3.590 \(\int (a+\frac{b}{x})^m \, dx\)

Optimal. Leaf size=40 \[ -\frac{b \left (a+\frac{b}{x}\right )^{m+1} \, _2F_1\left (2,m+1;m+2;\frac{b}{a x}+1\right )}{a^2 (m+1)} \]

[Out]

-((b*(a + b/x)^(1 + m)*Hypergeometric2F1[2, 1 + m, 2 + m, 1 + b/(a*x)])/(a^2*(1 + m)))

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Rubi [A]  time = 0.0101131, antiderivative size = 40, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 9, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {242, 65} \[ -\frac{b \left (a+\frac{b}{x}\right )^{m+1} \, _2F_1\left (2,m+1;m+2;\frac{b}{a x}+1\right )}{a^2 (m+1)} \]

Antiderivative was successfully verified.

[In]

Int[(a + b/x)^m,x]

[Out]

-((b*(a + b/x)^(1 + m)*Hypergeometric2F1[2, 1 + m, 2 + m, 1 + b/(a*x)])/(a^2*(1 + m)))

Rule 242

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + b/x^n)^p/x^2, x], x, 1/x] /; FreeQ[{a, b, p},
x] && ILtQ[n, 0]

Rule 65

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

Rubi steps

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

Mathematica [A]  time = 0.0173851, size = 50, normalized size = 1.25 \[ -\frac{x \left (a+\frac{b}{x}\right )^m \left (\frac{a x}{b}+1\right )^{-m} \, _2F_1\left (1-m,-m;2-m;-\frac{a x}{b}\right )}{m-1} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b/x)^m,x]

[Out]

-(((a + b/x)^m*x*Hypergeometric2F1[1 - m, -m, 2 - m, -((a*x)/b)])/((-1 + m)*(1 + (a*x)/b)^m))

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Maple [F]  time = 0.012, size = 0, normalized size = 0. \begin{align*} \int \left ( a+{\frac{b}{x}} \right ) ^{m}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b/x)^m,x)

[Out]

int((a+b/x)^m,x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x)^m,x, algorithm="maxima")

[Out]

integrate((a + b/x)^m, 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 )^{m}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x)^m,x, algorithm="fricas")

[Out]

integral(((a*x + b)/x)^m, x)

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Sympy [C]  time = 1.36163, size = 34, normalized size = 0.85 \begin{align*} \frac{b^{m} x x^{- m} \Gamma \left (1 - m\right ){{}_{2}F_{1}\left (\begin{matrix} - m, 1 - m \\ 2 - m \end{matrix}\middle |{\frac{a x e^{i \pi }}{b}} \right )}}{\Gamma \left (2 - m\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x)**m,x)

[Out]

b**m*x*x**(-m)*gamma(1 - m)*hyper((-m, 1 - m), (2 - m,), a*x*exp_polar(I*pi)/b)/gamma(2 - m)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x)^m,x, algorithm="giac")

[Out]

integrate((a + b/x)^m, x)