\(\int x (A+B x) \sqrt {a+b x+c x^2} \, dx\) [107]

Optimal result

Integrand size = 21, antiderivative size = 144 \[ \int x (A+B x) \sqrt {a+b x+c x^2} \, dx=\frac {\left (5 b^2 B-8 A b c-4 a B c\right ) (b+2 c x) \sqrt {a+b x+c x^2}}{64 c^3}-\frac {(5 b B-8 A c-6 B c x) \left (a+b x+c x^2\right )^{3/2}}{24 c^2}-\frac {\left (b^2-4 a c\right ) \left (5 b^2 B-8 A b c-4 a B c\right ) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{128 c^{7/2}} \] Output:

1/64*(-8*A*b*c-4*B*a*c+5*B*b^2)*(2*c*x+b)*(c*x^2+b*x+a)^(1/2)/c^3-1/24*(-6 
*B*c*x-8*A*c+5*B*b)*(c*x^2+b*x+a)^(3/2)/c^2-1/128*(-4*a*c+b^2)*(-8*A*b*c-4 
*B*a*c+5*B*b^2)*arctanh(1/2*(2*c*x+b)/c^(1/2)/(c*x^2+b*x+a)^(1/2))/c^(7/2)
 

Mathematica [A] (verified)

Time = 0.63 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.07 \[ \int x (A+B x) \sqrt {a+b x+c x^2} \, dx=\frac {2 \sqrt {c} \sqrt {a+x (b+c x)} \left (15 b^3 B-2 b^2 c (12 A+5 B x)+8 c^2 \left (8 a A+3 a B x+8 A c x^2+6 B c x^3\right )+4 b c (-13 a B+2 c x (2 A+B x))\right )+3 \left (b^2-4 a c\right ) \left (5 b^2 B-8 A b c-4 a B c\right ) \log \left (b+2 c x-2 \sqrt {c} \sqrt {a+x (b+c x)}\right )}{384 c^{7/2}} \] Input:

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

Output:

(2*Sqrt[c]*Sqrt[a + x*(b + c*x)]*(15*b^3*B - 2*b^2*c*(12*A + 5*B*x) + 8*c^ 
2*(8*a*A + 3*a*B*x + 8*A*c*x^2 + 6*B*c*x^3) + 4*b*c*(-13*a*B + 2*c*x*(2*A 
+ B*x))) + 3*(b^2 - 4*a*c)*(5*b^2*B - 8*A*b*c - 4*a*B*c)*Log[b + 2*c*x - 2 
*Sqrt[c]*Sqrt[a + x*(b + c*x)]])/(384*c^(7/2))
 

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.94, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {1225, 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.

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

\(\Big \downarrow \) 1225

\(\displaystyle \frac {\left (-4 a B c-8 A b c+5 b^2 B\right ) \int \sqrt {c x^2+b x+a}dx}{16 c^2}-\frac {\left (a+b x+c x^2\right )^{3/2} (-8 A c+5 b B-6 B c x)}{24 c^2}\)

\(\Big \downarrow \) 1087

\(\displaystyle \frac {\left (-4 a B c-8 A b c+5 b^2 B\right ) \left (\frac {(b+2 c x) \sqrt {a+b x+c x^2}}{4 c}-\frac {\left (b^2-4 a c\right ) \int \frac {1}{\sqrt {c x^2+b x+a}}dx}{8 c}\right )}{16 c^2}-\frac {\left (a+b x+c x^2\right )^{3/2} (-8 A c+5 b B-6 B c x)}{24 c^2}\)

\(\Big \downarrow \) 1092

\(\displaystyle \frac {\left (-4 a B c-8 A b c+5 b^2 B\right ) \left (\frac {(b+2 c x) \sqrt {a+b x+c x^2}}{4 c}-\frac {\left (b^2-4 a c\right ) \int \frac {1}{4 c-\frac {(b+2 c x)^2}{c x^2+b x+a}}d\frac {b+2 c x}{\sqrt {c x^2+b x+a}}}{4 c}\right )}{16 c^2}-\frac {\left (a+b x+c x^2\right )^{3/2} (-8 A c+5 b B-6 B c x)}{24 c^2}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\left (-4 a B c-8 A b c+5 b^2 B\right ) \left (\frac {(b+2 c x) \sqrt {a+b x+c x^2}}{4 c}-\frac {\left (b^2-4 a c\right ) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{8 c^{3/2}}\right )}{16 c^2}-\frac {\left (a+b x+c x^2\right )^{3/2} (-8 A c+5 b B-6 B c x)}{24 c^2}\)

Input:

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

Output:

-1/24*((5*b*B - 8*A*c - 6*B*c*x)*(a + b*x + c*x^2)^(3/2))/c^2 + ((5*b^2*B 
- 8*A*b*c - 4*a*B*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^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 = 1.19 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.16

method	result	size
risch	\(\frac {\left (48 B \,c^{3} x^{3}+64 A \,c^{3} x^{2}+8 B b \,c^{2} x^{2}+16 A b \,c^{2} x +24 B a \,c^{2} x -10 B \,b^{2} c x +64 A a \,c^{2}-24 A \,b^{2} c -52 B a b c +15 B \,b^{3}\right ) \sqrt {c \,x^{2}+b x +a}}{192 c^{3}}-\frac {\left (32 A a b \,c^{2}-8 A \,b^{3} c +16 B \,a^{2} c^{2}-24 B a \,b^{2} c +5 B \,b^{4}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{128 c^{\frac {7}{2}}}\)	\(167\)
default	\(A \left (\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}{3 c}-\frac {b \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x +a}}{4 c}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 c^{\frac {3}{2}}}\right )}{2 c}\right )+B \left (\frac {x \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}{4 c}-\frac {5 b \left (\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}{3 c}-\frac {b \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x +a}}{4 c}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 c^{\frac {3}{2}}}\right )}{2 c}\right )}{8 c}-\frac {a \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x +a}}{4 c}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 c^{\frac {3}{2}}}\right )}{4 c}\right )\)	\(277\)

Input:

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

Output:

1/192*(48*B*c^3*x^3+64*A*c^3*x^2+8*B*b*c^2*x^2+16*A*b*c^2*x+24*B*a*c^2*x-1 
0*B*b^2*c*x+64*A*a*c^2-24*A*b^2*c-52*B*a*b*c+15*B*b^3)/c^3*(c*x^2+b*x+a)^( 
1/2)-1/128*(32*A*a*b*c^2-8*A*b^3*c+16*B*a^2*c^2-24*B*a*b^2*c+5*B*b^4)/c^(7 
/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 393, normalized size of antiderivative = 2.73 \[ \int x (A+B x) \sqrt {a+b x+c x^2} \, dx=\left [\frac {3 \, {\left (5 \, B b^{4} + 16 \, {\left (B a^{2} + 2 \, A a b\right )} c^{2} - 8 \, {\left (3 \, B a b^{2} + A b^{3}\right )} c\right )} \sqrt {c} \log \left (-8 \, c^{2} x^{2} - 8 \, b c x - b^{2} + 4 \, \sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {c} - 4 \, a c\right ) + 4 \, {\left (48 \, B c^{4} x^{3} + 15 \, B b^{3} c + 64 \, A a c^{3} - 4 \, {\left (13 \, B a b + 6 \, A b^{2}\right )} c^{2} + 8 \, {\left (B b c^{3} + 8 \, A c^{4}\right )} x^{2} - 2 \, {\left (5 \, B b^{2} c^{2} - 4 \, {\left (3 \, B a + 2 \, A b\right )} c^{3}\right )} x\right )} \sqrt {c x^{2} + b x + a}}{768 \, c^{4}}, \frac {3 \, {\left (5 \, B b^{4} + 16 \, {\left (B a^{2} + 2 \, A a b\right )} c^{2} - 8 \, {\left (3 \, B a b^{2} + A b^{3}\right )} c\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{2} + b c x + a c\right )}}\right ) + 2 \, {\left (48 \, B c^{4} x^{3} + 15 \, B b^{3} c + 64 \, A a c^{3} - 4 \, {\left (13 \, B a b + 6 \, A b^{2}\right )} c^{2} + 8 \, {\left (B b c^{3} + 8 \, A c^{4}\right )} x^{2} - 2 \, {\left (5 \, B b^{2} c^{2} - 4 \, {\left (3 \, B a + 2 \, A b\right )} c^{3}\right )} x\right )} \sqrt {c x^{2} + b x + a}}{384 \, c^{4}}\right ] \] Input:

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

Output:

[1/768*(3*(5*B*b^4 + 16*(B*a^2 + 2*A*a*b)*c^2 - 8*(3*B*a*b^2 + A*b^3)*c)*s 
qrt(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*(48*B*c^4*x^3 + 15*B*b^3*c + 64*A*a*c^3 - 4*(13*B*a 
*b + 6*A*b^2)*c^2 + 8*(B*b*c^3 + 8*A*c^4)*x^2 - 2*(5*B*b^2*c^2 - 4*(3*B*a 
+ 2*A*b)*c^3)*x)*sqrt(c*x^2 + b*x + a))/c^4, 1/384*(3*(5*B*b^4 + 16*(B*a^2 
 + 2*A*a*b)*c^2 - 8*(3*B*a*b^2 + A*b^3)*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*(48*B*c^4*x^3 
 + 15*B*b^3*c + 64*A*a*c^3 - 4*(13*B*a*b + 6*A*b^2)*c^2 + 8*(B*b*c^3 + 8*A 
*c^4)*x^2 - 2*(5*B*b^2*c^2 - 4*(3*B*a + 2*A*b)*c^3)*x)*sqrt(c*x^2 + b*x + 
a))/c^4]
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 350 vs. \(2 (143) = 286\).

Time = 0.76 (sec) , antiderivative size = 350, normalized size of antiderivative = 2.43 \[ \int x (A+B x) \sqrt {a+b x+c x^2} \, dx=\begin {cases} \left (- \frac {a \left (A b + \frac {B a}{4} - \frac {5 b \left (A c + \frac {B b}{8}\right )}{6 c}\right )}{2 c} - \frac {b \left (A a - \frac {2 a \left (A c + \frac {B b}{8}\right )}{3 c} - \frac {3 b \left (A b + \frac {B a}{4} - \frac {5 b \left (A c + \frac {B b}{8}\right )}{6 c}\right )}{4 c}\right )}{2 c}\right ) \left (\begin {cases} \frac {\log {\left (b + 2 \sqrt {c} \sqrt {a + b x + c x^{2}} + 2 c x \right )}}{\sqrt {c}} & \text {for}\: a - \frac {b^{2}}{4 c} \neq 0 \\\frac {\left (\frac {b}{2 c} + x\right ) \log {\left (\frac {b}{2 c} + x \right )}}{\sqrt {c \left (\frac {b}{2 c} + x\right )^{2}}} & \text {otherwise} \end {cases}\right ) + \sqrt {a + b x + c x^{2}} \left (\frac {B x^{3}}{4} + \frac {x^{2} \left (A c + \frac {B b}{8}\right )}{3 c} + \frac {x \left (A b + \frac {B a}{4} - \frac {5 b \left (A c + \frac {B b}{8}\right )}{6 c}\right )}{2 c} + \frac {A a - \frac {2 a \left (A c + \frac {B b}{8}\right )}{3 c} - \frac {3 b \left (A b + \frac {B a}{4} - \frac {5 b \left (A c + \frac {B b}{8}\right )}{6 c}\right )}{4 c}}{c}\right ) & \text {for}\: c \neq 0 \\\frac {2 \left (\frac {B \left (a + b x\right )^{\frac {7}{2}}}{7 b} + \frac {\left (a + b x\right )^{\frac {5}{2}} \left (A b - 2 B a\right )}{5 b} + \frac {\left (a + b x\right )^{\frac {3}{2}} \left (- A a b + B a^{2}\right )}{3 b}\right )}{b^{2}} & \text {for}\: b \neq 0 \\\sqrt {a} \left (\frac {A x^{2}}{2} + \frac {B x^{3}}{3}\right ) & \text {otherwise} \end {cases} \] Input:

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

Output:

Piecewise(((-a*(A*b + B*a/4 - 5*b*(A*c + B*b/8)/(6*c))/(2*c) - b*(A*a - 2* 
a*(A*c + B*b/8)/(3*c) - 3*b*(A*b + B*a/4 - 5*b*(A*c + B*b/8)/(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*x**3/4 + x**2*(A*c + B*b/8) 
/(3*c) + x*(A*b + B*a/4 - 5*b*(A*c + B*b/8)/(6*c))/(2*c) + (A*a - 2*a*(A*c 
 + B*b/8)/(3*c) - 3*b*(A*b + B*a/4 - 5*b*(A*c + B*b/8)/(6*c))/(4*c))/c), N 
e(c, 0)), (2*(B*(a + b*x)**(7/2)/(7*b) + (a + b*x)**(5/2)*(A*b - 2*B*a)/(5 
*b) + (a + b*x)**(3/2)*(-A*a*b + B*a**2)/(3*b))/b**2, Ne(b, 0)), (sqrt(a)* 
(A*x**2/2 + B*x**3/3), True))
 

Maxima [F(-2)]

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

integrate(x*(B*x+A)*(c*x^2+b*x+a)^(1/2),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 = 176, normalized size of antiderivative = 1.22 \[ \int x (A+B x) \sqrt {a+b x+c x^2} \, dx=\frac {1}{192} \, \sqrt {c x^{2} + b x + a} {\left (2 \, {\left (4 \, {\left (6 \, B x + \frac {B b c^{2} + 8 \, A c^{3}}{c^{3}}\right )} x - \frac {5 \, B b^{2} c - 12 \, B a c^{2} - 8 \, A b c^{2}}{c^{3}}\right )} x + \frac {15 \, B b^{3} - 52 \, B a b c - 24 \, A b^{2} c + 64 \, A a c^{2}}{c^{3}}\right )} + \frac {{\left (5 \, B b^{4} - 24 \, B a b^{2} c - 8 \, A b^{3} c + 16 \, B a^{2} c^{2} + 32 \, A a b c^{2}\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} \sqrt {c} + b \right |}\right )}{128 \, c^{\frac {7}{2}}} \] Input:

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

Output:

1/192*sqrt(c*x^2 + b*x + a)*(2*(4*(6*B*x + (B*b*c^2 + 8*A*c^3)/c^3)*x - (5 
*B*b^2*c - 12*B*a*c^2 - 8*A*b*c^2)/c^3)*x + (15*B*b^3 - 52*B*a*b*c - 24*A* 
b^2*c + 64*A*a*c^2)/c^3) + 1/128*(5*B*b^4 - 24*B*a*b^2*c - 8*A*b^3*c + 16* 
B*a^2*c^2 + 32*A*a*b*c^2)*log(abs(2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*sq 
rt(c) + b))/c^(7/2)
 

Mupad [B] (verification not implemented)

Time = 10.90 (sec) , antiderivative size = 256, normalized size of antiderivative = 1.78 \[ \int x (A+B x) \sqrt {a+b x+c x^2} \, dx=\frac {A\,\ln \left (\frac {b+2\,c\,x}{\sqrt {c}}+2\,\sqrt {c\,x^2+b\,x+a}\right )\,\left (b^3-4\,a\,b\,c\right )}{16\,c^{5/2}}-\frac {B\,a\,\left (\left (\frac {x}{2}+\frac {b}{4\,c}\right )\,\sqrt {c\,x^2+b\,x+a}+\frac {\ln \left (\frac {\frac {b}{2}+c\,x}{\sqrt {c}}+\sqrt {c\,x^2+b\,x+a}\right )\,\left (a\,c-\frac {b^2}{4}\right )}{2\,c^{3/2}}\right )}{4\,c}-\frac {5\,B\,b\,\left (\frac {\ln \left (\frac {b+2\,c\,x}{\sqrt {c}}+2\,\sqrt {c\,x^2+b\,x+a}\right )\,\left (b^3-4\,a\,b\,c\right )}{16\,c^{5/2}}+\frac {\left (-3\,b^2+2\,c\,x\,b+8\,c\,\left (c\,x^2+a\right )\right )\,\sqrt {c\,x^2+b\,x+a}}{24\,c^2}\right )}{8\,c}+\frac {A\,\left (-3\,b^2+2\,c\,x\,b+8\,c\,\left (c\,x^2+a\right )\right )\,\sqrt {c\,x^2+b\,x+a}}{24\,c^2}+\frac {B\,x\,{\left (c\,x^2+b\,x+a\right )}^{3/2}}{4\,c} \] Input:

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

Output:

(A*log((b + 2*c*x)/c^(1/2) + 2*(a + b*x + c*x^2)^(1/2))*(b^3 - 4*a*b*c))/( 
16*c^(5/2)) - (B*a*((x/2 + b/(4*c))*(a + b*x + c*x^2)^(1/2) + (log((b/2 + 
c*x)/c^(1/2) + (a + b*x + c*x^2)^(1/2))*(a*c - b^2/4))/(2*c^(3/2))))/(4*c) 
 - (5*B*b*((log((b + 2*c*x)/c^(1/2) + 2*(a + b*x + c*x^2)^(1/2))*(b^3 - 4* 
a*b*c))/(16*c^(5/2)) + ((8*c*(a + c*x^2) - 3*b^2 + 2*b*c*x)*(a + b*x + c*x 
^2)^(1/2))/(24*c^2)))/(8*c) + (A*(8*c*(a + c*x^2) - 3*b^2 + 2*b*c*x)*(a + 
b*x + c*x^2)^(1/2))/(24*c^2) + (B*x*(a + b*x + c*x^2)^(3/2))/(4*c)
 

Reduce [F]

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

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

Output:

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