3.10.94 \(\int \frac {(a+\frac {b}{x^2})^p (c+\frac {d}{x^2})^q}{x^3} \, dx\) [994]

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

[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)

________________________________________________________________________________________

Rubi [A]
time = 0.04, 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 = {455, 72, 71} \begin {gather*} -\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)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 71

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c
- a*d))^n))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], 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 72

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 455

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} \text {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 ) \text {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*}

________________________________________________________________________________________

Mathematica [A]
time = 0.06, size = 110, normalized size = 1.29 \begin {gather*} -\frac {\left (a+\frac {b}{x^2}\right )^p \left (c+\frac {d}{x^2}\right )^q \left (1+\frac {a x^2}{b}\right )^{-p} \left (d+c x^2\right ) \left (1+\frac {c x^2}{d}\right )^p \, _2F_1\left (-p,-1-p-q;-p-q;\frac {(b c-a d) x^2}{b \left (d+c x^2\right )}\right )}{2 d (1+p+q) x^2} \end {gather*}

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)

________________________________________________________________________________________

Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {\left (\frac {b}{x^{2}}+a \right )^{p} \left (c +\frac {d}{x^{2}}\right )^{q}}{x^{3}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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/x^3,x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

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/x^3,x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

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/x**3,x)

[Out]

Timed out

________________________________________________________________________________________

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/x^3,x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

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}{x^3} \,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)/x^3,x)

[Out]

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

________________________________________________________________________________________