Integrand size = 29, antiderivative size = 107 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2}{x^{5/2}} \, dx=-\frac {2 a^4 A}{3 x^{3/2}}-\frac {2 a^3 (4 A b+a B)}{\sqrt {x}}+4 a^2 b (3 A b+2 a B) \sqrt {x}+\frac {4}{3} a b^2 (2 A b+3 a B) x^{3/2}+\frac {2}{5} b^3 (A b+4 a B) x^{5/2}+\frac {2}{7} b^4 B x^{7/2} \]
-2/3*a^4*A/x^(3/2)+4/3*a*b^2*(2*A*b+3*B*a)*x^(3/2)+2/5*b^3*(A*b+4*B*a)*x^( 5/2)+2/7*b^4*B*x^(7/2)-2*a^3*(4*A*b+B*a)/x^(1/2)+4*a^2*b*(3*A*b+2*B*a)*x^( 1/2)
Time = 0.07 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.80 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2}{x^{5/2}} \, dx=\frac {840 a^3 b x (-A+B x)+420 a^2 b^2 x^2 (3 A+B x)-70 a^4 (A+3 B x)+56 a b^3 x^3 (5 A+3 B x)+6 b^4 x^4 (7 A+5 B x)}{105 x^{3/2}} \]
(840*a^3*b*x*(-A + B*x) + 420*a^2*b^2*x^2*(3*A + B*x) - 70*a^4*(A + 3*B*x) + 56*a*b^3*x^3*(5*A + 3*B*x) + 6*b^4*x^4*(7*A + 5*B*x))/(105*x^(3/2))
Time = 0.24 (sec) , antiderivative size = 107, 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.
\(\displaystyle \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^2 (A+B x)}{x^{5/2}} \, dx\) |
\(\Big \downarrow \) 1184 |
\(\displaystyle \frac {\int \frac {b^4 (a+b x)^4 (A+B x)}{x^{5/2}}dx}{b^4}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \frac {(a+b x)^4 (A+B x)}{x^{5/2}}dx\) |
\(\Big \downarrow \) 85 |
\(\displaystyle \int \left (\frac {a^4 A}{x^{5/2}}+\frac {a^3 (a B+4 A b)}{x^{3/2}}+\frac {2 a^2 b (2 a B+3 A b)}{\sqrt {x}}+b^3 x^{3/2} (4 a B+A b)+2 a b^2 \sqrt {x} (3 a B+2 A b)+b^4 B x^{5/2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {2 a^4 A}{3 x^{3/2}}-\frac {2 a^3 (a B+4 A b)}{\sqrt {x}}+4 a^2 b \sqrt {x} (2 a B+3 A b)+\frac {2}{5} b^3 x^{5/2} (4 a B+A b)+\frac {4}{3} a b^2 x^{3/2} (3 a B+2 A b)+\frac {2}{7} b^4 B x^{7/2}\) |
(-2*a^4*A)/(3*x^(3/2)) - (2*a^3*(4*A*b + a*B))/Sqrt[x] + 4*a^2*b*(3*A*b + 2*a*B)*Sqrt[x] + (4*a*b^2*(2*A*b + 3*a*B)*x^(3/2))/3 + (2*b^3*(A*b + 4*a*B )*x^(5/2))/5 + (2*b^4*B*x^(7/2))/7
3.8.44.3.1 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 = 0.12 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.93
method | result | size |
derivativedivides | \(\frac {2 b^{4} B \,x^{\frac {7}{2}}}{7}+\frac {2 A \,b^{4} x^{\frac {5}{2}}}{5}+\frac {8 B a \,b^{3} x^{\frac {5}{2}}}{5}+\frac {8 A a \,b^{3} x^{\frac {3}{2}}}{3}+4 B \,a^{2} b^{2} x^{\frac {3}{2}}+12 A \,a^{2} b^{2} \sqrt {x}+8 B \,a^{3} b \sqrt {x}-\frac {2 a^{4} A}{3 x^{\frac {3}{2}}}-\frac {2 a^{3} \left (4 A b +B a \right )}{\sqrt {x}}\) | \(99\) |
default | \(\frac {2 b^{4} B \,x^{\frac {7}{2}}}{7}+\frac {2 A \,b^{4} x^{\frac {5}{2}}}{5}+\frac {8 B a \,b^{3} x^{\frac {5}{2}}}{5}+\frac {8 A a \,b^{3} x^{\frac {3}{2}}}{3}+4 B \,a^{2} b^{2} x^{\frac {3}{2}}+12 A \,a^{2} b^{2} \sqrt {x}+8 B \,a^{3} b \sqrt {x}-\frac {2 a^{4} A}{3 x^{\frac {3}{2}}}-\frac {2 a^{3} \left (4 A b +B a \right )}{\sqrt {x}}\) | \(99\) |
gosper | \(-\frac {2 \left (-15 b^{4} B \,x^{5}-21 A \,b^{4} x^{4}-84 x^{4} B \,b^{3} a -140 a A \,b^{3} x^{3}-210 x^{3} B \,a^{2} b^{2}-630 a^{2} A \,b^{2} x^{2}-420 x^{2} B \,a^{3} b +420 a^{3} A b x +105 a^{4} B x +35 A \,a^{4}\right )}{105 x^{\frac {3}{2}}}\) | \(100\) |
trager | \(-\frac {2 \left (-15 b^{4} B \,x^{5}-21 A \,b^{4} x^{4}-84 x^{4} B \,b^{3} a -140 a A \,b^{3} x^{3}-210 x^{3} B \,a^{2} b^{2}-630 a^{2} A \,b^{2} x^{2}-420 x^{2} B \,a^{3} b +420 a^{3} A b x +105 a^{4} B x +35 A \,a^{4}\right )}{105 x^{\frac {3}{2}}}\) | \(100\) |
risch | \(-\frac {2 \left (-15 b^{4} B \,x^{5}-21 A \,b^{4} x^{4}-84 x^{4} B \,b^{3} a -140 a A \,b^{3} x^{3}-210 x^{3} B \,a^{2} b^{2}-630 a^{2} A \,b^{2} x^{2}-420 x^{2} B \,a^{3} b +420 a^{3} A b x +105 a^{4} B x +35 A \,a^{4}\right )}{105 x^{\frac {3}{2}}}\) | \(100\) |
2/7*b^4*B*x^(7/2)+2/5*A*b^4*x^(5/2)+8/5*B*a*b^3*x^(5/2)+8/3*A*a*b^3*x^(3/2 )+4*B*a^2*b^2*x^(3/2)+12*A*a^2*b^2*x^(1/2)+8*B*a^3*b*x^(1/2)-2/3*a^4*A/x^( 3/2)-2*a^3*(4*A*b+B*a)/x^(1/2)
Time = 0.27 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.93 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2}{x^{5/2}} \, dx=\frac {2 \, {\left (15 \, B b^{4} x^{5} - 35 \, A a^{4} + 21 \, {\left (4 \, B a b^{3} + A b^{4}\right )} x^{4} + 70 \, {\left (3 \, B a^{2} b^{2} + 2 \, A a b^{3}\right )} x^{3} + 210 \, {\left (2 \, B a^{3} b + 3 \, A a^{2} b^{2}\right )} x^{2} - 105 \, {\left (B a^{4} + 4 \, A a^{3} b\right )} x\right )}}{105 \, x^{\frac {3}{2}}} \]
2/105*(15*B*b^4*x^5 - 35*A*a^4 + 21*(4*B*a*b^3 + A*b^4)*x^4 + 70*(3*B*a^2* b^2 + 2*A*a*b^3)*x^3 + 210*(2*B*a^3*b + 3*A*a^2*b^2)*x^2 - 105*(B*a^4 + 4* A*a^3*b)*x)/x^(3/2)
Time = 0.31 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.30 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2}{x^{5/2}} \, dx=- \frac {2 A a^{4}}{3 x^{\frac {3}{2}}} - \frac {8 A a^{3} b}{\sqrt {x}} + 12 A a^{2} b^{2} \sqrt {x} + \frac {8 A a b^{3} x^{\frac {3}{2}}}{3} + \frac {2 A b^{4} x^{\frac {5}{2}}}{5} - \frac {2 B a^{4}}{\sqrt {x}} + 8 B a^{3} b \sqrt {x} + 4 B a^{2} b^{2} x^{\frac {3}{2}} + \frac {8 B a b^{3} x^{\frac {5}{2}}}{5} + \frac {2 B b^{4} x^{\frac {7}{2}}}{7} \]
-2*A*a**4/(3*x**(3/2)) - 8*A*a**3*b/sqrt(x) + 12*A*a**2*b**2*sqrt(x) + 8*A *a*b**3*x**(3/2)/3 + 2*A*b**4*x**(5/2)/5 - 2*B*a**4/sqrt(x) + 8*B*a**3*b*s qrt(x) + 4*B*a**2*b**2*x**(3/2) + 8*B*a*b**3*x**(5/2)/5 + 2*B*b**4*x**(7/2 )/7
Time = 0.20 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.93 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2}{x^{5/2}} \, dx=\frac {2}{7} \, B b^{4} x^{\frac {7}{2}} + \frac {2}{5} \, {\left (4 \, B a b^{3} + A b^{4}\right )} x^{\frac {5}{2}} + \frac {4}{3} \, {\left (3 \, B a^{2} b^{2} + 2 \, A a b^{3}\right )} x^{\frac {3}{2}} + 4 \, {\left (2 \, B a^{3} b + 3 \, A a^{2} b^{2}\right )} \sqrt {x} - \frac {2 \, {\left (A a^{4} + 3 \, {\left (B a^{4} + 4 \, A a^{3} b\right )} x\right )}}{3 \, x^{\frac {3}{2}}} \]
2/7*B*b^4*x^(7/2) + 2/5*(4*B*a*b^3 + A*b^4)*x^(5/2) + 4/3*(3*B*a^2*b^2 + 2 *A*a*b^3)*x^(3/2) + 4*(2*B*a^3*b + 3*A*a^2*b^2)*sqrt(x) - 2/3*(A*a^4 + 3*( B*a^4 + 4*A*a^3*b)*x)/x^(3/2)
Time = 0.34 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.93 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2}{x^{5/2}} \, dx=\frac {2}{7} \, B b^{4} x^{\frac {7}{2}} + \frac {8}{5} \, B a b^{3} x^{\frac {5}{2}} + \frac {2}{5} \, A b^{4} x^{\frac {5}{2}} + 4 \, B a^{2} b^{2} x^{\frac {3}{2}} + \frac {8}{3} \, A a b^{3} x^{\frac {3}{2}} + 8 \, B a^{3} b \sqrt {x} + 12 \, A a^{2} b^{2} \sqrt {x} - \frac {2 \, {\left (3 \, B a^{4} x + 12 \, A a^{3} b x + A a^{4}\right )}}{3 \, x^{\frac {3}{2}}} \]
2/7*B*b^4*x^(7/2) + 8/5*B*a*b^3*x^(5/2) + 2/5*A*b^4*x^(5/2) + 4*B*a^2*b^2* x^(3/2) + 8/3*A*a*b^3*x^(3/2) + 8*B*a^3*b*sqrt(x) + 12*A*a^2*b^2*sqrt(x) - 2/3*(3*B*a^4*x + 12*A*a^3*b*x + A*a^4)/x^(3/2)
Time = 0.04 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.86 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2}{x^{5/2}} \, dx=x^{5/2}\,\left (\frac {2\,A\,b^4}{5}+\frac {8\,B\,a\,b^3}{5}\right )-\frac {x\,\left (2\,B\,a^4+8\,A\,b\,a^3\right )+\frac {2\,A\,a^4}{3}}{x^{3/2}}+\frac {2\,B\,b^4\,x^{7/2}}{7}+4\,a^2\,b\,\sqrt {x}\,\left (3\,A\,b+2\,B\,a\right )+\frac {4\,a\,b^2\,x^{3/2}\,\left (2\,A\,b+3\,B\,a\right )}{3} \]