Integrand size = 22, antiderivative size = 203 \[ \int x^2 (A+B x) \left (b x+c x^2\right )^{3/2} \, dx=\frac {b^4 (9 b B-14 A c) (b+2 c x) \sqrt {b x+c x^2}}{1024 c^5}-\frac {b^2 (9 b B-14 A c) (b+2 c x) \left (b x+c x^2\right )^{3/2}}{384 c^4}+\frac {b (9 b B-14 A c) \left (b x+c x^2\right )^{5/2}}{120 c^3}-\frac {(9 b B-14 A c) x \left (b x+c x^2\right )^{5/2}}{84 c^2}+\frac {B x^2 \left (b x+c x^2\right )^{5/2}}{7 c}-\frac {b^6 (9 b B-14 A c) \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{1024 c^{11/2}} \]
-1/384*b^2*(-14*A*c+9*B*b)*(2*c*x+b)*(c*x^2+b*x)^(3/2)/c^4+1/120*b*(-14*A* c+9*B*b)*(c*x^2+b*x)^(5/2)/c^3-1/84*(-14*A*c+9*B*b)*x*(c*x^2+b*x)^(5/2)/c^ 2+1/7*B*x^2*(c*x^2+b*x)^(5/2)/c-1/1024*b^6*(-14*A*c+9*B*b)*arctanh(x*c^(1/ 2)/(c*x^2+b*x)^(1/2))/c^(11/2)+1/1024*b^4*(-14*A*c+9*B*b)*(2*c*x+b)*(c*x^2 +b*x)^(1/2)/c^5
Time = 1.08 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.16 \[ \int x^2 (A+B x) \left (b x+c x^2\right )^{3/2} \, dx=\frac {(x (b+c x))^{3/2} \left (945 b^6 B-1470 A b^5 c-630 b^5 B c x+980 A b^4 c^2 x+504 b^4 B c^2 x^2-784 A b^3 c^3 x^2-432 b^3 B c^3 x^3+672 A b^2 c^4 x^3+384 b^2 B c^4 x^4+23296 A b c^5 x^4+19200 b B c^5 x^5+17920 A c^6 x^5+15360 B c^6 x^6\right )}{107520 c^5 x (b+c x)}-\frac {b^6 (9 b B-14 A c) (x (b+c x))^{3/2} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {x}}{-\sqrt {b}+\sqrt {b+c x}}\right )}{512 c^{11/2} x^{3/2} (b+c x)^{3/2}} \]
((x*(b + c*x))^(3/2)*(945*b^6*B - 1470*A*b^5*c - 630*b^5*B*c*x + 980*A*b^4 *c^2*x + 504*b^4*B*c^2*x^2 - 784*A*b^3*c^3*x^2 - 432*b^3*B*c^3*x^3 + 672*A *b^2*c^4*x^3 + 384*b^2*B*c^4*x^4 + 23296*A*b*c^5*x^4 + 19200*b*B*c^5*x^5 + 17920*A*c^6*x^5 + 15360*B*c^6*x^6))/(107520*c^5*x*(b + c*x)) - (b^6*(9*b* B - 14*A*c)*(x*(b + c*x))^(3/2)*ArcTanh[(Sqrt[c]*Sqrt[x])/(-Sqrt[b] + Sqrt [b + c*x])])/(512*c^(11/2)*x^(3/2)*(b + c*x)^(3/2))
Time = 0.35 (sec) , antiderivative size = 197, normalized size of antiderivative = 0.97, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {1221, 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.
| \(\displaystyle \int x^2 (A+B x) \left (b x+c x^2\right )^{3/2} \, dx\) |
| \(\Big \downarrow \) 1221 |
| \(\displaystyle \frac {B x^2 \left (b x+c x^2\right )^{5/2}}{7 c}-\frac {(9 b B-14 A c) \int x^2 \left (c x^2+b x\right )^{3/2}dx}{14 c}\) |
| \(\Big \downarrow \) 1134 |
| \(\displaystyle \frac {B x^2 \left (b x+c x^2\right )^{5/2}}{7 c}-\frac {(9 b B-14 A c) \left (\frac {x \left (b x+c x^2\right )^{5/2}}{6 c}-\frac {7 b \int x \left (c x^2+b x\right )^{3/2}dx}{12 c}\right )}{14 c}\) |
| \(\Big \downarrow \) 1160 |
| \(\displaystyle \frac {B x^2 \left (b x+c x^2\right )^{5/2}}{7 c}-\frac {(9 b B-14 A c) \left (\frac {x \left (b x+c x^2\right )^{5/2}}{6 c}-\frac {7 b \left (\frac {\left (b x+c x^2\right )^{5/2}}{5 c}-\frac {b \int \left (c x^2+b x\right )^{3/2}dx}{2 c}\right )}{12 c}\right )}{14 c}\) |
| \(\Big \downarrow \) 1087 |
| \(\displaystyle \frac {B x^2 \left (b x+c x^2\right )^{5/2}}{7 c}-\frac {(9 b B-14 A c) \left (\frac {x \left (b x+c x^2\right )^{5/2}}{6 c}-\frac {7 b \left (\frac {\left (b x+c x^2\right )^{5/2}}{5 c}-\frac {b \left (\frac {(b+2 c x) \left (b x+c x^2\right )^{3/2}}{8 c}-\frac {3 b^2 \int \sqrt {c x^2+b x}dx}{16 c}\right )}{2 c}\right )}{12 c}\right )}{14 c}\) |
| \(\Big \downarrow \) 1087 |
| \(\displaystyle \frac {B x^2 \left (b x+c x^2\right )^{5/2}}{7 c}-\frac {(9 b B-14 A c) \left (\frac {x \left (b x+c x^2\right )^{5/2}}{6 c}-\frac {7 b \left (\frac {\left (b x+c x^2\right )^{5/2}}{5 c}-\frac {b \left (\frac {(b+2 c x) \left (b x+c x^2\right )^{3/2}}{8 c}-\frac {3 b^2 \left (\frac {(b+2 c x) \sqrt {b x+c x^2}}{4 c}-\frac {b^2 \int \frac {1}{\sqrt {c x^2+b x}}dx}{8 c}\right )}{16 c}\right )}{2 c}\right )}{12 c}\right )}{14 c}\) |
| \(\Big \downarrow \) 1091 |
| \(\displaystyle \frac {B x^2 \left (b x+c x^2\right )^{5/2}}{7 c}-\frac {(9 b B-14 A c) \left (\frac {x \left (b x+c x^2\right )^{5/2}}{6 c}-\frac {7 b \left (\frac {\left (b x+c x^2\right )^{5/2}}{5 c}-\frac {b \left (\frac {(b+2 c x) \left (b x+c x^2\right )^{3/2}}{8 c}-\frac {3 b^2 \left (\frac {(b+2 c x) \sqrt {b x+c x^2}}{4 c}-\frac {b^2 \int \frac {1}{1-\frac {c x^2}{c x^2+b x}}d\frac {x}{\sqrt {c x^2+b x}}}{4 c}\right )}{16 c}\right )}{2 c}\right )}{12 c}\right )}{14 c}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {B x^2 \left (b x+c x^2\right )^{5/2}}{7 c}-\frac {(9 b B-14 A c) \left (\frac {x \left (b x+c x^2\right )^{5/2}}{6 c}-\frac {7 b \left (\frac {\left (b x+c x^2\right )^{5/2}}{5 c}-\frac {b \left (\frac {(b+2 c x) \left (b x+c x^2\right )^{3/2}}{8 c}-\frac {3 b^2 \left (\frac {(b+2 c x) \sqrt {b x+c x^2}}{4 c}-\frac {b^2 \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{4 c^{3/2}}\right )}{16 c}\right )}{2 c}\right )}{12 c}\right )}{14 c}\) |
(B*x^2*(b*x + c*x^2)^(5/2))/(7*c) - ((9*b*B - 14*A*c)*((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*Ar cTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/(4*c^(3/2))))/(16*c)))/(2*c)))/(12*c )))/(14*c)
3.1.80.3.1 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 = 0.22 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.73
| method | result | size |
| pseudoelliptic | \(-\frac {7 \left (-\frac {15 \left (A c -\frac {9 B b}{14}\right ) b^{6} \operatorname {arctanh}\left (\frac {\sqrt {x \left (c x +b \right )}}{x \sqrt {c}}\right )}{8}+\left (-\frac {208 \left (\frac {75 B x}{91}+A \right ) x^{4} b \,c^{\frac {11}{2}}}{7}-\frac {160 \left (\frac {6 B x}{7}+A \right ) x^{5} c^{\frac {13}{2}}}{7}+\left (\frac {15 \left (\frac {3 B x}{7}+A \right ) b^{3} c^{\frac {3}{2}}}{8}-\frac {5 x \left (\frac {18 B x}{35}+A \right ) b^{2} c^{\frac {5}{2}}}{4}+b \,x^{2} \left (\frac {27 B x}{49}+A \right ) c^{\frac {7}{2}}-\frac {6 x^{3} \left (\frac {4 B x}{7}+A \right ) c^{\frac {9}{2}}}{7}-\frac {135 B \,b^{4} \sqrt {c}}{112}\right ) b^{2}\right ) \sqrt {x \left (c x +b \right )}\right )}{960 c^{\frac {11}{2}}}\) | \(148\) |
| risch | \(-\frac {\left (-15360 B \,c^{6} x^{6}-17920 A \,c^{6} x^{5}-19200 B b \,c^{5} x^{5}-23296 A b \,c^{5} x^{4}-384 B \,b^{2} c^{4} x^{4}-672 A \,b^{2} c^{4} x^{3}+432 B \,b^{3} c^{3} x^{3}+784 A \,b^{3} c^{3} x^{2}-504 B \,b^{4} c^{2} x^{2}-980 A \,b^{4} c^{2} x +630 B \,b^{5} c x +1470 A \,b^{5} c -945 B \,b^{6}\right ) x \left (c x +b \right )}{107520 c^{5} \sqrt {x \left (c x +b \right )}}+\frac {b^{6} \left (14 A c -9 B b \right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{2048 c^{\frac {11}{2}}}\) | \(193\) |
| default | \(B \left (\frac {x^{2} \left (c \,x^{2}+b x \right )^{\frac {5}{2}}}{7 c}-\frac {9 b \left (\frac {x \left (c \,x^{2}+b x \right )^{\frac {5}{2}}}{6 c}-\frac {7 b \left (\frac {\left (c \,x^{2}+b x \right )^{\frac {5}{2}}}{5 c}-\frac {b \left (\frac {\left (2 c x +b \right ) \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{8 c}-\frac {3 b^{2} \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x}}{4 c}-\frac {b^{2} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{8 c^{\frac {3}{2}}}\right )}{16 c}\right )}{2 c}\right )}{12 c}\right )}{14 c}\right )+A \left (\frac {x \left (c \,x^{2}+b x \right )^{\frac {5}{2}}}{6 c}-\frac {7 b \left (\frac {\left (c \,x^{2}+b x \right )^{\frac {5}{2}}}{5 c}-\frac {b \left (\frac {\left (2 c x +b \right ) \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{8 c}-\frac {3 b^{2} \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x}}{4 c}-\frac {b^{2} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{8 c^{\frac {3}{2}}}\right )}{16 c}\right )}{2 c}\right )}{12 c}\right )\) | \(298\) |
-7/960*(-15/8*(A*c-9/14*B*b)*b^6*arctanh((x*(c*x+b))^(1/2)/x/c^(1/2))+(-20 8/7*(75/91*B*x+A)*x^4*b*c^(11/2)-160/7*(6/7*B*x+A)*x^5*c^(13/2)+(15/8*(3/7 *B*x+A)*b^3*c^(3/2)-5/4*x*(18/35*B*x+A)*b^2*c^(5/2)+b*x^2*(27/49*B*x+A)*c^ (7/2)-6/7*x^3*(4/7*B*x+A)*c^(9/2)-135/112*B*b^4*c^(1/2))*b^2)*(x*(c*x+b))^ (1/2))/c^(11/2)
Time = 0.28 (sec) , antiderivative size = 399, normalized size of antiderivative = 1.97 \[ \int x^2 (A+B x) \left (b x+c x^2\right )^{3/2} \, dx=\left [-\frac {105 \, {\left (9 \, B b^{7} - 14 \, A b^{6} c\right )} \sqrt {c} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) - 2 \, {\left (15360 \, B c^{7} x^{6} + 945 \, B b^{6} c - 1470 \, A b^{5} c^{2} + 1280 \, {\left (15 \, B b c^{6} + 14 \, A c^{7}\right )} x^{5} + 128 \, {\left (3 \, B b^{2} c^{5} + 182 \, A b c^{6}\right )} x^{4} - 48 \, {\left (9 \, B b^{3} c^{4} - 14 \, A b^{2} c^{5}\right )} x^{3} + 56 \, {\left (9 \, B b^{4} c^{3} - 14 \, A b^{3} c^{4}\right )} x^{2} - 70 \, {\left (9 \, B b^{5} c^{2} - 14 \, A b^{4} c^{3}\right )} x\right )} \sqrt {c x^{2} + b x}}{215040 \, c^{6}}, \frac {105 \, {\left (9 \, B b^{7} - 14 \, A b^{6} c\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) + {\left (15360 \, B c^{7} x^{6} + 945 \, B b^{6} c - 1470 \, A b^{5} c^{2} + 1280 \, {\left (15 \, B b c^{6} + 14 \, A c^{7}\right )} x^{5} + 128 \, {\left (3 \, B b^{2} c^{5} + 182 \, A b c^{6}\right )} x^{4} - 48 \, {\left (9 \, B b^{3} c^{4} - 14 \, A b^{2} c^{5}\right )} x^{3} + 56 \, {\left (9 \, B b^{4} c^{3} - 14 \, A b^{3} c^{4}\right )} x^{2} - 70 \, {\left (9 \, B b^{5} c^{2} - 14 \, A b^{4} c^{3}\right )} x\right )} \sqrt {c x^{2} + b x}}{107520 \, c^{6}}\right ] \]
[-1/215040*(105*(9*B*b^7 - 14*A*b^6*c)*sqrt(c)*log(2*c*x + b + 2*sqrt(c*x^ 2 + b*x)*sqrt(c)) - 2*(15360*B*c^7*x^6 + 945*B*b^6*c - 1470*A*b^5*c^2 + 12 80*(15*B*b*c^6 + 14*A*c^7)*x^5 + 128*(3*B*b^2*c^5 + 182*A*b*c^6)*x^4 - 48* (9*B*b^3*c^4 - 14*A*b^2*c^5)*x^3 + 56*(9*B*b^4*c^3 - 14*A*b^3*c^4)*x^2 - 7 0*(9*B*b^5*c^2 - 14*A*b^4*c^3)*x)*sqrt(c*x^2 + b*x))/c^6, 1/107520*(105*(9 *B*b^7 - 14*A*b^6*c)*sqrt(-c)*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) + ( 15360*B*c^7*x^6 + 945*B*b^6*c - 1470*A*b^5*c^2 + 1280*(15*B*b*c^6 + 14*A*c ^7)*x^5 + 128*(3*B*b^2*c^5 + 182*A*b*c^6)*x^4 - 48*(9*B*b^3*c^4 - 14*A*b^2 *c^5)*x^3 + 56*(9*B*b^4*c^3 - 14*A*b^3*c^4)*x^2 - 70*(9*B*b^5*c^2 - 14*A*b ^4*c^3)*x)*sqrt(c*x^2 + b*x))/c^6]
Leaf count of result is larger than twice the leaf count of optimal. 444 vs. \(2 (192) = 384\).
Time = 0.53 (sec) , antiderivative size = 444, normalized size of antiderivative = 2.19 \[ \int x^2 (A+B x) \left (b x+c x^2\right )^{3/2} \, dx=\begin {cases} \frac {35 b^{4} \left (A b^{2} - \frac {9 b \left (2 A b c + B b^{2} - \frac {11 b \left (A c^{2} + \frac {15 B b c}{14}\right )}{12 c}\right )}{10 c}\right ) \left (\begin {cases} \frac {\log {\left (b + 2 \sqrt {c} \sqrt {b x + c x^{2}} + 2 c x \right )}}{\sqrt {c}} & \text {for}\: \frac {b^{2}}{c} \neq 0 \\\frac {\left (\frac {b}{2 c} + x\right ) \log {\left (\frac {b}{2 c} + x \right )}}{\sqrt {c \left (\frac {b}{2 c} + x\right )^{2}}} & \text {otherwise} \end {cases}\right )}{128 c^{4}} + \sqrt {b x + c x^{2}} \left (\frac {B c x^{6}}{7} - \frac {35 b^{3} \left (A b^{2} - \frac {9 b \left (2 A b c + B b^{2} - \frac {11 b \left (A c^{2} + \frac {15 B b c}{14}\right )}{12 c}\right )}{10 c}\right )}{64 c^{4}} + \frac {35 b^{2} x \left (A b^{2} - \frac {9 b \left (2 A b c + B b^{2} - \frac {11 b \left (A c^{2} + \frac {15 B b c}{14}\right )}{12 c}\right )}{10 c}\right )}{96 c^{3}} - \frac {7 b x^{2} \left (A b^{2} - \frac {9 b \left (2 A b c + B b^{2} - \frac {11 b \left (A c^{2} + \frac {15 B b c}{14}\right )}{12 c}\right )}{10 c}\right )}{24 c^{2}} + \frac {x^{5} \left (A c^{2} + \frac {15 B b c}{14}\right )}{6 c} + \frac {x^{4} \cdot \left (2 A b c + B b^{2} - \frac {11 b \left (A c^{2} + \frac {15 B b c}{14}\right )}{12 c}\right )}{5 c} + \frac {x^{3} \left (A b^{2} - \frac {9 b \left (2 A b c + B b^{2} - \frac {11 b \left (A c^{2} + \frac {15 B b c}{14}\right )}{12 c}\right )}{10 c}\right )}{4 c}\right ) & \text {for}\: c \neq 0 \\\frac {2 \left (\frac {A \left (b x\right )^{\frac {9}{2}}}{9} + \frac {B \left (b x\right )^{\frac {11}{2}}}{11 b}\right )}{b^{3}} & \text {for}\: b \neq 0 \\0 & \text {otherwise} \end {cases} \]
Piecewise((35*b**4*(A*b**2 - 9*b*(2*A*b*c + B*b**2 - 11*b*(A*c**2 + 15*B*b *c/14)/(12*c))/(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*c*x**6/7 - 35*b** 3*(A*b**2 - 9*b*(2*A*b*c + B*b**2 - 11*b*(A*c**2 + 15*B*b*c/14)/(12*c))/(1 0*c))/(64*c**4) + 35*b**2*x*(A*b**2 - 9*b*(2*A*b*c + B*b**2 - 11*b*(A*c**2 + 15*B*b*c/14)/(12*c))/(10*c))/(96*c**3) - 7*b*x**2*(A*b**2 - 9*b*(2*A*b* c + B*b**2 - 11*b*(A*c**2 + 15*B*b*c/14)/(12*c))/(10*c))/(24*c**2) + x**5* (A*c**2 + 15*B*b*c/14)/(6*c) + x**4*(2*A*b*c + B*b**2 - 11*b*(A*c**2 + 15* B*b*c/14)/(12*c))/(5*c) + x**3*(A*b**2 - 9*b*(2*A*b*c + B*b**2 - 11*b*(A*c **2 + 15*B*b*c/14)/(12*c))/(10*c))/(4*c)), Ne(c, 0)), (2*(A*(b*x)**(9/2)/9 + B*(b*x)**(11/2)/(11*b))/b**3, Ne(b, 0)), (0, True))
Time = 0.19 (sec) , antiderivative size = 324, normalized size of antiderivative = 1.60 \[ \int x^2 (A+B x) \left (b x+c x^2\right )^{3/2} \, dx=\frac {{\left (c x^{2} + b x\right )}^{\frac {5}{2}} B x^{2}}{7 \, c} + \frac {9 \, \sqrt {c x^{2} + b x} B b^{5} x}{512 \, c^{4}} - \frac {3 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} B b^{3} x}{64 \, c^{3}} - \frac {7 \, \sqrt {c x^{2} + b x} A b^{4} x}{256 \, c^{3}} - \frac {3 \, {\left (c x^{2} + b x\right )}^{\frac {5}{2}} B b x}{28 \, c^{2}} + \frac {7 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} A b^{2} x}{96 \, c^{2}} + \frac {{\left (c x^{2} + b x\right )}^{\frac {5}{2}} A x}{6 \, c} - \frac {9 \, B b^{7} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{2048 \, c^{\frac {11}{2}}} + \frac {7 \, A b^{6} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{1024 \, c^{\frac {9}{2}}} + \frac {9 \, \sqrt {c x^{2} + b x} B b^{6}}{1024 \, c^{5}} - \frac {3 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} B b^{4}}{128 \, c^{4}} - \frac {7 \, \sqrt {c x^{2} + b x} A b^{5}}{512 \, c^{4}} + \frac {3 \, {\left (c x^{2} + b x\right )}^{\frac {5}{2}} B b^{2}}{40 \, c^{3}} + \frac {7 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} A b^{3}}{192 \, c^{3}} - \frac {7 \, {\left (c x^{2} + b x\right )}^{\frac {5}{2}} A b}{60 \, c^{2}} \]
1/7*(c*x^2 + b*x)^(5/2)*B*x^2/c + 9/512*sqrt(c*x^2 + b*x)*B*b^5*x/c^4 - 3/ 64*(c*x^2 + b*x)^(3/2)*B*b^3*x/c^3 - 7/256*sqrt(c*x^2 + b*x)*A*b^4*x/c^3 - 3/28*(c*x^2 + b*x)^(5/2)*B*b*x/c^2 + 7/96*(c*x^2 + b*x)^(3/2)*A*b^2*x/c^2 + 1/6*(c*x^2 + b*x)^(5/2)*A*x/c - 9/2048*B*b^7*log(2*c*x + b + 2*sqrt(c*x ^2 + b*x)*sqrt(c))/c^(11/2) + 7/1024*A*b^6*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/c^(9/2) + 9/1024*sqrt(c*x^2 + b*x)*B*b^6/c^5 - 3/128*(c*x^2 + b*x)^(3/2)*B*b^4/c^4 - 7/512*sqrt(c*x^2 + b*x)*A*b^5/c^4 + 3/40*(c*x^2 + b*x)^(5/2)*B*b^2/c^3 + 7/192*(c*x^2 + b*x)^(3/2)*A*b^3/c^3 - 7/60*(c*x^2 + b*x)^(5/2)*A*b/c^2
Time = 0.28 (sec) , antiderivative size = 220, normalized size of antiderivative = 1.08 \[ \int x^2 (A+B x) \left (b x+c x^2\right )^{3/2} \, dx=\frac {1}{107520} \, \sqrt {c x^{2} + b x} {\left (2 \, {\left (4 \, {\left (2 \, {\left (8 \, {\left (10 \, {\left (12 \, B c x + \frac {15 \, B b c^{6} + 14 \, A c^{7}}{c^{6}}\right )} x + \frac {3 \, B b^{2} c^{5} + 182 \, A b c^{6}}{c^{6}}\right )} x - \frac {3 \, {\left (9 \, B b^{3} c^{4} - 14 \, A b^{2} c^{5}\right )}}{c^{6}}\right )} x + \frac {7 \, {\left (9 \, B b^{4} c^{3} - 14 \, A b^{3} c^{4}\right )}}{c^{6}}\right )} x - \frac {35 \, {\left (9 \, B b^{5} c^{2} - 14 \, A b^{4} c^{3}\right )}}{c^{6}}\right )} x + \frac {105 \, {\left (9 \, B b^{6} c - 14 \, A b^{5} c^{2}\right )}}{c^{6}}\right )} + \frac {{\left (9 \, B b^{7} - 14 \, A b^{6} c\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} + b \right |}\right )}{2048 \, c^{\frac {11}{2}}} \]
1/107520*sqrt(c*x^2 + b*x)*(2*(4*(2*(8*(10*(12*B*c*x + (15*B*b*c^6 + 14*A* c^7)/c^6)*x + (3*B*b^2*c^5 + 182*A*b*c^6)/c^6)*x - 3*(9*B*b^3*c^4 - 14*A*b ^2*c^5)/c^6)*x + 7*(9*B*b^4*c^3 - 14*A*b^3*c^4)/c^6)*x - 35*(9*B*b^5*c^2 - 14*A*b^4*c^3)/c^6)*x + 105*(9*B*b^6*c - 14*A*b^5*c^2)/c^6) + 1/2048*(9*B* b^7 - 14*A*b^6*c)*log(abs(2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) + b))/ c^(11/2)
Timed out. \[ \int x^2 (A+B x) \left (b x+c x^2\right )^{3/2} \, dx=\int x^2\,{\left (c\,x^2+b\,x\right )}^{3/2}\,\left (A+B\,x\right ) \,d x \]