\(\int \frac {(a+b x^2)^p (c+d x^2)^q}{x^3} \, dx\) [1639]

Optimal result

Integrand size = 22, antiderivative size = 97 \[ \int \frac {\left (a+b x^2\right )^p \left (c+d x^2\right )^q}{x^3} \, dx=\frac {b \left (a+b x^2\right )^{1+p} \left (c+d x^2\right )^q \left (\frac {b \left (c+d x^2\right )}{b c-a d}\right )^{-q} \operatorname {AppellF1}\left (1+p,2,-q,2+p,1+\frac {b x^2}{a},-\frac {d \left (a+b x^2\right )}{b c-a d}\right )}{2 a^2 (1+p)} \] Output:

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

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.03 \[ \int \frac {\left (a+b x^2\right )^p \left (c+d x^2\right )^q}{x^3} \, dx=\frac {\left (1+\frac {a}{b x^2}\right )^{-p} \left (1+\frac {c}{d x^2}\right )^{-q} \left (a+b x^2\right )^p \left (c+d x^2\right )^q \operatorname {AppellF1}\left (1-p-q,-p,-q,2-p-q,-\frac {a}{b x^2},-\frac {c}{d x^2}\right )}{2 (-1+p+q) x^2} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.23 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.01, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {354, 154, 153}

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.

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

\(\Big \downarrow \) 354

\(\displaystyle \frac {1}{2} \int \frac {\left (b x^2+a\right )^p \left (d x^2+c\right )^q}{x^4}dx^2\)

\(\Big \downarrow \) 154

\(\displaystyle \frac {1}{2} \left (c+d x^2\right )^q \left (\frac {b \left (c+d x^2\right )}{b c-a d}\right )^{-q} \int \frac {\left (b x^2+a\right )^p \left (\frac {b d x^2}{b c-a d}+\frac {b c}{b c-a d}\right )^q}{x^4}dx^2\)

\(\Big \downarrow \) 153

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 153

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_)) 
^(p_), x_] :> Simp[(b*e - a*f)^p*((a + b*x)^(m + 1)/(b^(p + 1)*(m + 1)*Simp 
lify[b/(b*c - a*d)]^n))*AppellF1[m + 1, -n, -p, m + 2, (-d)*((a + b*x)/(b*c 
 - a*d)), (-f)*((a + b*x)/(b*e - a*f))], x] /; FreeQ[{a, b, c, d, e, f, m, 
n}, x] &&  !IntegerQ[m] &&  !IntegerQ[n] && IntegerQ[p] && GtQ[Simplify[b/( 
b*c - a*d)], 0] &&  !(GtQ[Simplify[d/(d*a - c*b)], 0] && SimplerQ[c + d*x, 
a + b*x])
 

rule 154

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_)) 
^(p_), x_] :> Simp[(c + d*x)^FracPart[n]/(Simplify[b/(b*c - a*d)]^IntPart[n 
]*(b*((c + d*x)/(b*c - a*d)))^FracPart[n])   Int[(a + b*x)^m*Simp[b*(c/(b*c 
 - a*d)) + b*d*(x/(b*c - a*d)), x]^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, 
 d, e, f, m, n}, x] &&  !IntegerQ[m] &&  !IntegerQ[n] && IntegerQ[p] &&  !G 
tQ[Simplify[b/(b*c - a*d)], 0] &&  !SimplerQ[c + d*x, a + b*x]
 

rule 354

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

Maple [F]

\[\int \frac {\left (b \,x^{2}+a \right )^{p} \left (x^{2} d +c \right )^{q}}{x^{3}}d x\]

Input:

int((b*x^2+a)^p*(d*x^2+c)^q/x^3,x)
 

Output:

int((b*x^2+a)^p*(d*x^2+c)^q/x^3,x)
 

Fricas [F]

\[ \int \frac {\left (a+b x^2\right )^p \left (c+d x^2\right )^q}{x^3} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{p} {\left (d x^{2} + c\right )}^{q}}{x^{3}} \,d x } \] Input:

integrate((b*x^2+a)^p*(d*x^2+c)^q/x^3,x, algorithm="fricas")
 

Output:

integral((b*x^2 + a)^p*(d*x^2 + c)^q/x^3, x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (a+b x^2\right )^p \left (c+d x^2\right )^q}{x^3} \, dx=\text {Timed out} \] Input:

integrate((b*x**2+a)**p*(d*x**2+c)**q/x**3,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {\left (a+b x^2\right )^p \left (c+d x^2\right )^q}{x^3} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{p} {\left (d x^{2} + c\right )}^{q}}{x^{3}} \,d x } \] Input:

integrate((b*x^2+a)^p*(d*x^2+c)^q/x^3,x, algorithm="maxima")
 

Output:

integrate((b*x^2 + a)^p*(d*x^2 + c)^q/x^3, x)
 

Giac [F]

\[ \int \frac {\left (a+b x^2\right )^p \left (c+d x^2\right )^q}{x^3} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{p} {\left (d x^{2} + c\right )}^{q}}{x^{3}} \,d x } \] Input:

integrate((b*x^2+a)^p*(d*x^2+c)^q/x^3,x, algorithm="giac")
 

Output:

integrate((b*x^2 + a)^p*(d*x^2 + c)^q/x^3, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b x^2\right )^p \left (c+d x^2\right )^q}{x^3} \, dx=\int \frac {{\left (b\,x^2+a\right )}^p\,{\left (d\,x^2+c\right )}^q}{x^3} \,d x \] Input:

int(((a + b*x^2)^p*(c + d*x^2)^q)/x^3,x)
 

Output:

int(((a + b*x^2)^p*(c + d*x^2)^q)/x^3, x)
 

Reduce [F]

\[ \int \frac {\left (a+b x^2\right )^p \left (c+d x^2\right )^q}{x^3} \, dx=\text {too large to display} \] Input:

int((b*x^2+a)^p*(d*x^2+c)^q/x^3,x)
 

Output:

( - (c + d*x**2)**q*(a + b*x**2)**p*a**2*d**2*q**2 + (c + d*x**2)**q*(a + 
b*x**2)**p*a**2*d**2*q - 2*(c + d*x**2)**q*(a + b*x**2)**p*a*b*c*d*p*q + ( 
c + d*x**2)**q*(a + b*x**2)**p*a*b*c*d*p + (c + d*x**2)**q*(a + b*x**2)**p 
*a*b*c*d*q - (c + d*x**2)**q*(a + b*x**2)**p*a*b*d**2*p*x**2 - (c + d*x**2 
)**q*(a + b*x**2)**p*b**2*c**2*p**2 + (c + d*x**2)**q*(a + b*x**2)**p*b**2 
*c**2*p - (c + d*x**2)**q*(a + b*x**2)**p*b**2*c*d*q*x**2 + 2*int(((c + d* 
x**2)**q*(a + b*x**2)**p*x**3)/(a**3*c*d**2*q**2 - a**3*c*d**2*q + a**3*d* 
*3*q**2*x**2 - a**3*d**3*q*x**2 + 2*a**2*b*c**2*d*p*q - a**2*b*c**2*d*p - 
a**2*b*c**2*d*q + 2*a**2*b*c*d**2*p*q*x**2 - a**2*b*c*d**2*p*x**2 + a**2*b 
*c*d**2*q**2*x**2 - 2*a**2*b*c*d**2*q*x**2 + a**2*b*d**3*q**2*x**4 - a**2* 
b*d**3*q*x**4 + a*b**2*c**3*p**2 - a*b**2*c**3*p + a*b**2*c**2*d*p**2*x**2 
 + 2*a*b**2*c**2*d*p*q*x**2 - 2*a*b**2*c**2*d*p*x**2 - a*b**2*c**2*d*q*x** 
2 + 2*a*b**2*c*d**2*p*q*x**4 - a*b**2*c*d**2*p*x**4 - a*b**2*c*d**2*q*x**4 
 + b**3*c**3*p**2*x**2 - b**3*c**3*p*x**2 + b**3*c**2*d*p**2*x**4 - b**3*c 
**2*d*p*x**4),x)*a**3*b**2*d**5*p**2*q**2*x**2 - 2*int(((c + d*x**2)**q*(a 
 + b*x**2)**p*x**3)/(a**3*c*d**2*q**2 - a**3*c*d**2*q + a**3*d**3*q**2*x** 
2 - a**3*d**3*q*x**2 + 2*a**2*b*c**2*d*p*q - a**2*b*c**2*d*p - a**2*b*c**2 
*d*q + 2*a**2*b*c*d**2*p*q*x**2 - a**2*b*c*d**2*p*x**2 + a**2*b*c*d**2*q** 
2*x**2 - 2*a**2*b*c*d**2*q*x**2 + a**2*b*d**3*q**2*x**4 - a**2*b*d**3*q*x* 
*4 + a*b**2*c**3*p**2 - a*b**2*c**3*p + a*b**2*c**2*d*p**2*x**2 + 2*a*b...