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

Optimal. Leaf size=91 \[ \frac{2 (e x)^{5/2} \left (a+\frac{b}{x^2}\right )^p \left (\frac{b}{a x^2}+1\right )^{-p} \left (c+\frac{d}{x^2}\right )^q \left (\frac{d}{c x^2}+1\right )^{-q} F_1\left (-\frac{5}{4};-p,-q;-\frac{1}{4};-\frac{b}{a x^2},-\frac{d}{c x^2}\right )}{5 e} \]

[Out]

(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)

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Rubi [A]  time = 0.121727, antiderivative size = 91, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.115, Rules used = {496, 511, 510} \[ \frac{2 (e x)^{5/2} \left (a+\frac{b}{x^2}\right )^p \left (\frac{b}{a x^2}+1\right )^{-p} \left (c+\frac{d}{x^2}\right )^q \left (\frac{d}{c x^2}+1\right )^{-q} F_1\left (-\frac{5}{4};-p,-q;-\frac{1}{4};-\frac{b}{a x^2},-\frac{d}{c x^2}\right )}{5 e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(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)

Rule 496

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> With[{g = Deno
minator[m]}, -Dist[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] && FractionQ[m]

Rule 511

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

Rule 510

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

Rubi steps

\begin{align*} \int \left (a+\frac{b}{x^2}\right )^p \left (c+\frac{d}{x^2}\right )^q (e x)^{3/2} \, dx &=-\frac{2 \operatorname{Subst}\left (\int \frac{\left (a+b e^2 x^4\right )^p \left (c+d e^2 x^4\right )^q}{x^6} \, dx,x,\frac{1}{\sqrt{e x}}\right )}{e}\\ &=-\frac{\left (2 \left (a+\frac{b}{x^2}\right )^p \left (1+\frac{b}{a x^2}\right )^{-p}\right ) \operatorname{Subst}\left (\int \frac{\left (1+\frac{b e^2 x^4}{a}\right )^p \left (c+d e^2 x^4\right )^q}{x^6} \, dx,x,\frac{1}{\sqrt{e x}}\right )}{e}\\ &=-\frac{\left (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}\right ) \operatorname{Subst}\left (\int \frac{\left (1+\frac{b e^2 x^4}{a}\right )^p \left (1+\frac{d e^2 x^4}{c}\right )^q}{x^6} \, dx,x,\frac{1}{\sqrt{e x}}\right )}{e}\\ &=\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} F_1\left (-\frac{5}{4};-p,-q;-\frac{1}{4};-\frac{b}{a x^2},-\frac{d}{c x^2}\right )}{5 e}\\ \end{align*}

Mathematica [A]  time = 0.112769, size = 111, normalized size = 1.22 \[ -\frac{2 x (e x)^{3/2} \left (a+\frac{b}{x^2}\right )^p \left (\frac{a x^2}{b}+1\right )^{-p} \left (c+\frac{d}{x^2}\right )^q \left (\frac{c x^2}{d}+1\right )^{-q} F_1\left (-p-q+\frac{5}{4};-p,-q;-p-q+\frac{9}{4};-\frac{a x^2}{b},-\frac{c x^2}{d}\right )}{4 p+4 q-5} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-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)

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Maple [F]  time = 0.025, size = 0, normalized size = 0. \begin{align*} \int \left ( a+{\frac{b}{{x}^{2}}} \right ) ^{p} \left ( c+{\frac{d}{{x}^{2}}} \right ) ^{q} \left ( ex \right ) ^{{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\sqrt{e x} e x \left (\frac{a x^{2} + b}{x^{2}}\right )^{p} \left (\frac{c x^{2} + d}{x^{2}}\right )^{q}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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