Integrand size = 23, antiderivative size = 133 \[ \int (A+B x) (d+e x)^2 \left (a+b x+c x^2\right ) \, dx=-\frac {(B d-A e) \left (c d^2-b d e+a e^2\right ) (d+e x)^3}{3 e^4}+\frac {\left (3 B c d^2-B e (2 b d-a e)-A e (2 c d-b e)\right ) (d+e x)^4}{4 e^4}-\frac {(3 B c d-b B e-A c e) (d+e x)^5}{5 e^4}+\frac {B c (d+e x)^6}{6 e^4} \] Output:
-1/3*(-A*e+B*d)*(a*e^2-b*d*e+c*d^2)*(e*x+d)^3/e^4+1/4*(3*B*c*d^2-B*e*(-a*e +2*b*d)-A*e*(-b*e+2*c*d))*(e*x+d)^4/e^4-1/5*(-A*c*e-B*b*e+3*B*c*d)*(e*x+d) ^5/e^4+1/6*B*c*(e*x+d)^6/e^4
Time = 0.06 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.03 \[ \int (A+B x) (d+e x)^2 \left (a+b x+c x^2\right ) \, dx=a A d^2 x+\frac {1}{2} d (A b d+a B d+2 a A e) x^2+\frac {1}{3} \left (A c d^2+2 a B d e+a A e^2+b d (B d+2 A e)\right ) x^3+\frac {1}{4} \left (B c d^2+B e (2 b d+a e)+A e (2 c d+b e)\right ) x^4+\frac {1}{5} e (2 B c d+b B e+A c e) x^5+\frac {1}{6} B c e^2 x^6 \] Input:
Integrate[(A + B*x)*(d + e*x)^2*(a + b*x + c*x^2),x]
Output:
a*A*d^2*x + (d*(A*b*d + a*B*d + 2*a*A*e)*x^2)/2 + ((A*c*d^2 + 2*a*B*d*e + a*A*e^2 + b*d*(B*d + 2*A*e))*x^3)/3 + ((B*c*d^2 + B*e*(2*b*d + a*e) + A*e* (2*c*d + b*e))*x^4)/4 + (e*(2*B*c*d + b*B*e + A*c*e)*x^5)/5 + (B*c*e^2*x^6 )/6
Time = 0.57 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {1195, 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.
| \(\displaystyle \int (A+B x) (d+e x)^2 \left (a+b x+c x^2\right ) \, dx\) |
| \(\Big \downarrow \) 1195 |
| \(\displaystyle \int \left (\frac {(d+e x)^3 \left (-B e (2 b d-a e)-A e (2 c d-b e)+3 B c d^2\right )}{e^3}+\frac {(d+e x)^2 (A e-B d) \left (a e^2-b d e+c d^2\right )}{e^3}+\frac {(d+e x)^4 (A c e+b B e-3 B c d)}{e^3}+\frac {B c (d+e x)^5}{e^3}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {(d+e x)^4 \left (-B e (2 b d-a e)-A e (2 c d-b e)+3 B c d^2\right )}{4 e^4}-\frac {(d+e x)^3 (B d-A e) \left (a e^2-b d e+c d^2\right )}{3 e^4}-\frac {(d+e x)^5 (-A c e-b B e+3 B c d)}{5 e^4}+\frac {B c (d+e x)^6}{6 e^4}\) |
Input:
Int[(A + B*x)*(d + e*x)^2*(a + b*x + c*x^2),x]
Output:
-1/3*((B*d - A*e)*(c*d^2 - b*d*e + a*e^2)*(d + e*x)^3)/e^4 + ((3*B*c*d^2 - B*e*(2*b*d - a*e) - A*e*(2*c*d - b*e))*(d + e*x)^4)/(4*e^4) - ((3*B*c*d - b*B*e - A*c*e)*(d + e*x)^5)/(5*e^4) + (B*c*(d + e*x)^6)/(6*e^4)
Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_.) + (b_.)*(x _) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n}, x ] && IGtQ[p, 0]
Time = 0.76 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.08
| method | result | size |
| norman | \(\frac {B \,e^{2} c \,x^{6}}{6}+\left (\frac {1}{5} A c \,e^{2}+\frac {1}{5} B \,e^{2} b +\frac {2}{5} B c d e \right ) x^{5}+\left (\frac {1}{4} A b \,e^{2}+\frac {1}{2} A c d e +\frac {1}{4} B a \,e^{2}+\frac {1}{2} B b d e +\frac {1}{4} B c \,d^{2}\right ) x^{4}+\left (\frac {1}{3} A a \,e^{2}+\frac {2}{3} A b d e +\frac {1}{3} A c \,d^{2}+\frac {2}{3} B a d e +\frac {1}{3} B b \,d^{2}\right ) x^{3}+\left (A a d e +\frac {1}{2} A b \,d^{2}+\frac {1}{2} a B \,d^{2}\right ) x^{2}+A a \,d^{2} x\) | \(144\) |
| default | \(\frac {B \,e^{2} c \,x^{6}}{6}+\frac {\left (\left (A \,e^{2}+2 B d e \right ) c +B \,e^{2} b \right ) x^{5}}{5}+\frac {\left (\left (2 A d e +B \,d^{2}\right ) c +\left (A \,e^{2}+2 B d e \right ) b +B a \,e^{2}\right ) x^{4}}{4}+\frac {\left (A c \,d^{2}+\left (2 A d e +B \,d^{2}\right ) b +\left (A \,e^{2}+2 B d e \right ) a \right ) x^{3}}{3}+\frac {\left (A b \,d^{2}+\left (2 A d e +B \,d^{2}\right ) a \right ) x^{2}}{2}+A a \,d^{2} x\) | \(145\) |
| orering | \(\frac {x \left (10 B \,e^{2} c \,x^{5}+12 A c \,e^{2} x^{4}+12 B \,e^{2} b \,x^{4}+24 B c d e \,x^{4}+15 x^{3} A b \,e^{2}+30 A c d e \,x^{3}+15 x^{3} B a \,e^{2}+30 x^{3} B b d e +15 B c \,d^{2} x^{3}+20 x^{2} A a \,e^{2}+40 x^{2} A b d e +20 A c \,d^{2} x^{2}+40 x^{2} B a d e +20 B b \,d^{2} x^{2}+60 x A a d e +30 A b \,d^{2} x +30 x a B \,d^{2}+60 A a \,d^{2}\right )}{60}\) | \(170\) |
| gosper | \(\frac {1}{6} B \,e^{2} c \,x^{6}+\frac {1}{5} x^{5} A c \,e^{2}+\frac {1}{5} x^{5} B \,e^{2} b +\frac {2}{5} x^{5} B c d e +\frac {1}{4} x^{4} A b \,e^{2}+\frac {1}{2} x^{4} A c d e +\frac {1}{4} x^{4} B a \,e^{2}+\frac {1}{2} x^{4} B b d e +\frac {1}{4} x^{4} B c \,d^{2}+\frac {1}{3} x^{3} A a \,e^{2}+\frac {2}{3} x^{3} A b d e +\frac {1}{3} x^{3} A c \,d^{2}+\frac {2}{3} x^{3} B a d e +\frac {1}{3} x^{3} B b \,d^{2}+x^{2} A a d e +\frac {1}{2} A b \,d^{2} x^{2}+\frac {1}{2} x^{2} a B \,d^{2}+A a \,d^{2} x\) | \(172\) |
| risch | \(\frac {1}{6} B \,e^{2} c \,x^{6}+\frac {1}{5} x^{5} A c \,e^{2}+\frac {1}{5} x^{5} B \,e^{2} b +\frac {2}{5} x^{5} B c d e +\frac {1}{4} x^{4} A b \,e^{2}+\frac {1}{2} x^{4} A c d e +\frac {1}{4} x^{4} B a \,e^{2}+\frac {1}{2} x^{4} B b d e +\frac {1}{4} x^{4} B c \,d^{2}+\frac {1}{3} x^{3} A a \,e^{2}+\frac {2}{3} x^{3} A b d e +\frac {1}{3} x^{3} A c \,d^{2}+\frac {2}{3} x^{3} B a d e +\frac {1}{3} x^{3} B b \,d^{2}+x^{2} A a d e +\frac {1}{2} A b \,d^{2} x^{2}+\frac {1}{2} x^{2} a B \,d^{2}+A a \,d^{2} x\) | \(172\) |
| parallelrisch | \(\frac {1}{6} B \,e^{2} c \,x^{6}+\frac {1}{5} x^{5} A c \,e^{2}+\frac {1}{5} x^{5} B \,e^{2} b +\frac {2}{5} x^{5} B c d e +\frac {1}{4} x^{4} A b \,e^{2}+\frac {1}{2} x^{4} A c d e +\frac {1}{4} x^{4} B a \,e^{2}+\frac {1}{2} x^{4} B b d e +\frac {1}{4} x^{4} B c \,d^{2}+\frac {1}{3} x^{3} A a \,e^{2}+\frac {2}{3} x^{3} A b d e +\frac {1}{3} x^{3} A c \,d^{2}+\frac {2}{3} x^{3} B a d e +\frac {1}{3} x^{3} B b \,d^{2}+x^{2} A a d e +\frac {1}{2} A b \,d^{2} x^{2}+\frac {1}{2} x^{2} a B \,d^{2}+A a \,d^{2} x\) | \(172\) |
Input:
int((B*x+A)*(e*x+d)^2*(c*x^2+b*x+a),x,method=_RETURNVERBOSE)
Output:
1/6*B*e^2*c*x^6+(1/5*A*c*e^2+1/5*B*e^2*b+2/5*B*c*d*e)*x^5+(1/4*A*b*e^2+1/2 *A*c*d*e+1/4*B*a*e^2+1/2*B*b*d*e+1/4*B*c*d^2)*x^4+(1/3*A*a*e^2+2/3*A*b*d*e +1/3*A*c*d^2+2/3*B*a*d*e+1/3*B*b*d^2)*x^3+(A*a*d*e+1/2*A*b*d^2+1/2*a*B*d^2 )*x^2+A*a*d^2*x
Time = 0.07 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.99 \[ \int (A+B x) (d+e x)^2 \left (a+b x+c x^2\right ) \, dx=\frac {1}{6} \, B c e^{2} x^{6} + \frac {1}{5} \, {\left (2 \, B c d e + {\left (B b + A c\right )} e^{2}\right )} x^{5} + A a d^{2} x + \frac {1}{4} \, {\left (B c d^{2} + 2 \, {\left (B b + A c\right )} d e + {\left (B a + A b\right )} e^{2}\right )} x^{4} + \frac {1}{3} \, {\left (A a e^{2} + {\left (B b + A c\right )} d^{2} + 2 \, {\left (B a + A b\right )} d e\right )} x^{3} + \frac {1}{2} \, {\left (2 \, A a d e + {\left (B a + A b\right )} d^{2}\right )} x^{2} \] Input:
integrate((B*x+A)*(e*x+d)^2*(c*x^2+b*x+a),x, algorithm="fricas")
Output:
1/6*B*c*e^2*x^6 + 1/5*(2*B*c*d*e + (B*b + A*c)*e^2)*x^5 + A*a*d^2*x + 1/4* (B*c*d^2 + 2*(B*b + A*c)*d*e + (B*a + A*b)*e^2)*x^4 + 1/3*(A*a*e^2 + (B*b + A*c)*d^2 + 2*(B*a + A*b)*d*e)*x^3 + 1/2*(2*A*a*d*e + (B*a + A*b)*d^2)*x^ 2
Time = 0.03 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.29 \[ \int (A+B x) (d+e x)^2 \left (a+b x+c x^2\right ) \, dx=A a d^{2} x + \frac {B c e^{2} x^{6}}{6} + x^{5} \left (\frac {A c e^{2}}{5} + \frac {B b e^{2}}{5} + \frac {2 B c d e}{5}\right ) + x^{4} \left (\frac {A b e^{2}}{4} + \frac {A c d e}{2} + \frac {B a e^{2}}{4} + \frac {B b d e}{2} + \frac {B c d^{2}}{4}\right ) + x^{3} \left (\frac {A a e^{2}}{3} + \frac {2 A b d e}{3} + \frac {A c d^{2}}{3} + \frac {2 B a d e}{3} + \frac {B b d^{2}}{3}\right ) + x^{2} \left (A a d e + \frac {A b d^{2}}{2} + \frac {B a d^{2}}{2}\right ) \] Input:
integrate((B*x+A)*(e*x+d)**2*(c*x**2+b*x+a),x)
Output:
A*a*d**2*x + B*c*e**2*x**6/6 + x**5*(A*c*e**2/5 + B*b*e**2/5 + 2*B*c*d*e/5 ) + x**4*(A*b*e**2/4 + A*c*d*e/2 + B*a*e**2/4 + B*b*d*e/2 + B*c*d**2/4) + x**3*(A*a*e**2/3 + 2*A*b*d*e/3 + A*c*d**2/3 + 2*B*a*d*e/3 + B*b*d**2/3) + x**2*(A*a*d*e + A*b*d**2/2 + B*a*d**2/2)
Time = 0.04 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.99 \[ \int (A+B x) (d+e x)^2 \left (a+b x+c x^2\right ) \, dx=\frac {1}{6} \, B c e^{2} x^{6} + \frac {1}{5} \, {\left (2 \, B c d e + {\left (B b + A c\right )} e^{2}\right )} x^{5} + A a d^{2} x + \frac {1}{4} \, {\left (B c d^{2} + 2 \, {\left (B b + A c\right )} d e + {\left (B a + A b\right )} e^{2}\right )} x^{4} + \frac {1}{3} \, {\left (A a e^{2} + {\left (B b + A c\right )} d^{2} + 2 \, {\left (B a + A b\right )} d e\right )} x^{3} + \frac {1}{2} \, {\left (2 \, A a d e + {\left (B a + A b\right )} d^{2}\right )} x^{2} \] Input:
integrate((B*x+A)*(e*x+d)^2*(c*x^2+b*x+a),x, algorithm="maxima")
Output:
1/6*B*c*e^2*x^6 + 1/5*(2*B*c*d*e + (B*b + A*c)*e^2)*x^5 + A*a*d^2*x + 1/4* (B*c*d^2 + 2*(B*b + A*c)*d*e + (B*a + A*b)*e^2)*x^4 + 1/3*(A*a*e^2 + (B*b + A*c)*d^2 + 2*(B*a + A*b)*d*e)*x^3 + 1/2*(2*A*a*d*e + (B*a + A*b)*d^2)*x^ 2
Time = 0.11 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.29 \[ \int (A+B x) (d+e x)^2 \left (a+b x+c x^2\right ) \, dx=\frac {1}{6} \, B c e^{2} x^{6} + \frac {2}{5} \, B c d e x^{5} + \frac {1}{5} \, B b e^{2} x^{5} + \frac {1}{5} \, A c e^{2} x^{5} + \frac {1}{4} \, B c d^{2} x^{4} + \frac {1}{2} \, B b d e x^{4} + \frac {1}{2} \, A c d e x^{4} + \frac {1}{4} \, B a e^{2} x^{4} + \frac {1}{4} \, A b e^{2} x^{4} + \frac {1}{3} \, B b d^{2} x^{3} + \frac {1}{3} \, A c d^{2} x^{3} + \frac {2}{3} \, B a d e x^{3} + \frac {2}{3} \, A b d e x^{3} + \frac {1}{3} \, A a e^{2} x^{3} + \frac {1}{2} \, B a d^{2} x^{2} + \frac {1}{2} \, A b d^{2} x^{2} + A a d e x^{2} + A a d^{2} x \] Input:
integrate((B*x+A)*(e*x+d)^2*(c*x^2+b*x+a),x, algorithm="giac")
Output:
1/6*B*c*e^2*x^6 + 2/5*B*c*d*e*x^5 + 1/5*B*b*e^2*x^5 + 1/5*A*c*e^2*x^5 + 1/ 4*B*c*d^2*x^4 + 1/2*B*b*d*e*x^4 + 1/2*A*c*d*e*x^4 + 1/4*B*a*e^2*x^4 + 1/4* A*b*e^2*x^4 + 1/3*B*b*d^2*x^3 + 1/3*A*c*d^2*x^3 + 2/3*B*a*d*e*x^3 + 2/3*A* b*d*e*x^3 + 1/3*A*a*e^2*x^3 + 1/2*B*a*d^2*x^2 + 1/2*A*b*d^2*x^2 + A*a*d*e* x^2 + A*a*d^2*x
Time = 0.06 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.08 \[ \int (A+B x) (d+e x)^2 \left (a+b x+c x^2\right ) \, dx=x^3\,\left (\frac {A\,a\,e^2}{3}+\frac {A\,c\,d^2}{3}+\frac {B\,b\,d^2}{3}+\frac {2\,A\,b\,d\,e}{3}+\frac {2\,B\,a\,d\,e}{3}\right )+x^4\,\left (\frac {A\,b\,e^2}{4}+\frac {B\,a\,e^2}{4}+\frac {B\,c\,d^2}{4}+\frac {A\,c\,d\,e}{2}+\frac {B\,b\,d\,e}{2}\right )+x^2\,\left (\frac {A\,b\,d^2}{2}+\frac {B\,a\,d^2}{2}+A\,a\,d\,e\right )+x^5\,\left (\frac {A\,c\,e^2}{5}+\frac {B\,b\,e^2}{5}+\frac {2\,B\,c\,d\,e}{5}\right )+A\,a\,d^2\,x+\frac {B\,c\,e^2\,x^6}{6} \] Input:
int((A + B*x)*(d + e*x)^2*(a + b*x + c*x^2),x)
Output:
x^3*((A*a*e^2)/3 + (A*c*d^2)/3 + (B*b*d^2)/3 + (2*A*b*d*e)/3 + (2*B*a*d*e) /3) + x^4*((A*b*e^2)/4 + (B*a*e^2)/4 + (B*c*d^2)/4 + (A*c*d*e)/2 + (B*b*d* e)/2) + x^2*((A*b*d^2)/2 + (B*a*d^2)/2 + A*a*d*e) + x^5*((A*c*e^2)/5 + (B* b*e^2)/5 + (2*B*c*d*e)/5) + A*a*d^2*x + (B*c*e^2*x^6)/6
Time = 0.23 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.11 \[ \int (A+B x) (d+e x)^2 \left (a+b x+c x^2\right ) \, dx=\frac {x \left (10 b c \,e^{2} x^{5}+12 a c \,e^{2} x^{4}+12 b^{2} e^{2} x^{4}+24 b c d e \,x^{4}+30 a b \,e^{2} x^{3}+30 a c d e \,x^{3}+30 b^{2} d e \,x^{3}+15 b c \,d^{2} x^{3}+20 a^{2} e^{2} x^{2}+80 a b d e \,x^{2}+20 a c \,d^{2} x^{2}+20 b^{2} d^{2} x^{2}+60 a^{2} d e x +60 a b \,d^{2} x +60 a^{2} d^{2}\right )}{60} \] Input:
int((B*x+A)*(e*x+d)^2*(c*x^2+b*x+a),x)
Output:
(x*(60*a**2*d**2 + 60*a**2*d*e*x + 20*a**2*e**2*x**2 + 60*a*b*d**2*x + 80* a*b*d*e*x**2 + 30*a*b*e**2*x**3 + 20*a*c*d**2*x**2 + 30*a*c*d*e*x**3 + 12* a*c*e**2*x**4 + 20*b**2*d**2*x**2 + 30*b**2*d*e*x**3 + 12*b**2*e**2*x**4 + 15*b*c*d**2*x**3 + 24*b*c*d*e*x**4 + 10*b*c*e**2*x**5))/60