\(\int \frac {(a+b x^2)^p (c+d x^2)^q}{(e x)^{5/2}} \, dx\) [1156]

Optimal result

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

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {525, 524} \[ \int \frac {\left (a+b x^2\right )^p \left (c+d x^2\right )^q}{(e x)^{5/2}} \, dx=-\frac {2 \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (c+d x^2\right )^q \left (\frac {d x^2}{c}+1\right )^{-q} \operatorname {AppellF1}\left (-\frac {3}{4},-p,-q,\frac {1}{4},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )}{3 e (e x)^{3/2}} \]

[In]

Int[((a + b*x^2)^p*(c + d*x^2)^q)/(e*x)^(5/2),x]

[Out]

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

Rule 524

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] /; Free
Q[{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 525

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[a^IntPar
t[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])

Rubi steps \begin{align*} \text {integral}& = \left (\left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p}\right ) \int \frac {\left (1+\frac {b x^2}{a}\right )^p \left (c+d x^2\right )^q}{(e x)^{5/2}} \, dx \\ & = \left (\left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \left (c+d x^2\right )^q \left (1+\frac {d x^2}{c}\right )^{-q}\right ) \int \frac {\left (1+\frac {b x^2}{a}\right )^p \left (1+\frac {d x^2}{c}\right )^q}{(e x)^{5/2}} \, dx \\ & = -\frac {2 \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \left (c+d x^2\right )^q \left (1+\frac {d x^2}{c}\right )^{-q} F_1\left (-\frac {3}{4};-p,-q;\frac {1}{4};-\frac {b x^2}{a},-\frac {d x^2}{c}\right )}{3 e (e x)^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.45 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.00 \[ \int \frac {\left (a+b x^2\right )^p \left (c+d x^2\right )^q}{(e x)^{5/2}} \, dx=-\frac {2 x \left (a+b x^2\right )^p \left (\frac {a+b x^2}{a}\right )^{-p} \left (c+d x^2\right )^q \left (\frac {c+d x^2}{c}\right )^{-q} \operatorname {AppellF1}\left (-\frac {3}{4},-p,-q,\frac {1}{4},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )}{3 (e x)^{5/2}} \]

[In]

Integrate[((a + b*x^2)^p*(c + d*x^2)^q)/(e*x)^(5/2),x]

[Out]

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

Maple [F]

\[\int \frac {\left (b \,x^{2}+a \right )^{p} \left (d \,x^{2}+c \right )^{q}}{\left (e x \right )^{\frac {5}{2}}}d x\]

[In]

int((b*x^2+a)^p*(d*x^2+c)^q/(e*x)^(5/2),x)

[Out]

int((b*x^2+a)^p*(d*x^2+c)^q/(e*x)^(5/2),x)

Fricas [F]

\[ \int \frac {\left (a+b x^2\right )^p \left (c+d x^2\right )^q}{(e x)^{5/2}} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{p} {\left (d x^{2} + c\right )}^{q}}{\left (e x\right )^{\frac {5}{2}}} \,d x } \]

[In]

integrate((b*x^2+a)^p*(d*x^2+c)^q/(e*x)^(5/2),x, algorithm="fricas")

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (a+b x^2\right )^p \left (c+d x^2\right )^q}{(e x)^{5/2}} \, dx=\text {Timed out} \]

[In]

integrate((b*x**2+a)**p*(d*x**2+c)**q/(e*x)**(5/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {\left (a+b x^2\right )^p \left (c+d x^2\right )^q}{(e x)^{5/2}} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{p} {\left (d x^{2} + c\right )}^{q}}{\left (e x\right )^{\frac {5}{2}}} \,d x } \]

[In]

integrate((b*x^2+a)^p*(d*x^2+c)^q/(e*x)^(5/2),x, algorithm="maxima")

[Out]

integrate((b*x^2 + a)^p*(d*x^2 + c)^q/(e*x)^(5/2), x)

Giac [F]

\[ \int \frac {\left (a+b x^2\right )^p \left (c+d x^2\right )^q}{(e x)^{5/2}} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{p} {\left (d x^{2} + c\right )}^{q}}{\left (e x\right )^{\frac {5}{2}}} \,d x } \]

[In]

integrate((b*x^2+a)^p*(d*x^2+c)^q/(e*x)^(5/2),x, algorithm="giac")

[Out]

integrate((b*x^2 + a)^p*(d*x^2 + c)^q/(e*x)^(5/2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b x^2\right )^p \left (c+d x^2\right )^q}{(e x)^{5/2}} \, dx=\int \frac {{\left (b\,x^2+a\right )}^p\,{\left (d\,x^2+c\right )}^q}{{\left (e\,x\right )}^{5/2}} \,d x \]

[In]

int(((a + b*x^2)^p*(c + d*x^2)^q)/(e*x)^(5/2),x)

[Out]

int(((a + b*x^2)^p*(c + d*x^2)^q)/(e*x)^(5/2), x)