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

Optimal. Leaf size=97 \[ \frac{\left (a+\frac{b}{x^2}\right )^{p+1} \left (c+\frac{d}{x^2}\right )^q \left (\frac{b \left (c+\frac{d}{x^2}\right )}{b c-a d}\right )^{-q} F_1\left (p+1;-q,1;p+2;-\frac{d \left (a+\frac{b}{x^2}\right )}{b c-a d},\frac{a+\frac{b}{x^2}}{a}\right )}{2 a (p+1)} \]

[Out]

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

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Rubi [A]  time = 0.0704423, antiderivative size = 97, normalized size of antiderivative = 1., 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 = {446, 137, 136} \[ \frac{\left (a+\frac{b}{x^2}\right )^{p+1} \left (c+\frac{d}{x^2}\right )^q \left (\frac{b \left (c+\frac{d}{x^2}\right )}{b c-a d}\right )^{-q} F_1\left (p+1;-q,1;p+2;-\frac{d \left (a+\frac{b}{x^2}\right )}{b c-a d},\frac{a+\frac{b}{x^2}}{a}\right )}{2 a (p+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(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] && IntegerQ[Simplify[(m + 1)/n]]

Rule 137

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), 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*((b*c)/(b*c
- a*d) + (b*d*x)/(b*c - a*d))^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] &&  !IntegerQ[m] &&
!IntegerQ[n] && IntegerQ[p] &&  !GtQ[b/(b*c - a*d), 0] &&  !SimplerQ[c + d*x, a + b*x]

Rule 136

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[((b*e - a*
f)^p*(a + b*x)^(m + 1)*AppellF1[m + 1, -n, -p, m + 2, -((d*(a + b*x))/(b*c - a*d)), -((f*(a + b*x))/(b*e - a*f
))])/(b^(p + 1)*(m + 1)*(b/(b*c - a*d))^n), x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] &&  !IntegerQ[m] &&  !Int
egerQ[n] && IntegerQ[p] && GtQ[b/(b*c - a*d), 0] &&  !(GtQ[d/(d*a - c*b), 0] && SimplerQ[c + d*x, a + b*x])

Rubi steps

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

Mathematica [A]  time = 0.0851773, size = 95, normalized size = 0.98 \[ -\frac{\left (a+\frac{b}{x^2}\right )^p \left (\frac{a x^2}{b}+1\right )^{-p} \left (c+\frac{d}{x^2}\right )^q \left (\frac{c x^2}{d}+1\right )^{-q} F_1\left (-p-q;-p,-q;-p-q+1;-\frac{a x^2}{b},-\frac{c x^2}{d}\right )}{2 (p+q)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [F]  time = 0.059, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{x} \left ( a+{\frac{b}{{x}^{2}}} \right ) ^{p} \left ( c+{\frac{d}{{x}^{2}}} \right ) ^{q}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (a + \frac{b}{x^{2}}\right )}^{p}{\left (c + \frac{d}{x^{2}}\right )}^{q}}{x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\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}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (a + \frac{b}{x^{2}}\right )}^{p}{\left (c + \frac{d}{x^{2}}\right )}^{q}}{x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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