\(\int \frac {x^2 (a+b x)^n}{(c+d x)^2} \, dx\) [562]

Optimal result

Integrand size = 18, antiderivative size = 122 \[ \int \frac {x^2 (a+b x)^n}{(c+d x)^2} \, dx=\frac {(a+b x)^{1+n}}{b d^2 (1+n)}+\frac {c^2 (a+b x)^{1+n}}{d^2 (b c-a d) (c+d x)}+\frac {c (2 a d-b c (2+n)) (a+b x)^{1+n} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,-\frac {d (a+b x)}{b c-a d}\right )}{d^2 (b c-a d)^2 (1+n)} \] Output:

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

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.94 \[ \int \frac {x^2 (a+b x)^n}{(c+d x)^2} \, dx=\frac {(a+b x)^{1+n} \left ((b c-a d) (-a d (c+d x)+b c (c (2+n)+d x))-b c (-2 a d+b c (2+n)) (c+d x) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {d (a+b x)}{-b c+a d}\right )\right )}{b d^2 (b c-a d)^2 (1+n) (c+d x)} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.14, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {100, 25, 90, 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.

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

\(\Big \downarrow \) 100

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 90

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

\(\Big \downarrow \) 78

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

Input:

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

Output:

(c^2*(a + b*x)^(1 + n))/(d^2*(b*c - a*d)*(c + d*x)) + (((b*c - a*d)*(a + b 
*x)^(1 + n))/(b*(1 + n)) + (c*(2*a*d - b*c*(2 + n))*(a + b*x)^(1 + n)*Hype 
rgeometric2F1[1, 1 + n, 2 + n, -((d*(a + b*x))/(b*c - a*d))])/((b*c - a*d) 
*(1 + n)))/(d^2*(b*c - a*d))
 

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 90

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

rule 100

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^( 
p_), x_] :> Simp[(b*c - a*d)^2*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d^2*(d 
*e - c*f)*(n + 1))), x] - Simp[1/(d^2*(d*e - c*f)*(n + 1))   Int[(c + d*x)^ 
(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*( 
p + 1)) - 2*a*b*d*(d*e*(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x 
, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ[n, -1] || (EqQ[n 
 + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 1])))
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-2)]

Exception generated. \[ \int \frac {x^2 (a+b x)^n}{(c+d x)^2} \, dx=\text {Exception raised: HeuristicGCDFailed} \] Input:

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

Output:

Exception raised: HeuristicGCDFailed >> no luck
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

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

Output:

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