3.290 \(\int \frac{(a+\frac{b}{x})^n}{x^3 (c+d x)} \, dx\)

Optimal. Leaf size=115 \[ \frac{(a c+b d) \left (a+\frac{b}{x}\right )^{n+1}}{b^2 c^2 (n+1)}-\frac{\left (a+\frac{b}{x}\right )^{n+2}}{b^2 c (n+2)}+\frac{d^2 \left (a+\frac{b}{x}\right )^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{c \left (a+\frac{b}{x}\right )}{a c-b d}\right )}{c^2 (n+1) (a c-b d)} \]

[Out]

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

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Rubi [A]  time = 0.0962083, antiderivative size = 115, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {514, 446, 88, 68} \[ \frac{(a c+b d) \left (a+\frac{b}{x}\right )^{n+1}}{b^2 c^2 (n+1)}-\frac{\left (a+\frac{b}{x}\right )^{n+2}}{b^2 c (n+2)}+\frac{d^2 \left (a+\frac{b}{x}\right )^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{c \left (a+\frac{b}{x}\right )}{a c-b d}\right )}{c^2 (n+1) (a c-b d)} \]

Antiderivative was successfully verified.

[In]

Int[(a + b/x)^n/(x^3*(c + d*x)),x]

[Out]

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

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 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

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]

Rubi steps

\begin{align*} \int \frac{\left (a+\frac{b}{x}\right )^n}{x^3 (c+d x)} \, dx &=\int \frac{\left (a+\frac{b}{x}\right )^n}{\left (d+\frac{c}{x}\right ) x^4} \, dx\\ &=-\operatorname{Subst}\left (\int \frac{x^2 (a+b x)^n}{d+c x} \, dx,x,\frac{1}{x}\right )\\ &=-\operatorname{Subst}\left (\int \left (\frac{(-a c-b d) (a+b x)^n}{b c^2}+\frac{(a+b x)^{1+n}}{b c}+\frac{d^2 (a+b x)^n}{c^2 (d+c x)}\right ) \, dx,x,\frac{1}{x}\right )\\ &=\frac{(a c+b d) \left (a+\frac{b}{x}\right )^{1+n}}{b^2 c^2 (1+n)}-\frac{\left (a+\frac{b}{x}\right )^{2+n}}{b^2 c (2+n)}-\frac{d^2 \operatorname{Subst}\left (\int \frac{(a+b x)^n}{d+c x} \, dx,x,\frac{1}{x}\right )}{c^2}\\ &=\frac{(a c+b d) \left (a+\frac{b}{x}\right )^{1+n}}{b^2 c^2 (1+n)}-\frac{\left (a+\frac{b}{x}\right )^{2+n}}{b^2 c (2+n)}+\frac{d^2 \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^2 (a c-b d) (1+n)}\\ \end{align*}

Mathematica [A]  time = 0.0939186, size = 112, normalized size = 0.97 \[ -\frac{(a x+b) \left (a+\frac{b}{x}\right )^n \left (b^2 d^2 (n+2) x \, _2F_1\left (1,n+1;n+2;\frac{c \left (a+\frac{b}{x}\right )}{a c-b d}\right )+(a c-b d) (a c x-b c (n+1)+b d (n+2) x)\right )}{b^2 c^2 (n+1) (n+2) x^2 (b d-a c)} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b/x)^n/(x^3*(c + d*x)),x]

[Out]

-(((a + b/x)^n*(b + a*x)*((a*c - b*d)*(-(b*c*(1 + n)) + a*c*x + b*d*(2 + n)*x) + b^2*d^2*(2 + n)*x*Hypergeomet
ric2F1[1, 1 + n, 2 + n, (c*(a + b/x))/(a*c - b*d)]))/(b^2*c^2*(-(a*c) + b*d)*(1 + n)*(2 + n)*x^2))

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Maple [F]  time = 0.52, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{x}^{3} \left ( dx+c \right ) } \left ( a+{\frac{b}{x}} \right ) ^{n}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (a + \frac{b}{x}\right )}^{n}}{{\left (d x + c\right )} x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x)^n/x^3/(d*x+c),x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\left (\frac{a x + b}{x}\right )^{n}}{d x^{4} + c x^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x)^n/x^3/(d*x+c),x, algorithm="fricas")

[Out]

integral(((a*x + b)/x)^n/(d*x^4 + c*x^3), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x)**n/x**3/(d*x+c),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (a + \frac{b}{x}\right )}^{n}}{{\left (d x + c\right )} x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x)^n/x^3/(d*x+c),x, algorithm="giac")

[Out]

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