Integrand size = 23, antiderivative size = 205 \[ \int x^2 (A+B x) \sqrt {a+b x+c x^2} \, dx=-\frac {\left (7 b^3 B-10 A b^2 c-12 a b B c+8 a A c^2\right ) (b+2 c x) \sqrt {a+b x+c x^2}}{128 c^4}+\frac {B x^2 \left (a+b x+c x^2\right )^{3/2}}{5 c}+\frac {\left (35 b^2 B-50 A b c-32 a B c-6 c (7 b B-10 A c) x\right ) \left (a+b x+c x^2\right )^{3/2}}{240 c^3}+\frac {\left (b^2-4 a c\right ) \left (7 b^3 B-10 A b^2 c-12 a b B c+8 a A c^2\right ) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{256 c^{9/2}} \]
1/5*B*x^2*(c*x^2+b*x+a)^(3/2)/c+1/240*(35*B*b^2-50*A*b*c-32*B*a*c-6*c*(-10 *A*c+7*B*b)*x)*(c*x^2+b*x+a)^(3/2)/c^3+1/256*(-4*a*c+b^2)*(8*A*a*c^2-10*A* b^2*c-12*B*a*b*c+7*B*b^3)*arctanh(1/2*(2*c*x+b)/c^(1/2)/(c*x^2+b*x+a)^(1/2 ))/c^(9/2)-1/128*(8*A*a*c^2-10*A*b^2*c-12*B*a*b*c+7*B*b^3)*(2*c*x+b)*(c*x^ 2+b*x+a)^(1/2)/c^4
Time = 0.60 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.02 \[ \int x^2 (A+B x) \sqrt {a+b x+c x^2} \, dx=\frac {2 \sqrt {c} \sqrt {a+x (b+c x)} \left (-105 b^4 B+10 b^3 c (15 A+7 B x)+16 c^2 \left (-16 a^2 B+6 c^2 x^3 (5 A+4 B x)+a c x (15 A+8 B x)\right )+4 b^2 c (115 a B-c x (25 A+14 B x))+8 b c^2 \left (2 c x^2 (5 A+3 B x)-a (65 A+29 B x)\right )\right )-15 \left (b^2-4 a c\right ) \left (7 b^3 B-10 A b^2 c-12 a b B c+8 a A c^2\right ) \log \left (b+2 c x-2 \sqrt {c} \sqrt {a+x (b+c x)}\right )}{3840 c^{9/2}} \]
(2*Sqrt[c]*Sqrt[a + x*(b + c*x)]*(-105*b^4*B + 10*b^3*c*(15*A + 7*B*x) + 1 6*c^2*(-16*a^2*B + 6*c^2*x^3*(5*A + 4*B*x) + a*c*x*(15*A + 8*B*x)) + 4*b^2 *c*(115*a*B - c*x*(25*A + 14*B*x)) + 8*b*c^2*(2*c*x^2*(5*A + 3*B*x) - a*(6 5*A + 29*B*x))) - 15*(b^2 - 4*a*c)*(7*b^3*B - 10*A*b^2*c - 12*a*b*B*c + 8* a*A*c^2)*Log[b + 2*c*x - 2*Sqrt[c]*Sqrt[a + x*(b + c*x)]])/(3840*c^(9/2))
Time = 0.37 (sec) , antiderivative size = 194, normalized size of antiderivative = 0.95, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {1236, 27, 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.
| \(\displaystyle \int x^2 (A+B x) \sqrt {a+b x+c x^2} \, dx\) |
| \(\Big \downarrow \) 1236 |
| \(\displaystyle \frac {\int -\frac {1}{2} x (4 a B+(7 b B-10 A c) x) \sqrt {c x^2+b x+a}dx}{5 c}+\frac {B x^2 \left (a+b x+c x^2\right )^{3/2}}{5 c}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {B x^2 \left (a+b x+c x^2\right )^{3/2}}{5 c}-\frac {\int x (4 a B+(7 b B-10 A c) x) \sqrt {c x^2+b x+a}dx}{10 c}\) |
| \(\Big \downarrow \) 1225 |
| \(\displaystyle \frac {B x^2 \left (a+b x+c x^2\right )^{3/2}}{5 c}-\frac {\frac {5 \left (8 a A c^2-12 a b B c-10 A b^2 c+7 b^3 B\right ) \int \sqrt {c x^2+b x+a}dx}{16 c^2}-\frac {\left (a+b x+c x^2\right )^{3/2} \left (-32 a B c-6 c x (7 b B-10 A c)-50 A b c+35 b^2 B\right )}{24 c^2}}{10 c}\) |
| \(\Big \downarrow \) 1087 |
| \(\displaystyle \frac {B x^2 \left (a+b x+c x^2\right )^{3/2}}{5 c}-\frac {\frac {5 \left (8 a A c^2-12 a b B c-10 A b^2 c+7 b^3 B\right ) \left (\frac {(b+2 c x) \sqrt {a+b x+c x^2}}{4 c}-\frac {\left (b^2-4 a c\right ) \int \frac {1}{\sqrt {c x^2+b x+a}}dx}{8 c}\right )}{16 c^2}-\frac {\left (a+b x+c x^2\right )^{3/2} \left (-32 a B c-6 c x (7 b B-10 A c)-50 A b c+35 b^2 B\right )}{24 c^2}}{10 c}\) |
| \(\Big \downarrow \) 1092 |
| \(\displaystyle \frac {B x^2 \left (a+b x+c x^2\right )^{3/2}}{5 c}-\frac {\frac {5 \left (8 a A c^2-12 a b B c-10 A b^2 c+7 b^3 B\right ) \left (\frac {(b+2 c x) \sqrt {a+b x+c x^2}}{4 c}-\frac {\left (b^2-4 a c\right ) \int \frac {1}{4 c-\frac {(b+2 c x)^2}{c x^2+b x+a}}d\frac {b+2 c x}{\sqrt {c x^2+b x+a}}}{4 c}\right )}{16 c^2}-\frac {\left (a+b x+c x^2\right )^{3/2} \left (-32 a B c-6 c x (7 b B-10 A c)-50 A b c+35 b^2 B\right )}{24 c^2}}{10 c}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {B x^2 \left (a+b x+c x^2\right )^{3/2}}{5 c}-\frac {\frac {5 \left (8 a A c^2-12 a b B c-10 A b^2 c+7 b^3 B\right ) \left (\frac {(b+2 c x) \sqrt {a+b x+c x^2}}{4 c}-\frac {\left (b^2-4 a c\right ) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{8 c^{3/2}}\right )}{16 c^2}-\frac {\left (a+b x+c x^2\right )^{3/2} \left (-32 a B c-6 c x (7 b B-10 A c)-50 A b c+35 b^2 B\right )}{24 c^2}}{10 c}\) |
(B*x^2*(a + b*x + c*x^2)^(3/2))/(5*c) - (-1/24*((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))/c^2 + (5*(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 ])/(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))/(10*c)
3.10.12.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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])
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])
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]
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]
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c _.)*(x_)^2)^(p_.), x_Symbol] :> Simp[g*(d + e*x)^m*((a + b*x + c*x^2)^(p + 1)/(c*(m + 2*p + 2))), x] + Simp[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] && GtQ[m, 0] && NeQ[m + 2*p + 2, 0] && (Intege rQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p]) && !(IGtQ[m, 0] && EqQ[f, 0])
Time = 0.50 (sec) , antiderivative size = 234, normalized size of antiderivative = 1.14
| method | result | size |
| risch | \(-\frac {\left (-384 B \,x^{4} c^{4}-480 A \,c^{4} x^{3}-48 B b \,c^{3} x^{3}-80 A b \,c^{3} x^{2}-128 B a \,c^{3} x^{2}+56 B \,b^{2} c^{2} x^{2}-240 A a \,c^{3} x +100 A \,b^{2} c^{2} x +232 B a b \,c^{2} x -70 B \,b^{3} c x +520 a A b \,c^{2}-150 A \,b^{3} c +256 a^{2} B \,c^{2}-460 a \,b^{2} B c +105 b^{4} B \right ) \sqrt {c \,x^{2}+b x +a}}{1920 c^{4}}-\frac {\left (32 A \,a^{2} c^{3}-48 A a \,b^{2} c^{2}+10 A \,b^{4} c -48 B \,a^{2} b \,c^{2}+40 B a \,b^{3} c -7 B \,b^{5}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{256 c^{\frac {9}{2}}}\) | \(234\) |
| default | \(B \left (\frac {x^{2} \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}{5 c}-\frac {7 b \left (\frac {x \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}{4 c}-\frac {5 b \left (\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}{3 c}-\frac {b \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x +a}}{4 c}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 c^{\frac {3}{2}}}\right )}{2 c}\right )}{8 c}-\frac {a \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x +a}}{4 c}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 c^{\frac {3}{2}}}\right )}{4 c}\right )}{10 c}-\frac {2 a \left (\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}{3 c}-\frac {b \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x +a}}{4 c}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 c^{\frac {3}{2}}}\right )}{2 c}\right )}{5 c}\right )+A \left (\frac {x \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}{4 c}-\frac {5 b \left (\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}{3 c}-\frac {b \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x +a}}{4 c}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 c^{\frac {3}{2}}}\right )}{2 c}\right )}{8 c}-\frac {a \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x +a}}{4 c}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 c^{\frac {3}{2}}}\right )}{4 c}\right )\) | \(493\) |
-1/1920*(-384*B*c^4*x^4-480*A*c^4*x^3-48*B*b*c^3*x^3-80*A*b*c^3*x^2-128*B* a*c^3*x^2+56*B*b^2*c^2*x^2-240*A*a*c^3*x+100*A*b^2*c^2*x+232*B*a*b*c^2*x-7 0*B*b^3*c*x+520*A*a*b*c^2-150*A*b^3*c+256*B*a^2*c^2-460*B*a*b^2*c+105*B*b^ 4)*(c*x^2+b*x+a)^(1/2)/c^4-1/256*(32*A*a^2*c^3-48*A*a*b^2*c^2+10*A*b^4*c-4 8*B*a^2*b*c^2+40*B*a*b^3*c-7*B*b^5)/c^(9/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+ b*x+a)^(1/2))
Time = 0.31 (sec) , antiderivative size = 517, normalized size of antiderivative = 2.52 \[ \int x^2 (A+B x) \sqrt {a+b x+c x^2} \, dx=\left [-\frac {15 \, {\left (7 \, B b^{5} - 32 \, A a^{2} c^{3} + 48 \, {\left (B a^{2} b + A a b^{2}\right )} c^{2} - 10 \, {\left (4 \, B a b^{3} + A b^{4}\right )} c\right )} \sqrt {c} \log \left (-8 \, c^{2} x^{2} - 8 \, b c x - b^{2} + 4 \, \sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {c} - 4 \, a c\right ) - 4 \, {\left (384 \, B c^{5} x^{4} - 105 \, B b^{4} c - 8 \, {\left (32 \, B a^{2} + 65 \, A a b\right )} c^{3} + 48 \, {\left (B b c^{4} + 10 \, A c^{5}\right )} x^{3} + 10 \, {\left (46 \, B a b^{2} + 15 \, A b^{3}\right )} c^{2} - 8 \, {\left (7 \, B b^{2} c^{3} - 2 \, {\left (8 \, B a + 5 \, A b\right )} c^{4}\right )} x^{2} + 2 \, {\left (35 \, B b^{3} c^{2} + 120 \, A a c^{4} - 2 \, {\left (58 \, B a b + 25 \, A b^{2}\right )} c^{3}\right )} x\right )} \sqrt {c x^{2} + b x + a}}{7680 \, c^{5}}, -\frac {15 \, {\left (7 \, B b^{5} - 32 \, A a^{2} c^{3} + 48 \, {\left (B a^{2} b + A a b^{2}\right )} c^{2} - 10 \, {\left (4 \, B a b^{3} + A b^{4}\right )} c\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{2} + b c x + a c\right )}}\right ) - 2 \, {\left (384 \, B c^{5} x^{4} - 105 \, B b^{4} c - 8 \, {\left (32 \, B a^{2} + 65 \, A a b\right )} c^{3} + 48 \, {\left (B b c^{4} + 10 \, A c^{5}\right )} x^{3} + 10 \, {\left (46 \, B a b^{2} + 15 \, A b^{3}\right )} c^{2} - 8 \, {\left (7 \, B b^{2} c^{3} - 2 \, {\left (8 \, B a + 5 \, A b\right )} c^{4}\right )} x^{2} + 2 \, {\left (35 \, B b^{3} c^{2} + 120 \, A a c^{4} - 2 \, {\left (58 \, B a b + 25 \, A b^{2}\right )} c^{3}\right )} x\right )} \sqrt {c x^{2} + b x + a}}{3840 \, c^{5}}\right ] \]
[-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)*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*(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]
Leaf count of result is larger than twice the leaf count of optimal. 539 vs. \(2 (214) = 428\).
Time = 0.62 (sec) , antiderivative size = 539, normalized size of antiderivative = 2.63 \[ \int x^2 (A+B x) \sqrt {a+b x+c x^2} \, dx=\begin {cases} \left (- \frac {a \left (A a - \frac {3 a \left (A c + \frac {B b}{10}\right )}{4 c} - \frac {5 b \left (A b + \frac {B a}{5} - \frac {7 b \left (A c + \frac {B b}{10}\right )}{8 c}\right )}{6 c}\right )}{2 c} - \frac {b \left (- \frac {2 a \left (A b + \frac {B a}{5} - \frac {7 b \left (A c + \frac {B b}{10}\right )}{8 c}\right )}{3 c} - \frac {3 b \left (A a - \frac {3 a \left (A c + \frac {B b}{10}\right )}{4 c} - \frac {5 b \left (A b + \frac {B a}{5} - \frac {7 b \left (A c + \frac {B b}{10}\right )}{8 c}\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^{4}}{5} + \frac {x^{3} \left (A c + \frac {B b}{10}\right )}{4 c} + \frac {x^{2} \left (A b + \frac {B a}{5} - \frac {7 b \left (A c + \frac {B b}{10}\right )}{8 c}\right )}{3 c} + \frac {x \left (A a - \frac {3 a \left (A c + \frac {B b}{10}\right )}{4 c} - \frac {5 b \left (A b + \frac {B a}{5} - \frac {7 b \left (A c + \frac {B b}{10}\right )}{8 c}\right )}{6 c}\right )}{2 c} + \frac {- \frac {2 a \left (A b + \frac {B a}{5} - \frac {7 b \left (A c + \frac {B b}{10}\right )}{8 c}\right )}{3 c} - \frac {3 b \left (A a - \frac {3 a \left (A c + \frac {B b}{10}\right )}{4 c} - \frac {5 b \left (A b + \frac {B a}{5} - \frac {7 b \left (A c + \frac {B b}{10}\right )}{8 c}\right )}{6 c}\right )}{4 c}}{c}\right ) & \text {for}\: c \neq 0 \\\frac {2 \left (\frac {B \left (a + b x\right )^{\frac {9}{2}}}{9 b} + \frac {\left (a + b x\right )^{\frac {7}{2}} \left (A b - 3 B a\right )}{7 b} + \frac {\left (a + b x\right )^{\frac {5}{2}} \left (- 2 A a b + 3 B a^{2}\right )}{5 b} + \frac {\left (a + b x\right )^{\frac {3}{2}} \left (A a^{2} b - B a^{3}\right )}{3 b}\right )}{b^{3}} & \text {for}\: b \neq 0 \\\sqrt {a} \left (\frac {A x^{3}}{3} + \frac {B x^{4}}{4}\right ) & \text {otherwise} \end {cases} \]
Piecewise(((-a*(A*a - 3*a*(A*c + B*b/10)/(4*c) - 5*b*(A*b + B*a/5 - 7*b*(A *c + B*b/10)/(8*c))/(6*c))/(2*c) - b*(-2*a*(A*b + B*a/5 - 7*b*(A*c + B*b/1 0)/(8*c))/(3*c) - 3*b*(A*a - 3*a*(A*c + B*b/10)/(4*c) - 5*b*(A*b + B*a/5 - 7*b*(A*c + B*b/10)/(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*x**4/5 + x**3*(A*c + B*b/10)/(4*c) + x**2*(A*b + B*a/5 - 7*b* (A*c + B*b/10)/(8*c))/(3*c) + x*(A*a - 3*a*(A*c + B*b/10)/(4*c) - 5*b*(A*b + B*a/5 - 7*b*(A*c + B*b/10)/(8*c))/(6*c))/(2*c) + (-2*a*(A*b + B*a/5 - 7 *b*(A*c + B*b/10)/(8*c))/(3*c) - 3*b*(A*a - 3*a*(A*c + B*b/10)/(4*c) - 5*b *(A*b + B*a/5 - 7*b*(A*c + B*b/10)/(8*c))/(6*c))/(4*c))/c), Ne(c, 0)), (2* (B*(a + b*x)**(9/2)/(9*b) + (a + b*x)**(7/2)*(A*b - 3*B*a)/(7*b) + (a + b* x)**(5/2)*(-2*A*a*b + 3*B*a**2)/(5*b) + (a + b*x)**(3/2)*(A*a**2*b - B*a** 3)/(3*b))/b**3, Ne(b, 0)), (sqrt(a)*(A*x**3/3 + B*x**4/4), True))
Exception generated. \[ \int x^2 (A+B x) \sqrt {a+b x+c x^2} \, dx=\text {Exception raised: ValueError} \]
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
Time = 0.30 (sec) , antiderivative size = 243, normalized size of antiderivative = 1.19 \[ \int x^2 (A+B x) \sqrt {a+b x+c x^2} \, dx=\frac {1}{1920} \, \sqrt {c x^{2} + b x + a} {\left (2 \, {\left (4 \, {\left (6 \, {\left (8 \, B x + \frac {B b c^{3} + 10 \, A c^{4}}{c^{4}}\right )} x - \frac {7 \, B b^{2} c^{2} - 16 \, B a c^{3} - 10 \, A b c^{3}}{c^{4}}\right )} x + \frac {35 \, B b^{3} c - 116 \, B a b c^{2} - 50 \, A b^{2} c^{2} + 120 \, A a c^{3}}{c^{4}}\right )} x - \frac {105 \, B b^{4} - 460 \, B a b^{2} c - 150 \, A b^{3} c + 256 \, B a^{2} c^{2} + 520 \, A a b c^{2}}{c^{4}}\right )} - \frac {{\left (7 \, B b^{5} - 40 \, B a b^{3} c - 10 \, A b^{4} c + 48 \, B a^{2} b c^{2} + 48 \, A a b^{2} c^{2} - 32 \, A a^{2} c^{3}\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} \sqrt {c} + b \right |}\right )}{256 \, c^{\frac {9}{2}}} \]
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)
Time = 10.30 (sec) , antiderivative size = 463, normalized size of antiderivative = 2.26 \[ \int x^2 (A+B x) \sqrt {a+b x+c x^2} \, dx=\frac {B\,x^2\,{\left (c\,x^2+b\,x+a\right )}^{3/2}}{5\,c}-\frac {A\,a\,\left (\left (\frac {x}{2}+\frac {b}{4\,c}\right )\,\sqrt {c\,x^2+b\,x+a}+\frac {\ln \left (\frac {\frac {b}{2}+c\,x}{\sqrt {c}}+\sqrt {c\,x^2+b\,x+a}\right )\,\left (a\,c-\frac {b^2}{4}\right )}{2\,c^{3/2}}\right )}{4\,c}-\frac {5\,A\,b\,\left (\frac {\ln \left (\frac {b+2\,c\,x}{\sqrt {c}}+2\,\sqrt {c\,x^2+b\,x+a}\right )\,\left (b^3-4\,a\,b\,c\right )}{16\,c^{5/2}}+\frac {\left (-3\,b^2+2\,c\,x\,b+8\,c\,\left (c\,x^2+a\right )\right )\,\sqrt {c\,x^2+b\,x+a}}{24\,c^2}\right )}{8\,c}-\frac {2\,B\,a\,\left (\frac {\ln \left (\frac {b+2\,c\,x}{\sqrt {c}}+2\,\sqrt {c\,x^2+b\,x+a}\right )\,\left (b^3-4\,a\,b\,c\right )}{16\,c^{5/2}}+\frac {\left (-3\,b^2+2\,c\,x\,b+8\,c\,\left (c\,x^2+a\right )\right )\,\sqrt {c\,x^2+b\,x+a}}{24\,c^2}\right )}{5\,c}+\frac {A\,x\,{\left (c\,x^2+b\,x+a\right )}^{3/2}}{4\,c}+\frac {7\,B\,b\,\left (\frac {5\,b\,\left (\frac {\ln \left (\frac {b+2\,c\,x}{\sqrt {c}}+2\,\sqrt {c\,x^2+b\,x+a}\right )\,\left (b^3-4\,a\,b\,c\right )}{16\,c^{5/2}}+\frac {\left (-3\,b^2+2\,c\,x\,b+8\,c\,\left (c\,x^2+a\right )\right )\,\sqrt {c\,x^2+b\,x+a}}{24\,c^2}\right )}{8\,c}-\frac {x\,{\left (c\,x^2+b\,x+a\right )}^{3/2}}{4\,c}+\frac {a\,\left (\left (\frac {x}{2}+\frac {b}{4\,c}\right )\,\sqrt {c\,x^2+b\,x+a}+\frac {\ln \left (\frac {\frac {b}{2}+c\,x}{\sqrt {c}}+\sqrt {c\,x^2+b\,x+a}\right )\,\left (a\,c-\frac {b^2}{4}\right )}{2\,c^{3/2}}\right )}{4\,c}\right )}{10\,c} \]
(B*x^2*(a + b*x + c*x^2)^(3/2))/(5*c) - (A*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*A*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) - (2*B*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)) + ((8*c*(a + c*x^2) - 3*b^2 + 2*b*c*x)*(a + b*x + c*x^2)^(1/2))/(24*c^2))) /(5*c) + (A*x*(a + b*x + c*x^2)^(3/2))/(4*c) + (7*B*b*((5*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) - (x*(a + b*x + c*x^2)^(3/2))/(4*c) + (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)))/(10*c)