\(\int \frac {(a+\frac {b}{x^2})^p (c+\frac {d}{x^2})^q}{x^4} \, dx\) [198]

Optimal result

Integrand size = 22, antiderivative size = 84 \[ \int \frac {\left (a+\frac {b}{x^2}\right )^p \left (c+\frac {d}{x^2}\right )^q}{x^4} \, dx=-\frac {\left (a+\frac {b}{x^2}\right )^p \left (1+\frac {b}{a x^2}\right )^{-p} \left (c+\frac {d}{x^2}\right )^q \left (1+\frac {d}{c x^2}\right )^{-q} \operatorname {AppellF1}\left (\frac {3}{2},-p,-q,\frac {5}{2},-\frac {b}{a x^2},-\frac {d}{c x^2}\right )}{3 x^3} \] Output:

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

Mathematica [A] (verified)

Time = 0.33 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.26 \[ \int \frac {\left (a+\frac {b}{x^2}\right )^p \left (c+\frac {d}{x^2}\right )^q}{x^4} \, dx=-\frac {\left (a+\frac {b}{x^2}\right )^p \left (c+\frac {d}{x^2}\right )^q \left (1+\frac {a x^2}{b}\right )^{-p} \left (1+\frac {c x^2}{d}\right )^{-q} \operatorname {AppellF1}\left (-\frac {3}{2}-p-q,-p,-q,-\frac {1}{2}-p-q,-\frac {a x^2}{b},-\frac {c x^2}{d}\right )}{(3+2 p+2 q) x^3} \] Input:

Integrate[((a + b/x^2)^p*(c + d/x^2)^q)/x^4,x]
 

Output:

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

Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {997, 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.

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

\(\Big \downarrow \) 997

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

\(\Big \downarrow \) 395

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

\(\Big \downarrow \) 395

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

\(\Big \downarrow \) 394

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

Input:

Int[((a + b/x^2)^p*(c + d/x^2)^q)/x^4,x]
 

Output:

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

Defintions of rubi rules used

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 997

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), 
 x_Symbol] :> -Subst[Int[(a + b/x^n)^p*((c + d/x^n)^q/x^(m + 2)), x], x, 1/ 
x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && ILtQ[n, 0] && In 
tegerQ[m]
 

Maple [F]

\[\int \frac {\left (a +\frac {b}{x^{2}}\right )^{p} \left (c +\frac {d}{x^{2}}\right )^{q}}{x^{4}}d x\]

Input:

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

Output:

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

Fricas [F]

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

integrate((a+b/x^2)^p*(c+d/x^2)^q/x^4,x, algorithm="fricas")
 

Output:

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

Sympy [F(-1)]

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

integrate((a+b/x**2)**p*(c+d/x**2)**q/x**4,x)
 

Output:

Timed out
 

Maxima [F]

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

integrate((a+b/x^2)^p*(c+d/x^2)^q/x^4,x, algorithm="maxima")
 

Output:

integrate((a + b/x^2)^p*(c + d/x^2)^q/x^4, x)
 

Giac [F]

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

integrate((a+b/x^2)^p*(c+d/x^2)^q/x^4,x, algorithm="giac")
 

Output:

integrate((a + b/x^2)^p*(c + d/x^2)^q/x^4, x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

( - (c*x**2 + d)**q*(a*x**2 + b)**p + 8*x**(2*p + 2*q)*int(((c*x**2 + d)** 
q*(a*x**2 + b)**p)/(4*x**(2*p + 2*q)*a**2*c*d*p*q*x**6 + 6*x**(2*p + 2*q)* 
a**2*c*d*p*x**6 + 4*x**(2*p + 2*q)*a**2*c*d*q**2*x**6 + 12*x**(2*p + 2*q)* 
a**2*c*d*q*x**6 + 9*x**(2*p + 2*q)*a**2*c*d*x**6 + 4*x**(2*p + 2*q)*a**2*d 
**2*p*q*x**4 + 6*x**(2*p + 2*q)*a**2*d**2*p*x**4 + 4*x**(2*p + 2*q)*a**2*d 
**2*q**2*x**4 + 12*x**(2*p + 2*q)*a**2*d**2*q*x**4 + 9*x**(2*p + 2*q)*a**2 
*d**2*x**4 + 4*x**(2*p + 2*q)*a*b*c**2*p**2*x**6 + 4*x**(2*p + 2*q)*a*b*c* 
*2*p*q*x**6 + 12*x**(2*p + 2*q)*a*b*c**2*p*x**6 + 6*x**(2*p + 2*q)*a*b*c** 
2*q*x**6 + 9*x**(2*p + 2*q)*a*b*c**2*x**6 + 4*x**(2*p + 2*q)*a*b*c*d*p**2* 
x**4 + 8*x**(2*p + 2*q)*a*b*c*d*p*q*x**4 + 18*x**(2*p + 2*q)*a*b*c*d*p*x** 
4 + 4*x**(2*p + 2*q)*a*b*c*d*q**2*x**4 + 18*x**(2*p + 2*q)*a*b*c*d*q*x**4 
+ 18*x**(2*p + 2*q)*a*b*c*d*x**4 + 4*x**(2*p + 2*q)*a*b*d**2*p*q*x**2 + 6* 
x**(2*p + 2*q)*a*b*d**2*p*x**2 + 4*x**(2*p + 2*q)*a*b*d**2*q**2*x**2 + 12* 
x**(2*p + 2*q)*a*b*d**2*q*x**2 + 9*x**(2*p + 2*q)*a*b*d**2*x**2 + 4*x**(2* 
p + 2*q)*b**2*c**2*p**2*x**4 + 4*x**(2*p + 2*q)*b**2*c**2*p*q*x**4 + 12*x* 
*(2*p + 2*q)*b**2*c**2*p*x**4 + 6*x**(2*p + 2*q)*b**2*c**2*q*x**4 + 9*x**( 
2*p + 2*q)*b**2*c**2*x**4 + 4*x**(2*p + 2*q)*b**2*c*d*p**2*x**2 + 4*x**(2* 
p + 2*q)*b**2*c*d*p*q*x**2 + 12*x**(2*p + 2*q)*b**2*c*d*p*x**2 + 6*x**(2*p 
 + 2*q)*b**2*c*d*q*x**2 + 9*x**(2*p + 2*q)*b**2*c*d*x**2),x)*a**2*d**2*p** 
2*q*x**3 + 12*x**(2*p + 2*q)*int(((c*x**2 + d)**q*(a*x**2 + b)**p)/(4*x...