{d}\right )}{e^3 (3+2 q) (5+2 q) (7+2 q)} \] Output:

-(A*e*(7+2*q)*(3*c*d-b*e*(5+2*q))-B*(15*c*d^2-e*(7+2*q)*(3*b*d-a*e*(5+2*q) 
)))*x*(e*x^2+d)^(1+q)/e^3/(3+2*q)/(5+2*q)/(7+2*q)+(A*c*e*(7+2*q)-B*(5*c*d- 
b*e*(7+2*q)))*x^3*(e*x^2+d)^(1+q)/e^2/(5+2*q)/(7+2*q)+B*c*x^5*(e*x^2+d)^(1 
+q)/e/(7+2*q)+(A*e*(7+2*q)*(3*c*d^2-e*(5+2*q)*(b*d-a*e*(3+2*q)))-B*d*(15*c 
*d^2-e*(7+2*q)*(3*b*d-a*e*(5+2*q))))*x*(e*x^2+d)^q*hypergeom([1/2, -q],[3/ 
2],-e*x^2/d)/e^3/(3+2*q)/(5+2*q)/(7+2*q)/((1+e*x^2/d)^q)
 

Mathematica [A] (verified)

Time = 0.42 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.44 \[ \int \left (A+B x^2\right ) \left (d+e x^2\right )^q \left (a+b x^2+c x^4\right ) \, dx=\frac {1}{105} x \left (d+e x^2\right )^q \left (1+\frac {e x^2}{d}\right )^{-q} \left (105 a A \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-q,\frac {3}{2},-\frac {e x^2}{d}\right )+35 (A b+a B) x^2 \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-q,\frac {5}{2},-\frac {e x^2}{d}\right )+21 (b B+A c) x^4 \operatorname {Hypergeometric2F1}\left (\frac {5}{2},-q,\frac {7}{2},-\frac {e x^2}{d}\right )+15 B c x^6 \operatorname {Hypergeometric2F1}\left (\frac {7}{2},-q,\frac {9}{2},-\frac {e x^2}{d}\right )\right ) \] Input:

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

Output:

(x*(d + e*x^2)^q*(105*a*A*Hypergeometric2F1[1/2, -q, 3/2, -((e*x^2)/d)] + 
35*(A*b + a*B)*x^2*Hypergeometric2F1[3/2, -q, 5/2, -((e*x^2)/d)] + 21*(b*B 
 + A*c)*x^4*Hypergeometric2F1[5/2, -q, 7/2, -((e*x^2)/d)] + 15*B*c*x^6*Hyp 
ergeometric2F1[7/2, -q, 9/2, -((e*x^2)/d)]))/(105*(1 + (e*x^2)/d)^q)
 

Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 210, normalized size of antiderivative = 0.66, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, 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.

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

\(\Big \downarrow \) 2256

\(\displaystyle \int \left (x^2 (a B+A b) \left (d+e x^2\right )^q+a A \left (d+e x^2\right )^q+x^4 (A c+b B) \left (d+e x^2\right )^q+B c x^6 \left (d+e x^2\right )^q\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{3} x^3 (a B+A b) \left (d+e x^2\right )^q \left (\frac {e x^2}{d}+1\right )^{-q} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-q,\frac {5}{2},-\frac {e x^2}{d}\right )+a A x \left (d+e x^2\right )^q \left (\frac {e x^2}{d}+1\right )^{-q} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-q,\frac {3}{2},-\frac {e x^2}{d}\right )+\frac {1}{5} x^5 (A c+b B) \left (d+e x^2\right )^q \left (\frac {e x^2}{d}+1\right )^{-q} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},-q,\frac {7}{2},-\frac {e x^2}{d}\right )+\frac {1}{7} B c x^7 \left (d+e x^2\right )^q \left (\frac {e x^2}{d}+1\right )^{-q} \operatorname {Hypergeometric2F1}\left (\frac {7}{2},-q,\frac {9}{2},-\frac {e x^2}{d}\right )\)

Input:

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

Output:

(a*A*x*(d + e*x^2)^q*Hypergeometric2F1[1/2, -q, 3/2, -((e*x^2)/d)])/(1 + ( 
e*x^2)/d)^q + ((A*b + a*B)*x^3*(d + e*x^2)^q*Hypergeometric2F1[3/2, -q, 5/ 
2, -((e*x^2)/d)])/(3*(1 + (e*x^2)/d)^q) + ((b*B + A*c)*x^5*(d + e*x^2)^q*H 
ypergeometric2F1[5/2, -q, 7/2, -((e*x^2)/d)])/(5*(1 + (e*x^2)/d)^q) + (B*c 
*x^7*(d + e*x^2)^q*Hypergeometric2F1[7/2, -q, 9/2, -((e*x^2)/d)])/(7*(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]

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

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]

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

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

Output:

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 25.06 (sec) , antiderivative size = 178, normalized size of antiderivative = 0.56 \[ \int \left (A+B x^2\right ) \left (d+e x^2\right )^q \left (a+b x^2+c x^4\right ) \, dx=A a d^{q} x {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, - q \\ \frac {3}{2} \end {matrix}\middle | {\frac {e x^{2} e^{i \pi }}{d}} \right )} + \frac {A b d^{q} x^{3} {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, - q \\ \frac {5}{2} \end {matrix}\middle | {\frac {e x^{2} e^{i \pi }}{d}} \right )}}{3} + \frac {A c d^{q} x^{5} {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{2}, - q \\ \frac {7}{2} \end {matrix}\middle | {\frac {e x^{2} e^{i \pi }}{d}} \right )}}{5} + \frac {B a d^{q} x^{3} {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, - q \\ \frac {5}{2} \end {matrix}\middle | {\frac {e x^{2} e^{i \pi }}{d}} \right )}}{3} + \frac {B b d^{q} x^{5} {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{2}, - q \\ \frac {7}{2} \end {matrix}\middle | {\frac {e x^{2} e^{i \pi }}{d}} \right )}}{5} + \frac {B c d^{q} x^{7} {{}_{2}F_{1}\left (\begin {matrix} \frac {7}{2}, - q \\ \frac {9}{2} \end {matrix}\middle | {\frac {e x^{2} e^{i \pi }}{d}} \right )}}{7} \] Input:

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

Output:

A*a*d**q*x*hyper((1/2, -q), (3/2,), e*x**2*exp_polar(I*pi)/d) + A*b*d**q*x 
**3*hyper((3/2, -q), (5/2,), e*x**2*exp_polar(I*pi)/d)/3 + A*c*d**q*x**5*h 
yper((5/2, -q), (7/2,), e*x**2*exp_polar(I*pi)/d)/5 + B*a*d**q*x**3*hyper( 
(3/2, -q), (5/2,), e*x**2*exp_polar(I*pi)/d)/3 + B*b*d**q*x**5*hyper((5/2, 
 -q), (7/2,), e*x**2*exp_polar(I*pi)/d)/5 + B*c*d**q*x**7*hyper((7/2, -q), 
 (9/2,), e*x**2*exp_polar(I*pi)/d)/7
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

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

Output:

(8*(d + e*x**2)**q*a**2*e**3*q**3*x + 60*(d + e*x**2)**q*a**2*e**3*q**2*x 
+ 142*(d + e*x**2)**q*a**2*e**3*q*x + 105*(d + e*x**2)**q*a**2*e**3*x + 16 
*(d + e*x**2)**q*a*b*d*e**2*q**3*x + 96*(d + e*x**2)**q*a*b*d*e**2*q**2*x 
+ 140*(d + e*x**2)**q*a*b*d*e**2*q*x + 16*(d + e*x**2)**q*a*b*e**3*q**3*x* 
*3 + 104*(d + e*x**2)**q*a*b*e**3*q**2*x**3 + 188*(d + e*x**2)**q*a*b*e**3 
*q*x**3 + 70*(d + e*x**2)**q*a*b*e**3*x**3 - 12*(d + e*x**2)**q*a*c*d**2*e 
*q**2*x - 42*(d + e*x**2)**q*a*c*d**2*e*q*x + 8*(d + e*x**2)**q*a*c*d*e**2 
*q**3*x**3 + 32*(d + e*x**2)**q*a*c*d*e**2*q**2*x**3 + 14*(d + e*x**2)**q* 
a*c*d*e**2*q*x**3 + 8*(d + e*x**2)**q*a*c*e**3*q**3*x**5 + 44*(d + e*x**2) 
**q*a*c*e**3*q**2*x**5 + 62*(d + e*x**2)**q*a*c*e**3*q*x**5 + 21*(d + e*x* 
*2)**q*a*c*e**3*x**5 - 12*(d + e*x**2)**q*b**2*d**2*e*q**2*x - 42*(d + e*x 
**2)**q*b**2*d**2*e*q*x + 8*(d + e*x**2)**q*b**2*d*e**2*q**3*x**3 + 32*(d 
+ e*x**2)**q*b**2*d*e**2*q**2*x**3 + 14*(d + e*x**2)**q*b**2*d*e**2*q*x**3 
 + 8*(d + e*x**2)**q*b**2*e**3*q**3*x**5 + 44*(d + e*x**2)**q*b**2*e**3*q* 
*2*x**5 + 62*(d + e*x**2)**q*b**2*e**3*q*x**5 + 21*(d + e*x**2)**q*b**2*e* 
*3*x**5 + 30*(d + e*x**2)**q*b*c*d**3*q*x - 20*(d + e*x**2)**q*b*c*d**2*e* 
q**2*x**3 - 10*(d + e*x**2)**q*b*c*d**2*e*q*x**3 + 8*(d + e*x**2)**q*b*c*d 
*e**2*q**3*x**5 + 16*(d + e*x**2)**q*b*c*d*e**2*q**2*x**5 + 6*(d + e*x**2) 
**q*b*c*d*e**2*q*x**5 + 8*(d + e*x**2)**q*b*c*e**3*q**3*x**7 + 36*(d + e*x 
**2)**q*b*c*e**3*q**2*x**7 + 46*(d + e*x**2)**q*b*c*e**3*q*x**7 + 15*(d...