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

Optimal. Leaf size=105 \[ \frac {\left (a+\frac {b}{x}\right )^{n+1} (a c-b d (n+1)) \, _2F_1\left (1,n+1;n+2;\frac {c \left (a+\frac {b}{x}\right )}{a c-b d}\right )}{c (n+1) (a c-b d)^2}-\frac {d \left (a+\frac {b}{x}\right )^{n+1}}{c \left (\frac {c}{x}+d\right ) (a c-b d)} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {514, 446, 78, 68} \[ \frac {\left (a+\frac {b}{x}\right )^{n+1} (a c-b d (n+1)) \, _2F_1\left (1,n+1;n+2;\frac {c \left (a+\frac {b}{x}\right )}{a c-b d}\right )}{c (n+1) (a c-b d)^2}-\frac {d \left (a+\frac {b}{x}\right )^{n+1}}{c \left (\frac {c}{x}+d\right ) (a c-b d)} \]

Antiderivative was successfully verified.

[In]

Int[(a + b/x)^n/(x*(c + d*x)^2),x]

[Out]

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

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 78

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

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 514

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

Rubi steps

\begin {align*} \int \frac {\left (a+\frac {b}{x}\right )^n}{x (c+d x)^2} \, dx &=\int \frac {\left (a+\frac {b}{x}\right )^n}{\left (d+\frac {c}{x}\right )^2 x^3} \, dx\\ &=-\operatorname {Subst}\left (\int \frac {x (a+b x)^n}{(d+c x)^2} \, dx,x,\frac {1}{x}\right )\\ &=-\frac {d \left (a+\frac {b}{x}\right )^{1+n}}{c (a c-b d) \left (d+\frac {c}{x}\right )}-\frac {(a c-b d (1+n)) \operatorname {Subst}\left (\int \frac {(a+b x)^n}{d+c x} \, dx,x,\frac {1}{x}\right )}{c (a c-b d)}\\ &=-\frac {d \left (a+\frac {b}{x}\right )^{1+n}}{c (a c-b d) \left (d+\frac {c}{x}\right )}+\frac {(a c-b d (1+n)) \left (a+\frac {b}{x}\right )^{1+n} \, _2F_1\left (1,1+n;2+n;\frac {c \left (a+\frac {b}{x}\right )}{a c-b d}\right )}{c (a c-b d)^2 (1+n)}\\ \end {align*}

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Mathematica [A]  time = 0.12, size = 88, normalized size = 0.84 \[ \frac {\left (a+\frac {b}{x}\right )^{n+1} \left (\frac {(a c-b d (n+1)) \, _2F_1\left (1,n+1;n+2;\frac {c \left (a+\frac {b}{x}\right )}{a c-b d}\right )}{n+1}+\frac {d x (b d-a c)}{c+d x}\right )}{c (a c-b d)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.95, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\left (\frac {a x + b}{x}\right )^{n}}{d^{2} x^{3} + 2 \, c d x^{2} + c^{2} x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(((a*x + b)/x)^n/(d^2*x^3 + 2*c*d*x^2 + c^2*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.56, size = 0, normalized size = 0.00 \[ \int \frac {\left (a +\frac {b}{x}\right )^{n}}{\left (d x +c \right )^{2} x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+\frac {b}{x}\right )}^n}{x\,{\left (c+d\,x\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b/x)^n/(x*(c + d*x)^2),x)

[Out]

int((a + b/x)^n/(x*(c + d*x)^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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