3.298 \(\int \frac {(a+\frac {b}{x})^n}{x^3 (c+d x)^2} \, dx\)

Optimal. Leaf size=217 \[ -\frac {\left (a+\frac {b}{x}\right )^{n+1} \left (d (b d (n+2) (a c+b d (n+3))-a c (a c+b d (3 n+5)))-\frac {c (a c-b d) (a c+b d (n+3))}{x}\right )}{b^2 c^3 (n+1) (n+2) \left (\frac {c}{x}+d\right ) (a c-b d)}+\frac {d^2 \left (a+\frac {b}{x}\right )^{n+1} (3 a c-b d (n+3)) \, _2F_1\left (1,n+1;n+2;\frac {c \left (a+\frac {b}{x}\right )}{a c-b d}\right )}{c^3 (n+1) (a c-b d)^2}-\frac {\left (a+\frac {b}{x}\right )^{n+1}}{b c (n+2) x^2 \left (\frac {c}{x}+d\right )} \]

[Out]

-(a+b/x)^(1+n)*(d*(b*d*(2+n)*(a*c+b*d*(3+n))-a*c*(a*c+b*d*(5+3*n)))-c*(a*c-b*d)*(a*c+b*d*(3+n))/x)/b^2/c^3/(a*
c-b*d)/(1+n)/(2+n)/(d+c/x)-(a+b/x)^(1+n)/b/c/(2+n)/(d+c/x)/x^2+d^2*(3*a*c-b*d*(3+n))*(a+b/x)^(1+n)*hypergeom([
1, 1+n],[2+n],c*(a+b/x)/(a*c-b*d))/c^3/(a*c-b*d)^2/(1+n)

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Rubi [A]  time = 0.26, antiderivative size = 217, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {514, 446, 100, 146, 68} \[ -\frac {\left (a+\frac {b}{x}\right )^{n+1} \left (d (b d (n+2) (a c+b d (n+3))-a c (a c+b d (3 n+5)))-\frac {c (a c-b d) (a c+b d (n+3))}{x}\right )}{b^2 c^3 (n+1) (n+2) \left (\frac {c}{x}+d\right ) (a c-b d)}+\frac {d^2 \left (a+\frac {b}{x}\right )^{n+1} (3 a c-b d (n+3)) \, _2F_1\left (1,n+1;n+2;\frac {c \left (a+\frac {b}{x}\right )}{a c-b d}\right )}{c^3 (n+1) (a c-b d)^2}-\frac {\left (a+\frac {b}{x}\right )^{n+1}}{b c (n+2) x^2 \left (\frac {c}{x}+d\right )} \]

Antiderivative was successfully verified.

[In]

Int[(a + b/x)^n/(x^3*(c + d*x)^2),x]

[Out]

-(((a + b/x)^(1 + n)*(d*(b*d*(2 + n)*(a*c + b*d*(3 + n)) - a*c*(a*c + b*d*(5 + 3*n))) - (c*(a*c - b*d)*(a*c +
b*d*(3 + n)))/x))/(b^2*c^3*(a*c - b*d)*(1 + n)*(2 + n)*(d + c/x))) - (a + b/x)^(1 + n)/(b*c*(2 + n)*(d + c/x)*
x^2) + (d^2*(3*a*c - b*d*(3 + n))*(a + b/x)^(1 + n)*Hypergeometric2F1[1, 1 + n, 2 + n, (c*(a + b/x))/(a*c - b*
d)])/(c^3*(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 100

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

Rule 146

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_))*((g_.) + (h_.)*(x_)), x_Symbol] :
> Simp[((a^2*d*f*h*(n + 2) + b^2*d*e*g*(m + n + 3) + a*b*(c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b*f*h*(
b*c - a*d)*(m + 1)*x)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1))/(b^2*d*(b*c - a*d)*(m + 1)*(m + n + 3)), x] - Dist[
(a^2*d^2*f*h*(n + 1)*(n + 2) + a*b*d*(n + 1)*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b^2*(c^2*f*h*(m +
1)*(m + 2) - c*d*(f*g + e*h)*(m + 1)*(m + n + 3) + d^2*e*g*(m + n + 2)*(m + n + 3)))/(b^2*d*(b*c - a*d)*(m +
1)*(m + n + 3)), Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && ((Ge
Q[m, -2] && LtQ[m, -1]) || SumSimplerQ[m, 1]) && NeQ[m, -1] && NeQ[m + n + 3, 0]

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^3 (c+d x)^2} \, dx &=\int \frac {\left (a+\frac {b}{x}\right )^n}{\left (d+\frac {c}{x}\right )^2 x^5} \, dx\\ &=-\operatorname {Subst}\left (\int \frac {x^3 (a+b x)^n}{(d+c x)^2} \, dx,x,\frac {1}{x}\right )\\ &=-\frac {\left (a+\frac {b}{x}\right )^{1+n}}{b c (2+n) \left (d+\frac {c}{x}\right ) x^2}-\frac {\operatorname {Subst}\left (\int \frac {x (a+b x)^n (-2 a d+(-a c-b d (3+n)) x)}{(d+c x)^2} \, dx,x,\frac {1}{x}\right )}{b c (2+n)}\\ &=-\frac {\left (a+\frac {b}{x}\right )^{1+n} \left (d (b d (2+n) (a c+b d (3+n))-a c (a c+b d (5+3 n)))-\frac {c (a c-b d) (a c+b d (3+n))}{x}\right )}{b^2 c^3 (a c-b d) (1+n) (2+n) \left (d+\frac {c}{x}\right )}-\frac {\left (a+\frac {b}{x}\right )^{1+n}}{b c (2+n) \left (d+\frac {c}{x}\right ) x^2}-\frac {\left (d^2 (3 a c-b d (3+n))\right ) \operatorname {Subst}\left (\int \frac {(a+b x)^n}{d+c x} \, dx,x,\frac {1}{x}\right )}{c^3 (a c-b d)}\\ &=-\frac {\left (a+\frac {b}{x}\right )^{1+n} \left (d (b d (2+n) (a c+b d (3+n))-a c (a c+b d (5+3 n)))-\frac {c (a c-b d) (a c+b d (3+n))}{x}\right )}{b^2 c^3 (a c-b d) (1+n) (2+n) \left (d+\frac {c}{x}\right )}-\frac {\left (a+\frac {b}{x}\right )^{1+n}}{b c (2+n) \left (d+\frac {c}{x}\right ) x^2}+\frac {d^2 (3 a c-b d (3+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^3 (a c-b d)^2 (1+n)}\\ \end {align*}

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Mathematica [A]  time = 0.35, size = 182, normalized size = 0.84 \[ \frac {\left (a+\frac {b}{x}\right )^{n+1} \left (\frac {-a^2 c^2 (c+d x)-a b c d (c (n+2)+d (2 n+3) x)+b^2 d^2 (n+3) (c+d (n+2) x)}{b c^2 (n+1) (c+d x) (b d-a c)}-\frac {b d^2 (n+2) (b d (n+3)-3 a c) \, _2F_1\left (1,n+1;n+2;\frac {c \left (a+\frac {b}{x}\right )}{a c-b d}\right )}{c^2 (n+1) (a c-b d)^2}-\frac {1}{x (c+d x)}\right )}{b c (n+2)} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b/x)^n/(x^3*(c + d*x)^2),x]

[Out]

((a + b/x)^(1 + n)*(-(1/(x*(c + d*x))) + (-(a^2*c^2*(c + d*x)) + b^2*d^2*(3 + n)*(c + d*(2 + n)*x) - a*b*c*d*(
c*(2 + n) + d*(3 + 2*n)*x))/(b*c^2*(-(a*c) + b*d)*(1 + n)*(c + d*x)) - (b*d^2*(2 + n)*(-3*a*c + b*d*(3 + n))*H
ypergeometric2F1[1, 1 + n, 2 + n, (c*(a + b/x))/(a*c - b*d)])/(c^2*(a*c - b*d)^2*(1 + n))))/(b*c*(2 + n))

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fricas [F]  time = 1.02, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\left (\frac {a x + b}{x}\right )^{n}}{d^{2} x^{5} + 2 \, c d x^{4} + c^{2} x^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x)^n/x^3/(d*x+c)^2,x, algorithm="fricas")

[Out]

integral(((a*x + b)/x)^n/(d^2*x^5 + 2*c*d*x^4 + c^2*x^3), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x)^n/x^3/(d*x+c)^2,x, algorithm="giac")

[Out]

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

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maple [F]  time = 0.54, size = 0, normalized size = 0.00 \[ \int \frac {\left (a +\frac {b}{x}\right )^{n}}{\left (d x +c \right )^{2} x^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x)^n/x^3/(d*x+c)^2,x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (a+\frac {b}{x}\right )}^n}{x^3\,{\left (c+d\,x\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b/x)^n/(x^3*(c + d*x)^2),x)

[Out]

int((a + b/x)^n/(x^3*(c + d*x)^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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