Integrand size = 29, antiderivative size = 153 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^3}{x^{5/2}} \, dx=-\frac {2 a^6 A}{3 x^{3/2}}-\frac {2 a^5 (6 A b+a B)}{\sqrt {x}}+6 a^4 b (5 A b+2 a B) \sqrt {x}+\frac {10}{3} a^3 b^2 (4 A b+3 a B) x^{3/2}+2 a^2 b^3 (3 A b+4 a B) x^{5/2}+\frac {6}{7} a b^4 (2 A b+5 a B) x^{7/2}+\frac {2}{9} b^5 (A b+6 a B) x^{9/2}+\frac {2}{11} b^6 B x^{11/2} \]
-2/3*a^6*A/x^(3/2)+10/3*a^3*b^2*(4*A*b+3*B*a)*x^(3/2)+2*a^2*b^3*(3*A*b+4*B *a)*x^(5/2)+6/7*a*b^4*(2*A*b+5*B*a)*x^(7/2)+2/9*b^5*(A*b+6*B*a)*x^(9/2)+2/ 11*b^6*B*x^(11/2)-2*a^5*(6*A*b+B*a)/x^(1/2)+6*a^4*b*(5*A*b+2*B*a)*x^(1/2)
Time = 0.08 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.81 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^3}{x^{5/2}} \, dx=\frac {2 \left (4158 a^5 b x (-A+B x)+3465 a^4 b^2 x^2 (3 A+B x)-231 a^6 (A+3 B x)+924 a^3 b^3 x^3 (5 A+3 B x)+297 a^2 b^4 x^4 (7 A+5 B x)+66 a b^5 x^5 (9 A+7 B x)+7 b^6 x^6 (11 A+9 B x)\right )}{693 x^{3/2}} \]
(2*(4158*a^5*b*x*(-A + B*x) + 3465*a^4*b^2*x^2*(3*A + B*x) - 231*a^6*(A + 3*B*x) + 924*a^3*b^3*x^3*(5*A + 3*B*x) + 297*a^2*b^4*x^4*(7*A + 5*B*x) + 6 6*a*b^5*x^5*(9*A + 7*B*x) + 7*b^6*x^6*(11*A + 9*B*x)))/(693*x^(3/2))
Time = 0.29 (sec) , antiderivative size = 153, 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.
\(\displaystyle \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^3 (A+B x)}{x^{5/2}} \, dx\) |
\(\Big \downarrow \) 1184 |
\(\displaystyle \frac {\int \frac {b^6 (a+b x)^6 (A+B x)}{x^{5/2}}dx}{b^6}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \frac {(a+b x)^6 (A+B x)}{x^{5/2}}dx\) |
\(\Big \downarrow \) 85 |
\(\displaystyle \int \left (\frac {a^6 A}{x^{5/2}}+\frac {a^5 (a B+6 A b)}{x^{3/2}}+\frac {3 a^4 b (2 a B+5 A b)}{\sqrt {x}}+5 a^3 b^2 \sqrt {x} (3 a B+4 A b)+5 a^2 b^3 x^{3/2} (4 a B+3 A b)+b^5 x^{7/2} (6 a B+A b)+3 a b^4 x^{5/2} (5 a B+2 A b)+b^6 B x^{9/2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {2 a^6 A}{3 x^{3/2}}-\frac {2 a^5 (a B+6 A b)}{\sqrt {x}}+6 a^4 b \sqrt {x} (2 a B+5 A b)+\frac {10}{3} a^3 b^2 x^{3/2} (3 a B+4 A b)+2 a^2 b^3 x^{5/2} (4 a B+3 A b)+\frac {2}{9} b^5 x^{9/2} (6 a B+A b)+\frac {6}{7} a b^4 x^{7/2} (5 a B+2 A b)+\frac {2}{11} b^6 B x^{11/2}\) |
(-2*a^6*A)/(3*x^(3/2)) - (2*a^5*(6*A*b + a*B))/Sqrt[x] + 6*a^4*b*(5*A*b + 2*a*B)*Sqrt[x] + (10*a^3*b^2*(4*A*b + 3*a*B)*x^(3/2))/3 + 2*a^2*b^3*(3*A*b + 4*a*B)*x^(5/2) + (6*a*b^4*(2*A*b + 5*a*B)*x^(7/2))/7 + (2*b^5*(A*b + 6* a*B)*x^(9/2))/9 + (2*b^6*B*x^(11/2))/11
3.8.53.3.1 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 = 0.13 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.96
method | result | size |
derivativedivides | \(\frac {2 b^{6} B \,x^{\frac {11}{2}}}{11}+\frac {2 A \,b^{6} x^{\frac {9}{2}}}{9}+\frac {4 B a \,b^{5} x^{\frac {9}{2}}}{3}+\frac {12 A a \,b^{5} x^{\frac {7}{2}}}{7}+\frac {30 B \,a^{2} b^{4} x^{\frac {7}{2}}}{7}+6 A \,a^{2} b^{4} x^{\frac {5}{2}}+8 B \,a^{3} b^{3} x^{\frac {5}{2}}+\frac {40 A \,a^{3} b^{3} x^{\frac {3}{2}}}{3}+10 B \,a^{4} b^{2} x^{\frac {3}{2}}+30 A \,a^{4} b^{2} \sqrt {x}+12 B \,a^{5} b \sqrt {x}-\frac {2 a^{6} A}{3 x^{\frac {3}{2}}}-\frac {2 a^{5} \left (6 A b +B a \right )}{\sqrt {x}}\) | \(147\) |
default | \(\frac {2 b^{6} B \,x^{\frac {11}{2}}}{11}+\frac {2 A \,b^{6} x^{\frac {9}{2}}}{9}+\frac {4 B a \,b^{5} x^{\frac {9}{2}}}{3}+\frac {12 A a \,b^{5} x^{\frac {7}{2}}}{7}+\frac {30 B \,a^{2} b^{4} x^{\frac {7}{2}}}{7}+6 A \,a^{2} b^{4} x^{\frac {5}{2}}+8 B \,a^{3} b^{3} x^{\frac {5}{2}}+\frac {40 A \,a^{3} b^{3} x^{\frac {3}{2}}}{3}+10 B \,a^{4} b^{2} x^{\frac {3}{2}}+30 A \,a^{4} b^{2} \sqrt {x}+12 B \,a^{5} b \sqrt {x}-\frac {2 a^{6} A}{3 x^{\frac {3}{2}}}-\frac {2 a^{5} \left (6 A b +B a \right )}{\sqrt {x}}\) | \(147\) |
gosper | \(-\frac {2 \left (-63 b^{6} B \,x^{7}-77 A \,b^{6} x^{6}-462 x^{6} B a \,b^{5}-594 a A \,b^{5} x^{5}-1485 x^{5} B \,b^{4} a^{2}-2079 a^{2} A \,b^{4} x^{4}-2772 x^{4} B \,a^{3} b^{3}-4620 a^{3} A \,b^{3} x^{3}-3465 x^{3} B \,a^{4} b^{2}-10395 a^{4} A \,b^{2} x^{2}-4158 x^{2} B \,a^{5} b +4158 a^{5} A b x +693 x B \,a^{6}+231 A \,a^{6}\right )}{693 x^{\frac {3}{2}}}\) | \(148\) |
trager | \(-\frac {2 \left (-63 b^{6} B \,x^{7}-77 A \,b^{6} x^{6}-462 x^{6} B a \,b^{5}-594 a A \,b^{5} x^{5}-1485 x^{5} B \,b^{4} a^{2}-2079 a^{2} A \,b^{4} x^{4}-2772 x^{4} B \,a^{3} b^{3}-4620 a^{3} A \,b^{3} x^{3}-3465 x^{3} B \,a^{4} b^{2}-10395 a^{4} A \,b^{2} x^{2}-4158 x^{2} B \,a^{5} b +4158 a^{5} A b x +693 x B \,a^{6}+231 A \,a^{6}\right )}{693 x^{\frac {3}{2}}}\) | \(148\) |
risch | \(-\frac {2 \left (-63 b^{6} B \,x^{7}-77 A \,b^{6} x^{6}-462 x^{6} B a \,b^{5}-594 a A \,b^{5} x^{5}-1485 x^{5} B \,b^{4} a^{2}-2079 a^{2} A \,b^{4} x^{4}-2772 x^{4} B \,a^{3} b^{3}-4620 a^{3} A \,b^{3} x^{3}-3465 x^{3} B \,a^{4} b^{2}-10395 a^{4} A \,b^{2} x^{2}-4158 x^{2} B \,a^{5} b +4158 a^{5} A b x +693 x B \,a^{6}+231 A \,a^{6}\right )}{693 x^{\frac {3}{2}}}\) | \(148\) |
2/11*b^6*B*x^(11/2)+2/9*A*b^6*x^(9/2)+4/3*B*a*b^5*x^(9/2)+12/7*A*a*b^5*x^( 7/2)+30/7*B*a^2*b^4*x^(7/2)+6*A*a^2*b^4*x^(5/2)+8*B*a^3*b^3*x^(5/2)+40/3*A *a^3*b^3*x^(3/2)+10*B*a^4*b^2*x^(3/2)+30*A*a^4*b^2*x^(1/2)+12*B*a^5*b*x^(1 /2)-2/3*a^6*A/x^(3/2)-2*a^5*(6*A*b+B*a)/x^(1/2)
Time = 0.25 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.96 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^3}{x^{5/2}} \, dx=\frac {2 \, {\left (63 \, B b^{6} x^{7} - 231 \, A a^{6} + 77 \, {\left (6 \, B a b^{5} + A b^{6}\right )} x^{6} + 297 \, {\left (5 \, B a^{2} b^{4} + 2 \, A a b^{5}\right )} x^{5} + 693 \, {\left (4 \, B a^{3} b^{3} + 3 \, A a^{2} b^{4}\right )} x^{4} + 1155 \, {\left (3 \, B a^{4} b^{2} + 4 \, A a^{3} b^{3}\right )} x^{3} + 2079 \, {\left (2 \, B a^{5} b + 5 \, A a^{4} b^{2}\right )} x^{2} - 693 \, {\left (B a^{6} + 6 \, A a^{5} b\right )} x\right )}}{693 \, x^{\frac {3}{2}}} \]
2/693*(63*B*b^6*x^7 - 231*A*a^6 + 77*(6*B*a*b^5 + A*b^6)*x^6 + 297*(5*B*a^ 2*b^4 + 2*A*a*b^5)*x^5 + 693*(4*B*a^3*b^3 + 3*A*a^2*b^4)*x^4 + 1155*(3*B*a ^4*b^2 + 4*A*a^3*b^3)*x^3 + 2079*(2*B*a^5*b + 5*A*a^4*b^2)*x^2 - 693*(B*a^ 6 + 6*A*a^5*b)*x)/x^(3/2)
Time = 0.47 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.33 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^3}{x^{5/2}} \, dx=- \frac {2 A a^{6}}{3 x^{\frac {3}{2}}} - \frac {12 A a^{5} b}{\sqrt {x}} + 30 A a^{4} b^{2} \sqrt {x} + \frac {40 A a^{3} b^{3} x^{\frac {3}{2}}}{3} + 6 A a^{2} b^{4} x^{\frac {5}{2}} + \frac {12 A a b^{5} x^{\frac {7}{2}}}{7} + \frac {2 A b^{6} x^{\frac {9}{2}}}{9} - \frac {2 B a^{6}}{\sqrt {x}} + 12 B a^{5} b \sqrt {x} + 10 B a^{4} b^{2} x^{\frac {3}{2}} + 8 B a^{3} b^{3} x^{\frac {5}{2}} + \frac {30 B a^{2} b^{4} x^{\frac {7}{2}}}{7} + \frac {4 B a b^{5} x^{\frac {9}{2}}}{3} + \frac {2 B b^{6} x^{\frac {11}{2}}}{11} \]
-2*A*a**6/(3*x**(3/2)) - 12*A*a**5*b/sqrt(x) + 30*A*a**4*b**2*sqrt(x) + 40 *A*a**3*b**3*x**(3/2)/3 + 6*A*a**2*b**4*x**(5/2) + 12*A*a*b**5*x**(7/2)/7 + 2*A*b**6*x**(9/2)/9 - 2*B*a**6/sqrt(x) + 12*B*a**5*b*sqrt(x) + 10*B*a**4 *b**2*x**(3/2) + 8*B*a**3*b**3*x**(5/2) + 30*B*a**2*b**4*x**(7/2)/7 + 4*B* a*b**5*x**(9/2)/3 + 2*B*b**6*x**(11/2)/11
Time = 0.21 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.96 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^3}{x^{5/2}} \, dx=\frac {2}{11} \, B b^{6} x^{\frac {11}{2}} + \frac {2}{9} \, {\left (6 \, B a b^{5} + A b^{6}\right )} x^{\frac {9}{2}} + \frac {6}{7} \, {\left (5 \, B a^{2} b^{4} + 2 \, A a b^{5}\right )} x^{\frac {7}{2}} + 2 \, {\left (4 \, B a^{3} b^{3} + 3 \, A a^{2} b^{4}\right )} x^{\frac {5}{2}} + \frac {10}{3} \, {\left (3 \, B a^{4} b^{2} + 4 \, A a^{3} b^{3}\right )} x^{\frac {3}{2}} + 6 \, {\left (2 \, B a^{5} b + 5 \, A a^{4} b^{2}\right )} \sqrt {x} - \frac {2 \, {\left (A a^{6} + 3 \, {\left (B a^{6} + 6 \, A a^{5} b\right )} x\right )}}{3 \, x^{\frac {3}{2}}} \]
2/11*B*b^6*x^(11/2) + 2/9*(6*B*a*b^5 + A*b^6)*x^(9/2) + 6/7*(5*B*a^2*b^4 + 2*A*a*b^5)*x^(7/2) + 2*(4*B*a^3*b^3 + 3*A*a^2*b^4)*x^(5/2) + 10/3*(3*B*a^ 4*b^2 + 4*A*a^3*b^3)*x^(3/2) + 6*(2*B*a^5*b + 5*A*a^4*b^2)*sqrt(x) - 2/3*( A*a^6 + 3*(B*a^6 + 6*A*a^5*b)*x)/x^(3/2)
Time = 0.29 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.96 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^3}{x^{5/2}} \, dx=\frac {2}{11} \, B b^{6} x^{\frac {11}{2}} + \frac {4}{3} \, B a b^{5} x^{\frac {9}{2}} + \frac {2}{9} \, A b^{6} x^{\frac {9}{2}} + \frac {30}{7} \, B a^{2} b^{4} x^{\frac {7}{2}} + \frac {12}{7} \, A a b^{5} x^{\frac {7}{2}} + 8 \, B a^{3} b^{3} x^{\frac {5}{2}} + 6 \, A a^{2} b^{4} x^{\frac {5}{2}} + 10 \, B a^{4} b^{2} x^{\frac {3}{2}} + \frac {40}{3} \, A a^{3} b^{3} x^{\frac {3}{2}} + 12 \, B a^{5} b \sqrt {x} + 30 \, A a^{4} b^{2} \sqrt {x} - \frac {2 \, {\left (3 \, B a^{6} x + 18 \, A a^{5} b x + A a^{6}\right )}}{3 \, x^{\frac {3}{2}}} \]
2/11*B*b^6*x^(11/2) + 4/3*B*a*b^5*x^(9/2) + 2/9*A*b^6*x^(9/2) + 30/7*B*a^2 *b^4*x^(7/2) + 12/7*A*a*b^5*x^(7/2) + 8*B*a^3*b^3*x^(5/2) + 6*A*a^2*b^4*x^ (5/2) + 10*B*a^4*b^2*x^(3/2) + 40/3*A*a^3*b^3*x^(3/2) + 12*B*a^5*b*sqrt(x) + 30*A*a^4*b^2*sqrt(x) - 2/3*(3*B*a^6*x + 18*A*a^5*b*x + A*a^6)/x^(3/2)
Time = 0.05 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.86 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^3}{x^{5/2}} \, dx=x^{9/2}\,\left (\frac {2\,A\,b^6}{9}+\frac {4\,B\,a\,b^5}{3}\right )-\frac {x\,\left (2\,B\,a^6+12\,A\,b\,a^5\right )+\frac {2\,A\,a^6}{3}}{x^{3/2}}+\frac {2\,B\,b^6\,x^{11/2}}{11}+\frac {10\,a^3\,b^2\,x^{3/2}\,\left (4\,A\,b+3\,B\,a\right )}{3}+2\,a^2\,b^3\,x^{5/2}\,\left (3\,A\,b+4\,B\,a\right )+6\,a^4\,b\,\sqrt {x}\,\left (5\,A\,b+2\,B\,a\right )+\frac {6\,a\,b^4\,x^{7/2}\,\left (2\,A\,b+5\,B\,a\right )}{7} \]