Integrand size = 26, antiderivative size = 119 \[ \int \left (A+B x+C x^2\right ) \left (2+3 x-5 x^2+x^3\right )^2 \, dx=4 A x+2 (3 A+B) x^2-\frac {1}{3} (11 A-4 (3 B+C)) x^3-\frac {1}{4} (26 A+11 B-12 C) x^4+\frac {1}{5} (31 A-26 B-11 C) x^5-\frac {1}{6} (10 A-31 B+26 C) x^6+\frac {1}{7} (A-10 B+31 C) x^7+\frac {1}{8} (B-10 C) x^8+\frac {C x^9}{9} \] Output:
4*A*x+2*(3*A+B)*x^2-1/3*(11*A-12*B-4*C)*x^3-1/4*(26*A+11*B-12*C)*x^4+1/5*( 31*A-26*B-11*C)*x^5-1/6*(10*A-31*B+26*C)*x^6+1/7*(A-10*B+31*C)*x^7+1/8*(B- 10*C)*x^8+1/9*C*x^9
Time = 0.03 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.99 \[ \int \left (A+B x+C x^2\right ) \left (2+3 x-5 x^2+x^3\right )^2 \, dx=4 A x+2 (3 A+B) x^2+\frac {1}{3} (-11 A+12 B+4 C) x^3+\frac {1}{4} (-26 A-11 B+12 C) x^4+\frac {1}{5} (31 A-26 B-11 C) x^5+\frac {1}{6} (-10 A+31 B-26 C) x^6+\frac {1}{7} (A-10 B+31 C) x^7+\frac {1}{8} (B-10 C) x^8+\frac {C x^9}{9} \] Input:
Integrate[(A + B*x + C*x^2)*(2 + 3*x - 5*x^2 + x^3)^2,x]
Output:
4*A*x + 2*(3*A + B)*x^2 + ((-11*A + 12*B + 4*C)*x^3)/3 + ((-26*A - 11*B + 12*C)*x^4)/4 + ((31*A - 26*B - 11*C)*x^5)/5 + ((-10*A + 31*B - 26*C)*x^6)/ 6 + ((A - 10*B + 31*C)*x^7)/7 + ((B - 10*C)*x^8)/8 + (C*x^9)/9
Time = 0.60 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {2188, 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.
\(\displaystyle \int \left (x^3-5 x^2+3 x+2\right )^2 \left (A+B x+C x^2\right ) \, dx\) |
\(\Big \downarrow \) 2188 |
\(\displaystyle \int \left (x^6 (A-10 B+31 C)-x^5 (10 A-31 B+26 C)+x^4 (31 A-26 B-11 C)-x^3 (26 A+11 B-12 C)-x^2 (11 A-12 B-4 C)+4 x (3 A+B)+4 A+x^7 (B-10 C)+C x^8\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{7} x^7 (A-10 B+31 C)-\frac {1}{6} x^6 (10 A-31 B+26 C)+\frac {1}{5} x^5 (31 A-26 B-11 C)-\frac {1}{4} x^4 (26 A+11 B-12 C)-\frac {1}{3} x^3 (11 A-4 (3 B+C))+2 x^2 (3 A+B)+4 A x+\frac {1}{8} x^8 (B-10 C)+\frac {C x^9}{9}\) |
Input:
Int[(A + B*x + C*x^2)*(2 + 3*x - 5*x^2 + x^3)^2,x]
Output:
4*A*x + 2*(3*A + B)*x^2 - ((11*A - 4*(3*B + C))*x^3)/3 - ((26*A + 11*B - 1 2*C)*x^4)/4 + ((31*A - 26*B - 11*C)*x^5)/5 - ((10*A - 31*B + 26*C)*x^6)/6 + ((A - 10*B + 31*C)*x^7)/7 + ((B - 10*C)*x^8)/8 + (C*x^9)/9
Int[(Pq_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[Expand Integrand[Pq*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c}, x] && PolyQ[Pq , x] && IGtQ[p, -2]
Time = 0.17 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.87
method | result | size |
norman | \(\frac {C \,x^{9}}{9}+\left (\frac {B}{8}-\frac {5 C}{4}\right ) x^{8}+\left (\frac {A}{7}-\frac {10 B}{7}+\frac {31 C}{7}\right ) x^{7}+\left (-\frac {5 A}{3}+\frac {31 B}{6}-\frac {13 C}{3}\right ) x^{6}+\left (\frac {31 A}{5}-\frac {26 B}{5}-\frac {11 C}{5}\right ) x^{5}+\left (-\frac {13 A}{2}-\frac {11 B}{4}+3 C \right ) x^{4}+\left (-\frac {11 A}{3}+4 B +\frac {4 C}{3}\right ) x^{3}+\left (6 A +2 B \right ) x^{2}+4 A x\) | \(104\) |
default | \(\frac {C \,x^{9}}{9}+\frac {\left (B -10 C \right ) x^{8}}{8}+\frac {\left (A -10 B +31 C \right ) x^{7}}{7}+\frac {\left (-10 A +31 B -26 C \right ) x^{6}}{6}+\frac {\left (31 A -26 B -11 C \right ) x^{5}}{5}+\frac {\left (-26 A -11 B +12 C \right ) x^{4}}{4}+\frac {\left (-11 A +12 B +4 C \right ) x^{3}}{3}+\frac {\left (12 A +4 B \right ) x^{2}}{2}+4 A x\) | \(107\) |
orering | \(\frac {x \left (280 x^{8} C +315 x^{7} B -3150 x^{7} C +360 x^{6} A -3600 x^{6} B +11160 C \,x^{6}-4200 x^{5} A +13020 B \,x^{5}-10920 x^{5} C +15624 x^{4} A -13104 x^{4} B -5544 C \,x^{4}-16380 x^{3} A -6930 B \,x^{3}+7560 C \,x^{3}-9240 A \,x^{2}+10080 B \,x^{2}+3360 C \,x^{2}+15120 A x +5040 B x +10080 A \right )}{2520}\) | \(124\) |
gosper | \(\frac {1}{9} C \,x^{9}+\frac {1}{8} B \,x^{8}-\frac {5}{4} x^{8} C +\frac {1}{7} x^{7} A -\frac {10}{7} x^{7} B +\frac {31}{7} x^{7} C -\frac {5}{3} x^{6} A +\frac {31}{6} x^{6} B -\frac {13}{3} C \,x^{6}+\frac {31}{5} x^{5} A -\frac {26}{5} B \,x^{5}-\frac {11}{5} x^{5} C -\frac {13}{2} x^{4} A -\frac {11}{4} x^{4} B +3 C \,x^{4}-\frac {11}{3} x^{3} A +4 B \,x^{3}+\frac {4}{3} C \,x^{3}+6 A \,x^{2}+2 B \,x^{2}+4 A x\) | \(126\) |
risch | \(\frac {1}{9} C \,x^{9}+\frac {1}{8} B \,x^{8}-\frac {5}{4} x^{8} C +\frac {1}{7} x^{7} A -\frac {10}{7} x^{7} B +\frac {31}{7} x^{7} C -\frac {5}{3} x^{6} A +\frac {31}{6} x^{6} B -\frac {13}{3} C \,x^{6}+\frac {31}{5} x^{5} A -\frac {26}{5} B \,x^{5}-\frac {11}{5} x^{5} C -\frac {13}{2} x^{4} A -\frac {11}{4} x^{4} B +3 C \,x^{4}-\frac {11}{3} x^{3} A +4 B \,x^{3}+\frac {4}{3} C \,x^{3}+6 A \,x^{2}+2 B \,x^{2}+4 A x\) | \(126\) |
parallelrisch | \(\frac {1}{9} C \,x^{9}+\frac {1}{8} B \,x^{8}-\frac {5}{4} x^{8} C +\frac {1}{7} x^{7} A -\frac {10}{7} x^{7} B +\frac {31}{7} x^{7} C -\frac {5}{3} x^{6} A +\frac {31}{6} x^{6} B -\frac {13}{3} C \,x^{6}+\frac {31}{5} x^{5} A -\frac {26}{5} B \,x^{5}-\frac {11}{5} x^{5} C -\frac {13}{2} x^{4} A -\frac {11}{4} x^{4} B +3 C \,x^{4}-\frac {11}{3} x^{3} A +4 B \,x^{3}+\frac {4}{3} C \,x^{3}+6 A \,x^{2}+2 B \,x^{2}+4 A x\) | \(126\) |
Input:
int((C*x^2+B*x+A)*(x^3-5*x^2+3*x+2)^2,x,method=_RETURNVERBOSE)
Output:
1/9*C*x^9+(1/8*B-5/4*C)*x^8+(1/7*A-10/7*B+31/7*C)*x^7+(-5/3*A+31/6*B-13/3* C)*x^6+(31/5*A-26/5*B-11/5*C)*x^5+(-13/2*A-11/4*B+3*C)*x^4+(-11/3*A+4*B+4/ 3*C)*x^3+(6*A+2*B)*x^2+4*A*x
Time = 0.06 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.87 \[ \int \left (A+B x+C x^2\right ) \left (2+3 x-5 x^2+x^3\right )^2 \, dx=\frac {1}{9} \, C x^{9} + \frac {1}{8} \, {\left (B - 10 \, C\right )} x^{8} + \frac {1}{7} \, {\left (A - 10 \, B + 31 \, C\right )} x^{7} - \frac {1}{6} \, {\left (10 \, A - 31 \, B + 26 \, C\right )} x^{6} + \frac {1}{5} \, {\left (31 \, A - 26 \, B - 11 \, C\right )} x^{5} - \frac {1}{4} \, {\left (26 \, A + 11 \, B - 12 \, C\right )} x^{4} - \frac {1}{3} \, {\left (11 \, A - 12 \, B - 4 \, C\right )} x^{3} + 2 \, {\left (3 \, A + B\right )} x^{2} + 4 \, A x \] Input:
integrate((C*x^2+B*x+A)*(x^3-5*x^2+3*x+2)^2,x, algorithm="fricas")
Output:
1/9*C*x^9 + 1/8*(B - 10*C)*x^8 + 1/7*(A - 10*B + 31*C)*x^7 - 1/6*(10*A - 3 1*B + 26*C)*x^6 + 1/5*(31*A - 26*B - 11*C)*x^5 - 1/4*(26*A + 11*B - 12*C)* x^4 - 1/3*(11*A - 12*B - 4*C)*x^3 + 2*(3*A + B)*x^2 + 4*A*x
Time = 0.03 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.02 \[ \int \left (A+B x+C x^2\right ) \left (2+3 x-5 x^2+x^3\right )^2 \, dx=4 A x + \frac {C x^{9}}{9} + x^{8} \left (\frac {B}{8} - \frac {5 C}{4}\right ) + x^{7} \left (\frac {A}{7} - \frac {10 B}{7} + \frac {31 C}{7}\right ) + x^{6} \left (- \frac {5 A}{3} + \frac {31 B}{6} - \frac {13 C}{3}\right ) + x^{5} \cdot \left (\frac {31 A}{5} - \frac {26 B}{5} - \frac {11 C}{5}\right ) + x^{4} \left (- \frac {13 A}{2} - \frac {11 B}{4} + 3 C\right ) + x^{3} \left (- \frac {11 A}{3} + 4 B + \frac {4 C}{3}\right ) + x^{2} \cdot \left (6 A + 2 B\right ) \] Input:
integrate((C*x**2+B*x+A)*(x**3-5*x**2+3*x+2)**2,x)
Output:
4*A*x + C*x**9/9 + x**8*(B/8 - 5*C/4) + x**7*(A/7 - 10*B/7 + 31*C/7) + x** 6*(-5*A/3 + 31*B/6 - 13*C/3) + x**5*(31*A/5 - 26*B/5 - 11*C/5) + x**4*(-13 *A/2 - 11*B/4 + 3*C) + x**3*(-11*A/3 + 4*B + 4*C/3) + x**2*(6*A + 2*B)
Time = 0.03 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.87 \[ \int \left (A+B x+C x^2\right ) \left (2+3 x-5 x^2+x^3\right )^2 \, dx=\frac {1}{9} \, C x^{9} + \frac {1}{8} \, {\left (B - 10 \, C\right )} x^{8} + \frac {1}{7} \, {\left (A - 10 \, B + 31 \, C\right )} x^{7} - \frac {1}{6} \, {\left (10 \, A - 31 \, B + 26 \, C\right )} x^{6} + \frac {1}{5} \, {\left (31 \, A - 26 \, B - 11 \, C\right )} x^{5} - \frac {1}{4} \, {\left (26 \, A + 11 \, B - 12 \, C\right )} x^{4} - \frac {1}{3} \, {\left (11 \, A - 12 \, B - 4 \, C\right )} x^{3} + 2 \, {\left (3 \, A + B\right )} x^{2} + 4 \, A x \] Input:
integrate((C*x^2+B*x+A)*(x^3-5*x^2+3*x+2)^2,x, algorithm="maxima")
Output:
1/9*C*x^9 + 1/8*(B - 10*C)*x^8 + 1/7*(A - 10*B + 31*C)*x^7 - 1/6*(10*A - 3 1*B + 26*C)*x^6 + 1/5*(31*A - 26*B - 11*C)*x^5 - 1/4*(26*A + 11*B - 12*C)* x^4 - 1/3*(11*A - 12*B - 4*C)*x^3 + 2*(3*A + B)*x^2 + 4*A*x
Time = 0.12 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.05 \[ \int \left (A+B x+C x^2\right ) \left (2+3 x-5 x^2+x^3\right )^2 \, dx=\frac {1}{9} \, C x^{9} + \frac {1}{8} \, B x^{8} - \frac {5}{4} \, C x^{8} + \frac {1}{7} \, A x^{7} - \frac {10}{7} \, B x^{7} + \frac {31}{7} \, C x^{7} - \frac {5}{3} \, A x^{6} + \frac {31}{6} \, B x^{6} - \frac {13}{3} \, C x^{6} + \frac {31}{5} \, A x^{5} - \frac {26}{5} \, B x^{5} - \frac {11}{5} \, C x^{5} - \frac {13}{2} \, A x^{4} - \frac {11}{4} \, B x^{4} + 3 \, C x^{4} - \frac {11}{3} \, A x^{3} + 4 \, B x^{3} + \frac {4}{3} \, C x^{3} + 6 \, A x^{2} + 2 \, B x^{2} + 4 \, A x \] Input:
integrate((C*x^2+B*x+A)*(x^3-5*x^2+3*x+2)^2,x, algorithm="giac")
Output:
1/9*C*x^9 + 1/8*B*x^8 - 5/4*C*x^8 + 1/7*A*x^7 - 10/7*B*x^7 + 31/7*C*x^7 - 5/3*A*x^6 + 31/6*B*x^6 - 13/3*C*x^6 + 31/5*A*x^5 - 26/5*B*x^5 - 11/5*C*x^5 - 13/2*A*x^4 - 11/4*B*x^4 + 3*C*x^4 - 11/3*A*x^3 + 4*B*x^3 + 4/3*C*x^3 + 6*A*x^2 + 2*B*x^2 + 4*A*x
Time = 0.06 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.89 \[ \int \left (A+B x+C x^2\right ) \left (2+3 x-5 x^2+x^3\right )^2 \, dx=\frac {C\,x^9}{9}+\left (\frac {B}{8}-\frac {5\,C}{4}\right )\,x^8+\left (\frac {A}{7}-\frac {10\,B}{7}+\frac {31\,C}{7}\right )\,x^7+\left (\frac {31\,B}{6}-\frac {5\,A}{3}-\frac {13\,C}{3}\right )\,x^6+\left (\frac {31\,A}{5}-\frac {26\,B}{5}-\frac {11\,C}{5}\right )\,x^5+\left (3\,C-\frac {11\,B}{4}-\frac {13\,A}{2}\right )\,x^4+\left (4\,B-\frac {11\,A}{3}+\frac {4\,C}{3}\right )\,x^3+\left (6\,A+2\,B\right )\,x^2+4\,A\,x \] Input:
int((A + B*x + C*x^2)*(3*x - 5*x^2 + x^3 + 2)^2,x)
Output:
4*A*x + (C*x^9)/9 + x^3*(4*B - (11*A)/3 + (4*C)/3) - x^4*((13*A)/2 + (11*B )/4 - 3*C) - x^6*((5*A)/3 - (31*B)/6 + (13*C)/3) + x^7*(A/7 - (10*B)/7 + ( 31*C)/7) - x^5*((26*B)/5 - (31*A)/5 + (11*C)/5) + x^2*(6*A + 2*B) + x^8*(B /8 - (5*C)/4)
Time = 0.16 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.03 \[ \int \left (A+B x+C x^2\right ) \left (2+3 x-5 x^2+x^3\right )^2 \, dx=\frac {x \left (280 c \,x^{8}+315 b \,x^{7}-3150 c \,x^{7}+360 a \,x^{6}-3600 b \,x^{6}+11160 c \,x^{6}-4200 a \,x^{5}+13020 b \,x^{5}-10920 c \,x^{5}+15624 a \,x^{4}-13104 b \,x^{4}-5544 c \,x^{4}-16380 a \,x^{3}-6930 b \,x^{3}+7560 c \,x^{3}-9240 a \,x^{2}+10080 b \,x^{2}+3360 c \,x^{2}+15120 a x +5040 b x +10080 a \right )}{2520} \] Input:
int((C*x^2+B*x+A)*(x^3-5*x^2+3*x+2)^2,x)
Output:
(x*(360*a*x**6 - 4200*a*x**5 + 15624*a*x**4 - 16380*a*x**3 - 9240*a*x**2 + 15120*a*x + 10080*a + 315*b*x**7 - 3600*b*x**6 + 13020*b*x**5 - 13104*b*x **4 - 6930*b*x**3 + 10080*b*x**2 + 5040*b*x + 280*c*x**8 - 3150*c*x**7 + 1 1160*c*x**6 - 10920*c*x**5 - 5544*c*x**4 + 7560*c*x**3 + 3360*c*x**2))/252 0