Integrand size = 29, antiderivative size = 73 \[ \int \frac {(A+B x) (d+e x)}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=-\frac {(A b-a B) (b d-a e)}{3 b^3 (a+b x)^3}-\frac {b B d+A b e-2 a B e}{2 b^3 (a+b x)^2}-\frac {B e}{b^3 (a+b x)} \] Output:
-1/3*(A*b-B*a)*(-a*e+b*d)/b^3/(b*x+a)^3-1/2*(A*b*e-2*B*a*e+B*b*d)/b^3/(b*x +a)^2-B*e/b^3/(b*x+a)
Time = 0.04 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.84 \[ \int \frac {(A+B x) (d+e x)}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=-\frac {A b (2 b d+a e+3 b e x)+B \left (2 a^2 e+3 b^2 x (d+2 e x)+a b (d+6 e x)\right )}{6 b^3 (a+b x)^3} \] Input:
Integrate[((A + B*x)*(d + e*x))/(a^2 + 2*a*b*x + b^2*x^2)^2,x]
Output:
-1/6*(A*b*(2*b*d + a*e + 3*b*e*x) + B*(2*a^2*e + 3*b^2*x*(d + 2*e*x) + a*b *(d + 6*e*x)))/(b^3*(a + b*x)^3)
Time = 0.42 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {1184, 27, 86, 2009}
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.
\(\displaystyle \int \frac {(A+B x) (d+e x)}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx\) |
\(\Big \downarrow \) 1184 |
\(\displaystyle b^4 \int \frac {(A+B x) (d+e x)}{b^4 (a+b x)^4}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \frac {(A+B x) (d+e x)}{(a+b x)^4}dx\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \int \left (\frac {-2 a B e+A b e+b B d}{b^2 (a+b x)^3}+\frac {(A b-a B) (b d-a e)}{b^2 (a+b x)^4}+\frac {B e}{b^2 (a+b x)^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {-2 a B e+A b e+b B d}{2 b^3 (a+b x)^2}-\frac {(A b-a B) (b d-a e)}{3 b^3 (a+b x)^3}-\frac {B e}{b^3 (a+b x)}\) |
Input:
Int[((A + B*x)*(d + e*x))/(a^2 + 2*a*b*x + b^2*x^2)^2,x]
Output:
-1/3*((A*b - a*B)*(b*d - a*e))/(b^3*(a + b*x)^3) - (b*B*d + A*b*e - 2*a*B* e)/(2*b^3*(a + b*x)^2) - (B*e)/(b^3*(a + b*x))
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_ .), x_] :> Int[ExpandIntegrand[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && ((ILtQ[n, 0] && ILtQ[p, 0]) || EqQ[p, 1 ] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))
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]
Time = 1.31 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.96
method | result | size |
norman | \(\frac {-\frac {B e \,x^{2}}{b}-\frac {\left (A b e +2 B a e +B b d \right ) x}{2 b^{2}}-\frac {A a b e +2 A \,b^{2} d +2 B e \,a^{2}+B a b d}{6 b^{3}}}{\left (b x +a \right )^{3}}\) | \(70\) |
default | \(-\frac {-A a b e +A \,b^{2} d +B e \,a^{2}-B a b d}{3 b^{3} \left (b x +a \right )^{3}}-\frac {B e}{b^{3} \left (b x +a \right )}-\frac {A b e -2 B a e +B b d}{2 b^{3} \left (b x +a \right )^{2}}\) | \(79\) |
orering | \(-\frac {\left (6 B e \,x^{2} b^{2}+3 A \,b^{2} e x +6 B a b e x +3 B \,b^{2} d x +A a b e +2 A \,b^{2} d +2 B e \,a^{2}+B a b d \right ) \left (b x +a \right )}{6 b^{3} \left (b^{2} x^{2}+2 a b x +a^{2}\right )^{2}}\) | \(87\) |
risch | \(\frac {-\frac {B e \,x^{2}}{b}-\frac {\left (A b e +2 B a e +B b d \right ) x}{2 b^{2}}-\frac {A a b e +2 A \,b^{2} d +2 B e \,a^{2}+B a b d}{6 b^{3}}}{\left (b x +a \right ) \left (b^{2} x^{2}+2 a b x +a^{2}\right )}\) | \(88\) |
gosper | \(-\frac {6 B e \,x^{2} b^{2}+3 A \,b^{2} e x +6 B a b e x +3 B \,b^{2} d x +A a b e +2 A \,b^{2} d +2 B e \,a^{2}+B a b d}{6 \left (b x +a \right ) \left (b^{2} x^{2}+2 a b x +a^{2}\right ) b^{3}}\) | \(89\) |
parallelrisch | \(-\frac {6 B e \,x^{2} b^{2}+3 A \,b^{2} e x +6 B a b e x +3 B \,b^{2} d x +A a b e +2 A \,b^{2} d +2 B e \,a^{2}+B a b d}{6 \left (b x +a \right ) \left (b^{2} x^{2}+2 a b x +a^{2}\right ) b^{3}}\) | \(89\) |
Input:
int((B*x+A)*(e*x+d)/(b^2*x^2+2*a*b*x+a^2)^2,x,method=_RETURNVERBOSE)
Output:
(-B*e/b*x^2-1/2*(A*b*e+2*B*a*e+B*b*d)/b^2*x-1/6*(A*a*b*e+2*A*b^2*d+2*B*a^2 *e+B*a*b*d)/b^3)/(b*x+a)^3
Time = 0.07 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.33 \[ \int \frac {(A+B x) (d+e x)}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=-\frac {6 \, B b^{2} e x^{2} + {\left (B a b + 2 \, A b^{2}\right )} d + {\left (2 \, B a^{2} + A a b\right )} e + 3 \, {\left (B b^{2} d + {\left (2 \, B a b + A b^{2}\right )} e\right )} x}{6 \, {\left (b^{6} x^{3} + 3 \, a b^{5} x^{2} + 3 \, a^{2} b^{4} x + a^{3} b^{3}\right )}} \] Input:
integrate((B*x+A)*(e*x+d)/(b^2*x^2+2*a*b*x+a^2)^2,x, algorithm="fricas")
Output:
-1/6*(6*B*b^2*e*x^2 + (B*a*b + 2*A*b^2)*d + (2*B*a^2 + A*a*b)*e + 3*(B*b^2 *d + (2*B*a*b + A*b^2)*e)*x)/(b^6*x^3 + 3*a*b^5*x^2 + 3*a^2*b^4*x + a^3*b^ 3)
Time = 0.76 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.47 \[ \int \frac {(A+B x) (d+e x)}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\frac {- A a b e - 2 A b^{2} d - 2 B a^{2} e - B a b d - 6 B b^{2} e x^{2} + x \left (- 3 A b^{2} e - 6 B a b e - 3 B b^{2} d\right )}{6 a^{3} b^{3} + 18 a^{2} b^{4} x + 18 a b^{5} x^{2} + 6 b^{6} x^{3}} \] Input:
integrate((B*x+A)*(e*x+d)/(b**2*x**2+2*a*b*x+a**2)**2,x)
Output:
(-A*a*b*e - 2*A*b**2*d - 2*B*a**2*e - B*a*b*d - 6*B*b**2*e*x**2 + x*(-3*A* b**2*e - 6*B*a*b*e - 3*B*b**2*d))/(6*a**3*b**3 + 18*a**2*b**4*x + 18*a*b** 5*x**2 + 6*b**6*x**3)
Time = 0.03 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.33 \[ \int \frac {(A+B x) (d+e x)}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=-\frac {6 \, B b^{2} e x^{2} + {\left (B a b + 2 \, A b^{2}\right )} d + {\left (2 \, B a^{2} + A a b\right )} e + 3 \, {\left (B b^{2} d + {\left (2 \, B a b + A b^{2}\right )} e\right )} x}{6 \, {\left (b^{6} x^{3} + 3 \, a b^{5} x^{2} + 3 \, a^{2} b^{4} x + a^{3} b^{3}\right )}} \] Input:
integrate((B*x+A)*(e*x+d)/(b^2*x^2+2*a*b*x+a^2)^2,x, algorithm="maxima")
Output:
-1/6*(6*B*b^2*e*x^2 + (B*a*b + 2*A*b^2)*d + (2*B*a^2 + A*a*b)*e + 3*(B*b^2 *d + (2*B*a*b + A*b^2)*e)*x)/(b^6*x^3 + 3*a*b^5*x^2 + 3*a^2*b^4*x + a^3*b^ 3)
Time = 0.14 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.96 \[ \int \frac {(A+B x) (d+e x)}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=-\frac {6 \, B b^{2} e x^{2} + 3 \, B b^{2} d x + 6 \, B a b e x + 3 \, A b^{2} e x + B a b d + 2 \, A b^{2} d + 2 \, B a^{2} e + A a b e}{6 \, {\left (b x + a\right )}^{3} b^{3}} \] Input:
integrate((B*x+A)*(e*x+d)/(b^2*x^2+2*a*b*x+a^2)^2,x, algorithm="giac")
Output:
-1/6*(6*B*b^2*e*x^2 + 3*B*b^2*d*x + 6*B*a*b*e*x + 3*A*b^2*e*x + B*a*b*d + 2*A*b^2*d + 2*B*a^2*e + A*a*b*e)/((b*x + a)^3*b^3)
Time = 11.67 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.25 \[ \int \frac {(A+B x) (d+e x)}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=-\frac {\frac {2\,A\,b^2\,d+2\,B\,a^2\,e+A\,a\,b\,e+B\,a\,b\,d}{6\,b^3}+\frac {x\,\left (A\,b\,e+2\,B\,a\,e+B\,b\,d\right )}{2\,b^2}+\frac {B\,e\,x^2}{b}}{a^3+3\,a^2\,b\,x+3\,a\,b^2\,x^2+b^3\,x^3} \] Input:
int(((A + B*x)*(d + e*x))/(a^2 + b^2*x^2 + 2*a*b*x)^2,x)
Output:
-((2*A*b^2*d + 2*B*a^2*e + A*a*b*e + B*a*b*d)/(6*b^3) + (x*(A*b*e + 2*B*a* e + B*b*d))/(2*b^2) + (B*e*x^2)/b)/(a^3 + b^3*x^3 + 3*a*b^2*x^2 + 3*a^2*b* x)
Time = 0.21 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.51 \[ \int \frac {(A+B x) (d+e x)}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\frac {b e \,x^{2}-a d}{2 a b \left (b^{2} x^{2}+2 a b x +a^{2}\right )} \] Input:
int((B*x+A)*(e*x+d)/(b^2*x^2+2*a*b*x+a^2)^2,x)
Output:
( - a*d + b*e*x**2)/(2*a*b*(a**2 + 2*a*b*x + b**2*x**2))