3.592 \(\int \frac {(a+\frac {b}{x})^m}{(c+d x)^2} \, dx\)

Optimal. Leaf size=56 \[ -\frac {b \left (a+\frac {b}{x}\right )^{m+1} \, _2F_1\left (2,m+1;m+2;\frac {c \left (a+\frac {b}{x}\right )}{a c-b d}\right )}{(m+1) (a c-b d)^2} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 56, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {434, 444, 68} \[ -\frac {b \left (a+\frac {b}{x}\right )^{m+1} \, _2F_1\left (2,m+1;m+2;\frac {c \left (a+\frac {b}{x}\right )}{a c-b d}\right )}{(m+1) (a c-b d)^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b/x)^m/(c + d*x)^2,x]

[Out]

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

Rule 68

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((b*c - a*d)^n*(a + b*x)^(m + 1)*Hype
rgeometric2F1[-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b^(n + 1)*(m + 1)), x] /; FreeQ[{a, b, c, d, m
}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

Rule 434

Int[((c_) + (d_.)*(x_)^(mn_.))^(q_.)*((a_) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[((a + b*x^n)^p*(d + c*x
^n)^q)/x^(n*q), x] /; FreeQ[{a, b, c, d, n, p}, x] && EqQ[mn, -n] && IntegerQ[q] && (PosQ[n] ||  !IntegerQ[p])

Rule 444

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

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Mathematica [A]  time = 0.02, size = 57, normalized size = 1.02 \[ -\frac {b \left (a+\frac {b}{x}\right )^{m+1} \, _2F_1\left (2,m+1;m+2;-\frac {c \left (a+\frac {b}{x}\right )}{b d-a c}\right )}{(m+1) (b d-a c)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b/x)^m/(c + d*x)^2,x]

[Out]

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

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fricas [F]  time = 0.41, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\left (\frac {a x + b}{x}\right )^{m}}{d^{2} x^{2} + 2 \, c d x + c^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (a + \frac {b}{x}\right )}^{m}}{{\left (d x + c\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((a + b/x)^m/(d*x + c)^2, x)

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maple [F]  time = 0.06, size = 0, normalized size = 0.00 \[ \int \frac {\left (a +\frac {b}{x}\right )^{m}}{\left (d x +c \right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (a + \frac {b}{x}\right )}^{m}}{{\left (d x + c\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((a + b/x)^m/(d*x + c)^2, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {{\left (a+\frac {b}{x}\right )}^m}{{\left (c+d\,x\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b/x)^m/(c + d*x)^2,x)

[Out]

int((a + b/x)^m/(c + d*x)^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b/x)**m/(c + d*x)**2, x)

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