3.9.40 \(\int \frac {x^m (A+B x)}{(a^2+2 a b x+b^2 x^2)^2} \, dx\) [840]

3.9.40.1 Optimal result

Integrand size = 27, antiderivative size = 81 \[ \int \frac {x^m (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\frac {(A b-a B) x^{1+m}}{3 a b (a+b x)^3}+\frac {(A b (2-m)+a B (1+m)) x^{1+m} \operatorname {Hypergeometric2F1}\left (3,1+m,2+m,-\frac {b x}{a}\right )}{3 a^4 b (1+m)} \]

output

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

3.9.40.2 Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.88 \[ \int \frac {x^m (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\frac {x^{1+m} \left (\frac {a^3 (A b-a B)}{(a+b x)^3}-\frac {(A b (-2+m)-a B (1+m)) \operatorname {Hypergeometric2F1}\left (3,1+m,2+m,-\frac {b x}{a}\right )}{1+m}\right )}{3 a^4 b} \]

input

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

output

(x^(1 + 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)
 

3.9.40.3 Rubi [A] (verified)

Time = 0.22 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, 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.

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

\(\Big \downarrow \) 1184

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 87

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

\(\Big \downarrow \) 74

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

input

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

output

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

3.9.40.3.1 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]
 

3.9.40.4 Maple [F]

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

input

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

output

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

3.9.40.5 Fricas [F]

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

input

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

output

integral((B*x + A)*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)
 

3.9.40.6 Sympy [F]

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

input

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

output

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

3.9.40.7 Maxima [F]

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

input

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

output

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

3.9.40.8 Giac [F]

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

input

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

output

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

3.9.40.9 Mupad [F(-1)]

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

input

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

output

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