Integrand size = 18, antiderivative size = 95 \[ \int x^2 (a+b x)^{3/2} (A+B x) \, dx=\frac {2 a^2 (A b-a B) (a+b x)^{5/2}}{5 b^4}-\frac {2 a (2 A b-3 a B) (a+b x)^{7/2}}{7 b^4}+\frac {2 (A b-3 a B) (a+b x)^{9/2}}{9 b^4}+\frac {2 B (a+b x)^{11/2}}{11 b^4} \]
2/5*a^2*(A*b-B*a)*(b*x+a)^(5/2)/b^4-2/7*a*(2*A*b-3*B*a)*(b*x+a)^(7/2)/b^4+ 2/9*(A*b-3*B*a)*(b*x+a)^(9/2)/b^4+2/11*B*(b*x+a)^(11/2)/b^4
Time = 0.05 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.72 \[ \int x^2 (a+b x)^{3/2} (A+B x) \, dx=\frac {2 (a+b x)^{5/2} \left (-48 a^3 B+35 b^3 x^2 (11 A+9 B x)+8 a^2 b (11 A+15 B x)-10 a b^2 x (22 A+21 B x)\right )}{3465 b^4} \]
(2*(a + b*x)^(5/2)*(-48*a^3*B + 35*b^3*x^2*(11*A + 9*B*x) + 8*a^2*b*(11*A + 15*B*x) - 10*a*b^2*x*(22*A + 21*B*x)))/(3465*b^4)
Time = 0.22 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {86, 2009}
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.
\(\displaystyle \int x^2 (a+b x)^{3/2} (A+B x) \, dx\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \int \left (-\frac {a^2 (a+b x)^{3/2} (a B-A b)}{b^3}+\frac {(a+b x)^{7/2} (A b-3 a B)}{b^3}+\frac {a (a+b x)^{5/2} (3 a B-2 A b)}{b^3}+\frac {B (a+b x)^{9/2}}{b^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2 a^2 (a+b x)^{5/2} (A b-a B)}{5 b^4}+\frac {2 (a+b x)^{9/2} (A b-3 a B)}{9 b^4}-\frac {2 a (a+b x)^{7/2} (2 A b-3 a B)}{7 b^4}+\frac {2 B (a+b x)^{11/2}}{11 b^4}\) |
(2*a^2*(A*b - a*B)*(a + b*x)^(5/2))/(5*b^4) - (2*a*(2*A*b - 3*a*B)*(a + b* x)^(7/2))/(7*b^4) + (2*(A*b - 3*a*B)*(a + b*x)^(9/2))/(9*b^4) + (2*B*(a + b*x)^(11/2))/(11*b^4)
3.4.100.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_ .), x_] :> Int[ExpandIntegrand[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && ((ILtQ[n, 0] && ILtQ[p, 0]) || EqQ[p, 1 ] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))
Time = 0.52 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.61
method | result | size |
pseudoelliptic | \(\frac {16 \left (b x +a \right )^{\frac {5}{2}} \left (\frac {35 x^{2} \left (\frac {9 B x}{11}+A \right ) b^{3}}{8}-\frac {5 x a \left (\frac {21 B x}{22}+A \right ) b^{2}}{2}+a^{2} \left (\frac {15 B x}{11}+A \right ) b -\frac {6 a^{3} B}{11}\right )}{315 b^{4}}\) | \(58\) |
gosper | \(\frac {2 \left (b x +a \right )^{\frac {5}{2}} \left (315 b^{3} B \,x^{3}+385 A \,b^{3} x^{2}-210 B a \,b^{2} x^{2}-220 a \,b^{2} A x +120 a^{2} b B x +88 a^{2} b A -48 a^{3} B \right )}{3465 b^{4}}\) | \(71\) |
derivativedivides | \(\frac {\frac {2 B \left (b x +a \right )^{\frac {11}{2}}}{11}+\frac {2 \left (A b -3 B a \right ) \left (b x +a \right )^{\frac {9}{2}}}{9}+\frac {2 \left (a^{2} B -2 a \left (A b -B a \right )\right ) \left (b x +a \right )^{\frac {7}{2}}}{7}+\frac {2 a^{2} \left (A b -B a \right ) \left (b x +a \right )^{\frac {5}{2}}}{5}}{b^{4}}\) | \(80\) |
default | \(\frac {\frac {2 B \left (b x +a \right )^{\frac {11}{2}}}{11}+\frac {2 \left (A b -3 B a \right ) \left (b x +a \right )^{\frac {9}{2}}}{9}+\frac {2 \left (a^{2} B -2 a \left (A b -B a \right )\right ) \left (b x +a \right )^{\frac {7}{2}}}{7}+\frac {2 a^{2} \left (A b -B a \right ) \left (b x +a \right )^{\frac {5}{2}}}{5}}{b^{4}}\) | \(80\) |
trager | \(\frac {2 \left (315 b^{5} B \,x^{5}+385 A \,b^{5} x^{4}+420 B a \,b^{4} x^{4}+550 A a \,b^{4} x^{3}+15 B \,a^{2} b^{3} x^{3}+33 A \,a^{2} b^{3} x^{2}-18 B \,a^{3} b^{2} x^{2}-44 a^{3} b^{2} A x +24 a^{4} b B x +88 a^{4} b A -48 a^{5} B \right ) \sqrt {b x +a}}{3465 b^{4}}\) | \(119\) |
risch | \(\frac {2 \left (315 b^{5} B \,x^{5}+385 A \,b^{5} x^{4}+420 B a \,b^{4} x^{4}+550 A a \,b^{4} x^{3}+15 B \,a^{2} b^{3} x^{3}+33 A \,a^{2} b^{3} x^{2}-18 B \,a^{3} b^{2} x^{2}-44 a^{3} b^{2} A x +24 a^{4} b B x +88 a^{4} b A -48 a^{5} B \right ) \sqrt {b x +a}}{3465 b^{4}}\) | \(119\) |
16/315*(b*x+a)^(5/2)*(35/8*x^2*(9/11*B*x+A)*b^3-5/2*x*a*(21/22*B*x+A)*b^2+ a^2*(15/11*B*x+A)*b-6/11*a^3*B)/b^4
Time = 0.22 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.26 \[ \int x^2 (a+b x)^{3/2} (A+B x) \, dx=\frac {2 \, {\left (315 \, B b^{5} x^{5} - 48 \, B a^{5} + 88 \, A a^{4} b + 35 \, {\left (12 \, B a b^{4} + 11 \, A b^{5}\right )} x^{4} + 5 \, {\left (3 \, B a^{2} b^{3} + 110 \, A a b^{4}\right )} x^{3} - 3 \, {\left (6 \, B a^{3} b^{2} - 11 \, A a^{2} b^{3}\right )} x^{2} + 4 \, {\left (6 \, B a^{4} b - 11 \, A a^{3} b^{2}\right )} x\right )} \sqrt {b x + a}}{3465 \, b^{4}} \]
2/3465*(315*B*b^5*x^5 - 48*B*a^5 + 88*A*a^4*b + 35*(12*B*a*b^4 + 11*A*b^5) *x^4 + 5*(3*B*a^2*b^3 + 110*A*a*b^4)*x^3 - 3*(6*B*a^3*b^2 - 11*A*a^2*b^3)* x^2 + 4*(6*B*a^4*b - 11*A*a^3*b^2)*x)*sqrt(b*x + a)/b^4
Time = 0.70 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.16 \[ \int x^2 (a+b x)^{3/2} (A+B x) \, dx=\begin {cases} \frac {2 \left (\frac {B \left (a + b x\right )^{\frac {11}{2}}}{11 b} + \frac {\left (a + b x\right )^{\frac {9}{2}} \left (A b - 3 B a\right )}{9 b} + \frac {\left (a + b x\right )^{\frac {7}{2}} \left (- 2 A a b + 3 B a^{2}\right )}{7 b} + \frac {\left (a + b x\right )^{\frac {5}{2}} \left (A a^{2} b - B a^{3}\right )}{5 b}\right )}{b^{3}} & \text {for}\: b \neq 0 \\a^{\frac {3}{2}} \left (\frac {A x^{3}}{3} + \frac {B x^{4}}{4}\right ) & \text {otherwise} \end {cases} \]
Piecewise((2*(B*(a + b*x)**(11/2)/(11*b) + (a + b*x)**(9/2)*(A*b - 3*B*a)/ (9*b) + (a + b*x)**(7/2)*(-2*A*a*b + 3*B*a**2)/(7*b) + (a + b*x)**(5/2)*(A *a**2*b - B*a**3)/(5*b))/b**3, Ne(b, 0)), (a**(3/2)*(A*x**3/3 + B*x**4/4), True))
Time = 0.19 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.81 \[ \int x^2 (a+b x)^{3/2} (A+B x) \, dx=\frac {2 \, {\left (315 \, {\left (b x + a\right )}^{\frac {11}{2}} B - 385 \, {\left (3 \, B a - A b\right )} {\left (b x + a\right )}^{\frac {9}{2}} + 495 \, {\left (3 \, B a^{2} - 2 \, A a b\right )} {\left (b x + a\right )}^{\frac {7}{2}} - 693 \, {\left (B a^{3} - A a^{2} b\right )} {\left (b x + a\right )}^{\frac {5}{2}}\right )}}{3465 \, b^{4}} \]
2/3465*(315*(b*x + a)^(11/2)*B - 385*(3*B*a - A*b)*(b*x + a)^(9/2) + 495*( 3*B*a^2 - 2*A*a*b)*(b*x + a)^(7/2) - 693*(B*a^3 - A*a^2*b)*(b*x + a)^(5/2) )/b^4
Leaf count of result is larger than twice the leaf count of optimal. 350 vs. \(2 (80) = 160\).
Time = 0.29 (sec) , antiderivative size = 350, normalized size of antiderivative = 3.68 \[ \int x^2 (a+b x)^{3/2} (A+B x) \, dx=\frac {2 \, {\left (\frac {231 \, {\left (3 \, {\left (b x + a\right )}^{\frac {5}{2}} - 10 \, {\left (b x + a\right )}^{\frac {3}{2}} a + 15 \, \sqrt {b x + a} a^{2}\right )} A a^{2}}{b^{2}} + \frac {99 \, {\left (5 \, {\left (b x + a\right )}^{\frac {7}{2}} - 21 \, {\left (b x + a\right )}^{\frac {5}{2}} a + 35 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{2} - 35 \, \sqrt {b x + a} a^{3}\right )} B a^{2}}{b^{3}} + \frac {198 \, {\left (5 \, {\left (b x + a\right )}^{\frac {7}{2}} - 21 \, {\left (b x + a\right )}^{\frac {5}{2}} a + 35 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{2} - 35 \, \sqrt {b x + a} a^{3}\right )} A a}{b^{2}} + \frac {22 \, {\left (35 \, {\left (b x + a\right )}^{\frac {9}{2}} - 180 \, {\left (b x + a\right )}^{\frac {7}{2}} a + 378 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{2} - 420 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{3} + 315 \, \sqrt {b x + a} a^{4}\right )} B a}{b^{3}} + \frac {11 \, {\left (35 \, {\left (b x + a\right )}^{\frac {9}{2}} - 180 \, {\left (b x + a\right )}^{\frac {7}{2}} a + 378 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{2} - 420 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{3} + 315 \, \sqrt {b x + a} a^{4}\right )} A}{b^{2}} + \frac {5 \, {\left (63 \, {\left (b x + a\right )}^{\frac {11}{2}} - 385 \, {\left (b x + a\right )}^{\frac {9}{2}} a + 990 \, {\left (b x + a\right )}^{\frac {7}{2}} a^{2} - 1386 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{3} + 1155 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{4} - 693 \, \sqrt {b x + a} a^{5}\right )} B}{b^{3}}\right )}}{3465 \, b} \]
2/3465*(231*(3*(b*x + a)^(5/2) - 10*(b*x + a)^(3/2)*a + 15*sqrt(b*x + a)*a ^2)*A*a^2/b^2 + 99*(5*(b*x + a)^(7/2) - 21*(b*x + a)^(5/2)*a + 35*(b*x + a )^(3/2)*a^2 - 35*sqrt(b*x + a)*a^3)*B*a^2/b^3 + 198*(5*(b*x + a)^(7/2) - 2 1*(b*x + a)^(5/2)*a + 35*(b*x + a)^(3/2)*a^2 - 35*sqrt(b*x + a)*a^3)*A*a/b ^2 + 22*(35*(b*x + a)^(9/2) - 180*(b*x + a)^(7/2)*a + 378*(b*x + a)^(5/2)* a^2 - 420*(b*x + a)^(3/2)*a^3 + 315*sqrt(b*x + a)*a^4)*B*a/b^3 + 11*(35*(b *x + a)^(9/2) - 180*(b*x + a)^(7/2)*a + 378*(b*x + a)^(5/2)*a^2 - 420*(b*x + a)^(3/2)*a^3 + 315*sqrt(b*x + a)*a^4)*A/b^2 + 5*(63*(b*x + a)^(11/2) - 385*(b*x + a)^(9/2)*a + 990*(b*x + a)^(7/2)*a^2 - 1386*(b*x + a)^(5/2)*a^3 + 1155*(b*x + a)^(3/2)*a^4 - 693*sqrt(b*x + a)*a^5)*B/b^3)/b
Time = 0.06 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.89 \[ \int x^2 (a+b x)^{3/2} (A+B x) \, dx=\frac {\left (6\,B\,a^2-4\,A\,a\,b\right )\,{\left (a+b\,x\right )}^{7/2}}{7\,b^4}+\frac {2\,B\,{\left (a+b\,x\right )}^{11/2}}{11\,b^4}+\frac {\left (2\,A\,b-6\,B\,a\right )\,{\left (a+b\,x\right )}^{9/2}}{9\,b^4}-\frac {\left (2\,B\,a^3-2\,A\,a^2\,b\right )\,{\left (a+b\,x\right )}^{5/2}}{5\,b^4} \]