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

Optimal. Leaf size=205 \[ \frac{\left (a+b x+c x^2\right )^{3/2} \left (-32 a B c-6 c x (7 b B-10 A c)-50 A b c+35 b^2 B\right )}{240 c^3}-\frac{(b+2 c x) \sqrt{a+b x+c x^2} \left (8 a A c^2-12 a b B c-10 A b^2 c+7 b^3 B\right )}{128 c^4}+\frac{\left (b^2-4 a c\right ) \left (8 a A c^2-12 a b B c-10 A b^2 c+7 b^3 B\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{256 c^{9/2}}+\frac{B x^2 \left (a+b x+c x^2\right )^{3/2}}{5 c} \]

[Out]

-((7*b^3*B - 10*A*b^2*c - 12*a*b*B*c + 8*a*A*c^2)*(b + 2*c*x)*Sqrt[a + b*x + c*x^2])/(128*c^4) + (B*x^2*(a + b
*x + c*x^2)^(3/2))/(5*c) + ((35*b^2*B - 50*A*b*c - 32*a*B*c - 6*c*(7*b*B - 10*A*c)*x)*(a + b*x + c*x^2)^(3/2))
/(240*c^3) + ((b^2 - 4*a*c)*(7*b^3*B - 10*A*b^2*c - 12*a*b*B*c + 8*a*A*c^2)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqr
t[a + b*x + c*x^2])])/(256*c^(9/2))

________________________________________________________________________________________

Rubi [A]  time = 0.196367, antiderivative size = 205, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {832, 779, 612, 621, 206} \[ \frac{\left (a+b x+c x^2\right )^{3/2} \left (-32 a B c-6 c x (7 b B-10 A c)-50 A b c+35 b^2 B\right )}{240 c^3}-\frac{(b+2 c x) \sqrt{a+b x+c x^2} \left (8 a A c^2-12 a b B c-10 A b^2 c+7 b^3 B\right )}{128 c^4}+\frac{\left (b^2-4 a c\right ) \left (8 a A c^2-12 a b B c-10 A b^2 c+7 b^3 B\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{256 c^{9/2}}+\frac{B x^2 \left (a+b x+c x^2\right )^{3/2}}{5 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((7*b^3*B - 10*A*b^2*c - 12*a*b*B*c + 8*a*A*c^2)*(b + 2*c*x)*Sqrt[a + b*x + c*x^2])/(128*c^4) + (B*x^2*(a + b
*x + c*x^2)^(3/2))/(5*c) + ((35*b^2*B - 50*A*b*c - 32*a*B*c - 6*c*(7*b*B - 10*A*c)*x)*(a + b*x + c*x^2)^(3/2))
/(240*c^3) + ((b^2 - 4*a*c)*(7*b^3*B - 10*A*b^2*c - 12*a*b*B*c + 8*a*A*c^2)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqr
t[a + b*x + c*x^2])])/(256*c^(9/2))

Rule 832

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Sim
p[(g*(d + e*x)^m*(a + b*x + c*x^2)^(p + 1))/(c*(m + 2*p + 2)), x] + Dist[1/(c*(m + 2*p + 2)), Int[(d + e*x)^(m
- 1)*(a + b*x + c*x^2)^p*Simp[m*(c*d*f - a*e*g) + d*(2*c*f - b*g)*(p + 1) + (m*(c*e*f + c*d*g - b*e*g) + e*(p
+ 1)*(2*c*f - b*g))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 -
b*d*e + a*e^2, 0] && GtQ[m, 0] && NeQ[m + 2*p + 2, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])
&&  !(IGtQ[m, 0] && EqQ[f, 0])

Rule 779

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] + Dist[(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] && NeQ[b^2 - 4*a*c, 0] &&  !LeQ[p, -1]

Rule 612

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] - Dist[(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]
&& NeQ[b^2 - 4*a*c, 0] && GtQ[p, 0] && IntegerQ[4*p]

Rule 621

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(4*c - x^2), x], x, (b + 2*c*x)
/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 206

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

Rubi steps

\begin{align*} \int x^2 (A+B x) \sqrt{a+b x+c x^2} \, dx &=\frac{B x^2 \left (a+b x+c x^2\right )^{3/2}}{5 c}+\frac{\int x \left (-2 a B-\frac{1}{2} (7 b B-10 A c) x\right ) \sqrt{a+b x+c x^2} \, dx}{5 c}\\ &=\frac{B x^2 \left (a+b x+c x^2\right )^{3/2}}{5 c}+\frac{\left (35 b^2 B-50 A b c-32 a B c-6 c (7 b B-10 A c) x\right ) \left (a+b x+c x^2\right )^{3/2}}{240 c^3}-\frac{\left (7 b^3 B-10 A b^2 c-12 a b B c+8 a A c^2\right ) \int \sqrt{a+b x+c x^2} \, dx}{32 c^3}\\ &=-\frac{\left (7 b^3 B-10 A b^2 c-12 a b B c+8 a A c^2\right ) (b+2 c x) \sqrt{a+b x+c x^2}}{128 c^4}+\frac{B x^2 \left (a+b x+c x^2\right )^{3/2}}{5 c}+\frac{\left (35 b^2 B-50 A b c-32 a B c-6 c (7 b B-10 A c) x\right ) \left (a+b x+c x^2\right )^{3/2}}{240 c^3}+\frac{\left (\left (b^2-4 a c\right ) \left (7 b^3 B-10 A b^2 c-12 a b B c+8 a A c^2\right )\right ) \int \frac{1}{\sqrt{a+b x+c x^2}} \, dx}{256 c^4}\\ &=-\frac{\left (7 b^3 B-10 A b^2 c-12 a b B c+8 a A c^2\right ) (b+2 c x) \sqrt{a+b x+c x^2}}{128 c^4}+\frac{B x^2 \left (a+b x+c x^2\right )^{3/2}}{5 c}+\frac{\left (35 b^2 B-50 A b c-32 a B c-6 c (7 b B-10 A c) x\right ) \left (a+b x+c x^2\right )^{3/2}}{240 c^3}+\frac{\left (\left (b^2-4 a c\right ) \left (7 b^3 B-10 A b^2 c-12 a b B c+8 a A c^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{4 c-x^2} \, dx,x,\frac{b+2 c x}{\sqrt{a+b x+c x^2}}\right )}{128 c^4}\\ &=-\frac{\left (7 b^3 B-10 A b^2 c-12 a b B c+8 a A c^2\right ) (b+2 c x) \sqrt{a+b x+c x^2}}{128 c^4}+\frac{B x^2 \left (a+b x+c x^2\right )^{3/2}}{5 c}+\frac{\left (35 b^2 B-50 A b c-32 a B c-6 c (7 b B-10 A c) x\right ) \left (a+b x+c x^2\right )^{3/2}}{240 c^3}+\frac{\left (b^2-4 a c\right ) \left (7 b^3 B-10 A b^2 c-12 a b B c+8 a A c^2\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{256 c^{9/2}}\\ \end{align*}

Mathematica [A]  time = 0.253531, size = 179, normalized size = 0.87 \[ \frac{\frac{(a+x (b+c x))^{3/2} \left (4 c (15 A c x-8 a B)-2 b c (25 A+21 B x)+35 b^2 B\right )}{48 c^2}+\frac{5 \left (8 a A c^2-12 a b B c-10 A b^2 c+7 b^3 B\right ) \left (\left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+x (b+c x)}}\right )-2 \sqrt{c} (b+2 c x) \sqrt{a+x (b+c x)}\right )}{256 c^{7/2}}+B x^2 (a+x (b+c x))^{3/2}}{5 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(B*x^2*(a + x*(b + c*x))^(3/2) + ((a + x*(b + c*x))^(3/2)*(35*b^2*B - 2*b*c*(25*A + 21*B*x) + 4*c*(-8*a*B + 15
*A*c*x)))/(48*c^2) + (5*(7*b^3*B - 10*A*b^2*c - 12*a*b*B*c + 8*a*A*c^2)*(-2*Sqrt[c]*(b + 2*c*x)*Sqrt[a + x*(b
+ c*x)] + (b^2 - 4*a*c)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + x*(b + c*x)])]))/(256*c^(7/2)))/(5*c)

________________________________________________________________________________________

Maple [B]  time = 0.01, size = 497, normalized size = 2.4 \begin{align*}{\frac{B{x}^{2}}{5\,c} \left ( c{x}^{2}+bx+a \right ) ^{{\frac{3}{2}}}}-{\frac{7\,bBx}{40\,{c}^{2}} \left ( c{x}^{2}+bx+a \right ) ^{{\frac{3}{2}}}}+{\frac{7\,{b}^{2}B}{48\,{c}^{3}} \left ( c{x}^{2}+bx+a \right ) ^{{\frac{3}{2}}}}-{\frac{7\,{b}^{3}Bx}{64\,{c}^{3}}\sqrt{c{x}^{2}+bx+a}}-{\frac{7\,{b}^{4}B}{128\,{c}^{4}}\sqrt{c{x}^{2}+bx+a}}-{\frac{5\,Ba{b}^{3}}{32}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx+a} \right ){c}^{-{\frac{7}{2}}}}+{\frac{7\,B{b}^{5}}{256}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx+a} \right ){c}^{-{\frac{9}{2}}}}+{\frac{3\,abBx}{16\,{c}^{2}}\sqrt{c{x}^{2}+bx+a}}+{\frac{3\,Ba{b}^{2}}{32\,{c}^{3}}\sqrt{c{x}^{2}+bx+a}}+{\frac{3\,B{a}^{2}b}{16}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx+a} \right ){c}^{-{\frac{5}{2}}}}-{\frac{2\,aB}{15\,{c}^{2}} \left ( c{x}^{2}+bx+a \right ) ^{{\frac{3}{2}}}}+{\frac{Ax}{4\,c} \left ( c{x}^{2}+bx+a \right ) ^{{\frac{3}{2}}}}-{\frac{5\,Ab}{24\,{c}^{2}} \left ( c{x}^{2}+bx+a \right ) ^{{\frac{3}{2}}}}+{\frac{5\,A{b}^{2}x}{32\,{c}^{2}}\sqrt{c{x}^{2}+bx+a}}+{\frac{5\,A{b}^{3}}{64\,{c}^{3}}\sqrt{c{x}^{2}+bx+a}}+{\frac{3\,A{b}^{2}a}{16}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx+a} \right ){c}^{-{\frac{5}{2}}}}-{\frac{5\,A{b}^{4}}{128}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx+a} \right ){c}^{-{\frac{7}{2}}}}-{\frac{aAx}{8\,c}\sqrt{c{x}^{2}+bx+a}}-{\frac{Aab}{16\,{c}^{2}}\sqrt{c{x}^{2}+bx+a}}-{\frac{A{a}^{2}}{8}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx+a} \right ){c}^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*B*x^2*(c*x^2+b*x+a)^(3/2)/c-7/40*B*b/c^2*x*(c*x^2+b*x+a)^(3/2)+7/48*B*b^2/c^3*(c*x^2+b*x+a)^(3/2)-7/64*B*b
^3/c^3*(c*x^2+b*x+a)^(1/2)*x-7/128*B*b^4/c^4*(c*x^2+b*x+a)^(1/2)-5/32*B*b^3/c^(7/2)*ln((1/2*b+c*x)/c^(1/2)+(c*
x^2+b*x+a)^(1/2))*a+7/256*B*b^5/c^(9/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))+3/16*B*b/c^2*a*(c*x^2+b*x+
a)^(1/2)*x+3/32*B*b^2/c^3*a*(c*x^2+b*x+a)^(1/2)+3/16*B*b/c^(5/2)*a^2*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2
))-2/15*B*a/c^2*(c*x^2+b*x+a)^(3/2)+1/4*A*x*(c*x^2+b*x+a)^(3/2)/c-5/24*A*b/c^2*(c*x^2+b*x+a)^(3/2)+5/32*A*b^2/
c^2*(c*x^2+b*x+a)^(1/2)*x+5/64*A*b^3/c^3*(c*x^2+b*x+a)^(1/2)+3/16*A*b^2/c^(5/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+
b*x+a)^(1/2))*a-5/128*A*b^4/c^(7/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))-1/8*A*a/c*(c*x^2+b*x+a)^(1/2)*
x-1/16*A*a/c^2*(c*x^2+b*x+a)^(1/2)*b-1/8*A*a^2/c^(3/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))

________________________________________________________________________________________

Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 1.91892, size = 1220, normalized size = 5.95 \begin{align*} \left [-\frac{15 \,{\left (7 \, B b^{5} - 32 \, A a^{2} c^{3} + 48 \,{\left (B a^{2} b + A a b^{2}\right )} c^{2} - 10 \,{\left (4 \, B a b^{3} + A b^{4}\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 (384 \, B c^{5} x^{4} - 105 \, B b^{4} c - 8 \,{\left (32 \, B a^{2} + 65 \, A a b\right )} c^{3} + 48 \,{\left (B b c^{4} + 10 \, A c^{5}\right )} x^{3} + 10 \,{\left (46 \, B a b^{2} + 15 \, A b^{3}\right )} c^{2} - 8 \,{\left (7 \, B b^{2} c^{3} - 2 \,{\left (8 \, B a + 5 \, A b\right )} c^{4}\right )} x^{2} + 2 \,{\left (35 \, B b^{3} c^{2} + 120 \, A a c^{4} - 2 \,{\left (58 \, B a b + 25 \, A b^{2}\right )} c^{3}\right )} x\right )} \sqrt{c x^{2} + b x + a}}{7680 \, c^{5}}, -\frac{15 \,{\left (7 \, B b^{5} - 32 \, A a^{2} c^{3} + 48 \,{\left (B a^{2} b + A a b^{2}\right )} c^{2} - 10 \,{\left (4 \, B a b^{3} + A b^{4}\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 (384 \, B c^{5} x^{4} - 105 \, B b^{4} c - 8 \,{\left (32 \, B a^{2} + 65 \, A a b\right )} c^{3} + 48 \,{\left (B b c^{4} + 10 \, A c^{5}\right )} x^{3} + 10 \,{\left (46 \, B a b^{2} + 15 \, A b^{3}\right )} c^{2} - 8 \,{\left (7 \, B b^{2} c^{3} - 2 \,{\left (8 \, B a + 5 \, A b\right )} c^{4}\right )} x^{2} + 2 \,{\left (35 \, B b^{3} c^{2} + 120 \, A a c^{4} - 2 \,{\left (58 \, B a b + 25 \, A b^{2}\right )} c^{3}\right )} x\right )} \sqrt{c x^{2} + b x + a}}{3840 \, c^{5}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/7680*(15*(7*B*b^5 - 32*A*a^2*c^3 + 48*(B*a^2*b + A*a*b^2)*c^2 - 10*(4*B*a*b^3 + A*b^4)*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*(384*B*c^5*x^4 - 105*B*b^4*c
- 8*(32*B*a^2 + 65*A*a*b)*c^3 + 48*(B*b*c^4 + 10*A*c^5)*x^3 + 10*(46*B*a*b^2 + 15*A*b^3)*c^2 - 8*(7*B*b^2*c^3
- 2*(8*B*a + 5*A*b)*c^4)*x^2 + 2*(35*B*b^3*c^2 + 120*A*a*c^4 - 2*(58*B*a*b + 25*A*b^2)*c^3)*x)*sqrt(c*x^2 + b
*x + a))/c^5, -1/3840*(15*(7*B*b^5 - 32*A*a^2*c^3 + 48*(B*a^2*b + A*a*b^2)*c^2 - 10*(4*B*a*b^3 + A*b^4)*c)*sqr
t(-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*(384*B*c^5*x^4 - 105*
B*b^4*c - 8*(32*B*a^2 + 65*A*a*b)*c^3 + 48*(B*b*c^4 + 10*A*c^5)*x^3 + 10*(46*B*a*b^2 + 15*A*b^3)*c^2 - 8*(7*B*
b^2*c^3 - 2*(8*B*a + 5*A*b)*c^4)*x^2 + 2*(35*B*b^3*c^2 + 120*A*a*c^4 - 2*(58*B*a*b + 25*A*b^2)*c^3)*x)*sqrt(c*
x^2 + b*x + a))/c^5]

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{2} \left (A + B x\right ) \sqrt{a + b x + c x^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**2*(A + B*x)*sqrt(a + b*x + c*x**2), x)

________________________________________________________________________________________

Giac [A]  time = 1.32766, size = 331, normalized size = 1.61 \begin{align*} \frac{1}{1920} \, \sqrt{c x^{2} + b x + a}{\left (2 \,{\left (4 \,{\left (6 \,{\left (8 \, B x + \frac{B b c^{3} + 10 \, A c^{4}}{c^{4}}\right )} x - \frac{7 \, B b^{2} c^{2} - 16 \, B a c^{3} - 10 \, A b c^{3}}{c^{4}}\right )} x + \frac{35 \, B b^{3} c - 116 \, B a b c^{2} - 50 \, A b^{2} c^{2} + 120 \, A a c^{3}}{c^{4}}\right )} x - \frac{105 \, B b^{4} - 460 \, B a b^{2} c - 150 \, A b^{3} c + 256 \, B a^{2} c^{2} + 520 \, A a b c^{2}}{c^{4}}\right )} - \frac{{\left (7 \, B b^{5} - 40 \, B a b^{3} c - 10 \, A b^{4} c + 48 \, B a^{2} b c^{2} + 48 \, A a b^{2} c^{2} - 32 \, A a^{2} c^{3}\right )} \log \left ({\left | -2 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )} \sqrt{c} - b \right |}\right )}{256 \, c^{\frac{9}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1920*sqrt(c*x^2 + b*x + a)*(2*(4*(6*(8*B*x + (B*b*c^3 + 10*A*c^4)/c^4)*x - (7*B*b^2*c^2 - 16*B*a*c^3 - 10*A*
b*c^3)/c^4)*x + (35*B*b^3*c - 116*B*a*b*c^2 - 50*A*b^2*c^2 + 120*A*a*c^3)/c^4)*x - (105*B*b^4 - 460*B*a*b^2*c
- 150*A*b^3*c + 256*B*a^2*c^2 + 520*A*a*b*c^2)/c^4) - 1/256*(7*B*b^5 - 40*B*a*b^3*c - 10*A*b^4*c + 48*B*a^2*b*
c^2 + 48*A*a*b^2*c^2 - 32*A*a^2*c^3)*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*sqrt(c) - b))/c^(9/2)