\(\int (a+\frac {b}{x^2})^p (c+\frac {d}{x^2})^q (e x)^{3/2} \, dx\) [205]

Optimal result

Integrand size = 26, antiderivative size = 91 \[ \int \left (a+\frac {b}{x^2}\right )^p \left (c+\frac {d}{x^2}\right )^q (e x)^{3/2} \, dx=\frac {2 \left (a+\frac {b}{x^2}\right )^p \left (1+\frac {b}{a x^2}\right )^{-p} \left (c+\frac {d}{x^2}\right )^q \left (1+\frac {d}{c x^2}\right )^{-q} (e x)^{5/2} \operatorname {AppellF1}\left (-\frac {5}{4},-p,-q,-\frac {1}{4},-\frac {b}{a x^2},-\frac {d}{c x^2}\right )}{5 e} \] Output:

2/5*(a+b/x^2)^p*(c+d/x^2)^q*(e*x)^(5/2)*AppellF1(-5/4,-p,-q,-1/4,-b/a/x^2, 
-d/c/x^2)/e/((1+b/a/x^2)^p)/((1+d/c/x^2)^q)
 

Mathematica [A] (verified)

Time = 0.41 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.22 \[ \int \left (a+\frac {b}{x^2}\right )^p \left (c+\frac {d}{x^2}\right )^q (e x)^{3/2} \, dx=-\frac {2 \left (a+\frac {b}{x^2}\right )^p \left (c+\frac {d}{x^2}\right )^q x (e x)^{3/2} \left (1+\frac {a x^2}{b}\right )^{-p} \left (1+\frac {c x^2}{d}\right )^{-q} \operatorname {AppellF1}\left (\frac {5}{4}-p-q,-p,-q,\frac {9}{4}-p-q,-\frac {a x^2}{b},-\frac {c x^2}{d}\right )}{-5+4 p+4 q} \] Input:

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

Output:

(-2*(a + b/x^2)^p*(c + d/x^2)^q*x*(e*x)^(3/2)*AppellF1[5/4 - p - q, -p, -q 
, 9/4 - p - q, -((a*x^2)/b), -((c*x^2)/d)])/((-5 + 4*p + 4*q)*(1 + (a*x^2) 
/b)^p*(1 + (c*x^2)/d)^q)
 

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {998, 1013, 1013, 1012}

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.

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

\(\Big \downarrow \) 998

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

\(\Big \downarrow \) 1013

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

\(\Big \downarrow \) 1013

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

\(\Big \downarrow \) 1012

\(\displaystyle \frac {2 (e x)^{5/2} \left (a+\frac {b}{x^2}\right )^p \left (\frac {b}{a x^2}+1\right )^{-p} \left (c+\frac {d}{x^2}\right )^q \left (\frac {d}{c x^2}+1\right )^{-q} \operatorname {AppellF1}\left (-\frac {5}{4},-p,-q,-\frac {1}{4},-\frac {b}{a x^2},-\frac {d}{c x^2}\right )}{5 e}\)

Input:

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

Output:

(2*(a + b/x^2)^p*(c + d/x^2)^q*(e*x)^(5/2)*AppellF1[-5/4, -p, -q, -1/4, -( 
b/(a*x^2)), -(d/(c*x^2))])/(5*e*(1 + b/(a*x^2))^p*(1 + d/(c*x^2))^q)
 

Defintions of rubi rules used

rule 998

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_) 
)^(q_), x_Symbol] :> With[{g = Denominator[m]}, Simp[-g/e   Subst[Int[(a + 
b/(e^n*x^(g*n)))^p*((c + d/(e^n*x^(g*n)))^q/x^(g*(m + 1) + 1)), x], x, 1/(e 
*x)^(1/g)], x]] /; FreeQ[{a, b, c, d, e, p, q}, x] && ILtQ[n, 0] && Fractio 
nQ[m]
 

rule 1012

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

rule 1013

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

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(2*sqrt(e)*e*(4*sqrt(x)*(c*x**2 + d)**q*(a*x**2 + b)**p*a*d*q*x**2 - sqrt( 
x)*(c*x**2 + d)**q*(a*x**2 + b)**p*a*d*x**2 + 4*sqrt(x)*(c*x**2 + d)**q*(a 
*x**2 + b)**p*b*c*p*x**2 - sqrt(x)*(c*x**2 + d)**q*(a*x**2 + b)**p*b*c*x** 
2 - 4*sqrt(x)*(c*x**2 + d)**q*(a*x**2 + b)**p*b*d*p - 4*sqrt(x)*(c*x**2 + 
d)**q*(a*x**2 + b)**p*b*d*q + 32*x**(2*p + 2*q)*int((sqrt(x)*(c*x**2 + d)* 
*q*(a*x**2 + b)**p*x**3)/(4*x**(2*p + 2*q)*a**2*c*d*q*x**4 - x**(2*p + 2*q 
)*a**2*c*d*x**4 + 4*x**(2*p + 2*q)*a**2*d**2*q*x**2 - x**(2*p + 2*q)*a**2* 
d**2*x**2 + 4*x**(2*p + 2*q)*a*b*c**2*p*x**4 - x**(2*p + 2*q)*a*b*c**2*x** 
4 + 4*x**(2*p + 2*q)*a*b*c*d*p*x**2 + 4*x**(2*p + 2*q)*a*b*c*d*q*x**2 - 2* 
x**(2*p + 2*q)*a*b*c*d*x**2 + 4*x**(2*p + 2*q)*a*b*d**2*q - x**(2*p + 2*q) 
*a*b*d**2 + 4*x**(2*p + 2*q)*b**2*c**2*p*x**2 - x**(2*p + 2*q)*b**2*c**2*x 
**2 + 4*x**(2*p + 2*q)*b**2*c*d*p - x**(2*p + 2*q)*b**2*c*d),x)*a**3*d**3* 
q**3 - 16*x**(2*p + 2*q)*int((sqrt(x)*(c*x**2 + d)**q*(a*x**2 + b)**p*x**3 
)/(4*x**(2*p + 2*q)*a**2*c*d*q*x**4 - x**(2*p + 2*q)*a**2*c*d*x**4 + 4*x** 
(2*p + 2*q)*a**2*d**2*q*x**2 - x**(2*p + 2*q)*a**2*d**2*x**2 + 4*x**(2*p + 
 2*q)*a*b*c**2*p*x**4 - x**(2*p + 2*q)*a*b*c**2*x**4 + 4*x**(2*p + 2*q)*a* 
b*c*d*p*x**2 + 4*x**(2*p + 2*q)*a*b*c*d*q*x**2 - 2*x**(2*p + 2*q)*a*b*c*d* 
x**2 + 4*x**(2*p + 2*q)*a*b*d**2*q - x**(2*p + 2*q)*a*b*d**2 + 4*x**(2*p + 
 2*q)*b**2*c**2*p*x**2 - x**(2*p + 2*q)*b**2*c**2*x**2 + 4*x**(2*p + 2*q)* 
b**2*c*d*p - x**(2*p + 2*q)*b**2*c*d),x)*a**3*d**3*q**2 + 2*x**(2*p + 2...