\(\int \sqrt {x} (A+B x) (a+b x+c x^2)^3 \, dx\) [77]

Optimal result

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

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

Mathematica [A] (verified)

Time = 0.24 (sec) , antiderivative size = 178, normalized size of antiderivative = 0.98 \[ \int \sqrt {x} (A+B x) \left (a+b x+c x^2\right )^3 \, dx=\frac {2 x^{3/2} \left (51051 a^3 (5 A+3 B x)+7293 a^2 x (9 A (7 b+5 c x)+5 B x (9 b+7 c x))+255 a x^2 \left (13 A \left (99 b^2+154 b c x+63 c^2 x^2\right )+7 B x \left (143 b^2+234 b c x+99 c^2 x^2\right )\right )+7 x^3 \left (17 A \left (715 b^3+1755 b^2 c x+1485 b c^2 x^2+429 c^3 x^3\right )+9 B x \left (1105 b^3+2805 b^2 c x+2431 b c^2 x^2+715 c^3 x^3\right )\right )\right )}{765765} \] Input:

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

Output:

(2*x^(3/2)*(51051*a^3*(5*A + 3*B*x) + 7293*a^2*x*(9*A*(7*b + 5*c*x) + 5*B* 
x*(9*b + 7*c*x)) + 255*a*x^2*(13*A*(99*b^2 + 154*b*c*x + 63*c^2*x^2) + 7*B 
*x*(143*b^2 + 234*b*c*x + 99*c^2*x^2)) + 7*x^3*(17*A*(715*b^3 + 1755*b^2*c 
*x + 1485*b*c^2*x^2 + 429*c^3*x^3) + 9*B*x*(1105*b^3 + 2805*b^2*c*x + 2431 
*b*c^2*x^2 + 715*c^3*x^3))))/765765
 

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.

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

\(\Big \downarrow \) 1195

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

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 = 1.07 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.05

method	result	size
gosper	\(\frac {2 x^{\frac {3}{2}} \left (45045 B \,c^{3} x^{7}+51051 A \,c^{3} x^{6}+153153 B b \,c^{2} x^{6}+176715 A b \,c^{2} x^{5}+176715 B a \,c^{2} x^{5}+176715 B \,b^{2} c \,x^{5}+208845 A a \,c^{2} x^{4}+208845 A \,b^{2} c \,x^{4}+417690 B a b c \,x^{4}+69615 B \,b^{3} x^{4}+510510 A a b c \,x^{3}+85085 A \,b^{3} x^{3}+255255 B \,a^{2} c \,x^{3}+255255 B a \,b^{2} x^{3}+328185 A \,a^{2} c \,x^{2}+328185 A a \,b^{2} x^{2}+328185 B \,a^{2} b \,x^{2}+459459 A \,a^{2} b x +153153 B \,a^{3} x +255255 a^{3} A \right )}{765765}\)	\(192\)
trager	\(\frac {2 x^{\frac {3}{2}} \left (45045 B \,c^{3} x^{7}+51051 A \,c^{3} x^{6}+153153 B b \,c^{2} x^{6}+176715 A b \,c^{2} x^{5}+176715 B a \,c^{2} x^{5}+176715 B \,b^{2} c \,x^{5}+208845 A a \,c^{2} x^{4}+208845 A \,b^{2} c \,x^{4}+417690 B a b c \,x^{4}+69615 B \,b^{3} x^{4}+510510 A a b c \,x^{3}+85085 A \,b^{3} x^{3}+255255 B \,a^{2} c \,x^{3}+255255 B a \,b^{2} x^{3}+328185 A \,a^{2} c \,x^{2}+328185 A a \,b^{2} x^{2}+328185 B \,a^{2} b \,x^{2}+459459 A \,a^{2} b x +153153 B \,a^{3} x +255255 a^{3} A \right )}{765765}\)	\(192\)
risch	\(\frac {2 x^{\frac {3}{2}} \left (45045 B \,c^{3} x^{7}+51051 A \,c^{3} x^{6}+153153 B b \,c^{2} x^{6}+176715 A b \,c^{2} x^{5}+176715 B a \,c^{2} x^{5}+176715 B \,b^{2} c \,x^{5}+208845 A a \,c^{2} x^{4}+208845 A \,b^{2} c \,x^{4}+417690 B a b c \,x^{4}+69615 B \,b^{3} x^{4}+510510 A a b c \,x^{3}+85085 A \,b^{3} x^{3}+255255 B \,a^{2} c \,x^{3}+255255 B a \,b^{2} x^{3}+328185 A \,a^{2} c \,x^{2}+328185 A a \,b^{2} x^{2}+328185 B \,a^{2} b \,x^{2}+459459 A \,a^{2} b x +153153 B \,a^{3} x +255255 a^{3} A \right )}{765765}\)	\(192\)
orering	\(\frac {2 x^{\frac {3}{2}} \left (45045 B \,c^{3} x^{7}+51051 A \,c^{3} x^{6}+153153 B b \,c^{2} x^{6}+176715 A b \,c^{2} x^{5}+176715 B a \,c^{2} x^{5}+176715 B \,b^{2} c \,x^{5}+208845 A a \,c^{2} x^{4}+208845 A \,b^{2} c \,x^{4}+417690 B a b c \,x^{4}+69615 B \,b^{3} x^{4}+510510 A a b c \,x^{3}+85085 A \,b^{3} x^{3}+255255 B \,a^{2} c \,x^{3}+255255 B a \,b^{2} x^{3}+328185 A \,a^{2} c \,x^{2}+328185 A a \,b^{2} x^{2}+328185 B \,a^{2} b \,x^{2}+459459 A \,a^{2} b x +153153 B \,a^{3} x +255255 a^{3} A \right )}{765765}\)	\(192\)
derivativedivides	\(\frac {2 B \,c^{3} x^{\frac {17}{2}}}{17}+\frac {2 \left (A \,c^{3}+3 B b \,c^{2}\right ) x^{\frac {15}{2}}}{15}+\frac {2 \left (3 A b \,c^{2}+B \left (a \,c^{2}+2 b^{2} c +c \left (2 a c +b^{2}\right )\right )\right ) x^{\frac {13}{2}}}{13}+\frac {2 \left (A \left (a \,c^{2}+2 b^{2} c +c \left (2 a c +b^{2}\right )\right )+B \left (4 a b c +b \left (2 a c +b^{2}\right )\right )\right ) x^{\frac {11}{2}}}{11}+\frac {2 \left (A \left (4 a b c +b \left (2 a c +b^{2}\right )\right )+B \left (a \left (2 a c +b^{2}\right )+2 a \,b^{2}+a^{2} c \right )\right ) x^{\frac {9}{2}}}{9}+\frac {2 \left (A \left (a \left (2 a c +b^{2}\right )+2 a \,b^{2}+a^{2} c \right )+3 B \,a^{2} b \right ) x^{\frac {7}{2}}}{7}+\frac {2 \left (3 A \,a^{2} b +B \,a^{3}\right ) x^{\frac {5}{2}}}{5}+\frac {2 a^{3} A \,x^{\frac {3}{2}}}{3}\)	\(226\)
default	\(\frac {2 B \,c^{3} x^{\frac {17}{2}}}{17}+\frac {2 \left (A \,c^{3}+3 B b \,c^{2}\right ) x^{\frac {15}{2}}}{15}+\frac {2 \left (3 A b \,c^{2}+B \left (a \,c^{2}+2 b^{2} c +c \left (2 a c +b^{2}\right )\right )\right ) x^{\frac {13}{2}}}{13}+\frac {2 \left (A \left (a \,c^{2}+2 b^{2} c +c \left (2 a c +b^{2}\right )\right )+B \left (4 a b c +b \left (2 a c +b^{2}\right )\right )\right ) x^{\frac {11}{2}}}{11}+\frac {2 \left (A \left (4 a b c +b \left (2 a c +b^{2}\right )\right )+B \left (a \left (2 a c +b^{2}\right )+2 a \,b^{2}+a^{2} c \right )\right ) x^{\frac {9}{2}}}{9}+\frac {2 \left (A \left (a \left (2 a c +b^{2}\right )+2 a \,b^{2}+a^{2} c \right )+3 B \,a^{2} b \right ) x^{\frac {7}{2}}}{7}+\frac {2 \left (3 A \,a^{2} b +B \,a^{3}\right ) x^{\frac {5}{2}}}{5}+\frac {2 a^{3} A \,x^{\frac {3}{2}}}{3}\)	\(226\)

Input:

int(x^(1/2)*(B*x+A)*(c*x^2+b*x+a)^3,x,method=_RETURNVERBOSE)
 

Output:

2/765765*x^(3/2)*(45045*B*c^3*x^7+51051*A*c^3*x^6+153153*B*b*c^2*x^6+17671 
5*A*b*c^2*x^5+176715*B*a*c^2*x^5+176715*B*b^2*c*x^5+208845*A*a*c^2*x^4+208 
845*A*b^2*c*x^4+417690*B*a*b*c*x^4+69615*B*b^3*x^4+510510*A*a*b*c*x^3+8508 
5*A*b^3*x^3+255255*B*a^2*c*x^3+255255*B*a*b^2*x^3+328185*A*a^2*c*x^2+32818 
5*A*a*b^2*x^2+328185*B*a^2*b*x^2+459459*A*a^2*b*x+153153*B*a^3*x+255255*A* 
a^3)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 169, normalized size of antiderivative = 0.93 \[ \int \sqrt {x} (A+B x) \left (a+b x+c x^2\right )^3 \, dx=\frac {2}{765765} \, {\left (45045 \, B c^{3} x^{8} + 51051 \, {\left (3 \, B b c^{2} + A c^{3}\right )} x^{7} + 176715 \, {\left (B b^{2} c + {\left (B a + A b\right )} c^{2}\right )} x^{6} + 69615 \, {\left (B b^{3} + 3 \, A a c^{2} + 3 \, {\left (2 \, B a b + A b^{2}\right )} c\right )} x^{5} + 255255 \, A a^{3} x + 85085 \, {\left (3 \, B a b^{2} + A b^{3} + 3 \, {\left (B a^{2} + 2 \, A a b\right )} c\right )} x^{4} + 328185 \, {\left (B a^{2} b + A a b^{2} + A a^{2} c\right )} x^{3} + 153153 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x^{2}\right )} \sqrt {x} \] Input:

integrate(x^(1/2)*(B*x+A)*(c*x^2+b*x+a)^3,x, algorithm="fricas")
 

Output:

2/765765*(45045*B*c^3*x^8 + 51051*(3*B*b*c^2 + A*c^3)*x^7 + 176715*(B*b^2* 
c + (B*a + A*b)*c^2)*x^6 + 69615*(B*b^3 + 3*A*a*c^2 + 3*(2*B*a*b + A*b^2)* 
c)*x^5 + 255255*A*a^3*x + 85085*(3*B*a*b^2 + A*b^3 + 3*(B*a^2 + 2*A*a*b)*c 
)*x^4 + 328185*(B*a^2*b + A*a*b^2 + A*a^2*c)*x^3 + 153153*(B*a^3 + 3*A*a^2 
*b)*x^2)*sqrt(x)
 

Sympy [A] (verification not implemented)

Time = 0.98 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.19 \[ \int \sqrt {x} (A+B x) \left (a+b x+c x^2\right )^3 \, dx=\frac {2 A a^{3} x^{\frac {3}{2}}}{3} + \frac {2 B c^{3} x^{\frac {17}{2}}}{17} + \frac {2 x^{\frac {15}{2}} \left (A c^{3} + 3 B b c^{2}\right )}{15} + \frac {2 x^{\frac {13}{2}} \cdot \left (3 A b c^{2} + 3 B a c^{2} + 3 B b^{2} c\right )}{13} + \frac {2 x^{\frac {11}{2}} \cdot \left (3 A a c^{2} + 3 A b^{2} c + 6 B a b c + B b^{3}\right )}{11} + \frac {2 x^{\frac {9}{2}} \cdot \left (6 A a b c + A b^{3} + 3 B a^{2} c + 3 B a b^{2}\right )}{9} + \frac {2 x^{\frac {7}{2}} \cdot \left (3 A a^{2} c + 3 A a b^{2} + 3 B a^{2} b\right )}{7} + \frac {2 x^{\frac {5}{2}} \cdot \left (3 A a^{2} b + B a^{3}\right )}{5} \] Input:

integrate(x**(1/2)*(B*x+A)*(c*x**2+b*x+a)**3,x)
 

Output:

2*A*a**3*x**(3/2)/3 + 2*B*c**3*x**(17/2)/17 + 2*x**(15/2)*(A*c**3 + 3*B*b* 
c**2)/15 + 2*x**(13/2)*(3*A*b*c**2 + 3*B*a*c**2 + 3*B*b**2*c)/13 + 2*x**(1 
1/2)*(3*A*a*c**2 + 3*A*b**2*c + 6*B*a*b*c + B*b**3)/11 + 2*x**(9/2)*(6*A*a 
*b*c + A*b**3 + 3*B*a**2*c + 3*B*a*b**2)/9 + 2*x**(7/2)*(3*A*a**2*c + 3*A* 
a*b**2 + 3*B*a**2*b)/7 + 2*x**(5/2)*(3*A*a**2*b + B*a**3)/5
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 166, normalized size of antiderivative = 0.91 \[ \int \sqrt {x} (A+B x) \left (a+b x+c x^2\right )^3 \, dx=\frac {2}{17} \, B c^{3} x^{\frac {17}{2}} + \frac {2}{15} \, {\left (3 \, B b c^{2} + A c^{3}\right )} x^{\frac {15}{2}} + \frac {6}{13} \, {\left (B b^{2} c + {\left (B a + A b\right )} c^{2}\right )} x^{\frac {13}{2}} + \frac {2}{11} \, {\left (B b^{3} + 3 \, A a c^{2} + 3 \, {\left (2 \, B a b + A b^{2}\right )} c\right )} x^{\frac {11}{2}} + \frac {2}{3} \, A a^{3} x^{\frac {3}{2}} + \frac {2}{9} \, {\left (3 \, B a b^{2} + A b^{3} + 3 \, {\left (B a^{2} + 2 \, A a b\right )} c\right )} x^{\frac {9}{2}} + \frac {6}{7} \, {\left (B a^{2} b + A a b^{2} + A a^{2} c\right )} x^{\frac {7}{2}} + \frac {2}{5} \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x^{\frac {5}{2}} \] Input:

integrate(x^(1/2)*(B*x+A)*(c*x^2+b*x+a)^3,x, algorithm="maxima")
 

Output:

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

Giac [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.06 \[ \int \sqrt {x} (A+B x) \left (a+b x+c x^2\right )^3 \, dx=\frac {2}{17} \, B c^{3} x^{\frac {17}{2}} + \frac {2}{5} \, B b c^{2} x^{\frac {15}{2}} + \frac {2}{15} \, A c^{3} x^{\frac {15}{2}} + \frac {6}{13} \, B b^{2} c x^{\frac {13}{2}} + \frac {6}{13} \, B a c^{2} x^{\frac {13}{2}} + \frac {6}{13} \, A b c^{2} x^{\frac {13}{2}} + \frac {2}{11} \, B b^{3} x^{\frac {11}{2}} + \frac {12}{11} \, B a b c x^{\frac {11}{2}} + \frac {6}{11} \, A b^{2} c x^{\frac {11}{2}} + \frac {6}{11} \, A a c^{2} x^{\frac {11}{2}} + \frac {2}{3} \, B a b^{2} x^{\frac {9}{2}} + \frac {2}{9} \, A b^{3} x^{\frac {9}{2}} + \frac {2}{3} \, B a^{2} c x^{\frac {9}{2}} + \frac {4}{3} \, A a b c x^{\frac {9}{2}} + \frac {6}{7} \, B a^{2} b x^{\frac {7}{2}} + \frac {6}{7} \, A a b^{2} x^{\frac {7}{2}} + \frac {6}{7} \, A a^{2} c x^{\frac {7}{2}} + \frac {2}{5} \, B a^{3} x^{\frac {5}{2}} + \frac {6}{5} \, A a^{2} b x^{\frac {5}{2}} + \frac {2}{3} \, A a^{3} x^{\frac {3}{2}} \] Input:

integrate(x^(1/2)*(B*x+A)*(c*x^2+b*x+a)^3,x, algorithm="giac")
                                                                                    
                                                                                    
 

Output:

2/17*B*c^3*x^(17/2) + 2/5*B*b*c^2*x^(15/2) + 2/15*A*c^3*x^(15/2) + 6/13*B* 
b^2*c*x^(13/2) + 6/13*B*a*c^2*x^(13/2) + 6/13*A*b*c^2*x^(13/2) + 2/11*B*b^ 
3*x^(11/2) + 12/11*B*a*b*c*x^(11/2) + 6/11*A*b^2*c*x^(11/2) + 6/11*A*a*c^2 
*x^(11/2) + 2/3*B*a*b^2*x^(9/2) + 2/9*A*b^3*x^(9/2) + 2/3*B*a^2*c*x^(9/2) 
+ 4/3*A*a*b*c*x^(9/2) + 6/7*B*a^2*b*x^(7/2) + 6/7*A*a*b^2*x^(7/2) + 6/7*A* 
a^2*c*x^(7/2) + 2/5*B*a^3*x^(5/2) + 6/5*A*a^2*b*x^(5/2) + 2/3*A*a^3*x^(3/2 
)
 

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 169, normalized size of antiderivative = 0.93 \[ \int \sqrt {x} (A+B x) \left (a+b x+c x^2\right )^3 \, dx=x^{9/2}\,\left (\frac {2\,B\,c\,a^2}{3}+\frac {2\,B\,a\,b^2}{3}+\frac {4\,A\,c\,a\,b}{3}+\frac {2\,A\,b^3}{9}\right )+x^{11/2}\,\left (\frac {2\,B\,b^3}{11}+\frac {6\,A\,b^2\,c}{11}+\frac {12\,B\,a\,b\,c}{11}+\frac {6\,A\,a\,c^2}{11}\right )+x^{5/2}\,\left (\frac {2\,B\,a^3}{5}+\frac {6\,A\,b\,a^2}{5}\right )+x^{15/2}\,\left (\frac {2\,A\,c^3}{15}+\frac {2\,B\,b\,c^2}{5}\right )+x^{7/2}\,\left (\frac {6\,B\,a^2\,b}{7}+\frac {6\,A\,c\,a^2}{7}+\frac {6\,A\,a\,b^2}{7}\right )+x^{13/2}\,\left (\frac {6\,B\,b^2\,c}{13}+\frac {6\,A\,b\,c^2}{13}+\frac {6\,B\,a\,c^2}{13}\right )+\frac {2\,A\,a^3\,x^{3/2}}{3}+\frac {2\,B\,c^3\,x^{17/2}}{17} \] Input:

int(x^(1/2)*(A + B*x)*(a + b*x + c*x^2)^3,x)
 

Output:

x^(9/2)*((2*A*b^3)/9 + (2*B*a*b^2)/3 + (2*B*a^2*c)/3 + (4*A*a*b*c)/3) + x^ 
(11/2)*((2*B*b^3)/11 + (6*A*a*c^2)/11 + (6*A*b^2*c)/11 + (12*B*a*b*c)/11) 
+ x^(5/2)*((2*B*a^3)/5 + (6*A*a^2*b)/5) + x^(15/2)*((2*A*c^3)/15 + (2*B*b* 
c^2)/5) + x^(7/2)*((6*A*a*b^2)/7 + (6*A*a^2*c)/7 + (6*B*a^2*b)/7) + x^(13/ 
2)*((6*A*b*c^2)/13 + (6*B*a*c^2)/13 + (6*B*b^2*c)/13) + (2*A*a^3*x^(3/2))/ 
3 + (2*B*c^3*x^(17/2))/17
 

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.74 \[ \int \sqrt {x} (A+B x) \left (a+b x+c x^2\right )^3 \, dx=\frac {2 \sqrt {x}\, x \left (45045 b \,c^{3} x^{7}+51051 a \,c^{3} x^{6}+153153 b^{2} c^{2} x^{6}+353430 a b \,c^{2} x^{5}+176715 b^{3} c \,x^{5}+208845 a^{2} c^{2} x^{4}+626535 a \,b^{2} c \,x^{4}+69615 b^{4} x^{4}+765765 a^{2} b c \,x^{3}+340340 a \,b^{3} x^{3}+328185 a^{3} c \,x^{2}+656370 a^{2} b^{2} x^{2}+612612 a^{3} b x +255255 a^{4}\right )}{765765} \] Input:

int(x^(1/2)*(B*x+A)*(c*x^2+b*x+a)^3,x)
 

Output:

(2*sqrt(x)*x*(255255*a**4 + 612612*a**3*b*x + 328185*a**3*c*x**2 + 656370* 
a**2*b**2*x**2 + 765765*a**2*b*c*x**3 + 208845*a**2*c**2*x**4 + 340340*a*b 
**3*x**3 + 626535*a*b**2*c*x**4 + 353430*a*b*c**2*x**5 + 51051*a*c**3*x**6 
 + 69615*b**4*x**4 + 176715*b**3*c*x**5 + 153153*b**2*c**2*x**6 + 45045*b* 
c**3*x**7))/765765