\(\int \frac {(a+b x^2)^p (c+d x^2)^q (e+f x^2)}{x^2} \, dx\) [74]

Optimal result

Integrand size = 29, antiderivative size = 244 \[ \int \frac {\left (a+b x^2\right )^p \left (c+d x^2\right )^q \left (e+f x^2\right )}{x^2} \, dx=-\frac {e \left (a+b x^2\right )^{1+p} \left (c+d x^2\right )^{1+q}}{a c x}+\frac {(a c f+b c e (1+2 p)+a d (e+2 e q)) x \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \left (c+d x^2\right )^q \left (1+\frac {d x^2}{c}\right )^{-q} \operatorname {AppellF1}\left (\frac {1}{2},-p,-q,\frac {3}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )}{a c}+\frac {b d e (3+2 p+2 q) x^3 \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \left (c+d x^2\right )^q \left (1+\frac {d x^2}{c}\right )^{-q} \operatorname {AppellF1}\left (\frac {3}{2},-p,-q,\frac {5}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )}{3 a c} \] Output:

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

Mathematica [A] (warning: unable to verify)

Time = 0.28 (sec) , antiderivative size = 242, normalized size of antiderivative = 0.99 \[ \int \frac {\left (a+b x^2\right )^p \left (c+d x^2\right )^q \left (e+f x^2\right )}{x^2} \, dx=\frac {\left (a+b x^2\right )^p \left (c+d x^2\right )^q \left (-e \left (1+\frac {b x^2}{a}\right )^{-p} \left (1+\frac {d x^2}{c}\right )^{-q} \operatorname {AppellF1}\left (-\frac {1}{2},-p,-q,\frac {1}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )+\frac {3 a c f x^2 \operatorname {AppellF1}\left (\frac {1}{2},-p,-q,\frac {3}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )}{3 a c \operatorname {AppellF1}\left (\frac {1}{2},-p,-q,\frac {3}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )+2 x^2 \left (b c p \operatorname {AppellF1}\left (\frac {3}{2},1-p,-q,\frac {5}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )+a d q \operatorname {AppellF1}\left (\frac {3}{2},-p,1-q,\frac {5}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )\right )}\right )}{x} \] Input:

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

Output:

((a + b*x^2)^p*(c + d*x^2)^q*(-((e*AppellF1[-1/2, -p, -q, 1/2, -((b*x^2)/a 
), -((d*x^2)/c)])/((1 + (b*x^2)/a)^p*(1 + (d*x^2)/c)^q)) + (3*a*c*f*x^2*Ap 
pellF1[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)]))))/x
 

Rubi [A] (verified)

Time = 0.76 (sec) , antiderivative size = 240, normalized size of antiderivative = 0.98, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.310, Rules used = {445, 25, 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.

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

\(\Big \downarrow \) 445

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 406

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

\(\Big \downarrow \) 334

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

\(\Big \downarrow \) 334

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

\(\Big \downarrow \) 333

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

\(\Big \downarrow \) 395

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

\(\Big \downarrow \) 395

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

\(\Big \downarrow \) 394

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 445

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

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

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

Output:

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

Reduce [F]

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

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

Output:

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