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

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Rubi [A]  time = 0.01, antiderivative size = 40, normalized size of antiderivative = 1.00, 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 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])

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]

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*}

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Mathematica [A]  time = 0.02, 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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fricas [F]  time = 0.41, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\left (\frac {a x + b}{x}\right )^{m}, x\right ) \]

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

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)

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maple [F]  time = 0.02, size = 0, normalized size = 0.00 \[ \int \left (a +\frac {b}{x}\right )^{m}\, dx \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (a + \frac {b}{x}\right )}^{m}\,{d x} \]

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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mupad [B]  time = 3.13, size = 51, normalized size = 1.28 \[ -\frac {x\,{\left (a+\frac {b}{x}\right )}^m\,{{}}_2{\mathrm {F}}_1\left (1-m,-m;\ 2-m;\ -\frac {a\,x}{b}\right )}{{\left (\frac {a\,x}{b}+1\right )}^m\,\left (m-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b/x)^m,x)

[Out]

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

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sympy [C]  time = 1.47, size = 34, normalized size = 0.85 \[ \frac {b^{m} x x^{- m} \Gamma \left (1 - m\right ) {{}_{2}F_{1}\left (\begin {matrix} - m, 1 - m \\ 2 - m \end {matrix}\middle | {\frac {a x e^{i \pi }}{b}} \right )}}{\Gamma \left (2 - m\right )} \]

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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