Integrand size = 26, antiderivative size = 89 \[ \int \frac {\left (a+\frac {b}{x^2}\right )^p \left (c+\frac {d}{x^2}\right )^q}{\sqrt {e x}} \, dx=\frac {2 \left (a+\frac {b}{x^2}\right )^p \left (1+\frac {b}{a x^2}\right )^{-p} \left (c+\frac {d}{x^2}\right )^q \left (1+\frac {d}{c x^2}\right )^{-q} \sqrt {e x} \operatorname {AppellF1}\left (-\frac {1}{4},-p,-q,\frac {3}{4},-\frac {b}{a x^2},-\frac {d}{c x^2}\right )}{e} \]
2*(a+b/x^2)^p*(c+d/x^2)^q*AppellF1(-1/4,-p,-q,3/4,-b/a/x^2,-d/c/x^2)*(e*x) ^(1/2)/e/((1+b/a/x^2)^p)/((1+d/c/x^2)^q)
Time = 0.46 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.25 \[ \int \frac {\left (a+\frac {b}{x^2}\right )^p \left (c+\frac {d}{x^2}\right )^q}{\sqrt {e x}} \, dx=-\frac {2 \left (a+\frac {b}{x^2}\right )^p \left (c+\frac {d}{x^2}\right )^q x \left (1+\frac {a x^2}{b}\right )^{-p} \left (1+\frac {c x^2}{d}\right )^{-q} \operatorname {AppellF1}\left (\frac {1}{4}-p-q,-p,-q,\frac {5}{4}-p-q,-\frac {a x^2}{b},-\frac {c x^2}{d}\right )}{(-1+4 p+4 q) \sqrt {e x}} \]
(-2*(a + b/x^2)^p*(c + d/x^2)^q*x*AppellF1[1/4 - p - q, -p, -q, 5/4 - p - q, -((a*x^2)/b), -((c*x^2)/d)])/((-1 + 4*p + 4*q)*Sqrt[e*x]*(1 + (a*x^2)/b )^p*(1 + (c*x^2)/d)^q)
Time = 0.28 (sec) , antiderivative size = 89, 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.
\(\displaystyle \int \frac {\left (a+\frac {b}{x^2}\right )^p \left (c+\frac {d}{x^2}\right )^q}{\sqrt {e x}} \, dx\) |
\(\Big \downarrow \) 998 |
\(\displaystyle -\frac {2 \int e \left (a+\frac {b}{x^2}\right )^p \left (c+\frac {d}{x^2}\right )^q xd\frac {1}{\sqrt {e x}}}{e}\) |
\(\Big \downarrow \) 1013 |
\(\displaystyle -\frac {2 \left (a+\frac {b}{x^2}\right )^p \left (\frac {b}{a x^2}+1\right )^{-p} \int e \left (\frac {b}{a x^2}+1\right )^p \left (c+\frac {d}{x^2}\right )^q xd\frac {1}{\sqrt {e x}}}{e}\) |
\(\Big \downarrow \) 1013 |
\(\displaystyle -\frac {2 \left (a+\frac {b}{x^2}\right )^p \left (\frac {b}{a x^2}+1\right )^{-p} \left (c+\frac {d}{x^2}\right )^q \left (\frac {d}{c x^2}+1\right )^{-q} \int e \left (\frac {b}{a x^2}+1\right )^p \left (\frac {d}{c x^2}+1\right )^q xd\frac {1}{\sqrt {e x}}}{e}\) |
\(\Big \downarrow \) 1012 |
\(\displaystyle \frac {2 \sqrt {e x} \left (a+\frac {b}{x^2}\right )^p \left (\frac {b}{a x^2}+1\right )^{-p} \left (c+\frac {d}{x^2}\right )^q \left (\frac {d}{c x^2}+1\right )^{-q} \operatorname {AppellF1}\left (-\frac {1}{4},-p,-q,\frac {3}{4},-\frac {b}{a x^2},-\frac {d}{c x^2}\right )}{e}\) |
(2*(a + b/x^2)^p*(c + d/x^2)^q*Sqrt[e*x]*AppellF1[-1/4, -p, -q, 3/4, -(b/( a*x^2)), -(d/(c*x^2))])/(e*(1 + b/(a*x^2))^p*(1 + d/(c*x^2))^q)
3.10.99.3.1 Defintions of rubi rules used
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]
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])
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])
\[\int \frac {\left (a +\frac {b}{x^{2}}\right )^{p} \left (c +\frac {d}{x^{2}}\right )^{q}}{\sqrt {e x}}d x\]
\[ \int \frac {\left (a+\frac {b}{x^2}\right )^p \left (c+\frac {d}{x^2}\right )^q}{\sqrt {e x}} \, dx=\int { \frac {{\left (a + \frac {b}{x^{2}}\right )}^{p} {\left (c + \frac {d}{x^{2}}\right )}^{q}}{\sqrt {e x}} \,d x } \]
Timed out. \[ \int \frac {\left (a+\frac {b}{x^2}\right )^p \left (c+\frac {d}{x^2}\right )^q}{\sqrt {e x}} \, dx=\text {Timed out} \]
\[ \int \frac {\left (a+\frac {b}{x^2}\right )^p \left (c+\frac {d}{x^2}\right )^q}{\sqrt {e x}} \, dx=\int { \frac {{\left (a + \frac {b}{x^{2}}\right )}^{p} {\left (c + \frac {d}{x^{2}}\right )}^{q}}{\sqrt {e x}} \,d x } \]
\[ \int \frac {\left (a+\frac {b}{x^2}\right )^p \left (c+\frac {d}{x^2}\right )^q}{\sqrt {e x}} \, dx=\int { \frac {{\left (a + \frac {b}{x^{2}}\right )}^{p} {\left (c + \frac {d}{x^{2}}\right )}^{q}}{\sqrt {e x}} \,d x } \]
Timed out. \[ \int \frac {\left (a+\frac {b}{x^2}\right )^p \left (c+\frac {d}{x^2}\right )^q}{\sqrt {e x}} \, dx=\int \frac {{\left (a+\frac {b}{x^2}\right )}^p\,{\left (c+\frac {d}{x^2}\right )}^q}{\sqrt {e\,x}} \,d x \]