\(\int (a+b x^2)^p (A+C x^2) (c+d x^2)^q \, dx\) [55]

Optimal result

Integrand size = 26, antiderivative size = 166 \[ \int \left (a+b x^2\right )^p \left (A+C x^2\right ) \left (c+d x^2\right )^q \, dx=A x \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \left (c+d x^2\right )^q \left (1+\frac {d x^2}{c}\right )^{-q} \operatorname {AppellF1}\left (\frac {1}{2},-p,-q,\frac {3}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )+\frac {1}{3} C x^3 \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \left (c+d x^2\right )^q \left (1+\frac {d x^2}{c}\right )^{-q} \operatorname {AppellF1}\left (\frac {3}{2},-p,-q,\frac {5}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right ) \] Output:

A*x*(b*x^2+a)^p*(d*x^2+c)^q*AppellF1(1/2,-p,-q,3/2,-b*x^2/a,-d*x^2/c)/((1+ 
b*x^2/a)^p)/((1+d*x^2/c)^q)+1/3*C*x^3*(b*x^2+a)^p*(d*x^2+c)^q*AppellF1(3/2 
,-p,-q,5/2,-b*x^2/a,-d*x^2/c)/((1+b*x^2/a)^p)/((1+d*x^2/c)^q)
 

Mathematica [A] (warning: unable to verify)

Time = 0.34 (sec) , antiderivative size = 242, normalized size of antiderivative = 1.46 \[ \int \left (a+b x^2\right )^p \left (A+C x^2\right ) \left (c+d x^2\right )^q \, dx=\frac {1}{3} x \left (a+b x^2\right )^p \left (c+d x^2\right )^q \left (\frac {9 a A c \operatorname {AppellF1}\left (\frac {1}{2},-p,-q,\frac {3}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )}{3 a c \operatorname {AppellF1}\left (\frac {1}{2},-p,-q,\frac {3}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )+2 x^2 \left (b c p \operatorname {AppellF1}\left (\frac {3}{2},1-p,-q,\frac {5}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )+a d q \operatorname {AppellF1}\left (\frac {3}{2},-p,1-q,\frac {5}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )\right )}+C x^2 \left (1+\frac {b x^2}{a}\right )^{-p} \left (1+\frac {d x^2}{c}\right )^{-q} \operatorname {AppellF1}\left (\frac {3}{2},-p,-q,\frac {5}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )\right ) \] Input:

Integrate[(a + b*x^2)^p*(A + C*x^2)*(c + d*x^2)^q,x]
 

Output:

(x*(a + b*x^2)^p*(c + d*x^2)^q*((9*a*A*c*AppellF1[1/2, -p, -q, 3/2, -((b*x 
^2)/a), -((d*x^2)/c)])/(3*a*c*AppellF1[1/2, -p, -q, 3/2, -((b*x^2)/a), -(( 
d*x^2)/c)] + 2*x^2*(b*c*p*AppellF1[3/2, 1 - p, -q, 5/2, -((b*x^2)/a), -((d 
*x^2)/c)] + a*d*q*AppellF1[3/2, -p, 1 - q, 5/2, -((b*x^2)/a), -((d*x^2)/c) 
])) + (C*x^2*AppellF1[3/2, -p, -q, 5/2, -((b*x^2)/a), -((d*x^2)/c)])/((1 + 
 (b*x^2)/a)^p*(1 + (d*x^2)/c)^q)))/3
 

Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {406, 334, 334, 333, 395, 395, 394}

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.

\(\displaystyle \int \left (A+C x^2\right ) \left (a+b x^2\right )^p \left (c+d x^2\right )^q \, dx\)

\(\Big \downarrow \) 406

\(\displaystyle A \int \left (b x^2+a\right )^p \left (d x^2+c\right )^qdx+C \int x^2 \left (b x^2+a\right )^p \left (d x^2+c\right )^qdx\)

\(\Big \downarrow \) 334

\(\displaystyle A \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \int \left (\frac {b x^2}{a}+1\right )^p \left (d x^2+c\right )^qdx+C \int x^2 \left (b x^2+a\right )^p \left (d x^2+c\right )^qdx\)

\(\Big \downarrow \) 334

\(\displaystyle A \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (c+d x^2\right )^q \left (\frac {d x^2}{c}+1\right )^{-q} \int \left (\frac {b x^2}{a}+1\right )^p \left (\frac {d x^2}{c}+1\right )^qdx+C \int x^2 \left (b x^2+a\right )^p \left (d x^2+c\right )^qdx\)

\(\Big \downarrow \) 333

\(\displaystyle C \int x^2 \left (b x^2+a\right )^p \left (d x^2+c\right )^qdx+A x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (c+d x^2\right )^q \left (\frac {d x^2}{c}+1\right )^{-q} \operatorname {AppellF1}\left (\frac {1}{2},-p,-q,\frac {3}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )\)

\(\Big \downarrow \) 395

\(\displaystyle C \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \int x^2 \left (\frac {b x^2}{a}+1\right )^p \left (d x^2+c\right )^qdx+A x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (c+d x^2\right )^q \left (\frac {d x^2}{c}+1\right )^{-q} \operatorname {AppellF1}\left (\frac {1}{2},-p,-q,\frac {3}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )\)

\(\Big \downarrow \) 395

\(\displaystyle C \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (c+d x^2\right )^q \left (\frac {d x^2}{c}+1\right )^{-q} \int x^2 \left (\frac {b x^2}{a}+1\right )^p \left (\frac {d x^2}{c}+1\right )^qdx+A x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (c+d x^2\right )^q \left (\frac {d x^2}{c}+1\right )^{-q} \operatorname {AppellF1}\left (\frac {1}{2},-p,-q,\frac {3}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )\)

\(\Big \downarrow \) 394

\(\displaystyle A x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (c+d x^2\right )^q \left (\frac {d x^2}{c}+1\right )^{-q} \operatorname {AppellF1}\left (\frac {1}{2},-p,-q,\frac {3}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )+\frac {1}{3} C x^3 \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (c+d x^2\right )^q \left (\frac {d x^2}{c}+1\right )^{-q} \operatorname {AppellF1}\left (\frac {3}{2},-p,-q,\frac {5}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )\)

Input:

Int[(a + b*x^2)^p*(A + C*x^2)*(c + d*x^2)^q,x]
 

Output:

(A*x*(a + b*x^2)^p*(c + d*x^2)^q*AppellF1[1/2, -p, -q, 3/2, -((b*x^2)/a), 
-((d*x^2)/c)])/((1 + (b*x^2)/a)^p*(1 + (d*x^2)/c)^q) + (C*x^3*(a + b*x^2)^ 
p*(c + d*x^2)^q*AppellF1[3/2, -p, -q, 5/2, -((b*x^2)/a), -((d*x^2)/c)])/(3 
*(1 + (b*x^2)/a)^p*(1 + (d*x^2)/c)^q)
 

Defintions of rubi rules used

rule 333

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim 
p[a^p*c^q*x*AppellF1[1/2, -p, -q, 3/2, (-b)*(x^2/a), (-d)*(x^2/c)], x] /; F 
reeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[p] || GtQ[a, 
0]) && (IntegerQ[q] || GtQ[c, 0])
 

rule 334

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim 
p[a^IntPart[p]*((a + b*x^2)^FracPart[p]/(1 + b*(x^2/a))^FracPart[p])   Int[ 
(1 + b*(x^2/a))^p*(c + d*x^2)^q, x], x] /; FreeQ[{a, b, c, d, p, q}, x] && 
NeQ[b*c - a*d, 0] &&  !(IntegerQ[p] || GtQ[a, 0])
 

rule 394

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ 
), x_Symbol] :> Simp[a^p*c^q*((e*x)^(m + 1)/(e*(m + 1)))*AppellF1[(m + 1)/2 
, -p, -q, 1 + (m + 1)/2, (-b)*(x^2/a), (-d)*(x^2/c)], x] /; FreeQ[{a, b, c, 
 d, e, m, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, 1] && (Int 
egerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])
 

rule 395

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ 
), x_Symbol] :> Simp[a^IntPart[p]*((a + b*x^2)^FracPart[p]/(1 + b*(x^2/a))^ 
FracPart[p])   Int[(e*x)^m*(1 + b*(x^2/a))^p*(c + d*x^2)^q, x], x] /; FreeQ 
[{a, b, c, d, e, m, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, 
1] &&  !(IntegerQ[p] || GtQ[a, 0])
 

rule 406

Int[((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.)*((e_) + (f_.)*( 
x_)^2), x_Symbol] :> Simp[e   Int[(a + b*x^2)^p*(c + d*x^2)^q, x], x] + Sim 
p[f   Int[x^2*(a + b*x^2)^p*(c + d*x^2)^q, x], x] /; FreeQ[{a, b, c, d, e, 
f, p, q}, x]
 

Maple [F]

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

Input:

int((b*x^2+a)^p*(C*x^2+A)*(d*x^2+c)^q,x)
 

Output:

int((b*x^2+a)^p*(C*x^2+A)*(d*x^2+c)^q,x)
 

Fricas [F]

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

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

Output:

integral((C*x^2 + A)*(b*x^2 + a)^p*(d*x^2 + c)^q, x)
 

Sympy [F(-1)]

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

integrate((b*x**2+a)**p*(C*x**2+A)*(d*x**2+c)**q,x)
 

Output:

Timed out
 

Maxima [F]

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

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

Output:

integrate((C*x^2 + A)*(b*x^2 + a)^p*(d*x^2 + c)^q, x)
 

Giac [F]

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

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

Output:

integrate((C*x^2 + A)*(b*x^2 + a)^p*(d*x^2 + c)^q, x)
 

Mupad [F(-1)]

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

int((A + C*x^2)*(a + b*x^2)^p*(c + d*x^2)^q,x)
 

Output:

int((A + C*x^2)*(a + b*x^2)^p*(c + d*x^2)^q, x)
 

Reduce [F]

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

int((b*x^2+a)^p*(C*x^2+A)*(d*x^2+c)^q,x)
 

Output:

(2*(c + d*x**2)**q*(a + b*x**2)**p*a*b*d*p*x + 2*(c + d*x**2)**q*(a + b*x* 
*2)**p*a*b*d*q*x + 3*(c + d*x**2)**q*(a + b*x**2)**p*a*b*d*x + 2*(c + d*x* 
*2)**q*(a + b*x**2)**p*a*c*d*p*x + 2*(c + d*x**2)**q*(a + b*x**2)**p*b*c** 
2*q*x + 2*(c + d*x**2)**q*(a + b*x**2)**p*b*c*d*p*x**3 + 2*(c + d*x**2)**q 
*(a + b*x**2)**p*b*c*d*q*x**3 + (c + d*x**2)**q*(a + b*x**2)**p*b*c*d*x**3 
 + 16*int(((c + d*x**2)**q*(a + b*x**2)**p*x**2)/(4*a*c*p**2 + 8*a*c*p*q + 
 8*a*c*p + 4*a*c*q**2 + 8*a*c*q + 3*a*c + 4*a*d*p**2*x**2 + 8*a*d*p*q*x**2 
 + 8*a*d*p*x**2 + 4*a*d*q**2*x**2 + 8*a*d*q*x**2 + 3*a*d*x**2 + 4*b*c*p**2 
*x**2 + 8*b*c*p*q*x**2 + 8*b*c*p*x**2 + 4*b*c*q**2*x**2 + 8*b*c*q*x**2 + 3 
*b*c*x**2 + 4*b*d*p**2*x**4 + 8*b*d*p*q*x**4 + 8*b*d*p*x**4 + 4*b*d*q**2*x 
**4 + 8*b*d*q*x**4 + 3*b*d*x**4),x)*a**2*b*d**2*p**4 + 48*int(((c + d*x**2 
)**q*(a + b*x**2)**p*x**2)/(4*a*c*p**2 + 8*a*c*p*q + 8*a*c*p + 4*a*c*q**2 
+ 8*a*c*q + 3*a*c + 4*a*d*p**2*x**2 + 8*a*d*p*q*x**2 + 8*a*d*p*x**2 + 4*a* 
d*q**2*x**2 + 8*a*d*q*x**2 + 3*a*d*x**2 + 4*b*c*p**2*x**2 + 8*b*c*p*q*x**2 
 + 8*b*c*p*x**2 + 4*b*c*q**2*x**2 + 8*b*c*q*x**2 + 3*b*c*x**2 + 4*b*d*p**2 
*x**4 + 8*b*d*p*q*x**4 + 8*b*d*p*x**4 + 4*b*d*q**2*x**4 + 8*b*d*q*x**4 + 3 
*b*d*x**4),x)*a**2*b*d**2*p**3*q + 56*int(((c + d*x**2)**q*(a + b*x**2)**p 
*x**2)/(4*a*c*p**2 + 8*a*c*p*q + 8*a*c*p + 4*a*c*q**2 + 8*a*c*q + 3*a*c + 
4*a*d*p**2*x**2 + 8*a*d*p*q*x**2 + 8*a*d*p*x**2 + 4*a*d*q**2*x**2 + 8*a*d* 
q*x**2 + 3*a*d*x**2 + 4*b*c*p**2*x**2 + 8*b*c*p*q*x**2 + 8*b*c*p*x**2 +...