Integrand size = 23, antiderivative size = 113 \[ \int x^{3/2} (A+B x) \left (a+b x+c x^2\right )^2 \, dx=\frac {2}{5} a^2 A x^{5/2}+\frac {2}{7} a (2 A b+a B) x^{7/2}+\frac {2}{9} \left (2 a b B+A \left (b^2+2 a c\right )\right ) x^{9/2}+\frac {2}{11} \left (b^2 B+2 A b c+2 a B c\right ) x^{11/2}+\frac {2}{13} c (2 b B+A c) x^{13/2}+\frac {2}{15} B c^2 x^{15/2} \] Output:
2/5*a^2*A*x^(5/2)+2/7*a*(2*A*b+B*a)*x^(7/2)+2/9*(2*a*b*B+A*(2*a*c+b^2))*x^ (9/2)+2/11*(2*A*b*c+2*B*a*c+B*b^2)*x^(11/2)+2/13*c*(A*c+2*B*b)*x^(13/2)+2/ 15*B*c^2*x^(15/2)
Time = 0.12 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.90 \[ \int x^{3/2} (A+B x) \left (a+b x+c x^2\right )^2 \, dx=\frac {2 x^{5/2} \left (1287 a^2 (7 A+5 B x)+130 a x (11 A (9 b+7 c x)+7 B x (11 b+9 c x))+7 x^2 \left (5 A \left (143 b^2+234 b c x+99 c^2 x^2\right )+3 B x \left (195 b^2+330 b c x+143 c^2 x^2\right )\right )\right )}{45045} \] Input:
Integrate[x^(3/2)*(A + B*x)*(a + b*x + c*x^2)^2,x]
Output:
(2*x^(5/2)*(1287*a^2*(7*A + 5*B*x) + 130*a*x*(11*A*(9*b + 7*c*x) + 7*B*x*( 11*b + 9*c*x)) + 7*x^2*(5*A*(143*b^2 + 234*b*c*x + 99*c^2*x^2) + 3*B*x*(19 5*b^2 + 330*b*c*x + 143*c^2*x^2))))/45045
Time = 0.26 (sec) , antiderivative size = 113, 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.
\(\displaystyle \int x^{3/2} (A+B x) \left (a+b x+c x^2\right )^2 \, dx\) |
\(\Big \downarrow \) 1195 |
\(\displaystyle \int \left (a^2 A x^{3/2}+x^{9/2} \left (2 a B c+2 A b c+b^2 B\right )+x^{7/2} \left (A \left (2 a c+b^2\right )+2 a b B\right )+a x^{5/2} (a B+2 A b)+c x^{11/2} (A c+2 b B)+B c^2 x^{13/2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2}{5} a^2 A x^{5/2}+\frac {2}{11} x^{11/2} \left (2 a B c+2 A b c+b^2 B\right )+\frac {2}{9} x^{9/2} \left (A \left (2 a c+b^2\right )+2 a b B\right )+\frac {2}{7} a x^{7/2} (a B+2 A b)+\frac {2}{13} c x^{13/2} (A c+2 b B)+\frac {2}{15} B c^2 x^{15/2}\) |
Input:
Int[x^(3/2)*(A + B*x)*(a + b*x + c*x^2)^2,x]
Output:
(2*a^2*A*x^(5/2))/5 + (2*a*(2*A*b + a*B)*x^(7/2))/7 + (2*(2*a*b*B + A*(b^2 + 2*a*c))*x^(9/2))/9 + (2*(b^2*B + 2*A*b*c + 2*a*B*c)*x^(11/2))/11 + (2*c *(2*b*B + A*c)*x^(13/2))/13 + (2*B*c^2*x^(15/2))/15
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 = 1.14 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.83
method | result | size |
derivativedivides | \(\frac {2 B \,c^{2} x^{\frac {15}{2}}}{15}+\frac {2 \left (A \,c^{2}+2 B b c \right ) x^{\frac {13}{2}}}{13}+\frac {2 \left (2 A b c +B \left (2 a c +b^{2}\right )\right ) x^{\frac {11}{2}}}{11}+\frac {2 \left (2 a b B +A \left (2 a c +b^{2}\right )\right ) x^{\frac {9}{2}}}{9}+\frac {2 \left (2 a b A +a^{2} B \right ) x^{\frac {7}{2}}}{7}+\frac {2 a^{2} A \,x^{\frac {5}{2}}}{5}\) | \(94\) |
default | \(\frac {2 B \,c^{2} x^{\frac {15}{2}}}{15}+\frac {2 \left (A \,c^{2}+2 B b c \right ) x^{\frac {13}{2}}}{13}+\frac {2 \left (2 A b c +B \left (2 a c +b^{2}\right )\right ) x^{\frac {11}{2}}}{11}+\frac {2 \left (2 a b B +A \left (2 a c +b^{2}\right )\right ) x^{\frac {9}{2}}}{9}+\frac {2 \left (2 a b A +a^{2} B \right ) x^{\frac {7}{2}}}{7}+\frac {2 a^{2} A \,x^{\frac {5}{2}}}{5}\) | \(94\) |
gosper | \(\frac {2 x^{\frac {5}{2}} \left (3003 B \,c^{2} x^{5}+3465 x^{4} A \,c^{2}+6930 x^{4} B b c +8190 x^{3} A b c +8190 B a c \,x^{3}+4095 x^{3} B \,b^{2}+10010 A a c \,x^{2}+5005 x^{2} b^{2} A +10010 B a \,x^{2} b +12870 a b A x +6435 a^{2} B x +9009 a^{2} A \right )}{45045}\) | \(102\) |
trager | \(\frac {2 x^{\frac {5}{2}} \left (3003 B \,c^{2} x^{5}+3465 x^{4} A \,c^{2}+6930 x^{4} B b c +8190 x^{3} A b c +8190 B a c \,x^{3}+4095 x^{3} B \,b^{2}+10010 A a c \,x^{2}+5005 x^{2} b^{2} A +10010 B a \,x^{2} b +12870 a b A x +6435 a^{2} B x +9009 a^{2} A \right )}{45045}\) | \(102\) |
risch | \(\frac {2 x^{\frac {5}{2}} \left (3003 B \,c^{2} x^{5}+3465 x^{4} A \,c^{2}+6930 x^{4} B b c +8190 x^{3} A b c +8190 B a c \,x^{3}+4095 x^{3} B \,b^{2}+10010 A a c \,x^{2}+5005 x^{2} b^{2} A +10010 B a \,x^{2} b +12870 a b A x +6435 a^{2} B x +9009 a^{2} A \right )}{45045}\) | \(102\) |
orering | \(\frac {2 x^{\frac {5}{2}} \left (3003 B \,c^{2} x^{5}+3465 x^{4} A \,c^{2}+6930 x^{4} B b c +8190 x^{3} A b c +8190 B a c \,x^{3}+4095 x^{3} B \,b^{2}+10010 A a c \,x^{2}+5005 x^{2} b^{2} A +10010 B a \,x^{2} b +12870 a b A x +6435 a^{2} B x +9009 a^{2} A \right )}{45045}\) | \(102\) |
Input:
int(x^(3/2)*(B*x+A)*(c*x^2+b*x+a)^2,x,method=_RETURNVERBOSE)
Output:
2/15*B*c^2*x^(15/2)+2/13*(A*c^2+2*B*b*c)*x^(13/2)+2/11*(2*A*b*c+B*(2*a*c+b ^2))*x^(11/2)+2/9*(2*a*b*B+A*(2*a*c+b^2))*x^(9/2)+2/7*(2*A*a*b+B*a^2)*x^(7 /2)+2/5*a^2*A*x^(5/2)
Time = 0.09 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.87 \[ \int x^{3/2} (A+B x) \left (a+b x+c x^2\right )^2 \, dx=\frac {2}{45045} \, {\left (3003 \, B c^{2} x^{7} + 3465 \, {\left (2 \, B b c + A c^{2}\right )} x^{6} + 4095 \, {\left (B b^{2} + 2 \, {\left (B a + A b\right )} c\right )} x^{5} + 9009 \, A a^{2} x^{2} + 5005 \, {\left (2 \, B a b + A b^{2} + 2 \, A a c\right )} x^{4} + 6435 \, {\left (B a^{2} + 2 \, A a b\right )} x^{3}\right )} \sqrt {x} \] Input:
integrate(x^(3/2)*(B*x+A)*(c*x^2+b*x+a)^2,x, algorithm="fricas")
Output:
2/45045*(3003*B*c^2*x^7 + 3465*(2*B*b*c + A*c^2)*x^6 + 4095*(B*b^2 + 2*(B* a + A*b)*c)*x^5 + 9009*A*a^2*x^2 + 5005*(2*B*a*b + A*b^2 + 2*A*a*c)*x^4 + 6435*(B*a^2 + 2*A*a*b)*x^3)*sqrt(x)
Time = 0.31 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.43 \[ \int x^{3/2} (A+B x) \left (a+b x+c x^2\right )^2 \, dx=\frac {2 A a^{2} x^{\frac {5}{2}}}{5} + \frac {4 A a b x^{\frac {7}{2}}}{7} + \frac {4 A a c x^{\frac {9}{2}}}{9} + \frac {2 A b^{2} x^{\frac {9}{2}}}{9} + \frac {4 A b c x^{\frac {11}{2}}}{11} + \frac {2 A c^{2} x^{\frac {13}{2}}}{13} + \frac {2 B a^{2} x^{\frac {7}{2}}}{7} + \frac {4 B a b x^{\frac {9}{2}}}{9} + \frac {4 B a c x^{\frac {11}{2}}}{11} + \frac {2 B b^{2} x^{\frac {11}{2}}}{11} + \frac {4 B b c x^{\frac {13}{2}}}{13} + \frac {2 B c^{2} x^{\frac {15}{2}}}{15} \] Input:
integrate(x**(3/2)*(B*x+A)*(c*x**2+b*x+a)**2,x)
Output:
2*A*a**2*x**(5/2)/5 + 4*A*a*b*x**(7/2)/7 + 4*A*a*c*x**(9/2)/9 + 2*A*b**2*x **(9/2)/9 + 4*A*b*c*x**(11/2)/11 + 2*A*c**2*x**(13/2)/13 + 2*B*a**2*x**(7/ 2)/7 + 4*B*a*b*x**(9/2)/9 + 4*B*a*c*x**(11/2)/11 + 2*B*b**2*x**(11/2)/11 + 4*B*b*c*x**(13/2)/13 + 2*B*c**2*x**(15/2)/15
Time = 0.03 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.82 \[ \int x^{3/2} (A+B x) \left (a+b x+c x^2\right )^2 \, dx=\frac {2}{15} \, B c^{2} x^{\frac {15}{2}} + \frac {2}{13} \, {\left (2 \, B b c + A c^{2}\right )} x^{\frac {13}{2}} + \frac {2}{11} \, {\left (B b^{2} + 2 \, {\left (B a + A b\right )} c\right )} x^{\frac {11}{2}} + \frac {2}{5} \, A a^{2} x^{\frac {5}{2}} + \frac {2}{9} \, {\left (2 \, B a b + A b^{2} + 2 \, A a c\right )} x^{\frac {9}{2}} + \frac {2}{7} \, {\left (B a^{2} + 2 \, A a b\right )} x^{\frac {7}{2}} \] Input:
integrate(x^(3/2)*(B*x+A)*(c*x^2+b*x+a)^2,x, algorithm="maxima")
Output:
2/15*B*c^2*x^(15/2) + 2/13*(2*B*b*c + A*c^2)*x^(13/2) + 2/11*(B*b^2 + 2*(B *a + A*b)*c)*x^(11/2) + 2/5*A*a^2*x^(5/2) + 2/9*(2*B*a*b + A*b^2 + 2*A*a*c )*x^(9/2) + 2/7*(B*a^2 + 2*A*a*b)*x^(7/2)
Time = 0.26 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.91 \[ \int x^{3/2} (A+B x) \left (a+b x+c x^2\right )^2 \, dx=\frac {2}{15} \, B c^{2} x^{\frac {15}{2}} + \frac {4}{13} \, B b c x^{\frac {13}{2}} + \frac {2}{13} \, A c^{2} x^{\frac {13}{2}} + \frac {2}{11} \, B b^{2} x^{\frac {11}{2}} + \frac {4}{11} \, B a c x^{\frac {11}{2}} + \frac {4}{11} \, A b c x^{\frac {11}{2}} + \frac {4}{9} \, B a b x^{\frac {9}{2}} + \frac {2}{9} \, A b^{2} x^{\frac {9}{2}} + \frac {4}{9} \, A a c x^{\frac {9}{2}} + \frac {2}{7} \, B a^{2} x^{\frac {7}{2}} + \frac {4}{7} \, A a b x^{\frac {7}{2}} + \frac {2}{5} \, A a^{2} x^{\frac {5}{2}} \] Input:
integrate(x^(3/2)*(B*x+A)*(c*x^2+b*x+a)^2,x, algorithm="giac")
Output:
2/15*B*c^2*x^(15/2) + 4/13*B*b*c*x^(13/2) + 2/13*A*c^2*x^(13/2) + 2/11*B*b ^2*x^(11/2) + 4/11*B*a*c*x^(11/2) + 4/11*A*b*c*x^(11/2) + 4/9*B*a*b*x^(9/2 ) + 2/9*A*b^2*x^(9/2) + 4/9*A*a*c*x^(9/2) + 2/7*B*a^2*x^(7/2) + 4/7*A*a*b* x^(7/2) + 2/5*A*a^2*x^(5/2)
Time = 0.04 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.82 \[ \int x^{3/2} (A+B x) \left (a+b x+c x^2\right )^2 \, dx=x^{7/2}\,\left (\frac {2\,B\,a^2}{7}+\frac {4\,A\,b\,a}{7}\right )+x^{13/2}\,\left (\frac {2\,A\,c^2}{13}+\frac {4\,B\,b\,c}{13}\right )+x^{9/2}\,\left (\frac {2\,A\,b^2}{9}+\frac {4\,B\,a\,b}{9}+\frac {4\,A\,a\,c}{9}\right )+x^{11/2}\,\left (\frac {2\,B\,b^2}{11}+\frac {4\,A\,c\,b}{11}+\frac {4\,B\,a\,c}{11}\right )+\frac {2\,A\,a^2\,x^{5/2}}{5}+\frac {2\,B\,c^2\,x^{15/2}}{15} \] Input:
int(x^(3/2)*(A + B*x)*(a + b*x + c*x^2)^2,x)
Output:
x^(7/2)*((2*B*a^2)/7 + (4*A*a*b)/7) + x^(13/2)*((2*A*c^2)/13 + (4*B*b*c)/1 3) + x^(9/2)*((2*A*b^2)/9 + (4*A*a*c)/9 + (4*B*a*b)/9) + x^(11/2)*((2*B*b^ 2)/11 + (4*A*b*c)/11 + (4*B*a*c)/11) + (2*A*a^2*x^(5/2))/5 + (2*B*c^2*x^(1 5/2))/15
Time = 0.23 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.72 \[ \int x^{3/2} (A+B x) \left (a+b x+c x^2\right )^2 \, dx=\frac {2 \sqrt {x}\, x^{2} \left (3003 b \,c^{2} x^{5}+3465 a \,c^{2} x^{4}+6930 b^{2} c \,x^{4}+16380 a b c \,x^{3}+4095 b^{3} x^{3}+10010 a^{2} c \,x^{2}+15015 a \,b^{2} x^{2}+19305 a^{2} b x +9009 a^{3}\right )}{45045} \] Input:
int(x^(3/2)*(B*x+A)*(c*x^2+b*x+a)^2,x)
Output:
(2*sqrt(x)*x**2*(9009*a**3 + 19305*a**2*b*x + 10010*a**2*c*x**2 + 15015*a* b**2*x**2 + 16380*a*b*c*x**3 + 3465*a*c**2*x**4 + 4095*b**3*x**3 + 6930*b* *2*c*x**4 + 3003*b*c**2*x**5))/45045