\(\int \frac {(g x)^m (A+B x)}{(a^2+2 a b x+b^2 x^2)^2} \, dx\) [469]

Optimal result

Integrand size = 29, antiderivative size = 81 \[ \int \frac {(g x)^m (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=-\frac {B (g x)^{1+m}}{b g (2-m) (a+b x)^3}+\frac {\left (\frac {A}{1+m}+\frac {a B}{2 b-b m}\right ) (g x)^{1+m} \operatorname {Hypergeometric2F1}\left (4,1+m,2+m,-\frac {b x}{a}\right )}{a^4 g} \] Output:

-B*(g*x)^(1+m)/b/g/(2-m)/(b*x+a)^3+(A/(1+m)+a*B/(-b*m+2*b))*(g*x)^(1+m)*hy 
pergeom([4, 1+m],[2+m],-b*x/a)/a^4/g
 

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.89 \[ \int \frac {(g x)^m (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\frac {x (g x)^m \left (\frac {a^3 (A b-a B)}{(a+b x)^3}-\frac {(A b (-2+m)-a B (1+m)) \operatorname {Hypergeometric2F1}\left (3,1+m,2+m,-\frac {b x}{a}\right )}{1+m}\right )}{3 a^4 b} \] Input:

Integrate[((g*x)^m*(A + B*x))/(a^2 + 2*a*b*x + b^2*x^2)^2,x]
 

Output:

(x*(g*x)^m*((a^3*(A*b - a*B))/(a + b*x)^3 - ((A*b*(-2 + m) - a*B*(1 + m))* 
Hypergeometric2F1[3, 1 + m, 2 + m, -((b*x)/a)])/(1 + m)))/(3*a^4*b)
 

Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.12, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {1184, 27, 87, 74}

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.

\(\displaystyle \int \frac {(A+B x) (g x)^m}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx\)

\(\Big \downarrow \) 1184

\(\displaystyle b^4 \int \frac {(g x)^m (A+B x)}{b^4 (a+b x)^4}dx\)

\(\Big \downarrow \) 27

\(\displaystyle \int \frac {(A+B x) (g x)^m}{(a+b x)^4}dx\)

\(\Big \downarrow \) 87

\(\displaystyle \frac {(a B (m+1)+A b (2-m)) \int \frac {(g x)^m}{(a+b x)^3}dx}{3 a b}+\frac {(g x)^{m+1} (A b-a B)}{3 a b g (a+b x)^3}\)

\(\Big \downarrow \) 74

\(\displaystyle \frac {(g x)^{m+1} (a B (m+1)+A b (2-m)) \operatorname {Hypergeometric2F1}\left (3,m+1,m+2,-\frac {b x}{a}\right )}{3 a^4 b g (m+1)}+\frac {(g x)^{m+1} (A b-a B)}{3 a b g (a+b x)^3}\)

Input:

Int[((g*x)^m*(A + B*x))/(a^2 + 2*a*b*x + b^2*x^2)^2,x]
 

Output:

((A*b - a*B)*(g*x)^(1 + m))/(3*a*b*g*(a + b*x)^3) + ((A*b*(2 - m) + a*B*(1 
 + m))*(g*x)^(1 + m)*Hypergeometric2F1[3, 1 + m, 2 + m, -((b*x)/a)])/(3*a^ 
4*b*g*(1 + m))
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 74

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

rule 87

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

rule 1184

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_) + (b_.)*(x_ 
) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[1/c^p   Int[(d + e*x)^m*(f + g*x 
)^n*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n}, x] && E 
qQ[b^2 - 4*a*c, 0] && IntegerQ[p]
 

Maple [F]

\[\int \frac {\left (g x \right )^{m} \left (B x +A \right )}{\left (b^{2} x^{2}+2 a b x +a^{2}\right )^{2}}d x\]

Input:

int((g*x)^m*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^2,x)
 

Output:

int((g*x)^m*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^2,x)
 

Fricas [F]

\[ \int \frac {(g x)^m (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\int { \frac {{\left (B x + A\right )} \left (g x\right )^{m}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{2}} \,d x } \] Input:

integrate((g*x)^m*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^2,x, algorithm="fricas")
 

Output:

integral((B*x + A)*(g*x)^m/(b^4*x^4 + 4*a*b^3*x^3 + 6*a^2*b^2*x^2 + 4*a^3* 
b*x + a^4), x)
 

Sympy [F]

\[ \int \frac {(g x)^m (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\int \frac {\left (g x\right )^{m} \left (A + B x\right )}{\left (a + b x\right )^{4}}\, dx \] Input:

integrate((g*x)**m*(B*x+A)/(b**2*x**2+2*a*b*x+a**2)**2,x)
 

Output:

Integral((g*x)**m*(A + B*x)/(a + b*x)**4, x)
 

Maxima [F]

\[ \int \frac {(g x)^m (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\int { \frac {{\left (B x + A\right )} \left (g x\right )^{m}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{2}} \,d x } \] Input:

integrate((g*x)^m*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^2,x, algorithm="maxima")
 

Output:

integrate((B*x + A)*(g*x)^m/(b^2*x^2 + 2*a*b*x + a^2)^2, x)
 

Giac [F]

\[ \int \frac {(g x)^m (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\int { \frac {{\left (B x + A\right )} \left (g x\right )^{m}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{2}} \,d x } \] Input:

integrate((g*x)^m*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^2,x, algorithm="giac")
 

Output:

integrate((B*x + A)*(g*x)^m/(b^2*x^2 + 2*a*b*x + a^2)^2, x)
 

Mupad [F(-1)]

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

int(((g*x)^m*(A + B*x))/(a^2 + b^2*x^2 + 2*a*b*x)^2,x)
 

Output:

int(((g*x)^m*(A + B*x))/(a^2 + b^2*x^2 + 2*a*b*x)^2, x)
 

Reduce [F]

\[ \int \frac {(g x)^m (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\frac {g^{m} \left (x^{m}-\left (\int \frac {x^{m}}{b^{3} m \,x^{4}+3 a \,b^{2} m \,x^{3}-2 b^{3} x^{4}+3 a^{2} b m \,x^{2}-6 a \,b^{2} x^{3}+a^{3} m x -6 a^{2} b \,x^{2}-2 a^{3} x}d x \right ) a^{3} m^{2}+2 \left (\int \frac {x^{m}}{b^{3} m \,x^{4}+3 a \,b^{2} m \,x^{3}-2 b^{3} x^{4}+3 a^{2} b m \,x^{2}-6 a \,b^{2} x^{3}+a^{3} m x -6 a^{2} b \,x^{2}-2 a^{3} x}d x \right ) a^{3} m -2 \left (\int \frac {x^{m}}{b^{3} m \,x^{4}+3 a \,b^{2} m \,x^{3}-2 b^{3} x^{4}+3 a^{2} b m \,x^{2}-6 a \,b^{2} x^{3}+a^{3} m x -6 a^{2} b \,x^{2}-2 a^{3} x}d x \right ) a^{2} b \,m^{2} x +4 \left (\int \frac {x^{m}}{b^{3} m \,x^{4}+3 a \,b^{2} m \,x^{3}-2 b^{3} x^{4}+3 a^{2} b m \,x^{2}-6 a \,b^{2} x^{3}+a^{3} m x -6 a^{2} b \,x^{2}-2 a^{3} x}d x \right ) a^{2} b m x -\left (\int \frac {x^{m}}{b^{3} m \,x^{4}+3 a \,b^{2} m \,x^{3}-2 b^{3} x^{4}+3 a^{2} b m \,x^{2}-6 a \,b^{2} x^{3}+a^{3} m x -6 a^{2} b \,x^{2}-2 a^{3} x}d x \right ) a \,b^{2} m^{2} x^{2}+2 \left (\int \frac {x^{m}}{b^{3} m \,x^{4}+3 a \,b^{2} m \,x^{3}-2 b^{3} x^{4}+3 a^{2} b m \,x^{2}-6 a \,b^{2} x^{3}+a^{3} m x -6 a^{2} b \,x^{2}-2 a^{3} x}d x \right ) a \,b^{2} m \,x^{2}\right )}{b \left (b^{2} m \,x^{2}+2 a b m x -2 b^{2} x^{2}+a^{2} m -4 a b x -2 a^{2}\right )} \] Input:

int((g*x)^m*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^2,x)
 

Output:

(g**m*(x**m - int(x**m/(a**3*m*x - 2*a**3*x + 3*a**2*b*m*x**2 - 6*a**2*b*x 
**2 + 3*a*b**2*m*x**3 - 6*a*b**2*x**3 + b**3*m*x**4 - 2*b**3*x**4),x)*a**3 
*m**2 + 2*int(x**m/(a**3*m*x - 2*a**3*x + 3*a**2*b*m*x**2 - 6*a**2*b*x**2 
+ 3*a*b**2*m*x**3 - 6*a*b**2*x**3 + b**3*m*x**4 - 2*b**3*x**4),x)*a**3*m - 
 2*int(x**m/(a**3*m*x - 2*a**3*x + 3*a**2*b*m*x**2 - 6*a**2*b*x**2 + 3*a*b 
**2*m*x**3 - 6*a*b**2*x**3 + b**3*m*x**4 - 2*b**3*x**4),x)*a**2*b*m**2*x + 
 4*int(x**m/(a**3*m*x - 2*a**3*x + 3*a**2*b*m*x**2 - 6*a**2*b*x**2 + 3*a*b 
**2*m*x**3 - 6*a*b**2*x**3 + b**3*m*x**4 - 2*b**3*x**4),x)*a**2*b*m*x - in 
t(x**m/(a**3*m*x - 2*a**3*x + 3*a**2*b*m*x**2 - 6*a**2*b*x**2 + 3*a*b**2*m 
*x**3 - 6*a*b**2*x**3 + b**3*m*x**4 - 2*b**3*x**4),x)*a*b**2*m**2*x**2 + 2 
*int(x**m/(a**3*m*x - 2*a**3*x + 3*a**2*b*m*x**2 - 6*a**2*b*x**2 + 3*a*b** 
2*m*x**3 - 6*a*b**2*x**3 + b**3*m*x**4 - 2*b**3*x**4),x)*a*b**2*m*x**2))/( 
b*(a**2*m - 2*a**2 + 2*a*b*m*x - 4*a*b*x + b**2*m*x**2 - 2*b**2*x**2))