Integrand size = 22, antiderivative size = 165 \[ \int x^2 (A+B x) \sqrt {b x+c x^2} \, dx=-\frac {b^2 (7 b B-10 A c) (b+2 c x) \sqrt {b x+c x^2}}{128 c^4}+\frac {b (7 b B-10 A c) \left (b x+c x^2\right )^{3/2}}{48 c^3}-\frac {(7 b B-10 A c) x \left (b x+c x^2\right )^{3/2}}{40 c^2}+\frac {B x^2 \left (b x+c x^2\right )^{3/2}}{5 c}+\frac {b^4 (7 b B-10 A c) \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{128 c^{9/2}} \]
1/48*b*(-10*A*c+7*B*b)*(c*x^2+b*x)^(3/2)/c^3-1/40*(-10*A*c+7*B*b)*x*(c*x^2 +b*x)^(3/2)/c^2+1/5*B*x^2*(c*x^2+b*x)^(3/2)/c+1/128*b^4*(-10*A*c+7*B*b)*ar ctanh(x*c^(1/2)/(c*x^2+b*x)^(1/2))/c^(9/2)-1/128*b^2*(-10*A*c+7*B*b)*(2*c* x+b)*(c*x^2+b*x)^(1/2)/c^4
Time = 0.68 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.22 \[ \int x^2 (A+B x) \sqrt {b x+c x^2} \, dx=\frac {\sqrt {x} \sqrt {b+c x} \left (\sqrt {c} \sqrt {x} \sqrt {b+c x} \left (-105 b^4 B+16 b c^3 x^2 (5 A+3 B x)+96 c^4 x^3 (5 A+4 B x)+10 b^3 c (15 A+7 B x)-4 b^2 c^2 x (25 A+14 B x)\right )+300 A b^4 c \text {arctanh}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}-\sqrt {b+c x}}\right )+210 b^5 B \text {arctanh}\left (\frac {\sqrt {c} \sqrt {x}}{-\sqrt {b}+\sqrt {b+c x}}\right )\right )}{1920 c^{9/2} \sqrt {x (b+c x)}} \]
(Sqrt[x]*Sqrt[b + c*x]*(Sqrt[c]*Sqrt[x]*Sqrt[b + c*x]*(-105*b^4*B + 16*b*c ^3*x^2*(5*A + 3*B*x) + 96*c^4*x^3*(5*A + 4*B*x) + 10*b^3*c*(15*A + 7*B*x) - 4*b^2*c^2*x*(25*A + 14*B*x)) + 300*A*b^4*c*ArcTanh[(Sqrt[c]*Sqrt[x])/(Sq rt[b] - Sqrt[b + c*x])] + 210*b^5*B*ArcTanh[(Sqrt[c]*Sqrt[x])/(-Sqrt[b] + Sqrt[b + c*x])]))/(1920*c^(9/2)*Sqrt[x*(b + c*x)])
Time = 0.31 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.97, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {1221, 1134, 1160, 1087, 1091, 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.
| \(\displaystyle \int x^2 (A+B x) \sqrt {b x+c x^2} \, dx\) |
| \(\Big \downarrow \) 1221 |
| \(\displaystyle \frac {B x^2 \left (b x+c x^2\right )^{3/2}}{5 c}-\frac {(7 b B-10 A c) \int x^2 \sqrt {c x^2+b x}dx}{10 c}\) |
| \(\Big \downarrow \) 1134 |
| \(\displaystyle \frac {B x^2 \left (b x+c x^2\right )^{3/2}}{5 c}-\frac {(7 b B-10 A c) \left (\frac {x \left (b x+c x^2\right )^{3/2}}{4 c}-\frac {5 b \int x \sqrt {c x^2+b x}dx}{8 c}\right )}{10 c}\) |
| \(\Big \downarrow \) 1160 |
| \(\displaystyle \frac {B x^2 \left (b x+c x^2\right )^{3/2}}{5 c}-\frac {(7 b B-10 A c) \left (\frac {x \left (b x+c x^2\right )^{3/2}}{4 c}-\frac {5 b \left (\frac {\left (b x+c x^2\right )^{3/2}}{3 c}-\frac {b \int \sqrt {c x^2+b x}dx}{2 c}\right )}{8 c}\right )}{10 c}\) |
| \(\Big \downarrow \) 1087 |
| \(\displaystyle \frac {B x^2 \left (b x+c x^2\right )^{3/2}}{5 c}-\frac {(7 b B-10 A c) \left (\frac {x \left (b x+c x^2\right )^{3/2}}{4 c}-\frac {5 b \left (\frac {\left (b x+c x^2\right )^{3/2}}{3 c}-\frac {b \left (\frac {(b+2 c x) \sqrt {b x+c x^2}}{4 c}-\frac {b^2 \int \frac {1}{\sqrt {c x^2+b x}}dx}{8 c}\right )}{2 c}\right )}{8 c}\right )}{10 c}\) |
| \(\Big \downarrow \) 1091 |
| \(\displaystyle \frac {B x^2 \left (b x+c x^2\right )^{3/2}}{5 c}-\frac {(7 b B-10 A c) \left (\frac {x \left (b x+c x^2\right )^{3/2}}{4 c}-\frac {5 b \left (\frac {\left (b x+c x^2\right )^{3/2}}{3 c}-\frac {b \left (\frac {(b+2 c x) \sqrt {b x+c x^2}}{4 c}-\frac {b^2 \int \frac {1}{1-\frac {c x^2}{c x^2+b x}}d\frac {x}{\sqrt {c x^2+b x}}}{4 c}\right )}{2 c}\right )}{8 c}\right )}{10 c}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {B x^2 \left (b x+c x^2\right )^{3/2}}{5 c}-\frac {(7 b B-10 A c) \left (\frac {x \left (b x+c x^2\right )^{3/2}}{4 c}-\frac {5 b \left (\frac {\left (b x+c x^2\right )^{3/2}}{3 c}-\frac {b \left (\frac {(b+2 c x) \sqrt {b x+c x^2}}{4 c}-\frac {b^2 \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{4 c^{3/2}}\right )}{2 c}\right )}{8 c}\right )}{10 c}\) |
(B*x^2*(b*x + c*x^2)^(3/2))/(5*c) - ((7*b*B - 10*A*c)*((x*(b*x + c*x^2)^(3 /2))/(4*c) - (5*b*((b*x + c*x^2)^(3/2)/(3*c) - (b*(((b + 2*c*x)*Sqrt[b*x + c*x^2])/(4*c) - (b^2*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/(4*c^(3/2))) )/(2*c)))/(8*c)))/(10*c)
3.1.68.3.1 Defintions of rubi rules used
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[(b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Simp[2 Subst[Int[1/(1 - c*x^2), x], x, x/Sqrt[b*x + c*x^2]], x] /; FreeQ[{b, c}, x]
Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S ymbol] :> Simp[e*(d + e*x)^(m - 1)*((a + b*x + c*x^2)^(p + 1)/(c*(m + 2*p + 1))), x] + Simp[(m + p)*((2*c*d - b*e)/(c*(m + 2*p + 1))) Int[(d + e*x)^ (m - 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[ c*d^2 - b*d*e + a*e^2, 0] && GtQ[m, 1] && NeQ[m + 2*p + 1, 0] && IntegerQ[2 *p]
Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol ] :> Simp[e*((a + b*x + c*x^2)^(p + 1)/(2*c*(p + 1))), x] + Simp[(2*c*d - b *e)/(2*c) Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[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[(m*(g*(c*d - b*e) + c*e*f) + e*(p + 1)*(2*c *f - b*g))/(c*e*(m + 2*p + 2)) Int[(d + e*x)^m*(a + b*x + c*x^2)^p, x], x ] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[m + 2*p + 2, 0]
Time = 0.18 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.70
| method | result | size |
| pseudoelliptic | \(-\frac {5 \left (\frac {3 \left (A \,b^{4} c -\frac {7}{10} B \,b^{5}\right ) \operatorname {arctanh}\left (\frac {\sqrt {x \left (c x +b \right )}}{x \sqrt {c}}\right )}{2}+\sqrt {x \left (c x +b \right )}\, \left (-\frac {3 \left (\frac {7 B x}{15}+A \right ) b^{3} c^{\frac {3}{2}}}{2}+b^{2} x \left (\frac {14 B x}{25}+A \right ) c^{\frac {5}{2}}-\frac {4 \left (\frac {3 B x}{5}+A \right ) x^{2} b \,c^{\frac {7}{2}}}{5}-\frac {24 \left (\frac {4 B x}{5}+A \right ) x^{3} c^{\frac {9}{2}}}{5}+\frac {21 B \,b^{4} \sqrt {c}}{20}\right )\right )}{96 c^{\frac {9}{2}}}\) | \(116\) |
| risch | \(\frac {\left (384 B \,c^{4} x^{4}+480 A \,c^{4} x^{3}+48 B b \,c^{3} x^{3}+80 A b \,c^{3} x^{2}-56 B \,b^{2} c^{2} x^{2}-100 A \,b^{2} c^{2} x +70 B \,b^{3} c x +150 A \,b^{3} c -105 b^{4} B \right ) x \left (c x +b \right )}{1920 c^{4} \sqrt {x \left (c x +b \right )}}-\frac {b^{4} \left (10 A c -7 B b \right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{256 c^{\frac {9}{2}}}\) | \(145\) |
| default | \(B \left (\frac {x^{2} \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{5 c}-\frac {7 b \left (\frac {x \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{4 c}-\frac {5 b \left (\frac {\left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{3 c}-\frac {b \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x}}{4 c}-\frac {b^{2} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{8 c^{\frac {3}{2}}}\right )}{2 c}\right )}{8 c}\right )}{10 c}\right )+A \left (\frac {x \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{4 c}-\frac {5 b \left (\frac {\left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{3 c}-\frac {b \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x}}{4 c}-\frac {b^{2} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{8 c^{\frac {3}{2}}}\right )}{2 c}\right )}{8 c}\right )\) | \(236\) |
-5/96/c^(9/2)*(3/2*(A*b^4*c-7/10*B*b^5)*arctanh((x*(c*x+b))^(1/2)/x/c^(1/2 ))+(x*(c*x+b))^(1/2)*(-3/2*(7/15*B*x+A)*b^3*c^(3/2)+b^2*x*(14/25*B*x+A)*c^ (5/2)-4/5*(3/5*B*x+A)*x^2*b*c^(7/2)-24/5*(4/5*B*x+A)*x^3*c^(9/2)+21/20*B*b ^4*c^(1/2)))
Time = 0.26 (sec) , antiderivative size = 302, normalized size of antiderivative = 1.83 \[ \int x^2 (A+B x) \sqrt {b x+c x^2} \, dx=\left [-\frac {15 \, {\left (7 \, B b^{5} - 10 \, A b^{4} c\right )} \sqrt {c} \log \left (2 \, c x + b - 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) - 2 \, {\left (384 \, B c^{5} x^{4} - 105 \, B b^{4} c + 150 \, A b^{3} c^{2} + 48 \, {\left (B b c^{4} + 10 \, A c^{5}\right )} x^{3} - 8 \, {\left (7 \, B b^{2} c^{3} - 10 \, A b c^{4}\right )} x^{2} + 10 \, {\left (7 \, B b^{3} c^{2} - 10 \, A b^{2} c^{3}\right )} x\right )} \sqrt {c x^{2} + b x}}{3840 \, c^{5}}, -\frac {15 \, {\left (7 \, B b^{5} - 10 \, A b^{4} c\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) - {\left (384 \, B c^{5} x^{4} - 105 \, B b^{4} c + 150 \, A b^{3} c^{2} + 48 \, {\left (B b c^{4} + 10 \, A c^{5}\right )} x^{3} - 8 \, {\left (7 \, B b^{2} c^{3} - 10 \, A b c^{4}\right )} x^{2} + 10 \, {\left (7 \, B b^{3} c^{2} - 10 \, A b^{2} c^{3}\right )} x\right )} \sqrt {c x^{2} + b x}}{1920 \, c^{5}}\right ] \]
[-1/3840*(15*(7*B*b^5 - 10*A*b^4*c)*sqrt(c)*log(2*c*x + b - 2*sqrt(c*x^2 + b*x)*sqrt(c)) - 2*(384*B*c^5*x^4 - 105*B*b^4*c + 150*A*b^3*c^2 + 48*(B*b* c^4 + 10*A*c^5)*x^3 - 8*(7*B*b^2*c^3 - 10*A*b*c^4)*x^2 + 10*(7*B*b^3*c^2 - 10*A*b^2*c^3)*x)*sqrt(c*x^2 + b*x))/c^5, -1/1920*(15*(7*B*b^5 - 10*A*b^4* c)*sqrt(-c)*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) - (384*B*c^5*x^4 - 10 5*B*b^4*c + 150*A*b^3*c^2 + 48*(B*b*c^4 + 10*A*c^5)*x^3 - 8*(7*B*b^2*c^3 - 10*A*b*c^4)*x^2 + 10*(7*B*b^3*c^2 - 10*A*b^2*c^3)*x)*sqrt(c*x^2 + b*x))/c ^5]
Time = 0.47 (sec) , antiderivative size = 238, normalized size of antiderivative = 1.44 \[ \int x^2 (A+B x) \sqrt {b x+c x^2} \, dx=\begin {cases} - \frac {5 b^{3} \left (A b - \frac {7 b \left (A c + \frac {B b}{10}\right )}{8 c}\right ) \left (\begin {cases} \frac {\log {\left (b + 2 \sqrt {c} \sqrt {b x + c x^{2}} + 2 c x \right )}}{\sqrt {c}} & \text {for}\: \frac {b^{2}}{c} \neq 0 \\\frac {\left (\frac {b}{2 c} + x\right ) \log {\left (\frac {b}{2 c} + x \right )}}{\sqrt {c \left (\frac {b}{2 c} + x\right )^{2}}} & \text {otherwise} \end {cases}\right )}{16 c^{3}} + \sqrt {b x + c x^{2}} \left (\frac {B x^{4}}{5} + \frac {5 b^{2} \left (A b - \frac {7 b \left (A c + \frac {B b}{10}\right )}{8 c}\right )}{8 c^{3}} - \frac {5 b x \left (A b - \frac {7 b \left (A c + \frac {B b}{10}\right )}{8 c}\right )}{12 c^{2}} + \frac {x^{3} \left (A c + \frac {B b}{10}\right )}{4 c} + \frac {x^{2} \left (A b - \frac {7 b \left (A c + \frac {B b}{10}\right )}{8 c}\right )}{3 c}\right ) & \text {for}\: c \neq 0 \\\frac {2 \left (\frac {A \left (b x\right )^{\frac {7}{2}}}{7} + \frac {B \left (b x\right )^{\frac {9}{2}}}{9 b}\right )}{b^{3}} & \text {for}\: b \neq 0 \\0 & \text {otherwise} \end {cases} \]
Piecewise((-5*b**3*(A*b - 7*b*(A*c + B*b/10)/(8*c))*Piecewise((log(b + 2*s qrt(c)*sqrt(b*x + c*x**2) + 2*c*x)/sqrt(c), Ne(b**2/c, 0)), ((b/(2*c) + x) *log(b/(2*c) + x)/sqrt(c*(b/(2*c) + x)**2), True))/(16*c**3) + sqrt(b*x + c*x**2)*(B*x**4/5 + 5*b**2*(A*b - 7*b*(A*c + B*b/10)/(8*c))/(8*c**3) - 5*b *x*(A*b - 7*b*(A*c + B*b/10)/(8*c))/(12*c**2) + x**3*(A*c + B*b/10)/(4*c) + x**2*(A*b - 7*b*(A*c + B*b/10)/(8*c))/(3*c)), Ne(c, 0)), (2*(A*(b*x)**(7 /2)/7 + B*(b*x)**(9/2)/(9*b))/b**3, Ne(b, 0)), (0, True))
Time = 0.18 (sec) , antiderivative size = 242, normalized size of antiderivative = 1.47 \[ \int x^2 (A+B x) \sqrt {b x+c x^2} \, dx=\frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}} B x^{2}}{5 \, c} - \frac {7 \, \sqrt {c x^{2} + b x} B b^{3} x}{64 \, c^{3}} - \frac {7 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} B b x}{40 \, c^{2}} + \frac {5 \, \sqrt {c x^{2} + b x} A b^{2} x}{32 \, c^{2}} + \frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}} A x}{4 \, c} + \frac {7 \, B b^{5} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{256 \, c^{\frac {9}{2}}} - \frac {5 \, A b^{4} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{128 \, c^{\frac {7}{2}}} - \frac {7 \, \sqrt {c x^{2} + b x} B b^{4}}{128 \, c^{4}} + \frac {7 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} B b^{2}}{48 \, c^{3}} + \frac {5 \, \sqrt {c x^{2} + b x} A b^{3}}{64 \, c^{3}} - \frac {5 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} A b}{24 \, c^{2}} \]
1/5*(c*x^2 + b*x)^(3/2)*B*x^2/c - 7/64*sqrt(c*x^2 + b*x)*B*b^3*x/c^3 - 7/4 0*(c*x^2 + b*x)^(3/2)*B*b*x/c^2 + 5/32*sqrt(c*x^2 + b*x)*A*b^2*x/c^2 + 1/4 *(c*x^2 + b*x)^(3/2)*A*x/c + 7/256*B*b^5*log(2*c*x + b + 2*sqrt(c*x^2 + b* x)*sqrt(c))/c^(9/2) - 5/128*A*b^4*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt (c))/c^(7/2) - 7/128*sqrt(c*x^2 + b*x)*B*b^4/c^4 + 7/48*(c*x^2 + b*x)^(3/2 )*B*b^2/c^3 + 5/64*sqrt(c*x^2 + b*x)*A*b^3/c^3 - 5/24*(c*x^2 + b*x)^(3/2)* A*b/c^2
Time = 0.27 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.96 \[ \int x^2 (A+B x) \sqrt {b x+c x^2} \, dx=\frac {1}{1920} \, \sqrt {c x^{2} + b x} {\left (2 \, {\left (4 \, {\left (6 \, {\left (8 \, B x + \frac {B b c^{3} + 10 \, A c^{4}}{c^{4}}\right )} x - \frac {7 \, B b^{2} c^{2} - 10 \, A b c^{3}}{c^{4}}\right )} x + \frac {5 \, {\left (7 \, B b^{3} c - 10 \, A b^{2} c^{2}\right )}}{c^{4}}\right )} x - \frac {15 \, {\left (7 \, B b^{4} - 10 \, A b^{3} c\right )}}{c^{4}}\right )} - \frac {{\left (7 \, B b^{5} - 10 \, A b^{4} c\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} + b \right |}\right )}{256 \, c^{\frac {9}{2}}} \]
1/1920*sqrt(c*x^2 + b*x)*(2*(4*(6*(8*B*x + (B*b*c^3 + 10*A*c^4)/c^4)*x - ( 7*B*b^2*c^2 - 10*A*b*c^3)/c^4)*x + 5*(7*B*b^3*c - 10*A*b^2*c^2)/c^4)*x - 1 5*(7*B*b^4 - 10*A*b^3*c)/c^4) - 1/256*(7*B*b^5 - 10*A*b^4*c)*log(abs(2*(sq rt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) + b))/c^(9/2)
Time = 10.49 (sec) , antiderivative size = 215, normalized size of antiderivative = 1.30 \[ \int x^2 (A+B x) \sqrt {b x+c x^2} \, dx=\frac {A\,x\,{\left (c\,x^2+b\,x\right )}^{3/2}}{4\,c}-\frac {5\,A\,b\,\left (\frac {b^3\,\ln \left (\frac {b+2\,c\,x}{\sqrt {c}}+2\,\sqrt {c\,x^2+b\,x}\right )}{16\,c^{5/2}}+\frac {\sqrt {c\,x^2+b\,x}\,\left (-3\,b^2+2\,b\,c\,x+8\,c^2\,x^2\right )}{24\,c^2}\right )}{8\,c}-\frac {7\,B\,b\,\left (\frac {x\,{\left (c\,x^2+b\,x\right )}^{3/2}}{4\,c}-\frac {5\,b\,\left (\frac {b^3\,\ln \left (\frac {b+2\,c\,x}{\sqrt {c}}+2\,\sqrt {c\,x^2+b\,x}\right )}{16\,c^{5/2}}+\frac {\sqrt {c\,x^2+b\,x}\,\left (-3\,b^2+2\,b\,c\,x+8\,c^2\,x^2\right )}{24\,c^2}\right )}{8\,c}\right )}{10\,c}+\frac {B\,x^2\,{\left (c\,x^2+b\,x\right )}^{3/2}}{5\,c} \]
(A*x*(b*x + c*x^2)^(3/2))/(4*c) - (5*A*b*((b^3*log((b + 2*c*x)/c^(1/2) + 2 *(b*x + c*x^2)^(1/2)))/(16*c^(5/2)) + ((b*x + c*x^2)^(1/2)*(8*c^2*x^2 - 3* b^2 + 2*b*c*x))/(24*c^2)))/(8*c) - (7*B*b*((x*(b*x + c*x^2)^(3/2))/(4*c) - (5*b*((b^3*log((b + 2*c*x)/c^(1/2) + 2*(b*x + c*x^2)^(1/2)))/(16*c^(5/2)) + ((b*x + c*x^2)^(1/2)*(8*c^2*x^2 - 3*b^2 + 2*b*c*x))/(24*c^2)))/(8*c)))/ (10*c) + (B*x^2*(b*x + c*x^2)^(3/2))/(5*c)