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

Optimal. Leaf size=96 \[ -\frac{b \left (a+\frac{b}{x}\right )^{p+1} \left (c+\frac{d}{x}\right )^q \left (\frac{b \left (c+\frac{d}{x}\right )}{b c-a d}\right )^{-q} F_1\left (p+1;-q,2;p+2;-\frac{d \left (a+\frac{b}{x}\right )}{b c-a d},\frac{a+\frac{b}{x}}{a}\right )}{a^2 (p+1)} \]

[Out]

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

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Rubi [A]  time = 0.0605418, antiderivative size = 96, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {375, 137, 136} \[ -\frac{b \left (a+\frac{b}{x}\right )^{p+1} \left (c+\frac{d}{x}\right )^q \left (\frac{b \left (c+\frac{d}{x}\right )}{b c-a d}\right )^{-q} F_1\left (p+1;-q,2;p+2;-\frac{d \left (a+\frac{b}{x}\right )}{b c-a d},\frac{a+\frac{b}{x}}{a}\right )}{a^2 (p+1)} \]

Antiderivative was successfully verified.

[In]

Int[(a + b/x)^p*(c + d/x)^q,x]

[Out]

-((b*(a + b/x)^(1 + p)*(c + d/x)^q*AppellF1[1 + p, -q, 2, 2 + p, -((d*(a + b/x))/(b*c - a*d)), (a + b/x)/a])/(
a^2*(1 + p)*((b*(c + d/x))/(b*c - a*d))^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 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 \left (a+\frac{b}{x}\right )^p \left (c+\frac{d}{x}\right )^q \, dx &=-\operatorname{Subst}\left (\int \frac{(a+b x)^p (c+d x)^q}{x^2} \, dx,x,\frac{1}{x}\right )\\ &=-\left (\left (\left (c+\frac{d}{x}\right )^q \left (\frac{b \left (c+\frac{d}{x}\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^2} \, dx,x,\frac{1}{x}\right )\right )\\ &=-\frac{b \left (a+\frac{b}{x}\right )^{1+p} \left (c+\frac{d}{x}\right )^q \left (\frac{b \left (c+\frac{d}{x}\right )}{b c-a d}\right )^{-q} F_1\left (1+p;-q,2;2+p;-\frac{d \left (a+\frac{b}{x}\right )}{b c-a d},\frac{a+\frac{b}{x}}{a}\right )}{a^2 (1+p)}\\ \end{align*}

Mathematica [B]  time = 0.302972, size = 206, normalized size = 2.15 \[ \frac{b d x (p+q-2) \left (a+\frac{b}{x}\right )^p \left (c+\frac{d}{x}\right )^q F_1\left (-p-q+1;-p,-q;-p-q+2;-\frac{a x}{b},-\frac{c x}{d}\right )}{(p+q-1) \left (x \left (a d p F_1\left (-p-q+2;1-p,-q;-p-q+3;-\frac{a x}{b},-\frac{c x}{d}\right )+b c q F_1\left (-p-q+2;-p,1-q;-p-q+3;-\frac{a x}{b},-\frac{c x}{d}\right )\right )-b d (p+q-2) F_1\left (-p-q+1;-p,-q;-p-q+2;-\frac{a x}{b},-\frac{c x}{d}\right )\right )} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b/x)^p*(c + d/x)^q,x]

[Out]

(b*d*(-2 + p + q)*(a + b/x)^p*(c + d/x)^q*x*AppellF1[1 - p - q, -p, -q, 2 - p - q, -((a*x)/b), -((c*x)/d)])/((
-1 + p + q)*(-(b*d*(-2 + p + q)*AppellF1[1 - p - q, -p, -q, 2 - p - q, -((a*x)/b), -((c*x)/d)]) + x*(a*d*p*App
ellF1[2 - p - q, 1 - p, -q, 3 - p - q, -((a*x)/b), -((c*x)/d)] + b*c*q*AppellF1[2 - p - q, -p, 1 - q, 3 - p -
q, -((a*x)/b), -((c*x)/d)])))

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Maple [F]  time = 0.121, size = 0, normalized size = 0. \begin{align*} \int \left ( a+{\frac{b}{x}} \right ) ^{p} \left ( c+{\frac{d}{x}} \right ) ^{q}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((a + b/x)^p*(c + d/x)^q, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((a + b/x)^p*(c + d/x)^q, x)