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

Optimal. Leaf size=79 \[ x \left (a+\frac {b}{x^2}\right )^p \left (\frac {b}{a x^2}+1\right )^{-p} \left (c+\frac {d}{x^2}\right )^q \left (\frac {d}{c x^2}+1\right )^{-q} F_1\left (-\frac {1}{2};-p,-q;\frac {1}{2};-\frac {b}{a x^2},-\frac {d}{c x^2}\right ) \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {375, 511, 510} \[ x \left (a+\frac {b}{x^2}\right )^p \left (\frac {b}{a x^2}+1\right )^{-p} \left (c+\frac {d}{x^2}\right )^q \left (\frac {d}{c x^2}+1\right )^{-q} F_1\left (-\frac {1}{2};-p,-q;\frac {1}{2};-\frac {b}{a x^2},-\frac {d}{c x^2}\right ) \]

Antiderivative was successfully verified.

[In]

Int[(a + b/x^2)^p*(c + d/x^2)^q,x]

[Out]

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

Rule 375

Int[((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^2, x], x, 1/x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && ILtQ[n, 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])

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

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Mathematica [A]  time = 0.10, size = 104, normalized size = 1.32 \[ -\frac {x \left (a+\frac {b}{x^2}\right )^p \left (\frac {a x^2}{b}+1\right )^{-p} \left (c+\frac {d}{x^2}\right )^q \left (\frac {c x^2}{d}+1\right )^{-q} F_1\left (-p-q+\frac {1}{2};-p,-q;-p-q+\frac {3}{2};-\frac {a x^2}{b},-\frac {c x^2}{d}\right )}{2 p+2 q-1} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [F]  time = 0.65, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\left (\frac {a x^{2} + b}{x^{2}}\right )^{p} \left (\frac {c x^{2} + d}{x^{2}}\right )^{q}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.04, size = 0, normalized size = 0.00 \[ \int \left (a +\frac {b}{x^{2}}\right )^{p} \left (c +\frac {d}{x^{2}}\right )^{q}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b/x^2)^p*(c + d/x^2)^q,x)

[Out]

int((a + b/x^2)^p*(c + d/x^2)^q, 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)

[Out]

Timed out

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