Integrand size = 29, antiderivative size = 151 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^3}{x^{9/2}} \, dx=-\frac {2 a^6 A}{7 x^{7/2}}-\frac {2 a^5 (6 A b+a B)}{5 x^{5/2}}-\frac {2 a^4 b (5 A b+2 a B)}{x^{3/2}}-\frac {10 a^3 b^2 (4 A b+3 a B)}{\sqrt {x}}+10 a^2 b^3 (3 A b+4 a B) \sqrt {x}+2 a b^4 (2 A b+5 a B) x^{3/2}+\frac {2}{5} b^5 (A b+6 a B) x^{5/2}+\frac {2}{7} b^6 B x^{7/2} \]
-2/7*a^6*A/x^(7/2)-2/5*a^5*(6*A*b+B*a)/x^(5/2)-2*a^4*b*(5*A*b+2*B*a)/x^(3/ 2)+2*a*b^4*(2*A*b+5*B*a)*x^(3/2)+2/5*b^5*(A*b+6*B*a)*x^(5/2)+2/7*b^6*B*x^( 7/2)-10*a^3*b^2*(4*A*b+3*B*a)/x^(1/2)+10*a^2*b^3*(3*A*b+4*B*a)*x^(1/2)
Time = 0.07 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.81 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^3}{x^{9/2}} \, dx=\frac {2 \left (700 a^3 b^3 x^3 (-A+B x)+175 a^2 b^4 x^4 (3 A+B x)-175 a^4 b^2 x^2 (A+3 B x)+14 a b^5 x^5 (5 A+3 B x)-14 a^5 b x (3 A+5 B x)+b^6 x^6 (7 A+5 B x)-a^6 (5 A+7 B x)\right )}{35 x^{7/2}} \]
(2*(700*a^3*b^3*x^3*(-A + B*x) + 175*a^2*b^4*x^4*(3*A + B*x) - 175*a^4*b^2 *x^2*(A + 3*B*x) + 14*a*b^5*x^5*(5*A + 3*B*x) - 14*a^5*b*x*(3*A + 5*B*x) + b^6*x^6*(7*A + 5*B*x) - a^6*(5*A + 7*B*x)))/(35*x^(7/2))
Time = 0.30 (sec) , antiderivative size = 151, 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.
\(\displaystyle \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^3 (A+B x)}{x^{9/2}} \, dx\) |
\(\Big \downarrow \) 1184 |
\(\displaystyle \frac {\int \frac {b^6 (a+b x)^6 (A+B x)}{x^{9/2}}dx}{b^6}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \frac {(a+b x)^6 (A+B x)}{x^{9/2}}dx\) |
\(\Big \downarrow \) 85 |
\(\displaystyle \int \left (\frac {a^6 A}{x^{9/2}}+\frac {a^5 (a B+6 A b)}{x^{7/2}}+\frac {3 a^4 b (2 a B+5 A b)}{x^{5/2}}+\frac {5 a^3 b^2 (3 a B+4 A b)}{x^{3/2}}+\frac {5 a^2 b^3 (4 a B+3 A b)}{\sqrt {x}}+b^5 x^{3/2} (6 a B+A b)+3 a b^4 \sqrt {x} (5 a B+2 A b)+b^6 B x^{5/2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {2 a^6 A}{7 x^{7/2}}-\frac {2 a^5 (a B+6 A b)}{5 x^{5/2}}-\frac {2 a^4 b (2 a B+5 A b)}{x^{3/2}}-\frac {10 a^3 b^2 (3 a B+4 A b)}{\sqrt {x}}+10 a^2 b^3 \sqrt {x} (4 a B+3 A b)+\frac {2}{5} b^5 x^{5/2} (6 a B+A b)+2 a b^4 x^{3/2} (5 a B+2 A b)+\frac {2}{7} b^6 B x^{7/2}\) |
(-2*a^6*A)/(7*x^(7/2)) - (2*a^5*(6*A*b + a*B))/(5*x^(5/2)) - (2*a^4*b*(5*A *b + 2*a*B))/x^(3/2) - (10*a^3*b^2*(4*A*b + 3*a*B))/Sqrt[x] + 10*a^2*b^3*( 3*A*b + 4*a*B)*Sqrt[x] + 2*a*b^4*(2*A*b + 5*a*B)*x^(3/2) + (2*b^5*(A*b + 6 *a*B)*x^(5/2))/5 + (2*b^6*B*x^(7/2))/7
3.8.55.3.1 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 = 139, normalized size of antiderivative = 0.92
method | result | size |
derivativedivides | \(\frac {2 b^{6} B \,x^{\frac {7}{2}}}{7}+\frac {2 A \,b^{6} x^{\frac {5}{2}}}{5}+\frac {12 B a \,b^{5} x^{\frac {5}{2}}}{5}+4 A a \,b^{5} x^{\frac {3}{2}}+10 B \,a^{2} b^{4} x^{\frac {3}{2}}+30 A \,a^{2} b^{4} \sqrt {x}+40 B \,a^{3} b^{3} \sqrt {x}-\frac {2 a^{4} b \left (5 A b +2 B a \right )}{x^{\frac {3}{2}}}-\frac {2 a^{5} \left (6 A b +B a \right )}{5 x^{\frac {5}{2}}}-\frac {2 a^{6} A}{7 x^{\frac {7}{2}}}-\frac {10 a^{3} b^{2} \left (4 A b +3 B a \right )}{\sqrt {x}}\) | \(139\) |
default | \(\frac {2 b^{6} B \,x^{\frac {7}{2}}}{7}+\frac {2 A \,b^{6} x^{\frac {5}{2}}}{5}+\frac {12 B a \,b^{5} x^{\frac {5}{2}}}{5}+4 A a \,b^{5} x^{\frac {3}{2}}+10 B \,a^{2} b^{4} x^{\frac {3}{2}}+30 A \,a^{2} b^{4} \sqrt {x}+40 B \,a^{3} b^{3} \sqrt {x}-\frac {2 a^{4} b \left (5 A b +2 B a \right )}{x^{\frac {3}{2}}}-\frac {2 a^{5} \left (6 A b +B a \right )}{5 x^{\frac {5}{2}}}-\frac {2 a^{6} A}{7 x^{\frac {7}{2}}}-\frac {10 a^{3} b^{2} \left (4 A b +3 B a \right )}{\sqrt {x}}\) | \(139\) |
gosper | \(-\frac {2 \left (-5 b^{6} B \,x^{7}-7 A \,b^{6} x^{6}-42 x^{6} B a \,b^{5}-70 a A \,b^{5} x^{5}-175 x^{5} B \,b^{4} a^{2}-525 a^{2} A \,b^{4} x^{4}-700 x^{4} B \,a^{3} b^{3}+700 a^{3} A \,b^{3} x^{3}+525 x^{3} B \,a^{4} b^{2}+175 a^{4} A \,b^{2} x^{2}+70 x^{2} B \,a^{5} b +42 a^{5} A b x +7 x B \,a^{6}+5 A \,a^{6}\right )}{35 x^{\frac {7}{2}}}\) | \(148\) |
trager | \(-\frac {2 \left (-5 b^{6} B \,x^{7}-7 A \,b^{6} x^{6}-42 x^{6} B a \,b^{5}-70 a A \,b^{5} x^{5}-175 x^{5} B \,b^{4} a^{2}-525 a^{2} A \,b^{4} x^{4}-700 x^{4} B \,a^{3} b^{3}+700 a^{3} A \,b^{3} x^{3}+525 x^{3} B \,a^{4} b^{2}+175 a^{4} A \,b^{2} x^{2}+70 x^{2} B \,a^{5} b +42 a^{5} A b x +7 x B \,a^{6}+5 A \,a^{6}\right )}{35 x^{\frac {7}{2}}}\) | \(148\) |
risch | \(-\frac {2 \left (-5 b^{6} B \,x^{7}-7 A \,b^{6} x^{6}-42 x^{6} B a \,b^{5}-70 a A \,b^{5} x^{5}-175 x^{5} B \,b^{4} a^{2}-525 a^{2} A \,b^{4} x^{4}-700 x^{4} B \,a^{3} b^{3}+700 a^{3} A \,b^{3} x^{3}+525 x^{3} B \,a^{4} b^{2}+175 a^{4} A \,b^{2} x^{2}+70 x^{2} B \,a^{5} b +42 a^{5} A b x +7 x B \,a^{6}+5 A \,a^{6}\right )}{35 x^{\frac {7}{2}}}\) | \(148\) |
2/7*b^6*B*x^(7/2)+2/5*A*b^6*x^(5/2)+12/5*B*a*b^5*x^(5/2)+4*A*a*b^5*x^(3/2) +10*B*a^2*b^4*x^(3/2)+30*A*a^2*b^4*x^(1/2)+40*B*a^3*b^3*x^(1/2)-2*a^4*b*(5 *A*b+2*B*a)/x^(3/2)-2/5*a^5*(6*A*b+B*a)/x^(5/2)-2/7*a^6*A/x^(7/2)-10*a^3*b ^2*(4*A*b+3*B*a)/x^(1/2)
Time = 0.27 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.97 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^3}{x^{9/2}} \, dx=\frac {2 \, {\left (5 \, B b^{6} x^{7} - 5 \, A a^{6} + 7 \, {\left (6 \, B a b^{5} + A b^{6}\right )} x^{6} + 35 \, {\left (5 \, B a^{2} b^{4} + 2 \, A a b^{5}\right )} x^{5} + 175 \, {\left (4 \, B a^{3} b^{3} + 3 \, A a^{2} b^{4}\right )} x^{4} - 175 \, {\left (3 \, B a^{4} b^{2} + 4 \, A a^{3} b^{3}\right )} x^{3} - 35 \, {\left (2 \, B a^{5} b + 5 \, A a^{4} b^{2}\right )} x^{2} - 7 \, {\left (B a^{6} + 6 \, A a^{5} b\right )} x\right )}}{35 \, x^{\frac {7}{2}}} \]
2/35*(5*B*b^6*x^7 - 5*A*a^6 + 7*(6*B*a*b^5 + A*b^6)*x^6 + 35*(5*B*a^2*b^4 + 2*A*a*b^5)*x^5 + 175*(4*B*a^3*b^3 + 3*A*a^2*b^4)*x^4 - 175*(3*B*a^4*b^2 + 4*A*a^3*b^3)*x^3 - 35*(2*B*a^5*b + 5*A*a^4*b^2)*x^2 - 7*(B*a^6 + 6*A*a^5 *b)*x)/x^(7/2)
Time = 0.69 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.34 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^3}{x^{9/2}} \, dx=- \frac {2 A a^{6}}{7 x^{\frac {7}{2}}} - \frac {12 A a^{5} b}{5 x^{\frac {5}{2}}} - \frac {10 A a^{4} b^{2}}{x^{\frac {3}{2}}} - \frac {40 A a^{3} b^{3}}{\sqrt {x}} + 30 A a^{2} b^{4} \sqrt {x} + 4 A a b^{5} x^{\frac {3}{2}} + \frac {2 A b^{6} x^{\frac {5}{2}}}{5} - \frac {2 B a^{6}}{5 x^{\frac {5}{2}}} - \frac {4 B a^{5} b}{x^{\frac {3}{2}}} - \frac {30 B a^{4} b^{2}}{\sqrt {x}} + 40 B a^{3} b^{3} \sqrt {x} + 10 B a^{2} b^{4} x^{\frac {3}{2}} + \frac {12 B a b^{5} x^{\frac {5}{2}}}{5} + \frac {2 B b^{6} x^{\frac {7}{2}}}{7} \]
-2*A*a**6/(7*x**(7/2)) - 12*A*a**5*b/(5*x**(5/2)) - 10*A*a**4*b**2/x**(3/2 ) - 40*A*a**3*b**3/sqrt(x) + 30*A*a**2*b**4*sqrt(x) + 4*A*a*b**5*x**(3/2) + 2*A*b**6*x**(5/2)/5 - 2*B*a**6/(5*x**(5/2)) - 4*B*a**5*b/x**(3/2) - 30*B *a**4*b**2/sqrt(x) + 40*B*a**3*b**3*sqrt(x) + 10*B*a**2*b**4*x**(3/2) + 12 *B*a*b**5*x**(5/2)/5 + 2*B*b**6*x**(7/2)/7
Time = 0.20 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.98 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^3}{x^{9/2}} \, dx=\frac {2}{7} \, B b^{6} x^{\frac {7}{2}} + \frac {2}{5} \, {\left (6 \, B a b^{5} + A b^{6}\right )} x^{\frac {5}{2}} + 2 \, {\left (5 \, B a^{2} b^{4} + 2 \, A a b^{5}\right )} x^{\frac {3}{2}} + 10 \, {\left (4 \, B a^{3} b^{3} + 3 \, A a^{2} b^{4}\right )} \sqrt {x} - \frac {2 \, {\left (5 \, A a^{6} + 175 \, {\left (3 \, B a^{4} b^{2} + 4 \, A a^{3} b^{3}\right )} x^{3} + 35 \, {\left (2 \, B a^{5} b + 5 \, A a^{4} b^{2}\right )} x^{2} + 7 \, {\left (B a^{6} + 6 \, A a^{5} b\right )} x\right )}}{35 \, x^{\frac {7}{2}}} \]
2/7*B*b^6*x^(7/2) + 2/5*(6*B*a*b^5 + A*b^6)*x^(5/2) + 2*(5*B*a^2*b^4 + 2*A *a*b^5)*x^(3/2) + 10*(4*B*a^3*b^3 + 3*A*a^2*b^4)*sqrt(x) - 2/35*(5*A*a^6 + 175*(3*B*a^4*b^2 + 4*A*a^3*b^3)*x^3 + 35*(2*B*a^5*b + 5*A*a^4*b^2)*x^2 + 7*(B*a^6 + 6*A*a^5*b)*x)/x^(7/2)
Time = 0.29 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.98 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^3}{x^{9/2}} \, dx=\frac {2}{7} \, B b^{6} x^{\frac {7}{2}} + \frac {12}{5} \, B a b^{5} x^{\frac {5}{2}} + \frac {2}{5} \, A b^{6} x^{\frac {5}{2}} + 10 \, B a^{2} b^{4} x^{\frac {3}{2}} + 4 \, A a b^{5} x^{\frac {3}{2}} + 40 \, B a^{3} b^{3} \sqrt {x} + 30 \, A a^{2} b^{4} \sqrt {x} - \frac {2 \, {\left (525 \, B a^{4} b^{2} x^{3} + 700 \, A a^{3} b^{3} x^{3} + 70 \, B a^{5} b x^{2} + 175 \, A a^{4} b^{2} x^{2} + 7 \, B a^{6} x + 42 \, A a^{5} b x + 5 \, A a^{6}\right )}}{35 \, x^{\frac {7}{2}}} \]
2/7*B*b^6*x^(7/2) + 12/5*B*a*b^5*x^(5/2) + 2/5*A*b^6*x^(5/2) + 10*B*a^2*b^ 4*x^(3/2) + 4*A*a*b^5*x^(3/2) + 40*B*a^3*b^3*sqrt(x) + 30*A*a^2*b^4*sqrt(x ) - 2/35*(525*B*a^4*b^2*x^3 + 700*A*a^3*b^3*x^3 + 70*B*a^5*b*x^2 + 175*A*a ^4*b^2*x^2 + 7*B*a^6*x + 42*A*a^5*b*x + 5*A*a^6)/x^(7/2)
Time = 9.99 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.91 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^3}{x^{9/2}} \, dx=x^{5/2}\,\left (\frac {2\,A\,b^6}{5}+\frac {12\,B\,a\,b^5}{5}\right )-\frac {x\,\left (\frac {2\,B\,a^6}{5}+\frac {12\,A\,b\,a^5}{5}\right )+\frac {2\,A\,a^6}{7}+x^2\,\left (4\,B\,a^5\,b+10\,A\,a^4\,b^2\right )+x^3\,\left (30\,B\,a^4\,b^2+40\,A\,a^3\,b^3\right )}{x^{7/2}}+\frac {2\,B\,b^6\,x^{7/2}}{7}+10\,a^2\,b^3\,\sqrt {x}\,\left (3\,A\,b+4\,B\,a\right )+2\,a\,b^4\,x^{3/2}\,\left (2\,A\,b+5\,B\,a\right ) \]