Integrand size = 22, antiderivative size = 200 \[ \int x^3 (A+B x) \sqrt {b x+c x^2} \, dx=\frac {7 b^3 (3 b B-4 A c) (b+2 c x) \sqrt {b x+c x^2}}{512 c^5}-\frac {7 b^2 (3 b B-4 A c) \left (b x+c x^2\right )^{3/2}}{192 c^4}+\frac {7 b (3 b B-4 A c) x \left (b x+c x^2\right )^{3/2}}{160 c^3}-\frac {(3 b B-4 A c) x^2 \left (b x+c x^2\right )^{3/2}}{20 c^2}+\frac {B x^3 \left (b x+c x^2\right )^{3/2}}{6 c}-\frac {7 b^5 (3 b B-4 A c) \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{512 c^{11/2}} \]
-7/192*b^2*(-4*A*c+3*B*b)*(c*x^2+b*x)^(3/2)/c^4+7/160*b*(-4*A*c+3*B*b)*x*( c*x^2+b*x)^(3/2)/c^3-1/20*(-4*A*c+3*B*b)*x^2*(c*x^2+b*x)^(3/2)/c^2+1/6*B*x ^3*(c*x^2+b*x)^(3/2)/c-7/512*b^5*(-4*A*c+3*B*b)*arctanh(x*c^(1/2)/(c*x^2+b *x)^(1/2))/c^(11/2)+7/512*b^3*(-4*A*c+3*B*b)*(2*c*x+b)*(c*x^2+b*x)^(1/2)/c ^5
Time = 0.84 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.88 \[ \int x^3 (A+B x) \sqrt {b x+c x^2} \, dx=\frac {\sqrt {x (b+c x)} \left (\sqrt {c} \left (315 b^5 B-210 b^4 c (2 A+B x)+64 b c^4 x^3 (3 A+2 B x)+56 b^3 c^2 x (5 A+3 B x)+256 c^5 x^4 (6 A+5 B x)-16 b^2 c^3 x^2 (14 A+9 B x)\right )+\frac {210 b^5 (3 b B-4 A c) \text {arctanh}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}-\sqrt {b+c x}}\right )}{\sqrt {x} \sqrt {b+c x}}\right )}{7680 c^{11/2}} \]
(Sqrt[x*(b + c*x)]*(Sqrt[c]*(315*b^5*B - 210*b^4*c*(2*A + B*x) + 64*b*c^4* x^3*(3*A + 2*B*x) + 56*b^3*c^2*x*(5*A + 3*B*x) + 256*c^5*x^4*(6*A + 5*B*x) - 16*b^2*c^3*x^2*(14*A + 9*B*x)) + (210*b^5*(3*b*B - 4*A*c)*ArcTanh[(Sqrt [c]*Sqrt[x])/(Sqrt[b] - Sqrt[b + c*x])])/(Sqrt[x]*Sqrt[b + c*x])))/(7680*c ^(11/2))
Time = 0.36 (sec) , antiderivative size = 192, normalized size of antiderivative = 0.96, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {1221, 1134, 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.
| \(\displaystyle \int x^3 (A+B x) \sqrt {b x+c x^2} \, dx\) |
| \(\Big \downarrow \) 1221 |
| \(\displaystyle \frac {B x^3 \left (b x+c x^2\right )^{3/2}}{6 c}-\frac {(3 b B-4 A c) \int x^3 \sqrt {c x^2+b x}dx}{4 c}\) |
| \(\Big \downarrow \) 1134 |
| \(\displaystyle \frac {B x^3 \left (b x+c x^2\right )^{3/2}}{6 c}-\frac {(3 b B-4 A c) \left (\frac {x^2 \left (b x+c x^2\right )^{3/2}}{5 c}-\frac {7 b \int x^2 \sqrt {c x^2+b x}dx}{10 c}\right )}{4 c}\) |
| \(\Big \downarrow \) 1134 |
| \(\displaystyle \frac {B x^3 \left (b x+c x^2\right )^{3/2}}{6 c}-\frac {(3 b B-4 A c) \left (\frac {x^2 \left (b x+c x^2\right )^{3/2}}{5 c}-\frac {7 b \left (\frac {x \left (b x+c x^2\right )^{3/2}}{4 c}-\frac {5 b \int x \sqrt {c x^2+b x}dx}{8 c}\right )}{10 c}\right )}{4 c}\) |
| \(\Big \downarrow \) 1160 |
| \(\displaystyle \frac {B x^3 \left (b x+c x^2\right )^{3/2}}{6 c}-\frac {(3 b B-4 A c) \left (\frac {x^2 \left (b x+c x^2\right )^{3/2}}{5 c}-\frac {7 b \left (\frac {x \left (b x+c x^2\right )^{3/2}}{4 c}-\frac {5 b \left (\frac {\left (b x+c x^2\right )^{3/2}}{3 c}-\frac {b \int \sqrt {c x^2+b x}dx}{2 c}\right )}{8 c}\right )}{10 c}\right )}{4 c}\) |
| \(\Big \downarrow \) 1087 |
| \(\displaystyle \frac {B x^3 \left (b x+c x^2\right )^{3/2}}{6 c}-\frac {(3 b B-4 A c) \left (\frac {x^2 \left (b x+c x^2\right )^{3/2}}{5 c}-\frac {7 b \left (\frac {x \left (b x+c x^2\right )^{3/2}}{4 c}-\frac {5 b \left (\frac {\left (b x+c x^2\right )^{3/2}}{3 c}-\frac {b \left (\frac {(b+2 c x) \sqrt {b x+c x^2}}{4 c}-\frac {b^2 \int \frac {1}{\sqrt {c x^2+b x}}dx}{8 c}\right )}{2 c}\right )}{8 c}\right )}{10 c}\right )}{4 c}\) |
| \(\Big \downarrow \) 1091 |
| \(\displaystyle \frac {B x^3 \left (b x+c x^2\right )^{3/2}}{6 c}-\frac {(3 b B-4 A c) \left (\frac {x^2 \left (b x+c x^2\right )^{3/2}}{5 c}-\frac {7 b \left (\frac {x \left (b x+c x^2\right )^{3/2}}{4 c}-\frac {5 b \left (\frac {\left (b x+c x^2\right )^{3/2}}{3 c}-\frac {b \left (\frac {(b+2 c x) \sqrt {b x+c x^2}}{4 c}-\frac {b^2 \int \frac {1}{1-\frac {c x^2}{c x^2+b x}}d\frac {x}{\sqrt {c x^2+b x}}}{4 c}\right )}{2 c}\right )}{8 c}\right )}{10 c}\right )}{4 c}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {B x^3 \left (b x+c x^2\right )^{3/2}}{6 c}-\frac {(3 b B-4 A c) \left (\frac {x^2 \left (b x+c x^2\right )^{3/2}}{5 c}-\frac {7 b \left (\frac {x \left (b x+c x^2\right )^{3/2}}{4 c}-\frac {5 b \left (\frac {\left (b x+c x^2\right )^{3/2}}{3 c}-\frac {b \left (\frac {(b+2 c x) \sqrt {b x+c x^2}}{4 c}-\frac {b^2 \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{4 c^{3/2}}\right )}{2 c}\right )}{8 c}\right )}{10 c}\right )}{4 c}\) |
(B*x^3*(b*x + c*x^2)^(3/2))/(6*c) - ((3*b*B - 4*A*c)*((x^2*(b*x + c*x^2)^( 3/2))/(5*c) - (7*b*((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)))/(4*c)
3.1.67.3.1 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 = 0.19 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.66
| method | result | size |
| pseudoelliptic | \(-\frac {7 \left (\left (-\frac {15}{8} A \,b^{5} c +\frac {45}{32} B \,b^{6}\right ) \operatorname {arctanh}\left (\frac {\sqrt {x \left (c x +b \right )}}{x \sqrt {c}}\right )+\sqrt {x \left (c x +b \right )}\, \left (-\frac {48 \left (\frac {5 B x}{6}+A \right ) x^{4} c^{\frac {11}{2}}}{7}+\left (\frac {15 \left (\frac {B x}{2}+A \right ) b^{3} c^{\frac {3}{2}}}{8}-\frac {5 \left (\frac {3 B x}{5}+A \right ) x \,b^{2} c^{\frac {5}{2}}}{4}+b \,x^{2} \left (\frac {9 B x}{14}+A \right ) c^{\frac {7}{2}}-\frac {6 x^{3} \left (\frac {2 B x}{3}+A \right ) c^{\frac {9}{2}}}{7}-\frac {45 B \,b^{4} \sqrt {c}}{32}\right ) b \right )\right )}{240 c^{\frac {11}{2}}}\) | \(133\) |
| risch | \(-\frac {\left (-1280 B \,c^{5} x^{5}-1536 A \,c^{5} x^{4}-128 B b \,c^{4} x^{4}-192 A b \,c^{4} x^{3}+144 B \,b^{2} c^{3} x^{3}+224 A \,b^{2} c^{3} x^{2}-168 B \,b^{3} c^{2} x^{2}-280 A \,b^{3} c^{2} x +210 B \,b^{4} c x +420 A \,b^{4} c -315 B \,b^{5}\right ) x \left (c x +b \right )}{7680 c^{5} \sqrt {x \left (c x +b \right )}}+\frac {7 b^{5} \left (4 A c -3 B b \right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{1024 c^{\frac {11}{2}}}\) | \(169\) |
| default | \(B \left (\frac {x^{3} \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{6 c}-\frac {3 b \left (\frac {x^{2} \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{5 c}-\frac {7 b \left (\frac {x \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{4 c}-\frac {5 b \left (\frac {\left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{3 c}-\frac {b \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x}}{4 c}-\frac {b^{2} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{8 c^{\frac {3}{2}}}\right )}{2 c}\right )}{8 c}\right )}{10 c}\right )}{4 c}\right )+A \left (\frac {x^{2} \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{5 c}-\frac {7 b \left (\frac {x \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{4 c}-\frac {5 b \left (\frac {\left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{3 c}-\frac {b \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x}}{4 c}-\frac {b^{2} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{8 c^{\frac {3}{2}}}\right )}{2 c}\right )}{8 c}\right )}{10 c}\right )\) | \(288\) |
-7/240/c^(11/2)*((-15/8*A*b^5*c+45/32*B*b^6)*arctanh((x*(c*x+b))^(1/2)/x/c ^(1/2))+(x*(c*x+b))^(1/2)*(-48/7*(5/6*B*x+A)*x^4*c^(11/2)+(15/8*(1/2*B*x+A )*b^3*c^(3/2)-5/4*(3/5*B*x+A)*x*b^2*c^(5/2)+b*x^2*(9/14*B*x+A)*c^(7/2)-6/7 *x^3*(2/3*B*x+A)*c^(9/2)-45/32*B*b^4*c^(1/2))*b))
Time = 0.27 (sec) , antiderivative size = 349, normalized size of antiderivative = 1.74 \[ \int x^3 (A+B x) \sqrt {b x+c x^2} \, dx=\left [-\frac {105 \, {\left (3 \, B b^{6} - 4 \, A b^{5} c\right )} \sqrt {c} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) - 2 \, {\left (1280 \, B c^{6} x^{5} + 315 \, B b^{5} c - 420 \, A b^{4} c^{2} + 128 \, {\left (B b c^{5} + 12 \, A c^{6}\right )} x^{4} - 48 \, {\left (3 \, B b^{2} c^{4} - 4 \, A b c^{5}\right )} x^{3} + 56 \, {\left (3 \, B b^{3} c^{3} - 4 \, A b^{2} c^{4}\right )} x^{2} - 70 \, {\left (3 \, B b^{4} c^{2} - 4 \, A b^{3} c^{3}\right )} x\right )} \sqrt {c x^{2} + b x}}{15360 \, c^{6}}, \frac {105 \, {\left (3 \, B b^{6} - 4 \, A b^{5} c\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) + {\left (1280 \, B c^{6} x^{5} + 315 \, B b^{5} c - 420 \, A b^{4} c^{2} + 128 \, {\left (B b c^{5} + 12 \, A c^{6}\right )} x^{4} - 48 \, {\left (3 \, B b^{2} c^{4} - 4 \, A b c^{5}\right )} x^{3} + 56 \, {\left (3 \, B b^{3} c^{3} - 4 \, A b^{2} c^{4}\right )} x^{2} - 70 \, {\left (3 \, B b^{4} c^{2} - 4 \, A b^{3} c^{3}\right )} x\right )} \sqrt {c x^{2} + b x}}{7680 \, c^{6}}\right ] \]
[-1/15360*(105*(3*B*b^6 - 4*A*b^5*c)*sqrt(c)*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c)) - 2*(1280*B*c^6*x^5 + 315*B*b^5*c - 420*A*b^4*c^2 + 128*(B *b*c^5 + 12*A*c^6)*x^4 - 48*(3*B*b^2*c^4 - 4*A*b*c^5)*x^3 + 56*(3*B*b^3*c^ 3 - 4*A*b^2*c^4)*x^2 - 70*(3*B*b^4*c^2 - 4*A*b^3*c^3)*x)*sqrt(c*x^2 + b*x) )/c^6, 1/7680*(105*(3*B*b^6 - 4*A*b^5*c)*sqrt(-c)*arctan(sqrt(c*x^2 + b*x) *sqrt(-c)/(c*x)) + (1280*B*c^6*x^5 + 315*B*b^5*c - 420*A*b^4*c^2 + 128*(B* b*c^5 + 12*A*c^6)*x^4 - 48*(3*B*b^2*c^4 - 4*A*b*c^5)*x^3 + 56*(3*B*b^3*c^3 - 4*A*b^2*c^4)*x^2 - 70*(3*B*b^4*c^2 - 4*A*b^3*c^3)*x)*sqrt(c*x^2 + b*x)) /c^6]
Time = 0.46 (sec) , antiderivative size = 270, normalized size of antiderivative = 1.35 \[ \int x^3 (A+B x) \sqrt {b x+c x^2} \, dx=\begin {cases} \frac {35 b^{4} \left (A b - \frac {9 b \left (A c + \frac {B b}{12}\right )}{10 c}\right ) \left (\begin {cases} \frac {\log {\left (b + 2 \sqrt {c} \sqrt {b x + c x^{2}} + 2 c x \right )}}{\sqrt {c}} & \text {for}\: \frac {b^{2}}{c} \neq 0 \\\frac {\left (\frac {b}{2 c} + x\right ) \log {\left (\frac {b}{2 c} + x \right )}}{\sqrt {c \left (\frac {b}{2 c} + x\right )^{2}}} & \text {otherwise} \end {cases}\right )}{128 c^{4}} + \sqrt {b x + c x^{2}} \left (\frac {B x^{5}}{6} - \frac {35 b^{3} \left (A b - \frac {9 b \left (A c + \frac {B b}{12}\right )}{10 c}\right )}{64 c^{4}} + \frac {35 b^{2} x \left (A b - \frac {9 b \left (A c + \frac {B b}{12}\right )}{10 c}\right )}{96 c^{3}} - \frac {7 b x^{2} \left (A b - \frac {9 b \left (A c + \frac {B b}{12}\right )}{10 c}\right )}{24 c^{2}} + \frac {x^{4} \left (A c + \frac {B b}{12}\right )}{5 c} + \frac {x^{3} \left (A b - \frac {9 b \left (A c + \frac {B b}{12}\right )}{10 c}\right )}{4 c}\right ) & \text {for}\: c \neq 0 \\\frac {2 \left (\frac {A \left (b x\right )^{\frac {9}{2}}}{9} + \frac {B \left (b x\right )^{\frac {11}{2}}}{11 b}\right )}{b^{4}} & \text {for}\: b \neq 0 \\0 & \text {otherwise} \end {cases} \]
Piecewise((35*b**4*(A*b - 9*b*(A*c + B*b/12)/(10*c))*Piecewise((log(b + 2* sqrt(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))/(128*c**4) + sqrt(b*x + c*x**2)*(B*x**5/6 - 35*b**3*(A*b - 9*b*(A*c + B*b/12)/(10*c))/(64*c**4) + 35*b**2*x*(A*b - 9*b*(A*c + B*b/12)/(10*c))/(96*c**3) - 7*b*x**2*(A*b - 9*b*(A*c + B*b/12)/(10*c))/(24*c**2) + x**4*(A*c + B*b/12)/(5*c) + x**3*(A *b - 9*b*(A*c + B*b/12)/(10*c))/(4*c)), Ne(c, 0)), (2*(A*(b*x)**(9/2)/9 + B*(b*x)**(11/2)/(11*b))/b**4, Ne(b, 0)), (0, True))
Time = 0.18 (sec) , antiderivative size = 288, normalized size of antiderivative = 1.44 \[ \int x^3 (A+B x) \sqrt {b x+c x^2} \, dx=\frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}} B x^{3}}{6 \, c} - \frac {3 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} B b x^{2}}{20 \, c^{2}} + \frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}} A x^{2}}{5 \, c} + \frac {21 \, \sqrt {c x^{2} + b x} B b^{4} x}{256 \, c^{4}} + \frac {21 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} B b^{2} x}{160 \, c^{3}} - \frac {7 \, \sqrt {c x^{2} + b x} A b^{3} x}{64 \, c^{3}} - \frac {7 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} A b x}{40 \, c^{2}} - \frac {21 \, B b^{6} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{1024 \, c^{\frac {11}{2}}} + \frac {7 \, A b^{5} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{256 \, c^{\frac {9}{2}}} + \frac {21 \, \sqrt {c x^{2} + b x} B b^{5}}{512 \, c^{5}} - \frac {7 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} B b^{3}}{64 \, c^{4}} - \frac {7 \, \sqrt {c x^{2} + b x} A b^{4}}{128 \, c^{4}} + \frac {7 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} A b^{2}}{48 \, c^{3}} \]
1/6*(c*x^2 + b*x)^(3/2)*B*x^3/c - 3/20*(c*x^2 + b*x)^(3/2)*B*b*x^2/c^2 + 1 /5*(c*x^2 + b*x)^(3/2)*A*x^2/c + 21/256*sqrt(c*x^2 + b*x)*B*b^4*x/c^4 + 21 /160*(c*x^2 + b*x)^(3/2)*B*b^2*x/c^3 - 7/64*sqrt(c*x^2 + b*x)*A*b^3*x/c^3 - 7/40*(c*x^2 + b*x)^(3/2)*A*b*x/c^2 - 21/1024*B*b^6*log(2*c*x + b + 2*sqr t(c*x^2 + b*x)*sqrt(c))/c^(11/2) + 7/256*A*b^5*log(2*c*x + b + 2*sqrt(c*x^ 2 + b*x)*sqrt(c))/c^(9/2) + 21/512*sqrt(c*x^2 + b*x)*B*b^5/c^5 - 7/64*(c*x ^2 + b*x)^(3/2)*B*b^3/c^4 - 7/128*sqrt(c*x^2 + b*x)*A*b^4/c^4 + 7/48*(c*x^ 2 + b*x)^(3/2)*A*b^2/c^3
Time = 0.27 (sec) , antiderivative size = 186, normalized size of antiderivative = 0.93 \[ \int x^3 (A+B x) \sqrt {b x+c x^2} \, dx=\frac {1}{7680} \, \sqrt {c x^{2} + b x} {\left (2 \, {\left (4 \, {\left (2 \, {\left (8 \, {\left (10 \, B x + \frac {B b c^{4} + 12 \, A c^{5}}{c^{5}}\right )} x - \frac {3 \, {\left (3 \, B b^{2} c^{3} - 4 \, A b c^{4}\right )}}{c^{5}}\right )} x + \frac {7 \, {\left (3 \, B b^{3} c^{2} - 4 \, A b^{2} c^{3}\right )}}{c^{5}}\right )} x - \frac {35 \, {\left (3 \, B b^{4} c - 4 \, A b^{3} c^{2}\right )}}{c^{5}}\right )} x + \frac {105 \, {\left (3 \, B b^{5} - 4 \, A b^{4} c\right )}}{c^{5}}\right )} + \frac {7 \, {\left (3 \, B b^{6} - 4 \, A b^{5} c\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} + b \right |}\right )}{1024 \, c^{\frac {11}{2}}} \]
1/7680*sqrt(c*x^2 + b*x)*(2*(4*(2*(8*(10*B*x + (B*b*c^4 + 12*A*c^5)/c^5)*x - 3*(3*B*b^2*c^3 - 4*A*b*c^4)/c^5)*x + 7*(3*B*b^3*c^2 - 4*A*b^2*c^3)/c^5) *x - 35*(3*B*b^4*c - 4*A*b^3*c^2)/c^5)*x + 105*(3*B*b^5 - 4*A*b^4*c)/c^5) + 7/1024*(3*B*b^6 - 4*A*b^5*c)*log(abs(2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*s qrt(c) + b))/c^(11/2)
Time = 10.74 (sec) , antiderivative size = 267, normalized size of antiderivative = 1.34 \[ \int x^3 (A+B x) \sqrt {b x+c x^2} \, dx=\frac {3\,B\,b\,\left (\frac {7\,b\,\left (\frac {x\,{\left (c\,x^2+b\,x\right )}^{3/2}}{4\,c}-\frac {5\,b\,\left (\frac {b^3\,\ln \left (\frac {b+2\,c\,x}{\sqrt {c}}+2\,\sqrt {c\,x^2+b\,x}\right )}{16\,c^{5/2}}+\frac {\sqrt {c\,x^2+b\,x}\,\left (-3\,b^2+2\,b\,c\,x+8\,c^2\,x^2\right )}{24\,c^2}\right )}{8\,c}\right )}{10\,c}-\frac {x^2\,{\left (c\,x^2+b\,x\right )}^{3/2}}{5\,c}\right )}{4\,c}-\frac {7\,A\,b\,\left (\frac {x\,{\left (c\,x^2+b\,x\right )}^{3/2}}{4\,c}-\frac {5\,b\,\left (\frac {b^3\,\ln \left (\frac {b+2\,c\,x}{\sqrt {c}}+2\,\sqrt {c\,x^2+b\,x}\right )}{16\,c^{5/2}}+\frac {\sqrt {c\,x^2+b\,x}\,\left (-3\,b^2+2\,b\,c\,x+8\,c^2\,x^2\right )}{24\,c^2}\right )}{8\,c}\right )}{10\,c}+\frac {A\,x^2\,{\left (c\,x^2+b\,x\right )}^{3/2}}{5\,c}+\frac {B\,x^3\,{\left (c\,x^2+b\,x\right )}^{3/2}}{6\,c} \]
(3*B*b*((7*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) - (x^2*(b*x + c*x^2)^( 3/2))/(5*c)))/(4*c) - (7*A*b*((x*(b*x + c*x^2)^(3/2))/(4*c) - (5*b*((b^3*l og((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) + (A*x ^2*(b*x + c*x^2)^(3/2))/(5*c) + (B*x^3*(b*x + c*x^2)^(3/2))/(6*c)