\(\int x^2 (a+b x^2)^p (c+d x^2)^q (e+f x^2) \, dx\) [72]

Optimal result

Integrand size = 29, antiderivative size = 273 \[ \int x^2 \left (a+b x^2\right )^p \left (c+d x^2\right )^q \left (e+f x^2\right ) \, dx=\frac {f x \left (a+b x^2\right )^{1+p} \left (c+d x^2\right )^{1+q}}{b d (5+2 p+2 q)}-\frac {a c f x \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \left (c+d x^2\right )^q \left (1+\frac {d x^2}{c}\right )^{-q} \operatorname {AppellF1}\left (\frac {1}{2},-p,-q,\frac {3}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )}{b d (5+2 p+2 q)}-\frac {(b c f (3+2 p)+a d f (3+2 q)-b d e (5+2 p+2 q)) x^3 \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \left (c+d x^2\right )^q \left (1+\frac {d x^2}{c}\right )^{-q} \operatorname {AppellF1}\left (\frac {3}{2},-p,-q,\frac {5}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )}{3 b d (5+2 p+2 q)} \] Output:

f*x*(b*x^2+a)^(p+1)*(d*x^2+c)^(1+q)/b/d/(5+2*p+2*q)-a*c*f*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)/b/d/(5+2*p+2*q)/((1+b* 
x^2/a)^p)/((1+d*x^2/c)^q)-1/3*(b*c*f*(3+2*p)+a*d*f*(3+2*q)-b*d*e*(5+2*p+2* 
q))*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)/ 
b/d/(5+2*p+2*q)/((1+b*x^2/a)^p)/((1+d*x^2/c)^q)
 

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.46 \[ \int x^2 \left (a+b x^2\right )^p \left (c+d x^2\right )^q \left (e+f x^2\right ) \, dx=\frac {1}{15} x^3 \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \left (c+d x^2\right )^q \left (1+\frac {d x^2}{c}\right )^{-q} \left (5 e \operatorname {AppellF1}\left (\frac {3}{2},-p,-q,\frac {5}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )+3 f x^2 \operatorname {AppellF1}\left (\frac {5}{2},-p,-q,\frac {7}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )\right ) \] Input:

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

Output:

(x^3*(a + b*x^2)^p*(c + d*x^2)^q*(5*e*AppellF1[3/2, -p, -q, 5/2, -((b*x^2) 
/a), -((d*x^2)/c)] + 3*f*x^2*AppellF1[5/2, -p, -q, 7/2, -((b*x^2)/a), -((d 
*x^2)/c)]))/(15*(1 + (b*x^2)/a)^p*(1 + (d*x^2)/c)^q)
 

Rubi [A] (verified)

Time = 0.79 (sec) , antiderivative size = 259, normalized size of antiderivative = 0.95, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.276, Rules used = {444, 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.

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

\(\Big \downarrow \) 444

\(\displaystyle \frac {f x \left (a+b x^2\right )^{p+1} \left (c+d x^2\right )^{q+1}}{b d (2 p+2 q+5)}-\frac {\int \left (b x^2+a\right )^p \left (d x^2+c\right )^q \left ((b c f (2 p+3)+a d f (2 q+3)-b d e (2 p+2 q+5)) x^2+a c f\right )dx}{b d (2 p+2 q+5)}\)

\(\Big \downarrow \) 406

\(\displaystyle \frac {f x \left (a+b x^2\right )^{p+1} \left (c+d x^2\right )^{q+1}}{b d (2 p+2 q+5)}-\frac {(a d f (2 q+3)+b c f (2 p+3)-b d e (2 p+2 q+5)) \int x^2 \left (b x^2+a\right )^p \left (d x^2+c\right )^qdx+a c f \int \left (b x^2+a\right )^p \left (d x^2+c\right )^qdx}{b d (2 p+2 q+5)}\)

\(\Big \downarrow \) 334

\(\displaystyle \frac {f x \left (a+b x^2\right )^{p+1} \left (c+d x^2\right )^{q+1}}{b d (2 p+2 q+5)}-\frac {(a d f (2 q+3)+b c f (2 p+3)-b d e (2 p+2 q+5)) \int x^2 \left (b x^2+a\right )^p \left (d x^2+c\right )^qdx+a c f \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \int \left (\frac {b x^2}{a}+1\right )^p \left (d x^2+c\right )^qdx}{b d (2 p+2 q+5)}\)

\(\Big \downarrow \) 334

\(\displaystyle \frac {f x \left (a+b x^2\right )^{p+1} \left (c+d x^2\right )^{q+1}}{b d (2 p+2 q+5)}-\frac {(a d f (2 q+3)+b c f (2 p+3)-b d e (2 p+2 q+5)) \int x^2 \left (b x^2+a\right )^p \left (d x^2+c\right )^qdx+a c f \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (c+d x^2\right )^q \left (\frac {d x^2}{c}+1\right )^{-q} \int \left (\frac {b x^2}{a}+1\right )^p \left (\frac {d x^2}{c}+1\right )^qdx}{b d (2 p+2 q+5)}\)

\(\Big \downarrow \) 333

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

\(\Big \downarrow \) 395

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

\(\Big \downarrow \) 395

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

\(\Big \downarrow \) 394

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

Input:

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

Output:

(f*x*(a + b*x^2)^(1 + p)*(c + d*x^2)^(1 + q))/(b*d*(5 + 2*p + 2*q)) - ((a* 
c*f*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) + ((b*c*f*(3 + 2*p) 
+ a*d*f*(3 + 2*q) - b*d*e*(5 + 2*p + 2*q))*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))/(b*d*(5 + 2*p + 2*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]
 

rule 444

Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q 
_.)*((e_) + (f_.)*(x_)^2), x_Symbol] :> Simp[f*g*(g*x)^(m - 1)*(a + b*x^2)^ 
(p + 1)*((c + d*x^2)^(q + 1)/(b*d*(m + 2*(p + q + 1) + 1))), x] - Simp[g^2/ 
(b*d*(m + 2*(p + q + 1) + 1))   Int[(g*x)^(m - 2)*(a + b*x^2)^p*(c + d*x^2) 
^q*Simp[a*f*c*(m - 1) + (a*f*d*(m + 2*q + 1) + b*(f*c*(m + 2*p + 1) - e*d*( 
m + 2*(p + q + 1) + 1)))*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p, 
q}, x] && GtQ[m, 1]
 

Maple [F]

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

Input:

int(x^2*(b*x^2+a)^p*(d*x^2+c)^q*(f*x^2+e),x)
 

Output:

int(x^2*(b*x^2+a)^p*(d*x^2+c)^q*(f*x^2+e),x)
 

Fricas [F]

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

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

Output:

integral((f*x^4 + e*x^2)*(b*x^2 + a)^p*(d*x^2 + c)^q, x)
 

Sympy [F(-1)]

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

integrate(x**2*(b*x**2+a)**p*(d*x**2+c)**q*(f*x**2+e),x)
 

Output:

Timed out
 

Maxima [F]

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

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

Output:

integrate((f*x^2 + e)*(b*x^2 + a)^p*(d*x^2 + c)^q*x^2, x)
 

Giac [F]

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

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

Output:

integrate((f*x^2 + e)*(b*x^2 + a)^p*(d*x^2 + c)^q*x^2, x)
 

Mupad [F(-1)]

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

int(x^2*(a + b*x^2)^p*(c + d*x^2)^q*(e + f*x^2),x)
 

Output:

int(x^2*(a + b*x^2)^p*(c + d*x^2)^q*(e + f*x^2), x)
 

Reduce [F]

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

int(x^2*(b*x^2+a)^p*(d*x^2+c)^q*(f*x^2+e),x)
                                                                                    
                                                                                    
 

Output:

( - 4*(c + d*x**2)**q*(a + b*x**2)**p*a**2*d**2*f*p*q*x - 6*(c + d*x**2)** 
q*(a + b*x**2)**p*a**2*d**2*f*p*x + 8*(c + d*x**2)**q*(a + b*x**2)**p*a*b* 
c*d*f*p*q*x + 4*(c + d*x**2)**q*(a + b*x**2)**p*a*b*d**2*e*p**2*x + 4*(c + 
 d*x**2)**q*(a + b*x**2)**p*a*b*d**2*e*p*q*x + 10*(c + d*x**2)**q*(a + b*x 
**2)**p*a*b*d**2*e*p*x + 4*(c + d*x**2)**q*(a + b*x**2)**p*a*b*d**2*f*p**2 
*x**3 + 4*(c + d*x**2)**q*(a + b*x**2)**p*a*b*d**2*f*p*q*x**3 + 2*(c + d*x 
**2)**q*(a + b*x**2)**p*a*b*d**2*f*p*x**3 - 4*(c + d*x**2)**q*(a + b*x**2) 
**p*b**2*c**2*f*p*q*x - 6*(c + d*x**2)**q*(a + b*x**2)**p*b**2*c**2*f*q*x 
+ 4*(c + d*x**2)**q*(a + b*x**2)**p*b**2*c*d*e*p*q*x + 4*(c + d*x**2)**q*( 
a + b*x**2)**p*b**2*c*d*e*q**2*x + 10*(c + d*x**2)**q*(a + b*x**2)**p*b**2 
*c*d*e*q*x + 4*(c + d*x**2)**q*(a + b*x**2)**p*b**2*c*d*f*p*q*x**3 + 4*(c 
+ d*x**2)**q*(a + b*x**2)**p*b**2*c*d*f*q**2*x**3 + 2*(c + d*x**2)**q*(a + 
 b*x**2)**p*b**2*c*d*f*q*x**3 + 4*(c + d*x**2)**q*(a + b*x**2)**p*b**2*d** 
2*e*p**2*x**3 + 8*(c + d*x**2)**q*(a + b*x**2)**p*b**2*d**2*e*p*q*x**3 + 1 
2*(c + d*x**2)**q*(a + b*x**2)**p*b**2*d**2*e*p*x**3 + 4*(c + d*x**2)**q*( 
a + b*x**2)**p*b**2*d**2*e*q**2*x**3 + 12*(c + d*x**2)**q*(a + b*x**2)**p* 
b**2*d**2*e*q*x**3 + 5*(c + d*x**2)**q*(a + b*x**2)**p*b**2*d**2*e*x**3 + 
4*(c + d*x**2)**q*(a + b*x**2)**p*b**2*d**2*f*p**2*x**5 + 8*(c + d*x**2)** 
q*(a + b*x**2)**p*b**2*d**2*f*p*q*x**5 + 8*(c + d*x**2)**q*(a + b*x**2)**p 
*b**2*d**2*f*p*x**5 + 4*(c + d*x**2)**q*(a + b*x**2)**p*b**2*d**2*f*q**...