\(\int x (A+B x) (a+b x+c x^2)^{3/2} \, dx\) [119]

Optimal result

Integrand size = 21, antiderivative size = 198 \[ \int x (A+B x) \left (a+b x+c x^2\right )^{3/2} \, dx=-\frac {\left (b^2-4 a c\right ) \left (7 b^2 B-12 A b c-4 a B c\right ) (b+2 c x) \sqrt {a+b x+c x^2}}{512 c^4}+\frac {\left (7 b^2 B-12 A b c-4 a B c\right ) (b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{192 c^3}-\frac {(7 b B-12 A c-10 B c x) \left (a+b x+c x^2\right )^{5/2}}{60 c^2}+\frac {\left (b^2-4 a c\right )^2 \left (7 b^2 B-12 A b c-4 a B c\right ) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{1024 c^{9/2}} \] Output:

-1/512*(-4*a*c+b^2)*(-12*A*b*c-4*B*a*c+7*B*b^2)*(2*c*x+b)*(c*x^2+b*x+a)^(1 
/2)/c^4+1/192*(-12*A*b*c-4*B*a*c+7*B*b^2)*(2*c*x+b)*(c*x^2+b*x+a)^(3/2)/c^ 
3-1/60*(-10*B*c*x-12*A*c+7*B*b)*(c*x^2+b*x+a)^(5/2)/c^2+1/1024*(-4*a*c+b^2 
)^2*(-12*A*b*c-4*B*a*c+7*B*b^2)*arctanh(1/2*(2*c*x+b)/c^(1/2)/(c*x^2+b*x+a 
)^(1/2))/c^(9/2)
 

Mathematica [A] (verified)

Time = 1.47 (sec) , antiderivative size = 254, normalized size of antiderivative = 1.28 \[ \int x (A+B x) \left (a+b x+c x^2\right )^{3/2} \, dx=\frac {2 \sqrt {c} \sqrt {a+x (b+c x)} \left (-105 b^5 B+10 b^4 c (18 A+7 B x)+8 b^3 c (95 a B-c x (15 A+7 B x))+48 b^2 c^2 \left (c x^2 (2 A+B x)-a (25 A+9 B x)\right )+16 b c^2 \left (-81 a^2 B+6 a c x (7 A+3 B x)+4 c^2 x^3 (33 A+26 B x)\right )+32 c^3 \left (8 c^2 x^4 (6 A+5 B x)+3 a^2 (16 A+5 B x)+2 a c x^2 (48 A+35 B x)\right )\right )-15 \left (b^2-4 a c\right )^2 \left (7 b^2 B-12 A b c-4 a B c\right ) \log \left (b+2 c x-2 \sqrt {c} \sqrt {a+x (b+c x)}\right )}{15360 c^{9/2}} \] Input:

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

Output:

(2*Sqrt[c]*Sqrt[a + x*(b + c*x)]*(-105*b^5*B + 10*b^4*c*(18*A + 7*B*x) + 8 
*b^3*c*(95*a*B - c*x*(15*A + 7*B*x)) + 48*b^2*c^2*(c*x^2*(2*A + B*x) - a*( 
25*A + 9*B*x)) + 16*b*c^2*(-81*a^2*B + 6*a*c*x*(7*A + 3*B*x) + 4*c^2*x^3*( 
33*A + 26*B*x)) + 32*c^3*(8*c^2*x^4*(6*A + 5*B*x) + 3*a^2*(16*A + 5*B*x) + 
 2*a*c*x^2*(48*A + 35*B*x))) - 15*(b^2 - 4*a*c)^2*(7*b^2*B - 12*A*b*c - 4* 
a*B*c)*Log[b + 2*c*x - 2*Sqrt[c]*Sqrt[a + x*(b + c*x)]])/(15360*c^(9/2))
 

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 178, normalized size of antiderivative = 0.90, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {1225, 1087, 1087, 1092, 219}

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.

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

\(\Big \downarrow \) 1225

\(\displaystyle \frac {\left (-4 a B c-12 A b c+7 b^2 B\right ) \int \left (c x^2+b x+a\right )^{3/2}dx}{24 c^2}-\frac {\left (a+b x+c x^2\right )^{5/2} (-12 A c+7 b B-10 B c x)}{60 c^2}\)

\(\Big \downarrow \) 1087

\(\displaystyle \frac {\left (-4 a B c-12 A b c+7 b^2 B\right ) \left (\frac {(b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{8 c}-\frac {3 \left (b^2-4 a c\right ) \int \sqrt {c x^2+b x+a}dx}{16 c}\right )}{24 c^2}-\frac {\left (a+b x+c x^2\right )^{5/2} (-12 A c+7 b B-10 B c x)}{60 c^2}\)

\(\Big \downarrow \) 1087

\(\displaystyle \frac {\left (-4 a B c-12 A b c+7 b^2 B\right ) \left (\frac {(b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{8 c}-\frac {3 \left (b^2-4 a c\right ) \left (\frac {(b+2 c x) \sqrt {a+b x+c x^2}}{4 c}-\frac {\left (b^2-4 a c\right ) \int \frac {1}{\sqrt {c x^2+b x+a}}dx}{8 c}\right )}{16 c}\right )}{24 c^2}-\frac {\left (a+b x+c x^2\right )^{5/2} (-12 A c+7 b B-10 B c x)}{60 c^2}\)

\(\Big \downarrow \) 1092

\(\displaystyle \frac {\left (-4 a B c-12 A b c+7 b^2 B\right ) \left (\frac {(b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{8 c}-\frac {3 \left (b^2-4 a c\right ) \left (\frac {(b+2 c x) \sqrt {a+b x+c x^2}}{4 c}-\frac {\left (b^2-4 a c\right ) \int \frac {1}{4 c-\frac {(b+2 c x)^2}{c x^2+b x+a}}d\frac {b+2 c x}{\sqrt {c x^2+b x+a}}}{4 c}\right )}{16 c}\right )}{24 c^2}-\frac {\left (a+b x+c x^2\right )^{5/2} (-12 A c+7 b B-10 B c x)}{60 c^2}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\left (-4 a B c-12 A b c+7 b^2 B\right ) \left (\frac {(b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{8 c}-\frac {3 \left (b^2-4 a c\right ) \left (\frac {(b+2 c x) \sqrt {a+b x+c x^2}}{4 c}-\frac {\left (b^2-4 a c\right ) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{8 c^{3/2}}\right )}{16 c}\right )}{24 c^2}-\frac {\left (a+b x+c x^2\right )^{5/2} (-12 A c+7 b B-10 B c x)}{60 c^2}\)

Input:

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

Output:

-1/60*((7*b*B - 12*A*c - 10*B*c*x)*(a + b*x + c*x^2)^(5/2))/c^2 + ((7*b^2* 
B - 12*A*b*c - 4*a*B*c)*(((b + 2*c*x)*(a + b*x + c*x^2)^(3/2))/(8*c) - (3* 
(b^2 - 4*a*c)*(((b + 2*c*x)*Sqrt[a + b*x + c*x^2])/(4*c) - ((b^2 - 4*a*c)* 
ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(8*c^(3/2))))/(16* 
c)))/(24*c^2)
 

Defintions of rubi rules used

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 1087

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(b + 2*c*x) 
*((a + b*x + c*x^2)^p/(2*c*(2*p + 1))), x] - Simp[p*((b^2 - 4*a*c)/(2*c*(2* 
p + 1)))   Int[(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && 
GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[3*p])
 

rule 1092

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Simp[2   Subst[I 
nt[1/(4*c - x^2), x], x, (b + 2*c*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a 
, b, c}, x]
 

rule 1225

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*( 
x_)^2)^(p_), x_Symbol] :> Simp[(-(b*e*g*(p + 2) - c*(e*f + d*g)*(2*p + 3) - 
 2*c*e*g*(p + 1)*x))*((a + b*x + c*x^2)^(p + 1)/(2*c^2*(p + 1)*(2*p + 3))), 
 x] + Simp[(b^2*e*g*(p + 2) - 2*a*c*e*g + c*(2*c*d*f - b*(e*f + d*g))*(2*p 
+ 3))/(2*c^2*(2*p + 3))   Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c 
, d, e, f, g, p}, x] &&  !LeQ[p, -1]
 

Maple [A] (verified)

Time = 1.16 (sec) , antiderivative size = 316, normalized size of antiderivative = 1.60

method	result	size
risch	\(\frac {\left (1280 B \,c^{5} x^{5}+1536 A \,c^{5} x^{4}+1664 B b \,c^{4} x^{4}+2112 A b \,c^{4} x^{3}+2240 B a \,c^{4} x^{3}+48 B \,b^{2} c^{3} x^{3}+3072 A a \,c^{4} x^{2}+96 A \,b^{2} c^{3} x^{2}+288 B a b \,c^{3} x^{2}-56 B \,b^{3} c^{2} x^{2}+672 A a b \,c^{3} x -120 A \,b^{3} c^{2} x +480 B \,a^{2} c^{3} x -432 B a \,b^{2} c^{2} x +70 B \,b^{4} c x +1536 a^{2} A \,c^{3}-1200 A a \,b^{2} c^{2}+180 A \,b^{4} c -1296 B \,a^{2} b \,c^{2}+760 B a \,b^{3} c -105 b^{5} B \right ) \sqrt {c \,x^{2}+b x +a}}{7680 c^{4}}-\frac {\left (192 a^{2} A b \,c^{3}-96 A a \,b^{3} c^{2}+12 A \,b^{5} c +64 B \,a^{3} c^{3}-144 B \,a^{2} b^{2} c^{2}+60 B a \,b^{4} c -7 B \,b^{6}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{1024 c^{\frac {9}{2}}}\)	\(316\)
default	\(A \left (\frac {\left (c \,x^{2}+b x +a \right )^{\frac {5}{2}}}{5 c}-\frac {b \left (\frac {\left (2 c x +b \right ) \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}{8 c}+\frac {3 \left (4 a c -b^{2}\right ) \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x +a}}{4 c}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 c^{\frac {3}{2}}}\right )}{16 c}\right )}{2 c}\right )+B \left (\frac {x \left (c \,x^{2}+b x +a \right )^{\frac {5}{2}}}{6 c}-\frac {7 b \left (\frac {\left (c \,x^{2}+b x +a \right )^{\frac {5}{2}}}{5 c}-\frac {b \left (\frac {\left (2 c x +b \right ) \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}{8 c}+\frac {3 \left (4 a c -b^{2}\right ) \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x +a}}{4 c}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 c^{\frac {3}{2}}}\right )}{16 c}\right )}{2 c}\right )}{12 c}-\frac {a \left (\frac {\left (2 c x +b \right ) \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}{8 c}+\frac {3 \left (4 a c -b^{2}\right ) \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x +a}}{4 c}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 c^{\frac {3}{2}}}\right )}{16 c}\right )}{6 c}\right )\)	\(394\)

Input:

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

Output:

1/7680/c^4*(1280*B*c^5*x^5+1536*A*c^5*x^4+1664*B*b*c^4*x^4+2112*A*b*c^4*x^ 
3+2240*B*a*c^4*x^3+48*B*b^2*c^3*x^3+3072*A*a*c^4*x^2+96*A*b^2*c^3*x^2+288* 
B*a*b*c^3*x^2-56*B*b^3*c^2*x^2+672*A*a*b*c^3*x-120*A*b^3*c^2*x+480*B*a^2*c 
^3*x-432*B*a*b^2*c^2*x+70*B*b^4*c*x+1536*A*a^2*c^3-1200*A*a*b^2*c^2+180*A* 
b^4*c-1296*B*a^2*b*c^2+760*B*a*b^3*c-105*B*b^5)*(c*x^2+b*x+a)^(1/2)-1/1024 
*(192*A*a^2*b*c^3-96*A*a*b^3*c^2+12*A*b^5*c+64*B*a^3*c^3-144*B*a^2*b^2*c^2 
+60*B*a*b^4*c-7*B*b^6)/c^(9/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))
 

Fricas [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 669, normalized size of antiderivative = 3.38 \[ \int x (A+B x) \left (a+b x+c x^2\right )^{3/2} \, dx =\text {Too large to display} \] Input:

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

Output:

[-1/30720*(15*(7*B*b^6 - 64*(B*a^3 + 3*A*a^2*b)*c^3 + 48*(3*B*a^2*b^2 + 2* 
A*a*b^3)*c^2 - 12*(5*B*a*b^4 + A*b^5)*c)*sqrt(c)*log(-8*c^2*x^2 - 8*b*c*x 
- b^2 + 4*sqrt(c*x^2 + b*x + a)*(2*c*x + b)*sqrt(c) - 4*a*c) - 4*(1280*B*c 
^6*x^5 - 105*B*b^5*c + 1536*A*a^2*c^4 + 128*(13*B*b*c^5 + 12*A*c^6)*x^4 - 
48*(27*B*a^2*b + 25*A*a*b^2)*c^3 + 16*(3*B*b^2*c^4 + 4*(35*B*a + 33*A*b)*c 
^5)*x^3 + 20*(38*B*a*b^3 + 9*A*b^4)*c^2 - 8*(7*B*b^3*c^3 - 384*A*a*c^5 - 1 
2*(3*B*a*b + A*b^2)*c^4)*x^2 + 2*(35*B*b^4*c^2 + 48*(5*B*a^2 + 7*A*a*b)*c^ 
4 - 12*(18*B*a*b^2 + 5*A*b^3)*c^3)*x)*sqrt(c*x^2 + b*x + a))/c^5, -1/15360 
*(15*(7*B*b^6 - 64*(B*a^3 + 3*A*a^2*b)*c^3 + 48*(3*B*a^2*b^2 + 2*A*a*b^3)* 
c^2 - 12*(5*B*a*b^4 + A*b^5)*c)*sqrt(-c)*arctan(1/2*sqrt(c*x^2 + b*x + a)* 
(2*c*x + b)*sqrt(-c)/(c^2*x^2 + b*c*x + a*c)) - 2*(1280*B*c^6*x^5 - 105*B* 
b^5*c + 1536*A*a^2*c^4 + 128*(13*B*b*c^5 + 12*A*c^6)*x^4 - 48*(27*B*a^2*b 
+ 25*A*a*b^2)*c^3 + 16*(3*B*b^2*c^4 + 4*(35*B*a + 33*A*b)*c^5)*x^3 + 20*(3 
8*B*a*b^3 + 9*A*b^4)*c^2 - 8*(7*B*b^3*c^3 - 384*A*a*c^5 - 12*(3*B*a*b + A* 
b^2)*c^4)*x^2 + 2*(35*B*b^4*c^2 + 48*(5*B*a^2 + 7*A*a*b)*c^4 - 12*(18*B*a* 
b^2 + 5*A*b^3)*c^3)*x)*sqrt(c*x^2 + b*x + a))/c^5]
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1175 vs. \(2 (199) = 398\).

Time = 0.90 (sec) , antiderivative size = 1175, normalized size of antiderivative = 5.93 \[ \int x (A+B x) \left (a+b x+c x^2\right )^{3/2} \, dx=\text {Too large to display} \] Input:

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

Output:

Piecewise(((-a*(2*A*a*b + B*a**2 - 3*a*(2*A*b*c + 7*B*a*c/6 + B*b**2 - 9*b 
*(A*c**2 + 13*B*b*c/12)/(10*c))/(4*c) - 5*b*(2*A*a*c + A*b**2 + 2*B*a*b - 
4*a*(A*c**2 + 13*B*b*c/12)/(5*c) - 7*b*(2*A*b*c + 7*B*a*c/6 + B*b**2 - 9*b 
*(A*c**2 + 13*B*b*c/12)/(10*c))/(8*c))/(6*c))/(2*c) - b*(A*a**2 - 2*a*(2*A 
*a*c + A*b**2 + 2*B*a*b - 4*a*(A*c**2 + 13*B*b*c/12)/(5*c) - 7*b*(2*A*b*c 
+ 7*B*a*c/6 + B*b**2 - 9*b*(A*c**2 + 13*B*b*c/12)/(10*c))/(8*c))/(3*c) - 3 
*b*(2*A*a*b + B*a**2 - 3*a*(2*A*b*c + 7*B*a*c/6 + B*b**2 - 9*b*(A*c**2 + 1 
3*B*b*c/12)/(10*c))/(4*c) - 5*b*(2*A*a*c + A*b**2 + 2*B*a*b - 4*a*(A*c**2 
+ 13*B*b*c/12)/(5*c) - 7*b*(2*A*b*c + 7*B*a*c/6 + B*b**2 - 9*b*(A*c**2 + 1 
3*B*b*c/12)/(10*c))/(8*c))/(6*c))/(4*c))/(2*c))*Piecewise((log(b + 2*sqrt( 
c)*sqrt(a + b*x + c*x**2) + 2*c*x)/sqrt(c), Ne(a - b**2/(4*c), 0)), ((b/(2 
*c) + x)*log(b/(2*c) + x)/sqrt(c*(b/(2*c) + x)**2), True)) + sqrt(a + b*x 
+ c*x**2)*(B*c*x**5/6 + x**4*(A*c**2 + 13*B*b*c/12)/(5*c) + x**3*(2*A*b*c 
+ 7*B*a*c/6 + B*b**2 - 9*b*(A*c**2 + 13*B*b*c/12)/(10*c))/(4*c) + x**2*(2* 
A*a*c + A*b**2 + 2*B*a*b - 4*a*(A*c**2 + 13*B*b*c/12)/(5*c) - 7*b*(2*A*b*c 
 + 7*B*a*c/6 + B*b**2 - 9*b*(A*c**2 + 13*B*b*c/12)/(10*c))/(8*c))/(3*c) + 
x*(2*A*a*b + B*a**2 - 3*a*(2*A*b*c + 7*B*a*c/6 + B*b**2 - 9*b*(A*c**2 + 13 
*B*b*c/12)/(10*c))/(4*c) - 5*b*(2*A*a*c + A*b**2 + 2*B*a*b - 4*a*(A*c**2 + 
 13*B*b*c/12)/(5*c) - 7*b*(2*A*b*c + 7*B*a*c/6 + B*b**2 - 9*b*(A*c**2 + 13 
*B*b*c/12)/(10*c))/(8*c))/(6*c))/(2*c) + (A*a**2 - 2*a*(2*A*a*c + A*b**...
 

Maxima [F(-2)]

Exception generated. \[ \int x (A+B x) \left (a+b x+c x^2\right )^{3/2} \, dx=\text {Exception raised: ValueError} \] Input:

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

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` for 
 more deta
 

Giac [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 330, normalized size of antiderivative = 1.67 \[ \int x (A+B x) \left (a+b x+c x^2\right )^{3/2} \, dx=\frac {1}{7680} \, \sqrt {c x^{2} + b x + a} {\left (2 \, {\left (4 \, {\left (2 \, {\left (8 \, {\left (10 \, B c x + \frac {13 \, B b c^{5} + 12 \, A c^{6}}{c^{5}}\right )} x + \frac {3 \, B b^{2} c^{4} + 140 \, B a c^{5} + 132 \, A b c^{5}}{c^{5}}\right )} x - \frac {7 \, B b^{3} c^{3} - 36 \, B a b c^{4} - 12 \, A b^{2} c^{4} - 384 \, A a c^{5}}{c^{5}}\right )} x + \frac {35 \, B b^{4} c^{2} - 216 \, B a b^{2} c^{3} - 60 \, A b^{3} c^{3} + 240 \, B a^{2} c^{4} + 336 \, A a b c^{4}}{c^{5}}\right )} x - \frac {105 \, B b^{5} c - 760 \, B a b^{3} c^{2} - 180 \, A b^{4} c^{2} + 1296 \, B a^{2} b c^{3} + 1200 \, A a b^{2} c^{3} - 1536 \, A a^{2} c^{4}}{c^{5}}\right )} - \frac {{\left (7 \, B b^{6} - 60 \, B a b^{4} c - 12 \, A b^{5} c + 144 \, B a^{2} b^{2} c^{2} + 96 \, A a b^{3} c^{2} - 64 \, B a^{3} c^{3} - 192 \, A a^{2} b c^{3}\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} \sqrt {c} + b \right |}\right )}{1024 \, c^{\frac {9}{2}}} \] Input:

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

Output:

1/7680*sqrt(c*x^2 + b*x + a)*(2*(4*(2*(8*(10*B*c*x + (13*B*b*c^5 + 12*A*c^ 
6)/c^5)*x + (3*B*b^2*c^4 + 140*B*a*c^5 + 132*A*b*c^5)/c^5)*x - (7*B*b^3*c^ 
3 - 36*B*a*b*c^4 - 12*A*b^2*c^4 - 384*A*a*c^5)/c^5)*x + (35*B*b^4*c^2 - 21 
6*B*a*b^2*c^3 - 60*A*b^3*c^3 + 240*B*a^2*c^4 + 336*A*a*b*c^4)/c^5)*x - (10 
5*B*b^5*c - 760*B*a*b^3*c^2 - 180*A*b^4*c^2 + 1296*B*a^2*b*c^3 + 1200*A*a* 
b^2*c^3 - 1536*A*a^2*c^4)/c^5) - 1/1024*(7*B*b^6 - 60*B*a*b^4*c - 12*A*b^5 
*c + 144*B*a^2*b^2*c^2 + 96*A*a*b^3*c^2 - 64*B*a^3*c^3 - 192*A*a^2*b*c^3)* 
log(abs(2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*sqrt(c) + b))/c^(9/2)
 

Mupad [F(-1)]

Timed out. \[ \int x (A+B x) \left (a+b x+c x^2\right )^{3/2} \, dx=\int x\,\left (A+B\,x\right )\,{\left (c\,x^2+b\,x+a\right )}^{3/2} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int x (A+B x) \left (a+b x+c x^2\right )^{3/2} \, dx=\int x \left (B x +A \right ) \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}d x \] Input:

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

Output:

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