Integrand size = 29, antiderivative size = 111 \[ \int \sqrt {x} (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2 \, dx=\frac {2}{3} a^4 A x^{3/2}+\frac {2}{5} a^3 (4 A b+a B) x^{5/2}+\frac {4}{7} a^2 b (3 A b+2 a B) x^{7/2}+\frac {4}{9} a b^2 (2 A b+3 a B) x^{9/2}+\frac {2}{11} b^3 (A b+4 a B) x^{11/2}+\frac {2}{13} b^4 B x^{13/2} \]
2/3*a^4*A*x^(3/2)+2/5*a^3*(4*A*b+B*a)*x^(5/2)+4/7*a^2*b*(3*A*b+2*B*a)*x^(7 /2)+4/9*a*b^2*(2*A*b+3*B*a)*x^(9/2)+2/11*b^3*(A*b+4*B*a)*x^(11/2)+2/13*b^4 *B*x^(13/2)
Time = 0.05 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.81 \[ \int \sqrt {x} (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2 \, dx=\frac {2 x^{3/2} \left (3003 a^4 (5 A+3 B x)+5148 a^3 b x (7 A+5 B x)+4290 a^2 b^2 x^2 (9 A+7 B x)+1820 a b^3 x^3 (11 A+9 B x)+315 b^4 x^4 (13 A+11 B x)\right )}{45045} \]
(2*x^(3/2)*(3003*a^4*(5*A + 3*B*x) + 5148*a^3*b*x*(7*A + 5*B*x) + 4290*a^2 *b^2*x^2*(9*A + 7*B*x) + 1820*a*b^3*x^3*(11*A + 9*B*x) + 315*b^4*x^4*(13*A + 11*B*x)))/45045
Time = 0.25 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {1184, 27, 85, 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.
\(\displaystyle \int \sqrt {x} \left (a^2+2 a b x+b^2 x^2\right )^2 (A+B x) \, dx\) |
\(\Big \downarrow \) 1184 |
\(\displaystyle \frac {\int b^4 \sqrt {x} (a+b x)^4 (A+B x)dx}{b^4}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \sqrt {x} (a+b x)^4 (A+B x)dx\) |
\(\Big \downarrow \) 85 |
\(\displaystyle \int \left (a^4 A \sqrt {x}+a^3 x^{3/2} (a B+4 A b)+2 a^2 b x^{5/2} (2 a B+3 A b)+b^3 x^{9/2} (4 a B+A b)+2 a b^2 x^{7/2} (3 a B+2 A b)+b^4 B x^{11/2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2}{3} a^4 A x^{3/2}+\frac {2}{5} a^3 x^{5/2} (a B+4 A b)+\frac {4}{7} a^2 b x^{7/2} (2 a B+3 A b)+\frac {2}{11} b^3 x^{11/2} (4 a B+A b)+\frac {4}{9} a b^2 x^{9/2} (3 a B+2 A b)+\frac {2}{13} b^4 B x^{13/2}\) |
(2*a^4*A*x^(3/2))/3 + (2*a^3*(4*A*b + a*B)*x^(5/2))/5 + (4*a^2*b*(3*A*b + 2*a*B)*x^(7/2))/7 + (4*a*b^2*(2*A*b + 3*a*B)*x^(9/2))/9 + (2*b^3*(A*b + 4* a*B)*x^(11/2))/11 + (2*b^4*B*x^(13/2))/13
3.8.41.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_] : > Int[ExpandIntegrand[(a + b*x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && (NeQ[n, -1] || EqQ[p, 1]) && NeQ[b*e + a* f, 0] && ( !IntegerQ[n] || LtQ[9*p + 5*n, 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, d, e, f])) && (NeQ[n + p + 3, 0] || EqQ[p, 1])
Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_) + (b_.)*(x_ ) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[1/c^p Int[(d + e*x)^m*(f + g*x )^n*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n}, x] && E qQ[b^2 - 4*a*c, 0] && IntegerQ[p]
Time = 0.18 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.90
method | result | size |
gosper | \(\frac {2 x^{\frac {3}{2}} \left (3465 b^{4} B \,x^{5}+4095 A \,b^{4} x^{4}+16380 x^{4} B \,b^{3} a +20020 a A \,b^{3} x^{3}+30030 x^{3} B \,a^{2} b^{2}+38610 a^{2} A \,b^{2} x^{2}+25740 x^{2} B \,a^{3} b +36036 a^{3} A b x +9009 a^{4} B x +15015 A \,a^{4}\right )}{45045}\) | \(100\) |
derivativedivides | \(\frac {2 b^{4} B \,x^{\frac {13}{2}}}{13}+\frac {2 \left (A \,b^{4}+4 B \,b^{3} a \right ) x^{\frac {11}{2}}}{11}+\frac {2 \left (4 A \,b^{3} a +6 B \,a^{2} b^{2}\right ) x^{\frac {9}{2}}}{9}+\frac {2 \left (6 A \,a^{2} b^{2}+4 B \,a^{3} b \right ) x^{\frac {7}{2}}}{7}+\frac {2 \left (4 A \,a^{3} b +B \,a^{4}\right ) x^{\frac {5}{2}}}{5}+\frac {2 a^{4} A \,x^{\frac {3}{2}}}{3}\) | \(100\) |
default | \(\frac {2 b^{4} B \,x^{\frac {13}{2}}}{13}+\frac {2 \left (A \,b^{4}+4 B \,b^{3} a \right ) x^{\frac {11}{2}}}{11}+\frac {2 \left (4 A \,b^{3} a +6 B \,a^{2} b^{2}\right ) x^{\frac {9}{2}}}{9}+\frac {2 \left (6 A \,a^{2} b^{2}+4 B \,a^{3} b \right ) x^{\frac {7}{2}}}{7}+\frac {2 \left (4 A \,a^{3} b +B \,a^{4}\right ) x^{\frac {5}{2}}}{5}+\frac {2 a^{4} A \,x^{\frac {3}{2}}}{3}\) | \(100\) |
trager | \(\frac {2 x^{\frac {3}{2}} \left (3465 b^{4} B \,x^{5}+4095 A \,b^{4} x^{4}+16380 x^{4} B \,b^{3} a +20020 a A \,b^{3} x^{3}+30030 x^{3} B \,a^{2} b^{2}+38610 a^{2} A \,b^{2} x^{2}+25740 x^{2} B \,a^{3} b +36036 a^{3} A b x +9009 a^{4} B x +15015 A \,a^{4}\right )}{45045}\) | \(100\) |
risch | \(\frac {2 x^{\frac {3}{2}} \left (3465 b^{4} B \,x^{5}+4095 A \,b^{4} x^{4}+16380 x^{4} B \,b^{3} a +20020 a A \,b^{3} x^{3}+30030 x^{3} B \,a^{2} b^{2}+38610 a^{2} A \,b^{2} x^{2}+25740 x^{2} B \,a^{3} b +36036 a^{3} A b x +9009 a^{4} B x +15015 A \,a^{4}\right )}{45045}\) | \(100\) |
2/45045*x^(3/2)*(3465*B*b^4*x^5+4095*A*b^4*x^4+16380*B*a*b^3*x^4+20020*A*a *b^3*x^3+30030*B*a^2*b^2*x^3+38610*A*a^2*b^2*x^2+25740*B*a^3*b*x^2+36036*A *a^3*b*x+9009*B*a^4*x+15015*A*a^4)
Time = 0.29 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.92 \[ \int \sqrt {x} (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2 \, dx=\frac {2}{45045} \, {\left (3465 \, B b^{4} x^{6} + 15015 \, A a^{4} x + 4095 \, {\left (4 \, B a b^{3} + A b^{4}\right )} x^{5} + 10010 \, {\left (3 \, B a^{2} b^{2} + 2 \, A a b^{3}\right )} x^{4} + 12870 \, {\left (2 \, B a^{3} b + 3 \, A a^{2} b^{2}\right )} x^{3} + 9009 \, {\left (B a^{4} + 4 \, A a^{3} b\right )} x^{2}\right )} \sqrt {x} \]
2/45045*(3465*B*b^4*x^6 + 15015*A*a^4*x + 4095*(4*B*a*b^3 + A*b^4)*x^5 + 1 0010*(3*B*a^2*b^2 + 2*A*a*b^3)*x^4 + 12870*(2*B*a^3*b + 3*A*a^2*b^2)*x^3 + 9009*(B*a^4 + 4*A*a^3*b)*x^2)*sqrt(x)
Time = 1.02 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.12 \[ \int \sqrt {x} (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2 \, dx=\frac {2 A a^{4} x^{\frac {3}{2}}}{3} + \frac {2 B b^{4} x^{\frac {13}{2}}}{13} + \frac {2 x^{\frac {11}{2}} \left (A b^{4} + 4 B a b^{3}\right )}{11} + \frac {2 x^{\frac {9}{2}} \cdot \left (4 A a b^{3} + 6 B a^{2} b^{2}\right )}{9} + \frac {2 x^{\frac {7}{2}} \cdot \left (6 A a^{2} b^{2} + 4 B a^{3} b\right )}{7} + \frac {2 x^{\frac {5}{2}} \cdot \left (4 A a^{3} b + B a^{4}\right )}{5} \]
2*A*a**4*x**(3/2)/3 + 2*B*b**4*x**(13/2)/13 + 2*x**(11/2)*(A*b**4 + 4*B*a* b**3)/11 + 2*x**(9/2)*(4*A*a*b**3 + 6*B*a**2*b**2)/9 + 2*x**(7/2)*(6*A*a** 2*b**2 + 4*B*a**3*b)/7 + 2*x**(5/2)*(4*A*a**3*b + B*a**4)/5
Time = 0.21 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.89 \[ \int \sqrt {x} (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2 \, dx=\frac {2}{13} \, B b^{4} x^{\frac {13}{2}} + \frac {2}{3} \, A a^{4} x^{\frac {3}{2}} + \frac {2}{11} \, {\left (4 \, B a b^{3} + A b^{4}\right )} x^{\frac {11}{2}} + \frac {4}{9} \, {\left (3 \, B a^{2} b^{2} + 2 \, A a b^{3}\right )} x^{\frac {9}{2}} + \frac {4}{7} \, {\left (2 \, B a^{3} b + 3 \, A a^{2} b^{2}\right )} x^{\frac {7}{2}} + \frac {2}{5} \, {\left (B a^{4} + 4 \, A a^{3} b\right )} x^{\frac {5}{2}} \]
2/13*B*b^4*x^(13/2) + 2/3*A*a^4*x^(3/2) + 2/11*(4*B*a*b^3 + A*b^4)*x^(11/2 ) + 4/9*(3*B*a^2*b^2 + 2*A*a*b^3)*x^(9/2) + 4/7*(2*B*a^3*b + 3*A*a^2*b^2)* x^(7/2) + 2/5*(B*a^4 + 4*A*a^3*b)*x^(5/2)
Time = 0.37 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.91 \[ \int \sqrt {x} (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2 \, dx=\frac {2}{13} \, B b^{4} x^{\frac {13}{2}} + \frac {8}{11} \, B a b^{3} x^{\frac {11}{2}} + \frac {2}{11} \, A b^{4} x^{\frac {11}{2}} + \frac {4}{3} \, B a^{2} b^{2} x^{\frac {9}{2}} + \frac {8}{9} \, A a b^{3} x^{\frac {9}{2}} + \frac {8}{7} \, B a^{3} b x^{\frac {7}{2}} + \frac {12}{7} \, A a^{2} b^{2} x^{\frac {7}{2}} + \frac {2}{5} \, B a^{4} x^{\frac {5}{2}} + \frac {8}{5} \, A a^{3} b x^{\frac {5}{2}} + \frac {2}{3} \, A a^{4} x^{\frac {3}{2}} \]
2/13*B*b^4*x^(13/2) + 8/11*B*a*b^3*x^(11/2) + 2/11*A*b^4*x^(11/2) + 4/3*B* a^2*b^2*x^(9/2) + 8/9*A*a*b^3*x^(9/2) + 8/7*B*a^3*b*x^(7/2) + 12/7*A*a^2*b ^2*x^(7/2) + 2/5*B*a^4*x^(5/2) + 8/5*A*a^3*b*x^(5/2) + 2/3*A*a^4*x^(3/2)
Time = 0.04 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.82 \[ \int \sqrt {x} (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2 \, dx=x^{5/2}\,\left (\frac {2\,B\,a^4}{5}+\frac {8\,A\,b\,a^3}{5}\right )+x^{11/2}\,\left (\frac {2\,A\,b^4}{11}+\frac {8\,B\,a\,b^3}{11}\right )+\frac {2\,A\,a^4\,x^{3/2}}{3}+\frac {2\,B\,b^4\,x^{13/2}}{13}+\frac {4\,a^2\,b\,x^{7/2}\,\left (3\,A\,b+2\,B\,a\right )}{7}+\frac {4\,a\,b^2\,x^{9/2}\,\left (2\,A\,b+3\,B\,a\right )}{9} \]