3.995 \(\int \frac {(a+\frac {b}{x^2})^p (c+\frac {d}{x^2})^q}{x^4} \, dx\)

Optimal. Leaf size=84 \[ -\frac {\left (a+\frac {b}{x^2}\right )^p \left (\frac {b}{a x^2}+1\right )^{-p} \left (c+\frac {d}{x^2}\right )^q \left (\frac {d}{c x^2}+1\right )^{-q} F_1\left (\frac {3}{2};-p,-q;\frac {5}{2};-\frac {b}{a x^2},-\frac {d}{c x^2}\right )}{3 x^3} \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {495, 511, 510} \[ -\frac {\left (a+\frac {b}{x^2}\right )^p \left (\frac {b}{a x^2}+1\right )^{-p} \left (c+\frac {d}{x^2}\right )^q \left (\frac {d}{c x^2}+1\right )^{-q} F_1\left (\frac {3}{2};-p,-q;\frac {5}{2};-\frac {b}{a x^2},-\frac {d}{c x^2}\right )}{3 x^3} \]

Antiderivative was successfully verified.

[In]

Int[((a + b/x^2)^p*(c + d/x^2)^q)/x^4,x]

[Out]

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

Rule 495

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Subst[Int[((a + b/x^
n)^p*(c + d/x^n)^q)/x^(m + 2), x], x, 1/x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && ILtQ[n, 0]
&& IntegerQ[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}{x^4} \, dx &=-\operatorname {Subst}\left (\int x^2 \left (a+b x^2\right )^p \left (c+d x^2\right )^q \, dx,x,\frac {1}{x}\right )\\ &=-\left (\left (\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 x^2}{a}\right )^p \left (c+d x^2\right )^q \, dx,x,\frac {1}{x}\right )\right )\\ &=-\left (\left (\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 x^2}{a}\right )^p \left (1+\frac {d x^2}{c}\right )^q \, dx,x,\frac {1}{x}\right )\right )\\ &=-\frac {\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}{2};-p,-q;\frac {5}{2};-\frac {b}{a x^2},-\frac {d}{c x^2}\right )}{3 x^3}\\ \end {align*}

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Mathematica [A]  time = 0.37, size = 106, normalized size = 1.26 \[ -\frac {\left (a+\frac {b}{x^2}\right )^p \left (\frac {a x^2}{b}+1\right )^{-p} \left (c+\frac {d}{x^2}\right )^q \left (\frac {c x^2}{d}+1\right )^{-q} F_1\left (-p-q-\frac {3}{2};-p,-q;-p-q-\frac {1}{2};-\frac {a x^2}{b},-\frac {c x^2}{d}\right )}{x^3 (2 p+2 q+3)} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[((a + b/x^2)^p*(c + d/x^2)^q)/x^4,x]

[Out]

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

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fricas [F]  time = 0.83, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\left (\frac {a x^{2} + b}{x^{2}}\right )^{p} \left (\frac {c x^{2} + d}{x^{2}}\right )^{q}}{x^{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x^2)^p*(c+d/x^2)^q/x^4,x, algorithm="fricas")

[Out]

integral(((a*x^2 + b)/x^2)^p*((c*x^2 + d)/x^2)^q/x^4, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (a + \frac {b}{x^{2}}\right )}^{p} {\left (c + \frac {d}{x^{2}}\right )}^{q}}{x^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x^2)^p*(c+d/x^2)^q/x^4,x, algorithm="giac")

[Out]

integrate((a + b/x^2)^p*(c + d/x^2)^q/x^4, x)

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maple [F]  time = 0.20, size = 0, normalized size = 0.00 \[ \int \frac {\left (a +\frac {b}{x^{2}}\right )^{p} \left (c +\frac {d}{x^{2}}\right )^{q}}{x^{4}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b/x^2)^p*(c+d/x^2)^q/x^4,x)

[Out]

int((a+b/x^2)^p*(c+d/x^2)^q/x^4,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (a + \frac {b}{x^{2}}\right )}^{p} {\left (c + \frac {d}{x^{2}}\right )}^{q}}{x^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x^2)^p*(c+d/x^2)^q/x^4,x, algorithm="maxima")

[Out]

integrate((a + b/x^2)^p*(c + d/x^2)^q/x^4, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+\frac {b}{x^2}\right )}^p\,{\left (c+\frac {d}{x^2}\right )}^q}{x^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b/x^2)^p*(c + d/x^2)^q)/x^4,x)

[Out]

int(((a + b/x^2)^p*(c + d/x^2)^q)/x^4, 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/x**4,x)

[Out]

Timed out

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