\(\int \frac {(a+\frac {b}{x})^n x^2}{(c+d x)^2} \, dx\) [291]

Optimal result

Integrand size = 20, antiderivative size = 202 \[ \int \frac {\left (a+\frac {b}{x}\right )^n x^2}{(c+d x)^2} \, dx=\frac {c (2 a c-b d) \left (a+\frac {b}{x}\right )^{1+n}}{a d^2 (a c-b d) \left (d+\frac {c}{x}\right )}+\frac {\left (a+\frac {b}{x}\right )^{1+n} x}{a d \left (d+\frac {c}{x}\right )}+\frac {c^2 (2 a c-b d (2-n)) \left (a+\frac {b}{x}\right )^{1+n} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {c \left (a+\frac {b}{x}\right )}{a c-b d}\right )}{d^3 (a c-b d)^2 (1+n)}-\frac {(2 a c-b d n) \left (a+\frac {b}{x}\right )^{1+n} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,1+\frac {b}{a x}\right )}{a^2 d^3 (1+n)} \] Output:

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

Mathematica [A] (verified)

Time = 0.25 (sec) , antiderivative size = 179, normalized size of antiderivative = 0.89 \[ \int \frac {\left (a+\frac {b}{x}\right )^n x^2}{(c+d x)^2} \, dx=\frac {\left (a+\frac {b}{x}\right )^{1+n} \left (a c d (a c-b d) (2 a c-b d) (1+n) x+a d^2 (a c-b d)^2 (1+n) x^2+(c+d x) \left (a^2 c^2 (2 a c+b d (-2+n)) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {c \left (a+\frac {b}{x}\right )}{a c-b d}\right )-(a c-b d)^2 (2 a c-b d n) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,1+\frac {b}{a x}\right )\right )\right )}{a^2 d^3 (a c-b d)^2 (1+n) (c+d x)} \] Input:

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

Output:

((a + b/x)^(1 + n)*(a*c*d*(a*c - b*d)*(2*a*c - b*d)*(1 + n)*x + a*d^2*(a*c 
 - b*d)^2*(1 + n)*x^2 + (c + d*x)*(a^2*c^2*(2*a*c + b*d*(-2 + n))*Hypergeo 
metric2F1[1, 1 + n, 2 + n, (c*(a + b/x))/(a*c - b*d)] - (a*c - b*d)^2*(2*a 
*c - b*d*n)*Hypergeometric2F1[1, 1 + n, 2 + n, 1 + b/(a*x)])))/(a^2*d^3*(a 
*c - b*d)^2*(1 + n)*(c + d*x))
 

Rubi [A] (verified)

Time = 0.68 (sec) , antiderivative size = 231, normalized size of antiderivative = 1.14, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {1016, 899, 114, 168, 174, 75, 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.

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

\(\Big \downarrow \) 1016

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

\(\Big \downarrow \) 899

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

\(\Big \downarrow \) 114

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

\(\Big \downarrow \) 168

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

\(\Big \downarrow \) 174

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

\(\Big \downarrow \) 75

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

\(\Big \downarrow \) 78

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

Input:

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

Output:

((a + b/x)^(1 + n)*x)/(a*d*(d + c/x)) + ((c*(2*a*c - b*d)*(a + b/x)^(1 + n 
))/(d*(a*c - b*d)*(d + c/x)) + ((a*c^2*(2*a*c - b*d*(2 - n))*(a + b/x)^(1 
+ n)*Hypergeometric2F1[1, 1 + n, 2 + n, (c*(a + b/x))/(a*c - b*d)])/(d*(a* 
c - b*d)*(1 + n)) - ((a*c - b*d)*(2*a*c - b*d*n)*(a + b/x)^(1 + n)*Hyperge 
ometric2F1[1, 1 + n, 2 + n, 1 + b/(a*x)])/(a*d*(1 + n)))/(d*(a*c - b*d)))/ 
(a*d)
 

Defintions of rubi rules used

rule 75

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((c + d*x 
)^(n + 1)/(d*(n + 1)*(-d/(b*c))^m))*Hypergeometric2F1[-m, n + 1, n + 2, 1 + 
 d*(x/c)], x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[n] && (IntegerQ[m] 
 || GtQ[-d/(b*c), 0])
 

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 114

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 
)/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + Simp[1/((m + 1)*(b*c - a*d)*(b*e 
 - a*f))   Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*(m + 1) 
 - b*(d*e*(m + n + 2) + c*f*(m + p + 2)) - b*d*f*(m + n + p + 3)*x, x], x], 
 x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && ILtQ[m, -1] && (IntegerQ[n] || 
 IntegersQ[2*n, 2*p] || ILtQ[m + n + p + 3, 0])
 

rule 168

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_)*((g_.) + (h_.)*(x_)), x_] :> Simp[(b*g - a*h)*(a + b*x)^(m + 1)*(c + 
 d*x)^(n + 1)*((e + f*x)^(p + 1)/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + S 
imp[1/((m + 1)*(b*c - a*d)*(b*e - a*f))   Int[(a + b*x)^(m + 1)*(c + d*x)^n 
*(e + f*x)^p*Simp[(a*d*f*g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a* 
h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p + 3)*x, x], x], 
 x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && ILtQ[m, -1]
 

rule 174

Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))* 
((c_.) + (d_.)*(x_))), x_] :> Simp[(b*g - a*h)/(b*c - a*d)   Int[(e + f*x)^ 
p/(a + b*x), x], x] - Simp[(d*g - c*h)/(b*c - a*d)   Int[(e + f*x)^p/(c + d 
*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x]
 

rule 899

Int[((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol 
] :> -Subst[Int[(a + b/x^n)^p*((c + d/x^n)^q/x^2), x], x, 1/x] /; FreeQ[{a, 
 b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && ILtQ[n, 0]
 

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])
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\left (a+\frac {b}{x}\right )^n x^2}{(c+d x)^2} \, dx=\text {too large to display} \] Input:

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

Output:

(2*(a*x + b)**n*a*c*d*n*x**2 - (a*x + b)**n*b*c**2*n**2 + 3*(a*x + b)**n*b 
*c**2*n - 2*(a*x + b)**n*b*c**2 + (a*x + b)**n*b*c*d*n**2*x - (a*x + b)**n 
*b*c*d*n*x - 2*(a*x + b)**n*b*c*d*x - (a*x + b)**n*b*d**2*n**2*x**2 - (a*x 
 + b)**n*b*d**2*n*x**2 - 2*x**n*int((a*x + b)**n/(2*x**n*a**2*c**3*x**2 + 
4*x**n*a**2*c**2*d*x**3 + 2*x**n*a**2*c*d**2*x**4 + 2*x**n*a*b*c**3*x - x* 
*n*a*b*c**2*d*n*x**2 + 3*x**n*a*b*c**2*d*x**2 - 2*x**n*a*b*c*d**2*n*x**3 - 
 x**n*a*b*d**3*n*x**4 - x**n*a*b*d**3*x**4 - x**n*b**2*c**2*d*n*x - x**n*b 
**2*c**2*d*x - 2*x**n*b**2*c*d**2*n*x**2 - 2*x**n*b**2*c*d**2*x**2 - x**n* 
b**2*d**3*n*x**3 - x**n*b**2*d**3*x**3),x)*a*b**2*c**5*n**3 + 6*x**n*int(( 
a*x + b)**n/(2*x**n*a**2*c**3*x**2 + 4*x**n*a**2*c**2*d*x**3 + 2*x**n*a**2 
*c*d**2*x**4 + 2*x**n*a*b*c**3*x - x**n*a*b*c**2*d*n*x**2 + 3*x**n*a*b*c** 
2*d*x**2 - 2*x**n*a*b*c*d**2*n*x**3 - x**n*a*b*d**3*n*x**4 - x**n*a*b*d**3 
*x**4 - x**n*b**2*c**2*d*n*x - x**n*b**2*c**2*d*x - 2*x**n*b**2*c*d**2*n*x 
**2 - 2*x**n*b**2*c*d**2*x**2 - x**n*b**2*d**3*n*x**3 - x**n*b**2*d**3*x** 
3),x)*a*b**2*c**5*n**2 - 4*x**n*int((a*x + b)**n/(2*x**n*a**2*c**3*x**2 + 
4*x**n*a**2*c**2*d*x**3 + 2*x**n*a**2*c*d**2*x**4 + 2*x**n*a*b*c**3*x - x* 
*n*a*b*c**2*d*n*x**2 + 3*x**n*a*b*c**2*d*x**2 - 2*x**n*a*b*c*d**2*n*x**3 - 
 x**n*a*b*d**3*n*x**4 - x**n*a*b*d**3*x**4 - x**n*b**2*c**2*d*n*x - x**n*b 
**2*c**2*d*x - 2*x**n*b**2*c*d**2*n*x**2 - 2*x**n*b**2*c*d**2*x**2 - x**n* 
b**2*d**3*n*x**3 - x**n*b**2*d**3*x**3),x)*a*b**2*c**5*n - 2*x**n*int((...