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

Optimal. Leaf size=85 \[ -\frac {\left (a+\frac {b}{x^2}\right )^{p+1} \left (c+\frac {d}{x^2}\right )^q \left (\frac {b \left (c+\frac {d}{x^2}\right )}{b c-a d}\right )^{-q} \, _2F_1\left (p+1,-q;p+2;-\frac {d \left (a+\frac {b}{x^2}\right )}{b c-a d}\right )}{2 b (p+1)} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {444, 70, 69} \[ -\frac {\left (a+\frac {b}{x^2}\right )^{p+1} \left (c+\frac {d}{x^2}\right )^q \left (\frac {b \left (c+\frac {d}{x^2}\right )}{b c-a d}\right )^{-q} \, _2F_1\left (p+1,-q;p+2;-\frac {d \left (a+\frac {b}{x^2}\right )}{b c-a d}\right )}{2 b (p+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a + b/x^2)^(1 + p)*(c + d/x^2)^q*Hypergeometric2F1[1 + p, -q, 2 + p, -((d*(a + b/x^2))/(b*c - a*d))])/(2*b*
(1 + p)*((b*(c + d/x^2))/(b*c - a*d))^q)

Rule 69

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*Hypergeometric2F1[
-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b*(m + 1)*(b/(b*c - a*d))^n), x] /; FreeQ[{a, b, c, d, m, n}
, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] ||  !(Ra
tionalQ[n] && GtQ[-(d/(b*c - a*d)), 0]))

Rule 70

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Dist[(c + d*x)^FracPart[n]/((b/(b*c - a*d)
)^IntPart[n]*((b*(c + d*x))/(b*c - a*d))^FracPart[n]), Int[(a + b*x)^m*Simp[(b*c)/(b*c - a*d) + (b*d*x)/(b*c -
a*d), x]^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] &&
(RationalQ[m] ||  !SimplerQ[n + 1, m + 1])

Rule 444

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && EqQ[m
- n + 1, 0]

Rubi steps

\begin {align*} \int \frac {\left (a+\frac {b}{x^2}\right )^p \left (c+\frac {d}{x^2}\right )^q}{x^3} \, dx &=-\left (\frac {1}{2} \operatorname {Subst}\left (\int (a+b x)^p (c+d x)^q \, dx,x,\frac {1}{x^2}\right )\right )\\ &=-\left (\frac {1}{2} \left (\left (c+\frac {d}{x^2}\right )^q \left (\frac {b \left (c+\frac {d}{x^2}\right )}{b c-a d}\right )^{-q}\right ) \operatorname {Subst}\left (\int (a+b x)^p \left (\frac {b c}{b c-a d}+\frac {b d x}{b c-a d}\right )^q \, dx,x,\frac {1}{x^2}\right )\right )\\ &=-\frac {\left (a+\frac {b}{x^2}\right )^{1+p} \left (c+\frac {d}{x^2}\right )^q \left (\frac {b \left (c+\frac {d}{x^2}\right )}{b c-a d}\right )^{-q} \, _2F_1\left (1+p,-q;2+p;-\frac {d \left (a+\frac {b}{x^2}\right )}{b c-a d}\right )}{2 b (1+p)}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 110, normalized size = 1.29 \[ -\frac {\left (c x^2+d\right ) \left (a+\frac {b}{x^2}\right )^p \left (\frac {a x^2}{b}+1\right )^{-p} \left (\frac {c x^2}{d}+1\right )^p \left (c+\frac {d}{x^2}\right )^q \, _2F_1\left (-p,-p-q-1;-p-q;\frac {(b c-a d) x^2}{b \left (c x^2+d\right )}\right )}{2 d x^2 (p+q+1)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-1/2*((a + b/x^2)^p*(c + d/x^2)^q*(d + c*x^2)*(1 + (c*x^2)/d)^p*Hypergeometric2F1[-p, -1 - p - q, -p - q, ((b*
c - a*d)*x^2)/(b*(d + c*x^2))])/(d*(1 + p + q)*x^2*(1 + (a*x^2)/b)^p)

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fricas [F]  time = 0.89, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\left (\frac {a x^{2} + b}{x^{2}}\right )^{p} \left (\frac {c x^{2} + d}{x^{2}}\right )^{q}}{x^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (a + \frac {b}{x^{2}}\right )}^{p} {\left (c + \frac {d}{x^{2}}\right )}^{q}}{x^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.19, size = 0, normalized size = 0.00 \[ \int \frac {\left (a +\frac {b}{x^{2}}\right )^{p} \left (c +\frac {d}{x^{2}}\right )^{q}}{x^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (a + \frac {b}{x^{2}}\right )}^{p} {\left (c + \frac {d}{x^{2}}\right )}^{q}}{x^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+\frac {b}{x^2}\right )}^p\,{\left (c+\frac {d}{x^2}\right )}^q}{x^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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