\(\int x^{3/2} (A+B x) (a+b x+c x^2)^3 \, dx\) [76]

Optimal result

Integrand size = 23, antiderivative size = 182 \[ \int x^{3/2} (A+B x) \left (a+b x+c x^2\right )^3 \, dx=\frac {2}{5} a^3 A x^{5/2}+\frac {2}{7} a^2 (3 A b+a B) x^{7/2}+\frac {2}{3} a \left (a b B+A \left (b^2+a c\right )\right ) x^{9/2}+\frac {2}{11} \left (3 a B \left (b^2+a c\right )+A \left (b^3+6 a b c\right )\right ) x^{11/2}+\frac {2}{13} \left (b^3 B+3 A b^2 c+6 a b B c+3 a A c^2\right ) x^{13/2}+\frac {2}{5} c \left (b^2 B+A b c+a B c\right ) x^{15/2}+\frac {2}{17} c^2 (3 b B+A c) x^{17/2}+\frac {2}{19} B c^3 x^{19/2} \] Output:

2/5*a^3*A*x^(5/2)+2/7*a^2*(3*A*b+B*a)*x^(7/2)+2/3*a*(a*b*B+A*(a*c+b^2))*x^ 
(9/2)+2/11*(3*a*B*(a*c+b^2)+A*(6*a*b*c+b^3))*x^(11/2)+2/13*(3*A*a*c^2+3*A* 
b^2*c+6*B*a*b*c+B*b^3)*x^(13/2)+2/5*c*(A*b*c+B*a*c+B*b^2)*x^(15/2)+2/17*c^ 
2*(A*c+3*B*b)*x^(17/2)+2/19*B*c^3*x^(19/2)
 

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 178, normalized size of antiderivative = 0.98 \[ \int x^{3/2} (A+B x) \left (a+b x+c x^2\right )^3 \, dx=\frac {2 x^{5/2} \left (138567 a^3 (7 A+5 B x)+20995 a^2 x (11 A (9 b+7 c x)+7 B x (11 b+9 c x))+2261 a x^2 \left (5 A \left (143 b^2+234 b c x+99 c^2 x^2\right )+3 B x \left (195 b^2+330 b c x+143 c^2 x^2\right )\right )+21 x^3 \left (19 A \left (1105 b^3+2805 b^2 c x+2431 b c^2 x^2+715 c^3 x^3\right )+11 B x \left (1615 b^3+4199 b^2 c x+3705 b c^2 x^2+1105 c^3 x^3\right )\right )\right )}{4849845} \] Input:

Integrate[x^(3/2)*(A + B*x)*(a + b*x + c*x^2)^3,x]
 

Output:

(2*x^(5/2)*(138567*a^3*(7*A + 5*B*x) + 20995*a^2*x*(11*A*(9*b + 7*c*x) + 7 
*B*x*(11*b + 9*c*x)) + 2261*a*x^2*(5*A*(143*b^2 + 234*b*c*x + 99*c^2*x^2) 
+ 3*B*x*(195*b^2 + 330*b*c*x + 143*c^2*x^2)) + 21*x^3*(19*A*(1105*b^3 + 28 
05*b^2*c*x + 2431*b*c^2*x^2 + 715*c^3*x^3) + 11*B*x*(1615*b^3 + 4199*b^2*c 
*x + 3705*b*c^2*x^2 + 1105*c^3*x^3))))/4849845
 

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 182, 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.

\(\displaystyle \int x^{3/2} (A+B x) \left (a+b x+c x^2\right )^3 \, dx\)

\(\Big \downarrow \) 1195

\(\displaystyle \int \left (a^3 A x^{3/2}+a^2 x^{5/2} (a B+3 A b)+3 c x^{13/2} \left (a B c+A b c+b^2 B\right )+3 a x^{7/2} \left (A \left (a c+b^2\right )+a b B\right )+x^{11/2} \left (3 a A c^2+6 a b B c+3 A b^2 c+b^3 B\right )+x^{9/2} \left (A \left (6 a b c+b^3\right )+3 a B \left (a c+b^2\right )\right )+c^2 x^{15/2} (A c+3 b B)+B c^3 x^{17/2}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {2}{5} a^3 A x^{5/2}+\frac {2}{7} a^2 x^{7/2} (a B+3 A b)+\frac {2}{5} c x^{15/2} \left (a B c+A b c+b^2 B\right )+\frac {2}{3} a x^{9/2} \left (A \left (a c+b^2\right )+a b B\right )+\frac {2}{13} x^{13/2} \left (3 a A c^2+6 a b B c+3 A b^2 c+b^3 B\right )+\frac {2}{11} x^{11/2} \left (A \left (6 a b c+b^3\right )+3 a B \left (a c+b^2\right )\right )+\frac {2}{17} c^2 x^{17/2} (A c+3 b B)+\frac {2}{19} B c^3 x^{19/2}\)

Input:

Int[x^(3/2)*(A + B*x)*(a + b*x + c*x^2)^3,x]
 

Output:

(2*a^3*A*x^(5/2))/5 + (2*a^2*(3*A*b + a*B)*x^(7/2))/7 + (2*a*(a*b*B + A*(b 
^2 + a*c))*x^(9/2))/3 + (2*(3*a*B*(b^2 + a*c) + A*(b^3 + 6*a*b*c))*x^(11/2 
))/11 + (2*(b^3*B + 3*A*b^2*c + 6*a*b*B*c + 3*a*A*c^2)*x^(13/2))/13 + (2*c 
*(b^2*B + A*b*c + a*B*c)*x^(15/2))/5 + (2*c^2*(3*b*B + A*c)*x^(17/2))/17 + 
 (2*B*c^3*x^(19/2))/19
 

Defintions of rubi rules used

rule 1195

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]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

Maple [A] (verified)

Time = 1.03 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.05

method	result	size
gosper	\(\frac {2 x^{\frac {5}{2}} \left (255255 B \,c^{3} x^{7}+285285 A \,c^{3} x^{6}+855855 B b \,c^{2} x^{6}+969969 A b \,c^{2} x^{5}+969969 B a \,c^{2} x^{5}+969969 B \,b^{2} c \,x^{5}+1119195 A a \,c^{2} x^{4}+1119195 A \,b^{2} c \,x^{4}+2238390 B a b c \,x^{4}+373065 B \,b^{3} x^{4}+2645370 A a b c \,x^{3}+440895 A \,b^{3} x^{3}+1322685 B \,a^{2} c \,x^{3}+1322685 B a \,b^{2} x^{3}+1616615 A \,a^{2} c \,x^{2}+1616615 A a \,b^{2} x^{2}+1616615 B \,a^{2} b \,x^{2}+2078505 A \,a^{2} b x +692835 B \,a^{3} x +969969 a^{3} A \right )}{4849845}\)	\(192\)
trager	\(\frac {2 x^{\frac {5}{2}} \left (255255 B \,c^{3} x^{7}+285285 A \,c^{3} x^{6}+855855 B b \,c^{2} x^{6}+969969 A b \,c^{2} x^{5}+969969 B a \,c^{2} x^{5}+969969 B \,b^{2} c \,x^{5}+1119195 A a \,c^{2} x^{4}+1119195 A \,b^{2} c \,x^{4}+2238390 B a b c \,x^{4}+373065 B \,b^{3} x^{4}+2645370 A a b c \,x^{3}+440895 A \,b^{3} x^{3}+1322685 B \,a^{2} c \,x^{3}+1322685 B a \,b^{2} x^{3}+1616615 A \,a^{2} c \,x^{2}+1616615 A a \,b^{2} x^{2}+1616615 B \,a^{2} b \,x^{2}+2078505 A \,a^{2} b x +692835 B \,a^{3} x +969969 a^{3} A \right )}{4849845}\)	\(192\)
risch	\(\frac {2 x^{\frac {5}{2}} \left (255255 B \,c^{3} x^{7}+285285 A \,c^{3} x^{6}+855855 B b \,c^{2} x^{6}+969969 A b \,c^{2} x^{5}+969969 B a \,c^{2} x^{5}+969969 B \,b^{2} c \,x^{5}+1119195 A a \,c^{2} x^{4}+1119195 A \,b^{2} c \,x^{4}+2238390 B a b c \,x^{4}+373065 B \,b^{3} x^{4}+2645370 A a b c \,x^{3}+440895 A \,b^{3} x^{3}+1322685 B \,a^{2} c \,x^{3}+1322685 B a \,b^{2} x^{3}+1616615 A \,a^{2} c \,x^{2}+1616615 A a \,b^{2} x^{2}+1616615 B \,a^{2} b \,x^{2}+2078505 A \,a^{2} b x +692835 B \,a^{3} x +969969 a^{3} A \right )}{4849845}\)	\(192\)
orering	\(\frac {2 x^{\frac {5}{2}} \left (255255 B \,c^{3} x^{7}+285285 A \,c^{3} x^{6}+855855 B b \,c^{2} x^{6}+969969 A b \,c^{2} x^{5}+969969 B a \,c^{2} x^{5}+969969 B \,b^{2} c \,x^{5}+1119195 A a \,c^{2} x^{4}+1119195 A \,b^{2} c \,x^{4}+2238390 B a b c \,x^{4}+373065 B \,b^{3} x^{4}+2645370 A a b c \,x^{3}+440895 A \,b^{3} x^{3}+1322685 B \,a^{2} c \,x^{3}+1322685 B a \,b^{2} x^{3}+1616615 A \,a^{2} c \,x^{2}+1616615 A a \,b^{2} x^{2}+1616615 B \,a^{2} b \,x^{2}+2078505 A \,a^{2} b x +692835 B \,a^{3} x +969969 a^{3} A \right )}{4849845}\)	\(192\)
derivativedivides	\(\frac {2 B \,c^{3} x^{\frac {19}{2}}}{19}+\frac {2 \left (A \,c^{3}+3 B b \,c^{2}\right ) x^{\frac {17}{2}}}{17}+\frac {2 \left (3 A b \,c^{2}+B \left (a \,c^{2}+2 b^{2} c +c \left (2 a c +b^{2}\right )\right )\right ) x^{\frac {15}{2}}}{15}+\frac {2 \left (A \left (a \,c^{2}+2 b^{2} c +c \left (2 a c +b^{2}\right )\right )+B \left (4 a b c +b \left (2 a c +b^{2}\right )\right )\right ) x^{\frac {13}{2}}}{13}+\frac {2 \left (A \left (4 a b c +b \left (2 a c +b^{2}\right )\right )+B \left (a \left (2 a c +b^{2}\right )+2 a \,b^{2}+a^{2} c \right )\right ) x^{\frac {11}{2}}}{11}+\frac {2 \left (A \left (a \left (2 a c +b^{2}\right )+2 a \,b^{2}+a^{2} c \right )+3 B \,a^{2} b \right ) x^{\frac {9}{2}}}{9}+\frac {2 \left (3 A \,a^{2} b +B \,a^{3}\right ) x^{\frac {7}{2}}}{7}+\frac {2 a^{3} A \,x^{\frac {5}{2}}}{5}\)	\(226\)
default	\(\frac {2 B \,c^{3} x^{\frac {19}{2}}}{19}+\frac {2 \left (A \,c^{3}+3 B b \,c^{2}\right ) x^{\frac {17}{2}}}{17}+\frac {2 \left (3 A b \,c^{2}+B \left (a \,c^{2}+2 b^{2} c +c \left (2 a c +b^{2}\right )\right )\right ) x^{\frac {15}{2}}}{15}+\frac {2 \left (A \left (a \,c^{2}+2 b^{2} c +c \left (2 a c +b^{2}\right )\right )+B \left (4 a b c +b \left (2 a c +b^{2}\right )\right )\right ) x^{\frac {13}{2}}}{13}+\frac {2 \left (A \left (4 a b c +b \left (2 a c +b^{2}\right )\right )+B \left (a \left (2 a c +b^{2}\right )+2 a \,b^{2}+a^{2} c \right )\right ) x^{\frac {11}{2}}}{11}+\frac {2 \left (A \left (a \left (2 a c +b^{2}\right )+2 a \,b^{2}+a^{2} c \right )+3 B \,a^{2} b \right ) x^{\frac {9}{2}}}{9}+\frac {2 \left (3 A \,a^{2} b +B \,a^{3}\right ) x^{\frac {7}{2}}}{7}+\frac {2 a^{3} A \,x^{\frac {5}{2}}}{5}\)	\(226\)

Input:

int(x^(3/2)*(B*x+A)*(c*x^2+b*x+a)^3,x,method=_RETURNVERBOSE)
 

Output:

2/4849845*x^(5/2)*(255255*B*c^3*x^7+285285*A*c^3*x^6+855855*B*b*c^2*x^6+96 
9969*A*b*c^2*x^5+969969*B*a*c^2*x^5+969969*B*b^2*c*x^5+1119195*A*a*c^2*x^4 
+1119195*A*b^2*c*x^4+2238390*B*a*b*c*x^4+373065*B*b^3*x^4+2645370*A*a*b*c* 
x^3+440895*A*b^3*x^3+1322685*B*a^2*c*x^3+1322685*B*a*b^2*x^3+1616615*A*a^2 
*c*x^2+1616615*A*a*b^2*x^2+1616615*B*a^2*b*x^2+2078505*A*a^2*b*x+692835*B* 
a^3*x+969969*A*a^3)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 171, normalized size of antiderivative = 0.94 \[ \int x^{3/2} (A+B x) \left (a+b x+c x^2\right )^3 \, dx=\frac {2}{4849845} \, {\left (255255 \, B c^{3} x^{9} + 285285 \, {\left (3 \, B b c^{2} + A c^{3}\right )} x^{8} + 969969 \, {\left (B b^{2} c + {\left (B a + A b\right )} c^{2}\right )} x^{7} + 373065 \, {\left (B b^{3} + 3 \, A a c^{2} + 3 \, {\left (2 \, B a b + A b^{2}\right )} c\right )} x^{6} + 969969 \, A a^{3} x^{2} + 440895 \, {\left (3 \, B a b^{2} + A b^{3} + 3 \, {\left (B a^{2} + 2 \, A a b\right )} c\right )} x^{5} + 1616615 \, {\left (B a^{2} b + A a b^{2} + A a^{2} c\right )} x^{4} + 692835 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x^{3}\right )} \sqrt {x} \] Input:

integrate(x^(3/2)*(B*x+A)*(c*x^2+b*x+a)^3,x, algorithm="fricas")
 

Output:

2/4849845*(255255*B*c^3*x^9 + 285285*(3*B*b*c^2 + A*c^3)*x^8 + 969969*(B*b 
^2*c + (B*a + A*b)*c^2)*x^7 + 373065*(B*b^3 + 3*A*a*c^2 + 3*(2*B*a*b + A*b 
^2)*c)*x^6 + 969969*A*a^3*x^2 + 440895*(3*B*a*b^2 + A*b^3 + 3*(B*a^2 + 2*A 
*a*b)*c)*x^5 + 1616615*(B*a^2*b + A*a*b^2 + A*a^2*c)*x^4 + 692835*(B*a^3 + 
 3*A*a^2*b)*x^3)*sqrt(x)
 

Sympy [A] (verification not implemented)

Time = 0.64 (sec) , antiderivative size = 294, normalized size of antiderivative = 1.62 \[ \int x^{3/2} (A+B x) \left (a+b x+c x^2\right )^3 \, dx=\frac {2 A a^{3} x^{\frac {5}{2}}}{5} + \frac {6 A a^{2} b x^{\frac {7}{2}}}{7} + \frac {2 A a^{2} c x^{\frac {9}{2}}}{3} + \frac {2 A a b^{2} x^{\frac {9}{2}}}{3} + \frac {12 A a b c x^{\frac {11}{2}}}{11} + \frac {6 A a c^{2} x^{\frac {13}{2}}}{13} + \frac {2 A b^{3} x^{\frac {11}{2}}}{11} + \frac {6 A b^{2} c x^{\frac {13}{2}}}{13} + \frac {2 A b c^{2} x^{\frac {15}{2}}}{5} + \frac {2 A c^{3} x^{\frac {17}{2}}}{17} + \frac {2 B a^{3} x^{\frac {7}{2}}}{7} + \frac {2 B a^{2} b x^{\frac {9}{2}}}{3} + \frac {6 B a^{2} c x^{\frac {11}{2}}}{11} + \frac {6 B a b^{2} x^{\frac {11}{2}}}{11} + \frac {12 B a b c x^{\frac {13}{2}}}{13} + \frac {2 B a c^{2} x^{\frac {15}{2}}}{5} + \frac {2 B b^{3} x^{\frac {13}{2}}}{13} + \frac {2 B b^{2} c x^{\frac {15}{2}}}{5} + \frac {6 B b c^{2} x^{\frac {17}{2}}}{17} + \frac {2 B c^{3} x^{\frac {19}{2}}}{19} \] Input:

integrate(x**(3/2)*(B*x+A)*(c*x**2+b*x+a)**3,x)
 

Output:

2*A*a**3*x**(5/2)/5 + 6*A*a**2*b*x**(7/2)/7 + 2*A*a**2*c*x**(9/2)/3 + 2*A* 
a*b**2*x**(9/2)/3 + 12*A*a*b*c*x**(11/2)/11 + 6*A*a*c**2*x**(13/2)/13 + 2* 
A*b**3*x**(11/2)/11 + 6*A*b**2*c*x**(13/2)/13 + 2*A*b*c**2*x**(15/2)/5 + 2 
*A*c**3*x**(17/2)/17 + 2*B*a**3*x**(7/2)/7 + 2*B*a**2*b*x**(9/2)/3 + 6*B*a 
**2*c*x**(11/2)/11 + 6*B*a*b**2*x**(11/2)/11 + 12*B*a*b*c*x**(13/2)/13 + 2 
*B*a*c**2*x**(15/2)/5 + 2*B*b**3*x**(13/2)/13 + 2*B*b**2*c*x**(15/2)/5 + 6 
*B*b*c**2*x**(17/2)/17 + 2*B*c**3*x**(19/2)/19
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 166, normalized size of antiderivative = 0.91 \[ \int x^{3/2} (A+B x) \left (a+b x+c x^2\right )^3 \, dx=\frac {2}{19} \, B c^{3} x^{\frac {19}{2}} + \frac {2}{17} \, {\left (3 \, B b c^{2} + A c^{3}\right )} x^{\frac {17}{2}} + \frac {2}{5} \, {\left (B b^{2} c + {\left (B a + A b\right )} c^{2}\right )} x^{\frac {15}{2}} + \frac {2}{13} \, {\left (B b^{3} + 3 \, A a c^{2} + 3 \, {\left (2 \, B a b + A b^{2}\right )} c\right )} x^{\frac {13}{2}} + \frac {2}{5} \, A a^{3} x^{\frac {5}{2}} + \frac {2}{11} \, {\left (3 \, B a b^{2} + A b^{3} + 3 \, {\left (B a^{2} + 2 \, A a b\right )} c\right )} x^{\frac {11}{2}} + \frac {2}{3} \, {\left (B a^{2} b + A a b^{2} + A a^{2} c\right )} x^{\frac {9}{2}} + \frac {2}{7} \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x^{\frac {7}{2}} \] Input:

integrate(x^(3/2)*(B*x+A)*(c*x^2+b*x+a)^3,x, algorithm="maxima")
 

Output:

2/19*B*c^3*x^(19/2) + 2/17*(3*B*b*c^2 + A*c^3)*x^(17/2) + 2/5*(B*b^2*c + ( 
B*a + A*b)*c^2)*x^(15/2) + 2/13*(B*b^3 + 3*A*a*c^2 + 3*(2*B*a*b + A*b^2)*c 
)*x^(13/2) + 2/5*A*a^3*x^(5/2) + 2/11*(3*B*a*b^2 + A*b^3 + 3*(B*a^2 + 2*A* 
a*b)*c)*x^(11/2) + 2/3*(B*a^2*b + A*a*b^2 + A*a^2*c)*x^(9/2) + 2/7*(B*a^3 
+ 3*A*a^2*b)*x^(7/2)
 

Giac [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.06 \[ \int x^{3/2} (A+B x) \left (a+b x+c x^2\right )^3 \, dx=\frac {2}{19} \, B c^{3} x^{\frac {19}{2}} + \frac {6}{17} \, B b c^{2} x^{\frac {17}{2}} + \frac {2}{17} \, A c^{3} x^{\frac {17}{2}} + \frac {2}{5} \, B b^{2} c x^{\frac {15}{2}} + \frac {2}{5} \, B a c^{2} x^{\frac {15}{2}} + \frac {2}{5} \, A b c^{2} x^{\frac {15}{2}} + \frac {2}{13} \, B b^{3} x^{\frac {13}{2}} + \frac {12}{13} \, B a b c x^{\frac {13}{2}} + \frac {6}{13} \, A b^{2} c x^{\frac {13}{2}} + \frac {6}{13} \, A a c^{2} x^{\frac {13}{2}} + \frac {6}{11} \, B a b^{2} x^{\frac {11}{2}} + \frac {2}{11} \, A b^{3} x^{\frac {11}{2}} + \frac {6}{11} \, B a^{2} c x^{\frac {11}{2}} + \frac {12}{11} \, A a b c x^{\frac {11}{2}} + \frac {2}{3} \, B a^{2} b x^{\frac {9}{2}} + \frac {2}{3} \, A a b^{2} x^{\frac {9}{2}} + \frac {2}{3} \, A a^{2} c x^{\frac {9}{2}} + \frac {2}{7} \, B a^{3} x^{\frac {7}{2}} + \frac {6}{7} \, A a^{2} b x^{\frac {7}{2}} + \frac {2}{5} \, A a^{3} x^{\frac {5}{2}} \] Input:

integrate(x^(3/2)*(B*x+A)*(c*x^2+b*x+a)^3,x, algorithm="giac")
                                                                                    
                                                                                    
 

Output:

2/19*B*c^3*x^(19/2) + 6/17*B*b*c^2*x^(17/2) + 2/17*A*c^3*x^(17/2) + 2/5*B* 
b^2*c*x^(15/2) + 2/5*B*a*c^2*x^(15/2) + 2/5*A*b*c^2*x^(15/2) + 2/13*B*b^3* 
x^(13/2) + 12/13*B*a*b*c*x^(13/2) + 6/13*A*b^2*c*x^(13/2) + 6/13*A*a*c^2*x 
^(13/2) + 6/11*B*a*b^2*x^(11/2) + 2/11*A*b^3*x^(11/2) + 6/11*B*a^2*c*x^(11 
/2) + 12/11*A*a*b*c*x^(11/2) + 2/3*B*a^2*b*x^(9/2) + 2/3*A*a*b^2*x^(9/2) + 
 2/3*A*a^2*c*x^(9/2) + 2/7*B*a^3*x^(7/2) + 6/7*A*a^2*b*x^(7/2) + 2/5*A*a^3 
*x^(5/2)
 

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 169, normalized size of antiderivative = 0.93 \[ \int x^{3/2} (A+B x) \left (a+b x+c x^2\right )^3 \, dx=x^{11/2}\,\left (\frac {6\,B\,c\,a^2}{11}+\frac {6\,B\,a\,b^2}{11}+\frac {12\,A\,c\,a\,b}{11}+\frac {2\,A\,b^3}{11}\right )+x^{13/2}\,\left (\frac {2\,B\,b^3}{13}+\frac {6\,A\,b^2\,c}{13}+\frac {12\,B\,a\,b\,c}{13}+\frac {6\,A\,a\,c^2}{13}\right )+x^{7/2}\,\left (\frac {2\,B\,a^3}{7}+\frac {6\,A\,b\,a^2}{7}\right )+x^{17/2}\,\left (\frac {2\,A\,c^3}{17}+\frac {6\,B\,b\,c^2}{17}\right )+x^{9/2}\,\left (\frac {2\,B\,a^2\,b}{3}+\frac {2\,A\,c\,a^2}{3}+\frac {2\,A\,a\,b^2}{3}\right )+x^{15/2}\,\left (\frac {2\,B\,b^2\,c}{5}+\frac {2\,A\,b\,c^2}{5}+\frac {2\,B\,a\,c^2}{5}\right )+\frac {2\,A\,a^3\,x^{5/2}}{5}+\frac {2\,B\,c^3\,x^{19/2}}{19} \] Input:

int(x^(3/2)*(A + B*x)*(a + b*x + c*x^2)^3,x)
 

Output:

x^(11/2)*((2*A*b^3)/11 + (6*B*a*b^2)/11 + (6*B*a^2*c)/11 + (12*A*a*b*c)/11 
) + x^(13/2)*((2*B*b^3)/13 + (6*A*a*c^2)/13 + (6*A*b^2*c)/13 + (12*B*a*b*c 
)/13) + x^(7/2)*((2*B*a^3)/7 + (6*A*a^2*b)/7) + x^(17/2)*((2*A*c^3)/17 + ( 
6*B*b*c^2)/17) + x^(9/2)*((2*A*a*b^2)/3 + (2*A*a^2*c)/3 + (2*B*a^2*b)/3) + 
 x^(15/2)*((2*A*b*c^2)/5 + (2*B*a*c^2)/5 + (2*B*b^2*c)/5) + (2*A*a^3*x^(5/ 
2))/5 + (2*B*c^3*x^(19/2))/19
 

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.75 \[ \int x^{3/2} (A+B x) \left (a+b x+c x^2\right )^3 \, dx=\frac {2 \sqrt {x}\, x^{2} \left (255255 b \,c^{3} x^{7}+285285 a \,c^{3} x^{6}+855855 b^{2} c^{2} x^{6}+1939938 a b \,c^{2} x^{5}+969969 b^{3} c \,x^{5}+1119195 a^{2} c^{2} x^{4}+3357585 a \,b^{2} c \,x^{4}+373065 b^{4} x^{4}+3968055 a^{2} b c \,x^{3}+1763580 a \,b^{3} x^{3}+1616615 a^{3} c \,x^{2}+3233230 a^{2} b^{2} x^{2}+2771340 a^{3} b x +969969 a^{4}\right )}{4849845} \] Input:

int(x^(3/2)*(B*x+A)*(c*x^2+b*x+a)^3,x)
 

Output:

(2*sqrt(x)*x**2*(969969*a**4 + 2771340*a**3*b*x + 1616615*a**3*c*x**2 + 32 
33230*a**2*b**2*x**2 + 3968055*a**2*b*c*x**3 + 1119195*a**2*c**2*x**4 + 17 
63580*a*b**3*x**3 + 3357585*a*b**2*c*x**4 + 1939938*a*b*c**2*x**5 + 285285 
*a*c**3*x**6 + 373065*b**4*x**4 + 969969*b**3*c*x**5 + 855855*b**2*c**2*x* 
*6 + 255255*b*c**3*x**7))/4849845