Integrand size = 27, antiderivative size = 75 \[ \int (A+B x) (d+e x) \left (a^2+2 a b x+b^2 x^2\right ) \, dx=\frac {(A b-a B) (b d-a e) (a+b x)^3}{3 b^3}+\frac {(b B d+A b e-2 a B e) (a+b x)^4}{4 b^3}+\frac {B e (a+b x)^5}{5 b^3} \]
1/3*(A*b-B*a)*(-a*e+b*d)*(b*x+a)^3/b^3+1/4*(A*b*e-2*B*a*e+B*b*d)*(b*x+a)^4 /b^3+1/5*B*e*(b*x+a)^5/b^3
Time = 0.02 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.28 \[ \int (A+B x) (d+e x) \left (a^2+2 a b x+b^2 x^2\right ) \, dx=a^2 A d x+\frac {1}{2} a (2 A b d+a B d+a A e) x^2+\frac {1}{3} \left (A b^2 d+2 a b B d+2 a A b e+a^2 B e\right ) x^3+\frac {1}{4} b (b B d+A b e+2 a B e) x^4+\frac {1}{5} b^2 B e x^5 \]
a^2*A*d*x + (a*(2*A*b*d + a*B*d + a*A*e)*x^2)/2 + ((A*b^2*d + 2*a*b*B*d + 2*a*A*b*e + a^2*B*e)*x^3)/3 + (b*(b*B*d + A*b*e + 2*a*B*e)*x^4)/4 + (b^2*B *e*x^5)/5
Time = 0.28 (sec) , antiderivative size = 75, 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, 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.
\(\displaystyle \int \left (a^2+2 a b x+b^2 x^2\right ) (A+B x) (d+e x) \, dx\) |
\(\Big \downarrow \) 1184 |
\(\displaystyle \frac {\int b^2 (a+b x)^2 (A+B x) (d+e x)dx}{b^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int (a+b x)^2 (A+B x) (d+e x)dx\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \int \left (\frac {(a+b x)^3 (-2 a B e+A b e+b B d)}{b^2}+\frac {(a+b x)^2 (A b-a B) (b d-a e)}{b^2}+\frac {B e (a+b x)^4}{b^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {(a+b x)^4 (-2 a B e+A b e+b B d)}{4 b^3}+\frac {(a+b x)^3 (A b-a B) (b d-a e)}{3 b^3}+\frac {B e (a+b x)^5}{5 b^3}\) |
((A*b - a*B)*(b*d - a*e)*(a + b*x)^3)/(3*b^3) + ((b*B*d + A*b*e - 2*a*B*e) *(a + b*x)^4)/(4*b^3) + (B*e*(a + b*x)^5)/(5*b^3)
3.17.63.3.1 Defintions of rubi rules used
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 = 0.04 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.25
method | result | size |
default | \(\frac {b^{2} B e \,x^{5}}{5}+\frac {\left (b^{2} \left (A e +B d \right )+2 B e b a \right ) x^{4}}{4}+\frac {\left (A d \,b^{2}+2 \left (A e +B d \right ) b a +B e \,a^{2}\right ) x^{3}}{3}+\frac {\left (2 A a b d +\left (A e +B d \right ) a^{2}\right ) x^{2}}{2}+a^{2} A d x\) | \(94\) |
norman | \(\frac {b^{2} B e \,x^{5}}{5}+\left (\frac {1}{4} A \,b^{2} e +\frac {1}{2} B e b a +\frac {1}{4} B \,b^{2} d \right ) x^{4}+\left (\frac {2}{3} A a b e +\frac {1}{3} A d \,b^{2}+\frac {1}{3} B e \,a^{2}+\frac {2}{3} B a b d \right ) x^{3}+\left (\frac {1}{2} A \,a^{2} e +A a b d +\frac {1}{2} B \,a^{2} d \right ) x^{2}+a^{2} A d x\) | \(99\) |
gosper | \(\frac {x \left (12 b^{2} B e \,x^{4}+15 x^{3} A \,b^{2} e +30 x^{3} B e b a +15 x^{3} B \,b^{2} d +40 x^{2} A a b e +20 x^{2} A d \,b^{2}+20 x^{2} B e \,a^{2}+40 x^{2} B a b d +30 x A \,a^{2} e +60 x A a b d +30 x B \,a^{2} d +60 d A \,a^{2}\right )}{60}\) | \(112\) |
risch | \(\frac {1}{5} b^{2} B e \,x^{5}+\frac {1}{4} x^{4} A \,b^{2} e +\frac {1}{2} x^{4} B e b a +\frac {1}{4} x^{4} B \,b^{2} d +\frac {2}{3} x^{3} A a b e +\frac {1}{3} A \,b^{2} d \,x^{3}+\frac {1}{3} x^{3} B e \,a^{2}+\frac {2}{3} x^{3} B a b d +\frac {1}{2} x^{2} A \,a^{2} e +x^{2} A a b d +\frac {1}{2} x^{2} B \,a^{2} d +a^{2} A d x\) | \(114\) |
parallelrisch | \(\frac {1}{5} b^{2} B e \,x^{5}+\frac {1}{4} x^{4} A \,b^{2} e +\frac {1}{2} x^{4} B e b a +\frac {1}{4} x^{4} B \,b^{2} d +\frac {2}{3} x^{3} A a b e +\frac {1}{3} A \,b^{2} d \,x^{3}+\frac {1}{3} x^{3} B e \,a^{2}+\frac {2}{3} x^{3} B a b d +\frac {1}{2} x^{2} A \,a^{2} e +x^{2} A a b d +\frac {1}{2} x^{2} B \,a^{2} d +a^{2} A d x\) | \(114\) |
1/5*b^2*B*e*x^5+1/4*(b^2*(A*e+B*d)+2*B*e*b*a)*x^4+1/3*(A*d*b^2+2*(A*e+B*d) *b*a+B*e*a^2)*x^3+1/2*(2*A*a*b*d+(A*e+B*d)*a^2)*x^2+a^2*A*d*x
Time = 0.31 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.33 \[ \int (A+B x) (d+e x) \left (a^2+2 a b x+b^2 x^2\right ) \, dx=\frac {1}{5} \, B b^{2} e x^{5} + A a^{2} d x + \frac {1}{4} \, {\left (B b^{2} d + {\left (2 \, B a b + A b^{2}\right )} e\right )} x^{4} + \frac {1}{3} \, {\left ({\left (2 \, B a b + A b^{2}\right )} d + {\left (B a^{2} + 2 \, A a b\right )} e\right )} x^{3} + \frac {1}{2} \, {\left (A a^{2} e + {\left (B a^{2} + 2 \, A a b\right )} d\right )} x^{2} \]
1/5*B*b^2*e*x^5 + A*a^2*d*x + 1/4*(B*b^2*d + (2*B*a*b + A*b^2)*e)*x^4 + 1/ 3*((2*B*a*b + A*b^2)*d + (B*a^2 + 2*A*a*b)*e)*x^3 + 1/2*(A*a^2*e + (B*a^2 + 2*A*a*b)*d)*x^2
Time = 0.02 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.55 \[ \int (A+B x) (d+e x) \left (a^2+2 a b x+b^2 x^2\right ) \, dx=A a^{2} d x + \frac {B b^{2} e x^{5}}{5} + x^{4} \left (\frac {A b^{2} e}{4} + \frac {B a b e}{2} + \frac {B b^{2} d}{4}\right ) + x^{3} \cdot \left (\frac {2 A a b e}{3} + \frac {A b^{2} d}{3} + \frac {B a^{2} e}{3} + \frac {2 B a b d}{3}\right ) + x^{2} \left (\frac {A a^{2} e}{2} + A a b d + \frac {B a^{2} d}{2}\right ) \]
A*a**2*d*x + B*b**2*e*x**5/5 + x**4*(A*b**2*e/4 + B*a*b*e/2 + B*b**2*d/4) + x**3*(2*A*a*b*e/3 + A*b**2*d/3 + B*a**2*e/3 + 2*B*a*b*d/3) + x**2*(A*a** 2*e/2 + A*a*b*d + B*a**2*d/2)
Time = 0.19 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.33 \[ \int (A+B x) (d+e x) \left (a^2+2 a b x+b^2 x^2\right ) \, dx=\frac {1}{5} \, B b^{2} e x^{5} + A a^{2} d x + \frac {1}{4} \, {\left (B b^{2} d + {\left (2 \, B a b + A b^{2}\right )} e\right )} x^{4} + \frac {1}{3} \, {\left ({\left (2 \, B a b + A b^{2}\right )} d + {\left (B a^{2} + 2 \, A a b\right )} e\right )} x^{3} + \frac {1}{2} \, {\left (A a^{2} e + {\left (B a^{2} + 2 \, A a b\right )} d\right )} x^{2} \]
1/5*B*b^2*e*x^5 + A*a^2*d*x + 1/4*(B*b^2*d + (2*B*a*b + A*b^2)*e)*x^4 + 1/ 3*((2*B*a*b + A*b^2)*d + (B*a^2 + 2*A*a*b)*e)*x^3 + 1/2*(A*a^2*e + (B*a^2 + 2*A*a*b)*d)*x^2
Time = 0.28 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.51 \[ \int (A+B x) (d+e x) \left (a^2+2 a b x+b^2 x^2\right ) \, dx=\frac {1}{5} \, B b^{2} e x^{5} + \frac {1}{4} \, B b^{2} d x^{4} + \frac {1}{2} \, B a b e x^{4} + \frac {1}{4} \, A b^{2} e x^{4} + \frac {2}{3} \, B a b d x^{3} + \frac {1}{3} \, A b^{2} d x^{3} + \frac {1}{3} \, B a^{2} e x^{3} + \frac {2}{3} \, A a b e x^{3} + \frac {1}{2} \, B a^{2} d x^{2} + A a b d x^{2} + \frac {1}{2} \, A a^{2} e x^{2} + A a^{2} d x \]
1/5*B*b^2*e*x^5 + 1/4*B*b^2*d*x^4 + 1/2*B*a*b*e*x^4 + 1/4*A*b^2*e*x^4 + 2/ 3*B*a*b*d*x^3 + 1/3*A*b^2*d*x^3 + 1/3*B*a^2*e*x^3 + 2/3*A*a*b*e*x^3 + 1/2* B*a^2*d*x^2 + A*a*b*d*x^2 + 1/2*A*a^2*e*x^2 + A*a^2*d*x
Time = 0.05 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.31 \[ \int (A+B x) (d+e x) \left (a^2+2 a b x+b^2 x^2\right ) \, dx=x^3\,\left (\frac {A\,b^2\,d}{3}+\frac {B\,a^2\,e}{3}+\frac {2\,A\,a\,b\,e}{3}+\frac {2\,B\,a\,b\,d}{3}\right )+x^2\,\left (\frac {A\,a^2\,e}{2}+\frac {B\,a^2\,d}{2}+A\,a\,b\,d\right )+x^4\,\left (\frac {A\,b^2\,e}{4}+\frac {B\,b^2\,d}{4}+\frac {B\,a\,b\,e}{2}\right )+A\,a^2\,d\,x+\frac {B\,b^2\,e\,x^5}{5} \]