Integrand size = 22, antiderivative size = 99 \[ \int \left (a+\frac {b}{x^2}\right )^p \left (c+\frac {d}{x^2}\right )^q x^3 \, dx=\frac {b^2 \left (a+\frac {b}{x^2}\right )^{1+p} \left (c+\frac {d}{x^2}\right )^q \left (\frac {b \left (c+\frac {d}{x^2}\right )}{b c-a d}\right )^{-q} \operatorname {AppellF1}\left (1+p,3,-q,2+p,1+\frac {b}{a x^2},-\frac {d \left (a+\frac {b}{x^2}\right )}{b c-a d}\right )}{2 a^3 (1+p)} \] Output:
1/2*b^2*(a+b/x^2)^(p+1)*(c+d/x^2)^q*AppellF1(p+1,-q,3,2+p,-d*(a+b/x^2)/(-a *d+b*c),1+b/a/x^2)/a^3/(p+1)/((b*(c+d/x^2)/(-a*d+b*c))^q)
Time = 0.18 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.01 \[ \int \left (a+\frac {b}{x^2}\right )^p \left (c+\frac {d}{x^2}\right )^q x^3 \, dx=-\frac {\left (a+\frac {b}{x^2}\right )^p \left (c+\frac {d}{x^2}\right )^q x^4 \left (1+\frac {a x^2}{b}\right )^{-p} \left (1+\frac {c x^2}{d}\right )^{-q} \operatorname {AppellF1}\left (2-p-q,-p,-q,3-p-q,-\frac {a x^2}{b},-\frac {c x^2}{d}\right )}{2 (-2+p+q)} \] Input:
Integrate[(a + b/x^2)^p*(c + d/x^2)^q*x^3,x]
Output:
-1/2*((a + b/x^2)^p*(c + d/x^2)^q*x^4*AppellF1[2 - p - q, -p, -q, 3 - p - q, -((a*x^2)/b), -((c*x^2)/d)])/((-2 + p + q)*(1 + (a*x^2)/b)^p*(1 + (c*x^ 2)/d)^q)
Time = 0.39 (sec) , antiderivative size = 100, 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 = {948, 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.
\(\displaystyle \int x^3 \left (a+\frac {b}{x^2}\right )^p \left (c+\frac {d}{x^2}\right )^q \, dx\) |
\(\Big \downarrow \) 948 |
\(\displaystyle -\frac {1}{2} \int \left (a+\frac {b}{x^2}\right )^p \left (c+\frac {d}{x^2}\right )^q x^6d\frac {1}{x^2}\) |
\(\Big \downarrow \) 154 |
\(\displaystyle -\frac {1}{2} \left (c+\frac {d}{x^2}\right )^q \left (\frac {b \left (c+\frac {d}{x^2}\right )}{b c-a d}\right )^{-q} \int \left (a+\frac {b}{x^2}\right )^p \left (\frac {b c}{b c-a d}+\frac {b d}{(b c-a d) x^2}\right )^q x^6d\frac {1}{x^2}\) |
\(\Big \downarrow \) 153 |
\(\displaystyle \frac {b^2 \left (a+\frac {b}{x^2}\right )^{p+1} \left (c+\frac {d}{x^2}\right )^q \left (\frac {b \left (c+\frac {d}{x^2}\right )}{b c-a d}\right )^{-q} \operatorname {AppellF1}\left (p+1,-q,3,p+2,-\frac {d \left (a+\frac {b}{x^2}\right )}{b c-a d},\frac {a+\frac {b}{x^2}}{a}\right )}{2 a^3 (p+1)}\) |
Input:
Int[(a + b/x^2)^p*(c + d/x^2)^q*x^3,x]
Output:
(b^2*(a + b/x^2)^(1 + p)*(c + d/x^2)^q*AppellF1[1 + p, -q, 3, 2 + p, -((d* (a + b/x^2))/(b*c - a*d)), (a + b/x^2)/a])/(2*a^3*(1 + p)*((b*(c + d/x^2)) /(b*c - a*d))^q)
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])
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]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_. ), x_Symbol] :> Simp[1/n Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^ p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ [b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]
\[\int \left (a +\frac {b}{x^{2}}\right )^{p} \left (c +\frac {d}{x^{2}}\right )^{q} x^{3}d x\]
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)
\[ \int \left (a+\frac {b}{x^2}\right )^p \left (c+\frac {d}{x^2}\right )^q x^3 \, dx=\int { {\left (a + \frac {b}{x^{2}}\right )}^{p} {\left (c + \frac {d}{x^{2}}\right )}^{q} x^{3} \,d x } \] Input:
integrate((a+b/x^2)^p*(c+d/x^2)^q*x^3,x, algorithm="fricas")
Output:
integral(x^3*((a*x^2 + b)/x^2)^p*((c*x^2 + d)/x^2)^q, x)
Timed out. \[ \int \left (a+\frac {b}{x^2}\right )^p \left (c+\frac {d}{x^2}\right )^q x^3 \, dx=\text {Timed out} \] Input:
integrate((a+b/x**2)**p*(c+d/x**2)**q*x**3,x)
Output:
Timed out
\[ \int \left (a+\frac {b}{x^2}\right )^p \left (c+\frac {d}{x^2}\right )^q x^3 \, dx=\int { {\left (a + \frac {b}{x^{2}}\right )}^{p} {\left (c + \frac {d}{x^{2}}\right )}^{q} x^{3} \,d x } \] Input:
integrate((a+b/x^2)^p*(c+d/x^2)^q*x^3,x, algorithm="maxima")
Output:
integrate((a + b/x^2)^p*(c + d/x^2)^q*x^3, x)
\[ \int \left (a+\frac {b}{x^2}\right )^p \left (c+\frac {d}{x^2}\right )^q x^3 \, dx=\int { {\left (a + \frac {b}{x^{2}}\right )}^{p} {\left (c + \frac {d}{x^{2}}\right )}^{q} x^{3} \,d x } \] Input:
integrate((a+b/x^2)^p*(c+d/x^2)^q*x^3,x, algorithm="giac")
Output:
integrate((a + b/x^2)^p*(c + d/x^2)^q*x^3, x)
Timed out. \[ \int \left (a+\frac {b}{x^2}\right )^p \left (c+\frac {d}{x^2}\right )^q x^3 \, dx=\int x^3\,{\left (a+\frac {b}{x^2}\right )}^p\,{\left (c+\frac {d}{x^2}\right )}^q \,d x \] Input:
int(x^3*(a + b/x^2)^p*(c + d/x^2)^q,x)
Output:
int(x^3*(a + b/x^2)^p*(c + d/x^2)^q, x)
\[ \int \left (a+\frac {b}{x^2}\right )^p \left (c+\frac {d}{x^2}\right )^q x^3 \, dx =\text {Too large to display} \] Input:
int((a+b/x^2)^p*(c+d/x^2)^q*x^3,x)
Output:
((c*x**2 + d)**q*(a*x**2 + b)**p*a*c*x**4 + (c*x**2 + d)**q*(a*x**2 + b)** p*a*d*q*x**2 + (c*x**2 + d)**q*(a*x**2 + b)**p*b*c*p*x**2 - (c*x**2 + d)** q*(a*x**2 + b)**p*b*d*p - (c*x**2 + d)**q*(a*x**2 + b)**p*b*d*q + (c*x**2 + d)**q*(a*x**2 + b)**p*b*d + 2*x**(2*p + 2*q)*int(((c*x**2 + d)**q*(a*x** 2 + b)**p*x**3)/(x**(2*p + 2*q)*a*c*x**4 + x**(2*p + 2*q)*a*d*x**2 + x**(2 *p + 2*q)*b*c*x**2 + x**(2*p + 2*q)*b*d),x)*a**2*d**2*q**2 - 2*x**(2*p + 2 *q)*int(((c*x**2 + d)**q*(a*x**2 + b)**p*x**3)/(x**(2*p + 2*q)*a*c*x**4 + x**(2*p + 2*q)*a*d*x**2 + x**(2*p + 2*q)*b*c*x**2 + x**(2*p + 2*q)*b*d),x) *a**2*d**2*q + 4*x**(2*p + 2*q)*int(((c*x**2 + d)**q*(a*x**2 + b)**p*x**3) /(x**(2*p + 2*q)*a*c*x**4 + x**(2*p + 2*q)*a*d*x**2 + x**(2*p + 2*q)*b*c*x **2 + x**(2*p + 2*q)*b*d),x)*a*b*c*d*p*q + 2*x**(2*p + 2*q)*int(((c*x**2 + d)**q*(a*x**2 + b)**p*x**3)/(x**(2*p + 2*q)*a*c*x**4 + x**(2*p + 2*q)*a*d *x**2 + x**(2*p + 2*q)*b*c*x**2 + x**(2*p + 2*q)*b*d),x)*b**2*c**2*p**2 - 2*x**(2*p + 2*q)*int(((c*x**2 + d)**q*(a*x**2 + b)**p*x**3)/(x**(2*p + 2*q )*a*c*x**4 + x**(2*p + 2*q)*a*d*x**2 + x**(2*p + 2*q)*b*c*x**2 + x**(2*p + 2*q)*b*d),x)*b**2*c**2*p - 2*x**(2*p + 2*q)*int(((c*x**2 + d)**q*(a*x**2 + b)**p)/(x**(2*p + 2*q)*a*c*x**5 + x**(2*p + 2*q)*a*d*x**3 + x**(2*p + 2* q)*b*c*x**3 + x**(2*p + 2*q)*b*d*x),x)*b**2*d**2*p**2 - 4*x**(2*p + 2*q)*i nt(((c*x**2 + d)**q*(a*x**2 + b)**p)/(x**(2*p + 2*q)*a*c*x**5 + x**(2*p + 2*q)*a*d*x**3 + x**(2*p + 2*q)*b*c*x**3 + x**(2*p + 2*q)*b*d*x),x)*b**2...