3.3.98 \(\int \frac {(a+\frac {b}{x})^n}{x^3 (c+d x)^2} \, dx\) [298]

3.3.98.1 Optimal result

Integrand size = 20, antiderivative size = 217 \[ \int \frac {\left (a+\frac {b}{x}\right )^n}{x^3 (c+d x)^2} \, dx=-\frac {\left (a+\frac {b}{x}\right )^{1+n} \left (d (b d (2+n) (a c+b d (3+n))-a c (a c+b d (5+3 n)))-\frac {c (a c-b d) (a c+b d (3+n))}{x}\right )}{b^2 c^3 (a c-b d) (1+n) (2+n) \left (d+\frac {c}{x}\right )}-\frac {\left (a+\frac {b}{x}\right )^{1+n}}{b c (2+n) \left (d+\frac {c}{x}\right ) x^2}+\frac {d^2 (3 a c-b d (3+n)) \left (a+\frac {b}{x}\right )^{1+n} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {c \left (a+\frac {b}{x}\right )}{a c-b d}\right )}{c^3 (a c-b d)^2 (1+n)} \]

output

-(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)
 

3.3.98.2 Mathematica [A] (verified)

Time = 0.38 (sec) , antiderivative size = 182, normalized size of antiderivative = 0.84 \[ \int \frac {\left (a+\frac {b}{x}\right )^n}{x^3 (c+d x)^2} \, dx=\frac {\left (a+\frac {b}{x}\right )^{1+n} \left (-\frac {1}{x (c+d x)}+\frac {-a^2 c^2 (c+d x)+b^2 d^2 (3+n) (c+d (2+n) x)-a b c d (c (2+n)+d (3+2 n) x)}{b c^2 (-a c+b d) (1+n) (c+d x)}-\frac {b d^2 (2+n) (-3 a c+b d (3+n)) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {c \left (a+\frac {b}{x}\right )}{a c-b d}\right )}{c^2 (a c-b d)^2 (1+n)}\right )}{b c (2+n)} \]

input

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

output

((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))*Hyp 
ergeometric2F1[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))
 

3.3.98.3 Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.06, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {1016, 948, 111, 25, 163, 78}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 1016

\(\displaystyle \int \frac {\left (a+\frac {b}{x}\right )^n}{x^5 \left (\frac {c}{x}+d\right )^2}dx\)

\(\Big \downarrow \) 948

\(\displaystyle -\int \frac {\left (a+\frac {b}{x}\right )^n}{\left (\frac {c}{x}+d\right )^2 x^3}d\frac {1}{x}\)

\(\Big \downarrow \) 111

\(\displaystyle -\frac {\int -\frac {\left (a+\frac {b}{x}\right )^n \left (2 a d+\frac {a c+b d (n+3)}{x}\right )}{\left (\frac {c}{x}+d\right )^2 x}d\frac {1}{x}}{b c (n+2)}-\frac {\left (a+\frac {b}{x}\right )^{n+1}}{b c (n+2) x^2 \left (\frac {c}{x}+d\right )}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\int \frac {\left (a+\frac {b}{x}\right )^n \left (2 a d+\frac {a c+b d (n+3)}{x}\right )}{\left (\frac {c}{x}+d\right )^2 x}d\frac {1}{x}}{b c (n+2)}-\frac {\left (a+\frac {b}{x}\right )^{n+1}}{b c (n+2) x^2 \left (\frac {c}{x}+d\right )}\)

\(\Big \downarrow \) 163

\(\displaystyle \frac {-\frac {b d^2 (n+2) (3 a c-b d (n+3)) \int \frac {\left (a+\frac {b}{x}\right )^n}{\frac {c}{x}+d}d\frac {1}{x}}{c^2 (a c-b d)}-\frac {\left (a+\frac {b}{x}\right )^{n+1} \left (d (b d (n+2) (a c+b d (n+3))-a c (a c+b d (3 n+5)))-\frac {c (a c-b d) (a c+b d (n+3))}{x}\right )}{b c^2 (n+1) \left (\frac {c}{x}+d\right ) (a c-b d)}}{b c (n+2)}-\frac {\left (a+\frac {b}{x}\right )^{n+1}}{b c (n+2) x^2 \left (\frac {c}{x}+d\right )}\)

\(\Big \downarrow \) 78

\(\displaystyle \frac {\frac {b d^2 (n+2) \left (a+\frac {b}{x}\right )^{n+1} (3 a c-b d (n+3)) \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {c \left (a+\frac {b}{x}\right )}{a c-b d}\right )}{c^2 (n+1) (a c-b d)^2}-\frac {\left (a+\frac {b}{x}\right )^{n+1} \left (d (b d (n+2) (a c+b d (n+3))-a c (a c+b d (3 n+5)))-\frac {c (a c-b d) (a c+b d (n+3))}{x}\right )}{b c^2 (n+1) \left (\frac {c}{x}+d\right ) (a c-b d)}}{b c (n+2)}-\frac {\left (a+\frac {b}{x}\right )^{n+1}}{b c (n+2) x^2 \left (\frac {c}{x}+d\right )}\)

input

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

output

-((a + b/x)^(1 + n)/(b*c*(2 + n)*(d + c/x)*x^2)) + (-(((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*c^2*(a*c - b*d)*(1 + n)*(d + c/x))) + ( 
b*d^2*(2 + n)*(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^2*(a*c - b*d)^2*(1 + n)))/(b 
*c*(2 + n))
 

3.3.98.3.1 Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 78

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(b 
*c - a*d)^n*((a + b*x)^(m + 1)/(b^(n + 1)*(m + 1)))*Hypergeometric2F1[-n, m 
 + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m}, x] 
 &&  !IntegerQ[m] && IntegerQ[n]
 

rule 111

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Simp[b*(a + b*x)^(m - 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1 
)/(d*f*(m + n + p + 1))), x] + Simp[1/(d*f*(m + n + p + 1))   Int[(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 163

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_) 
)*((g_.) + (h_.)*(x_)), x_] :> 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)/(b^2*d*(b*c - a*d)*(m + 1)*(m + n + 3)))*(a + b*x)^(m + 1)*(c + 
d*x)^(n + 1), x] - Simp[(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] && ((GeQ[m, -2] && LtQ[m, - 
1]) || SumSimplerQ[m, 1]) && NeQ[m, -1] && NeQ[m + n + 3, 0]
 

rule 948

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_. 
), x_Symbol] :> Simp[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 1016

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] ||  !I 
ntegerQ[p])
 

3.3.98.4 Maple [F]

\[\int \frac {\left (a +\frac {b}{x}\right )^{n}}{x^{3} \left (d x +c \right )^{2}}d x\]

input

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

output

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

3.3.98.5 Fricas [F]

\[ \int \frac {\left (a+\frac {b}{x}\right )^n}{x^3 (c+d x)^2} \, dx=\int { \frac {{\left (a + \frac {b}{x}\right )}^{n}}{{\left (d x + c\right )}^{2} x^{3}} \,d x } \]

input

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

output

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

3.3.98.6 Sympy [F]

\[ \int \frac {\left (a+\frac {b}{x}\right )^n}{x^3 (c+d x)^2} \, dx=\int \frac {\left (a + \frac {b}{x}\right )^{n}}{x^{3} \left (c + d x\right )^{2}}\, dx \]

input

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

output

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

3.3.98.7 Maxima [F]

\[ \int \frac {\left (a+\frac {b}{x}\right )^n}{x^3 (c+d x)^2} \, dx=\int { \frac {{\left (a + \frac {b}{x}\right )}^{n}}{{\left (d x + c\right )}^{2} x^{3}} \,d x } \]

input

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

output

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

3.3.98.8 Giac [F]

\[ \int \frac {\left (a+\frac {b}{x}\right )^n}{x^3 (c+d x)^2} \, dx=\int { \frac {{\left (a + \frac {b}{x}\right )}^{n}}{{\left (d x + c\right )}^{2} x^{3}} \,d x } \]

input

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

output

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

3.3.98.9 Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+\frac {b}{x}\right )^n}{x^3 (c+d x)^2} \, dx=\int \frac {{\left (a+\frac {b}{x}\right )}^n}{x^3\,{\left (c+d\,x\right )}^2} \,d x \]

input

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

output

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