3.11.1 \(\int \frac {(a+\frac {b}{x^2})^p (c+\frac {d}{x^2})^q}{(e x)^{5/2}} \, dx\) [1001]

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

[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 = 0.07, 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 = {510, 525, 524} \begin {gather*} -\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}} \end {gather*}

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 510

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 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*} \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 \text {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 ) \text {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 ) \text {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 = 0.13, size = 111, normalized size = 1.22 \begin {gather*} -\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} F_1\left (-\frac {3}{4}-p-q;-p,-q;\frac {1}{4}-p-q;-\frac {a x^2}{b},-\frac {c x^2}{d}\right )}{(3+4 p+4 q) (e x)^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[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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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {\left (\frac {b}{x^{2}}+a \right )^{p} \left (c +\frac {d}{x^{2}}\right )^{q}}{\left (e x \right )^{\frac {5}{2}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((b/x^2+a)^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 \begin {gather*} \text {Failed to integrate} \end {gather*}

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]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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^(-5/2)/x^(5/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \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 \end {gather*}

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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