3.1001 \(\int \frac {(a+\frac {b}{x^2})^p (c+\frac {d}{x^2})^q}{(e x)^{5/2}} \, dx\)

Optimal. Leaf size=91 \[ -\frac {2 \left (a+\frac {b}{x^2}\right )^p \left (\frac {b}{a x^2}+1\right )^{-p} \left (c+\frac {d}{x^2}\right )^q \left (\frac {d}{c x^2}+1\right )^{-q} F_1\left (\frac {3}{4};-p,-q;\frac {7}{4};-\frac {b}{a x^2},-\frac {d}{c x^2}\right )}{3 e (e x)^{3/2}} \]

[Out]

-2/3*(a+b/x^2)^p*(c+d/x^2)^q*AppellF1(3/4,-p,-q,7/4,-b/a/x^2,-d/c/x^2)/e/((1+b/a/x^2)^p)/((1+d/c/x^2)^q)/(e*x)
^(3/2)

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Rubi [A]  time = 0.12, antiderivative size = 91, normalized size of antiderivative = 1.00, 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} \[ -\frac {2 \left (a+\frac {b}{x^2}\right )^p \left (\frac {b}{a x^2}+1\right )^{-p} \left (c+\frac {d}{x^2}\right )^q \left (\frac {d}{c x^2}+1\right )^{-q} F_1\left (\frac {3}{4};-p,-q;\frac {7}{4};-\frac {b}{a x^2},-\frac {d}{c x^2}\right )}{3 e (e x)^{3/2}} \]

Antiderivative was successfully verified.

[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, 7/4, -(b/(a*x^2)), -(d/(c*x^2))])/(3*e*(1 + b/(a*x^2))^p
*(1 + d/(c*x^2))^q*(e*x)^(3/2))

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 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])

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])

Rubi steps

\begin {align*} \int \frac {\left (a+\frac {b}{x^2}\right )^p \left (c+\frac {d}{x^2}\right )^q}{(e x)^{5/2}} \, dx &=-\frac {2 \operatorname {Subst}\left (\int x^2 \left (a+b e^2 x^4\right )^p \left (c+d e^2 x^4\right )^q \, 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 x^2 \left (1+\frac {b e^2 x^4}{a}\right )^p \left (c+d e^2 x^4\right )^q \, 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 x^2 \left (1+\frac {b e^2 x^4}{a}\right )^p \left (1+\frac {d e^2 x^4}{c}\right )^q \, 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} F_1\left (\frac {3}{4};-p,-q;\frac {7}{4};-\frac {b}{a x^2},-\frac {d}{c x^2}\right )}{3 e (e x)^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.18, size = 111, normalized size = 1.22 \[ -\frac {2 x \left (a+\frac {b}{x^2}\right )^p \left (\frac {a x^2}{b}+1\right )^{-p} \left (c+\frac {d}{x^2}\right )^q \left (\frac {c x^2}{d}+1\right )^{-q} F_1\left (-p-q-\frac {3}{4};-p,-q;-p-q+\frac {1}{4};-\frac {a x^2}{b},-\frac {c x^2}{d}\right )}{(e x)^{5/2} (4 p+4 q+3)} \]

Warning: Unable to verify antiderivative.

[In]

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

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fricas [F]  time = 0.79, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {e x} \left (\frac {a x^{2} + b}{x^{2}}\right )^{p} \left (\frac {c x^{2} + d}{x^{2}}\right )^{q}}{e^{3} x^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (a + \frac {b}{x^{2}}\right )}^{p} {\left (c + \frac {d}{x^{2}}\right )}^{q}}{\left (e x\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.08, size = 0, normalized size = 0.00 \[ \int \frac {\left (a +\frac {b}{x^{2}}\right )^{p} \left (c +\frac {d}{x^{2}}\right )^{q}}{\left (e x \right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (a + \frac {b}{x^{2}}\right )}^{p} {\left (c + \frac {d}{x^{2}}\right )}^{q}}{\left (e x\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+\frac {b}{x^2}\right )}^p\,{\left (c+\frac {d}{x^2}\right )}^q}{{\left (e\,x\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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