Integrand size = 22, antiderivative size = 238 \[ \int x^3 (A+B x) \left (b x+c x^2\right )^{3/2} \, dx=-\frac {9 b^5 (11 b B-16 A c) (b+2 c x) \sqrt {b x+c x^2}}{16384 c^6}+\frac {3 b^3 (11 b B-16 A c) (b+2 c x) \left (b x+c x^2\right )^{3/2}}{2048 c^5}-\frac {3 b^2 (11 b B-16 A c) \left (b x+c x^2\right )^{5/2}}{640 c^4}+\frac {3 b (11 b B-16 A c) x \left (b x+c x^2\right )^{5/2}}{448 c^3}-\frac {(11 b B-16 A c) x^2 \left (b x+c x^2\right )^{5/2}}{112 c^2}+\frac {B x^3 \left (b x+c x^2\right )^{5/2}}{8 c}+\frac {9 b^7 (11 b B-16 A c) \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{16384 c^{13/2}} \]
3/2048*b^3*(-16*A*c+11*B*b)*(2*c*x+b)*(c*x^2+b*x)^(3/2)/c^5-3/640*b^2*(-16 *A*c+11*B*b)*(c*x^2+b*x)^(5/2)/c^4+3/448*b*(-16*A*c+11*B*b)*x*(c*x^2+b*x)^ (5/2)/c^3-1/112*(-16*A*c+11*B*b)*x^2*(c*x^2+b*x)^(5/2)/c^2+1/8*B*x^3*(c*x^ 2+b*x)^(5/2)/c+9/16384*b^7*(-16*A*c+11*B*b)*arctanh(x*c^(1/2)/(c*x^2+b*x)^ (1/2))/c^(13/2)-9/16384*b^5*(-16*A*c+11*B*b)*(2*c*x+b)*(c*x^2+b*x)^(1/2)/c ^6
Time = 1.32 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.08 \[ \int x^3 (A+B x) \left (b x+c x^2\right )^{3/2} \, dx=\frac {\sqrt {x} \sqrt {b+c x} \left (\sqrt {c} \sqrt {x} \sqrt {b+c x} \left (-3465 b^7 B+256 b^2 c^5 x^4 (8 A+5 B x)+10240 c^7 x^6 (8 A+7 B x)-128 b^3 c^4 x^3 (18 A+11 B x)-168 b^5 c^2 x (20 A+11 B x)+210 b^6 c (24 A+11 B x)+5120 b c^6 x^5 (20 A+17 B x)+48 b^4 c^3 x^2 (56 A+33 B x)\right )+10080 A b^7 c \text {arctanh}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}-\sqrt {b+c x}}\right )+6930 b^8 B \text {arctanh}\left (\frac {\sqrt {c} \sqrt {x}}{-\sqrt {b}+\sqrt {b+c x}}\right )\right )}{573440 c^{13/2} \sqrt {x (b+c x)}} \]
(Sqrt[x]*Sqrt[b + c*x]*(Sqrt[c]*Sqrt[x]*Sqrt[b + c*x]*(-3465*b^7*B + 256*b ^2*c^5*x^4*(8*A + 5*B*x) + 10240*c^7*x^6*(8*A + 7*B*x) - 128*b^3*c^4*x^3*( 18*A + 11*B*x) - 168*b^5*c^2*x*(20*A + 11*B*x) + 210*b^6*c*(24*A + 11*B*x) + 5120*b*c^6*x^5*(20*A + 17*B*x) + 48*b^4*c^3*x^2*(56*A + 33*B*x)) + 1008 0*A*b^7*c*ArcTanh[(Sqrt[c]*Sqrt[x])/(Sqrt[b] - Sqrt[b + c*x])] + 6930*b^8* B*ArcTanh[(Sqrt[c]*Sqrt[x])/(-Sqrt[b] + Sqrt[b + c*x])]))/(573440*c^(13/2) *Sqrt[x*(b + c*x)])
Time = 0.39 (sec) , antiderivative size = 229, normalized size of antiderivative = 0.96, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {1221, 1134, 1134, 1160, 1087, 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.
| \(\displaystyle \int x^3 (A+B x) \left (b x+c x^2\right )^{3/2} \, dx\) |
| \(\Big \downarrow \) 1221 |
| \(\displaystyle \frac {B x^3 \left (b x+c x^2\right )^{5/2}}{8 c}-\frac {(11 b B-16 A c) \int x^3 \left (c x^2+b x\right )^{3/2}dx}{16 c}\) |
| \(\Big \downarrow \) 1134 |
| \(\displaystyle \frac {B x^3 \left (b x+c x^2\right )^{5/2}}{8 c}-\frac {(11 b B-16 A c) \left (\frac {x^2 \left (b x+c x^2\right )^{5/2}}{7 c}-\frac {9 b \int x^2 \left (c x^2+b x\right )^{3/2}dx}{14 c}\right )}{16 c}\) |
| \(\Big \downarrow \) 1134 |
| \(\displaystyle \frac {B x^3 \left (b x+c x^2\right )^{5/2}}{8 c}-\frac {(11 b B-16 A c) \left (\frac {x^2 \left (b x+c x^2\right )^{5/2}}{7 c}-\frac {9 b \left (\frac {x \left (b x+c x^2\right )^{5/2}}{6 c}-\frac {7 b \int x \left (c x^2+b x\right )^{3/2}dx}{12 c}\right )}{14 c}\right )}{16 c}\) |
| \(\Big \downarrow \) 1160 |
| \(\displaystyle \frac {B x^3 \left (b x+c x^2\right )^{5/2}}{8 c}-\frac {(11 b B-16 A c) \left (\frac {x^2 \left (b x+c x^2\right )^{5/2}}{7 c}-\frac {9 b \left (\frac {x \left (b x+c x^2\right )^{5/2}}{6 c}-\frac {7 b \left (\frac {\left (b x+c x^2\right )^{5/2}}{5 c}-\frac {b \int \left (c x^2+b x\right )^{3/2}dx}{2 c}\right )}{12 c}\right )}{14 c}\right )}{16 c}\) |
| \(\Big \downarrow \) 1087 |
| \(\displaystyle \frac {B x^3 \left (b x+c x^2\right )^{5/2}}{8 c}-\frac {(11 b B-16 A c) \left (\frac {x^2 \left (b x+c x^2\right )^{5/2}}{7 c}-\frac {9 b \left (\frac {x \left (b x+c x^2\right )^{5/2}}{6 c}-\frac {7 b \left (\frac {\left (b x+c x^2\right )^{5/2}}{5 c}-\frac {b \left (\frac {(b+2 c x) \left (b x+c x^2\right )^{3/2}}{8 c}-\frac {3 b^2 \int \sqrt {c x^2+b x}dx}{16 c}\right )}{2 c}\right )}{12 c}\right )}{14 c}\right )}{16 c}\) |
| \(\Big \downarrow \) 1087 |
| \(\displaystyle \frac {B x^3 \left (b x+c x^2\right )^{5/2}}{8 c}-\frac {(11 b B-16 A c) \left (\frac {x^2 \left (b x+c x^2\right )^{5/2}}{7 c}-\frac {9 b \left (\frac {x \left (b x+c x^2\right )^{5/2}}{6 c}-\frac {7 b \left (\frac {\left (b x+c x^2\right )^{5/2}}{5 c}-\frac {b \left (\frac {(b+2 c x) \left (b x+c x^2\right )^{3/2}}{8 c}-\frac {3 b^2 \left (\frac {(b+2 c x) \sqrt {b x+c x^2}}{4 c}-\frac {b^2 \int \frac {1}{\sqrt {c x^2+b x}}dx}{8 c}\right )}{16 c}\right )}{2 c}\right )}{12 c}\right )}{14 c}\right )}{16 c}\) |
| \(\Big \downarrow \) 1091 |
| \(\displaystyle \frac {B x^3 \left (b x+c x^2\right )^{5/2}}{8 c}-\frac {(11 b B-16 A c) \left (\frac {x^2 \left (b x+c x^2\right )^{5/2}}{7 c}-\frac {9 b \left (\frac {x \left (b x+c x^2\right )^{5/2}}{6 c}-\frac {7 b \left (\frac {\left (b x+c x^2\right )^{5/2}}{5 c}-\frac {b \left (\frac {(b+2 c x) \left (b x+c x^2\right )^{3/2}}{8 c}-\frac {3 b^2 \left (\frac {(b+2 c x) \sqrt {b x+c x^2}}{4 c}-\frac {b^2 \int \frac {1}{1-\frac {c x^2}{c x^2+b x}}d\frac {x}{\sqrt {c x^2+b x}}}{4 c}\right )}{16 c}\right )}{2 c}\right )}{12 c}\right )}{14 c}\right )}{16 c}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {B x^3 \left (b x+c x^2\right )^{5/2}}{8 c}-\frac {(11 b B-16 A c) \left (\frac {x^2 \left (b x+c x^2\right )^{5/2}}{7 c}-\frac {9 b \left (\frac {x \left (b x+c x^2\right )^{5/2}}{6 c}-\frac {7 b \left (\frac {\left (b x+c x^2\right )^{5/2}}{5 c}-\frac {b \left (\frac {(b+2 c x) \left (b x+c x^2\right )^{3/2}}{8 c}-\frac {3 b^2 \left (\frac {(b+2 c x) \sqrt {b x+c x^2}}{4 c}-\frac {b^2 \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{4 c^{3/2}}\right )}{16 c}\right )}{2 c}\right )}{12 c}\right )}{14 c}\right )}{16 c}\) |
(B*x^3*(b*x + c*x^2)^(5/2))/(8*c) - ((11*b*B - 16*A*c)*((x^2*(b*x + c*x^2) ^(5/2))/(7*c) - (9*b*((x*(b*x + c*x^2)^(5/2))/(6*c) - (7*b*((b*x + c*x^2)^ (5/2)/(5*c) - (b*(((b + 2*c*x)*(b*x + c*x^2)^(3/2))/(8*c) - (3*b^2*(((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))))/(16*c)))/(2*c)))/(12*c)))/(14*c)))/(16*c)
3.1.79.3.1 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 = 0.23 (sec) , antiderivative size = 167, normalized size of antiderivative = 0.70
| method | result | size |
| pseudoelliptic | \(\frac {\frac {3 \left (-\frac {15}{8} A \,b^{7} c +\frac {165}{128} B \,b^{8}\right ) \operatorname {arctanh}\left (\frac {\sqrt {x \left (c x +b \right )}}{x \sqrt {c}}\right )}{640}+\frac {3 \sqrt {x \left (c x +b \right )}\, \left (\frac {16 \left (\frac {5 B x}{8}+A \right ) x^{4} b^{2} c^{\frac {11}{2}}}{21}+\frac {800 \left (\frac {17 B x}{20}+A \right ) x^{5} b \,c^{\frac {13}{2}}}{21}+\frac {640 x^{6} \left (\frac {7 B x}{8}+A \right ) c^{\frac {15}{2}}}{21}+\left (\frac {15 \left (\frac {11 B x}{24}+A \right ) b^{3} c^{\frac {3}{2}}}{8}-\frac {5 x \left (\frac {11 B x}{20}+A \right ) b^{2} c^{\frac {5}{2}}}{4}+b \,x^{2} \left (\frac {33 B x}{56}+A \right ) c^{\frac {7}{2}}-\frac {6 x^{3} \left (\frac {11 B x}{18}+A \right ) c^{\frac {9}{2}}}{7}-\frac {165 B \,b^{4} \sqrt {c}}{128}\right ) b^{3}\right )}{640}}{c^{\frac {13}{2}}}\) | \(167\) |
| risch | \(\frac {\left (71680 B \,c^{7} x^{7}+81920 A \,c^{7} x^{6}+87040 B b \,c^{6} x^{6}+102400 A b \,c^{6} x^{5}+1280 B \,b^{2} c^{5} x^{5}+2048 A \,b^{2} c^{5} x^{4}-1408 B \,b^{3} c^{4} x^{4}-2304 A \,b^{3} c^{4} x^{3}+1584 B \,b^{4} c^{3} x^{3}+2688 A \,b^{4} c^{3} x^{2}-1848 B \,b^{5} c^{2} x^{2}-3360 A \,b^{5} c^{2} x +2310 B \,b^{6} c x +5040 A \,b^{6} c -3465 B \,b^{7}\right ) x \left (c x +b \right )}{573440 c^{6} \sqrt {x \left (c x +b \right )}}-\frac {9 b^{7} \left (16 A c -11 B b \right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{32768 c^{\frac {13}{2}}}\) | \(217\) |
| default | \(B \left (\frac {x^{3} \left (c \,x^{2}+b x \right )^{\frac {5}{2}}}{8 c}-\frac {11 b \left (\frac {x^{2} \left (c \,x^{2}+b x \right )^{\frac {5}{2}}}{7 c}-\frac {9 b \left (\frac {x \left (c \,x^{2}+b x \right )^{\frac {5}{2}}}{6 c}-\frac {7 b \left (\frac {\left (c \,x^{2}+b x \right )^{\frac {5}{2}}}{5 c}-\frac {b \left (\frac {\left (2 c x +b \right ) \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{8 c}-\frac {3 b^{2} \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x}}{4 c}-\frac {b^{2} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{8 c^{\frac {3}{2}}}\right )}{16 c}\right )}{2 c}\right )}{12 c}\right )}{14 c}\right )}{16 c}\right )+A \left (\frac {x^{2} \left (c \,x^{2}+b x \right )^{\frac {5}{2}}}{7 c}-\frac {9 b \left (\frac {x \left (c \,x^{2}+b x \right )^{\frac {5}{2}}}{6 c}-\frac {7 b \left (\frac {\left (c \,x^{2}+b x \right )^{\frac {5}{2}}}{5 c}-\frac {b \left (\frac {\left (2 c x +b \right ) \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{8 c}-\frac {3 b^{2} \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x}}{4 c}-\frac {b^{2} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{8 c^{\frac {3}{2}}}\right )}{16 c}\right )}{2 c}\right )}{12 c}\right )}{14 c}\right )\) | \(350\) |
3/640*((-15/8*A*b^7*c+165/128*B*b^8)*arctanh((x*(c*x+b))^(1/2)/x/c^(1/2))+ (x*(c*x+b))^(1/2)*(16/21*(5/8*B*x+A)*x^4*b^2*c^(11/2)+800/21*(17/20*B*x+A) *x^5*b*c^(13/2)+640/21*x^6*(7/8*B*x+A)*c^(15/2)+(15/8*(11/24*B*x+A)*b^3*c^ (3/2)-5/4*x*(11/20*B*x+A)*b^2*c^(5/2)+b*x^2*(33/56*B*x+A)*c^(7/2)-6/7*x^3* (11/18*B*x+A)*c^(9/2)-165/128*B*b^4*c^(1/2))*b^3))/c^(13/2)
Time = 0.27 (sec) , antiderivative size = 446, normalized size of antiderivative = 1.87 \[ \int x^3 (A+B x) \left (b x+c x^2\right )^{3/2} \, dx=\left [-\frac {315 \, {\left (11 \, B b^{8} - 16 \, A b^{7} c\right )} \sqrt {c} \log \left (2 \, c x + b - 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) - 2 \, {\left (71680 \, B c^{8} x^{7} - 3465 \, B b^{7} c + 5040 \, A b^{6} c^{2} + 5120 \, {\left (17 \, B b c^{7} + 16 \, A c^{8}\right )} x^{6} + 1280 \, {\left (B b^{2} c^{6} + 80 \, A b c^{7}\right )} x^{5} - 128 \, {\left (11 \, B b^{3} c^{5} - 16 \, A b^{2} c^{6}\right )} x^{4} + 144 \, {\left (11 \, B b^{4} c^{4} - 16 \, A b^{3} c^{5}\right )} x^{3} - 168 \, {\left (11 \, B b^{5} c^{3} - 16 \, A b^{4} c^{4}\right )} x^{2} + 210 \, {\left (11 \, B b^{6} c^{2} - 16 \, A b^{5} c^{3}\right )} x\right )} \sqrt {c x^{2} + b x}}{1146880 \, c^{7}}, -\frac {315 \, {\left (11 \, B b^{8} - 16 \, A b^{7} c\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) - {\left (71680 \, B c^{8} x^{7} - 3465 \, B b^{7} c + 5040 \, A b^{6} c^{2} + 5120 \, {\left (17 \, B b c^{7} + 16 \, A c^{8}\right )} x^{6} + 1280 \, {\left (B b^{2} c^{6} + 80 \, A b c^{7}\right )} x^{5} - 128 \, {\left (11 \, B b^{3} c^{5} - 16 \, A b^{2} c^{6}\right )} x^{4} + 144 \, {\left (11 \, B b^{4} c^{4} - 16 \, A b^{3} c^{5}\right )} x^{3} - 168 \, {\left (11 \, B b^{5} c^{3} - 16 \, A b^{4} c^{4}\right )} x^{2} + 210 \, {\left (11 \, B b^{6} c^{2} - 16 \, A b^{5} c^{3}\right )} x\right )} \sqrt {c x^{2} + b x}}{573440 \, c^{7}}\right ] \]
[-1/1146880*(315*(11*B*b^8 - 16*A*b^7*c)*sqrt(c)*log(2*c*x + b - 2*sqrt(c* x^2 + b*x)*sqrt(c)) - 2*(71680*B*c^8*x^7 - 3465*B*b^7*c + 5040*A*b^6*c^2 + 5120*(17*B*b*c^7 + 16*A*c^8)*x^6 + 1280*(B*b^2*c^6 + 80*A*b*c^7)*x^5 - 12 8*(11*B*b^3*c^5 - 16*A*b^2*c^6)*x^4 + 144*(11*B*b^4*c^4 - 16*A*b^3*c^5)*x^ 3 - 168*(11*B*b^5*c^3 - 16*A*b^4*c^4)*x^2 + 210*(11*B*b^6*c^2 - 16*A*b^5*c ^3)*x)*sqrt(c*x^2 + b*x))/c^7, -1/573440*(315*(11*B*b^8 - 16*A*b^7*c)*sqrt (-c)*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) - (71680*B*c^8*x^7 - 3465*B* b^7*c + 5040*A*b^6*c^2 + 5120*(17*B*b*c^7 + 16*A*c^8)*x^6 + 1280*(B*b^2*c^ 6 + 80*A*b*c^7)*x^5 - 128*(11*B*b^3*c^5 - 16*A*b^2*c^6)*x^4 + 144*(11*B*b^ 4*c^4 - 16*A*b^3*c^5)*x^3 - 168*(11*B*b^5*c^3 - 16*A*b^4*c^4)*x^2 + 210*(1 1*B*b^6*c^2 - 16*A*b^5*c^3)*x)*sqrt(c*x^2 + b*x))/c^7]
Leaf count of result is larger than twice the leaf count of optimal. 502 vs. \(2 (235) = 470\).
Time = 0.53 (sec) , antiderivative size = 502, normalized size of antiderivative = 2.11 \[ \int x^3 (A+B x) \left (b x+c x^2\right )^{3/2} \, dx=\begin {cases} - \frac {63 b^{5} \left (A b^{2} - \frac {11 b \left (2 A b c + B b^{2} - \frac {13 b \left (A c^{2} + \frac {17 B b c}{16}\right )}{14 c}\right )}{12 c}\right ) \left (\begin {cases} \frac {\log {\left (b + 2 \sqrt {c} \sqrt {b x + c x^{2}} + 2 c x \right )}}{\sqrt {c}} & \text {for}\: \frac {b^{2}}{c} \neq 0 \\\frac {\left (\frac {b}{2 c} + x\right ) \log {\left (\frac {b}{2 c} + x \right )}}{\sqrt {c \left (\frac {b}{2 c} + x\right )^{2}}} & \text {otherwise} \end {cases}\right )}{256 c^{5}} + \sqrt {b x + c x^{2}} \left (\frac {B c x^{7}}{8} + \frac {63 b^{4} \left (A b^{2} - \frac {11 b \left (2 A b c + B b^{2} - \frac {13 b \left (A c^{2} + \frac {17 B b c}{16}\right )}{14 c}\right )}{12 c}\right )}{128 c^{5}} - \frac {21 b^{3} x \left (A b^{2} - \frac {11 b \left (2 A b c + B b^{2} - \frac {13 b \left (A c^{2} + \frac {17 B b c}{16}\right )}{14 c}\right )}{12 c}\right )}{64 c^{4}} + \frac {21 b^{2} x^{2} \left (A b^{2} - \frac {11 b \left (2 A b c + B b^{2} - \frac {13 b \left (A c^{2} + \frac {17 B b c}{16}\right )}{14 c}\right )}{12 c}\right )}{80 c^{3}} - \frac {9 b x^{3} \left (A b^{2} - \frac {11 b \left (2 A b c + B b^{2} - \frac {13 b \left (A c^{2} + \frac {17 B b c}{16}\right )}{14 c}\right )}{12 c}\right )}{40 c^{2}} + \frac {x^{6} \left (A c^{2} + \frac {17 B b c}{16}\right )}{7 c} + \frac {x^{5} \cdot \left (2 A b c + B b^{2} - \frac {13 b \left (A c^{2} + \frac {17 B b c}{16}\right )}{14 c}\right )}{6 c} + \frac {x^{4} \left (A b^{2} - \frac {11 b \left (2 A b c + B b^{2} - \frac {13 b \left (A c^{2} + \frac {17 B b c}{16}\right )}{14 c}\right )}{12 c}\right )}{5 c}\right ) & \text {for}\: c \neq 0 \\\frac {2 \left (\frac {A \left (b x\right )^{\frac {11}{2}}}{11} + \frac {B \left (b x\right )^{\frac {13}{2}}}{13 b}\right )}{b^{4}} & \text {for}\: b \neq 0 \\0 & \text {otherwise} \end {cases} \]
Piecewise((-63*b**5*(A*b**2 - 11*b*(2*A*b*c + B*b**2 - 13*b*(A*c**2 + 17*B *b*c/16)/(14*c))/(12*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))/(256*c**5) + sqrt(b*x + c*x**2)*(B*c*x**7/8 + 63*b **4*(A*b**2 - 11*b*(2*A*b*c + B*b**2 - 13*b*(A*c**2 + 17*B*b*c/16)/(14*c)) /(12*c))/(128*c**5) - 21*b**3*x*(A*b**2 - 11*b*(2*A*b*c + B*b**2 - 13*b*(A *c**2 + 17*B*b*c/16)/(14*c))/(12*c))/(64*c**4) + 21*b**2*x**2*(A*b**2 - 11 *b*(2*A*b*c + B*b**2 - 13*b*(A*c**2 + 17*B*b*c/16)/(14*c))/(12*c))/(80*c** 3) - 9*b*x**3*(A*b**2 - 11*b*(2*A*b*c + B*b**2 - 13*b*(A*c**2 + 17*B*b*c/1 6)/(14*c))/(12*c))/(40*c**2) + x**6*(A*c**2 + 17*B*b*c/16)/(7*c) + x**5*(2 *A*b*c + B*b**2 - 13*b*(A*c**2 + 17*B*b*c/16)/(14*c))/(6*c) + x**4*(A*b**2 - 11*b*(2*A*b*c + B*b**2 - 13*b*(A*c**2 + 17*B*b*c/16)/(14*c))/(12*c))/(5 *c)), Ne(c, 0)), (2*(A*(b*x)**(11/2)/11 + B*(b*x)**(13/2)/(13*b))/b**4, Ne (b, 0)), (0, True))
Time = 0.18 (sec) , antiderivative size = 370, normalized size of antiderivative = 1.55 \[ \int x^3 (A+B x) \left (b x+c x^2\right )^{3/2} \, dx=\frac {{\left (c x^{2} + b x\right )}^{\frac {5}{2}} B x^{3}}{8 \, c} - \frac {11 \, {\left (c x^{2} + b x\right )}^{\frac {5}{2}} B b x^{2}}{112 \, c^{2}} + \frac {{\left (c x^{2} + b x\right )}^{\frac {5}{2}} A x^{2}}{7 \, c} - \frac {99 \, \sqrt {c x^{2} + b x} B b^{6} x}{8192 \, c^{5}} + \frac {33 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} B b^{4} x}{1024 \, c^{4}} + \frac {9 \, \sqrt {c x^{2} + b x} A b^{5} x}{512 \, c^{4}} + \frac {33 \, {\left (c x^{2} + b x\right )}^{\frac {5}{2}} B b^{2} x}{448 \, c^{3}} - \frac {3 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} A b^{3} x}{64 \, c^{3}} - \frac {3 \, {\left (c x^{2} + b x\right )}^{\frac {5}{2}} A b x}{28 \, c^{2}} + \frac {99 \, B b^{8} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{32768 \, c^{\frac {13}{2}}} - \frac {9 \, A b^{7} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{2048 \, c^{\frac {11}{2}}} - \frac {99 \, \sqrt {c x^{2} + b x} B b^{7}}{16384 \, c^{6}} + \frac {33 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} B b^{5}}{2048 \, c^{5}} + \frac {9 \, \sqrt {c x^{2} + b x} A b^{6}}{1024 \, c^{5}} - \frac {33 \, {\left (c x^{2} + b x\right )}^{\frac {5}{2}} B b^{3}}{640 \, c^{4}} - \frac {3 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} A b^{4}}{128 \, c^{4}} + \frac {3 \, {\left (c x^{2} + b x\right )}^{\frac {5}{2}} A b^{2}}{40 \, c^{3}} \]
1/8*(c*x^2 + b*x)^(5/2)*B*x^3/c - 11/112*(c*x^2 + b*x)^(5/2)*B*b*x^2/c^2 + 1/7*(c*x^2 + b*x)^(5/2)*A*x^2/c - 99/8192*sqrt(c*x^2 + b*x)*B*b^6*x/c^5 + 33/1024*(c*x^2 + b*x)^(3/2)*B*b^4*x/c^4 + 9/512*sqrt(c*x^2 + b*x)*A*b^5*x /c^4 + 33/448*(c*x^2 + b*x)^(5/2)*B*b^2*x/c^3 - 3/64*(c*x^2 + b*x)^(3/2)*A *b^3*x/c^3 - 3/28*(c*x^2 + b*x)^(5/2)*A*b*x/c^2 + 99/32768*B*b^8*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/c^(13/2) - 9/2048*A*b^7*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/c^(11/2) - 99/16384*sqrt(c*x^2 + b*x)*B*b^7 /c^6 + 33/2048*(c*x^2 + b*x)^(3/2)*B*b^5/c^5 + 9/1024*sqrt(c*x^2 + b*x)*A* b^6/c^5 - 33/640*(c*x^2 + b*x)^(5/2)*B*b^3/c^4 - 3/128*(c*x^2 + b*x)^(3/2) *A*b^4/c^4 + 3/40*(c*x^2 + b*x)^(5/2)*A*b^2/c^3
Time = 0.28 (sec) , antiderivative size = 247, normalized size of antiderivative = 1.04 \[ \int x^3 (A+B x) \left (b x+c x^2\right )^{3/2} \, dx=\frac {1}{573440} \, \sqrt {c x^{2} + b x} {\left (2 \, {\left (4 \, {\left (2 \, {\left (8 \, {\left (10 \, {\left (4 \, {\left (14 \, B c x + \frac {17 \, B b c^{7} + 16 \, A c^{8}}{c^{7}}\right )} x + \frac {B b^{2} c^{6} + 80 \, A b c^{7}}{c^{7}}\right )} x - \frac {11 \, B b^{3} c^{5} - 16 \, A b^{2} c^{6}}{c^{7}}\right )} x + \frac {9 \, {\left (11 \, B b^{4} c^{4} - 16 \, A b^{3} c^{5}\right )}}{c^{7}}\right )} x - \frac {21 \, {\left (11 \, B b^{5} c^{3} - 16 \, A b^{4} c^{4}\right )}}{c^{7}}\right )} x + \frac {105 \, {\left (11 \, B b^{6} c^{2} - 16 \, A b^{5} c^{3}\right )}}{c^{7}}\right )} x - \frac {315 \, {\left (11 \, B b^{7} c - 16 \, A b^{6} c^{2}\right )}}{c^{7}}\right )} - \frac {9 \, {\left (11 \, B b^{8} - 16 \, A b^{7} c\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} + b \right |}\right )}{32768 \, c^{\frac {13}{2}}} \]
1/573440*sqrt(c*x^2 + b*x)*(2*(4*(2*(8*(10*(4*(14*B*c*x + (17*B*b*c^7 + 16 *A*c^8)/c^7)*x + (B*b^2*c^6 + 80*A*b*c^7)/c^7)*x - (11*B*b^3*c^5 - 16*A*b^ 2*c^6)/c^7)*x + 9*(11*B*b^4*c^4 - 16*A*b^3*c^5)/c^7)*x - 21*(11*B*b^5*c^3 - 16*A*b^4*c^4)/c^7)*x + 105*(11*B*b^6*c^2 - 16*A*b^5*c^3)/c^7)*x - 315*(1 1*B*b^7*c - 16*A*b^6*c^2)/c^7) - 9/32768*(11*B*b^8 - 16*A*b^7*c)*log(abs(2 *(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) + b))/c^(13/2)
Timed out. \[ \int x^3 (A+B x) \left (b x+c x^2\right )^{3/2} \, dx=\int x^3\,{\left (c\,x^2+b\,x\right )}^{3/2}\,\left (A+B\,x\right ) \,d x \]