Integrand size = 27, antiderivative size = 131 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^3}{x^3} \, dx=-\frac {a^6 A}{2 x^2}-\frac {a^5 (6 A b+a B)}{x}+5 a^3 b^2 (4 A b+3 a B) x+\frac {5}{2} a^2 b^3 (3 A b+4 a B) x^2+a b^4 (2 A b+5 a B) x^3+\frac {1}{4} b^5 (A b+6 a B) x^4+\frac {1}{5} b^6 B x^5+3 a^4 b (5 A b+2 a B) \log (x) \]
-1/2*a^6*A/x^2-a^5*(6*A*b+B*a)/x+5*a^3*b^2*(4*A*b+3*B*a)*x+5/2*a^2*b^3*(3* A*b+4*B*a)*x^2+a*b^4*(2*A*b+5*B*a)*x^3+1/4*b^5*(A*b+6*B*a)*x^4+1/5*b^6*B*x ^5+3*a^4*b*(5*A*b+2*B*a)*ln(x)
Time = 0.04 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.98 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^3}{x^3} \, dx=-\frac {6 a^5 A b}{x}+15 a^4 b^2 B x+10 a^3 b^3 x (2 A+B x)-\frac {a^6 (A+2 B x)}{2 x^2}+\frac {5}{2} a^2 b^4 x^2 (3 A+2 B x)+\frac {1}{2} a b^5 x^3 (4 A+3 B x)+\frac {1}{20} b^6 x^4 (5 A+4 B x)+3 a^4 b (5 A b+2 a B) \log (x) \]
(-6*a^5*A*b)/x + 15*a^4*b^2*B*x + 10*a^3*b^3*x*(2*A + B*x) - (a^6*(A + 2*B *x))/(2*x^2) + (5*a^2*b^4*x^2*(3*A + 2*B*x))/2 + (a*b^5*x^3*(4*A + 3*B*x)) /2 + (b^6*x^4*(5*A + 4*B*x))/20 + 3*a^4*b*(5*A*b + 2*a*B)*Log[x]
Time = 0.32 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, 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.
\(\displaystyle \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^3 (A+B x)}{x^3} \, dx\) |
\(\Big \downarrow \) 1184 |
\(\displaystyle \frac {\int \frac {b^6 (a+b x)^6 (A+B x)}{x^3}dx}{b^6}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \frac {(a+b x)^6 (A+B x)}{x^3}dx\) |
\(\Big \downarrow \) 85 |
\(\displaystyle \int \left (\frac {a^6 A}{x^3}+\frac {a^5 (a B+6 A b)}{x^2}+\frac {3 a^4 b (2 a B+5 A b)}{x}+5 a^3 b^2 (3 a B+4 A b)+5 a^2 b^3 x (4 a B+3 A b)+b^5 x^3 (6 a B+A b)+3 a b^4 x^2 (5 a B+2 A b)+b^6 B x^4\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {a^6 A}{2 x^2}-\frac {a^5 (a B+6 A b)}{x}+3 a^4 b \log (x) (2 a B+5 A b)+5 a^3 b^2 x (3 a B+4 A b)+\frac {5}{2} a^2 b^3 x^2 (4 a B+3 A b)+\frac {1}{4} b^5 x^4 (6 a B+A b)+a b^4 x^3 (5 a B+2 A b)+\frac {1}{5} b^6 B x^5\) |
-1/2*(a^6*A)/x^2 - (a^5*(6*A*b + a*B))/x + 5*a^3*b^2*(4*A*b + 3*a*B)*x + ( 5*a^2*b^3*(3*A*b + 4*a*B)*x^2)/2 + a*b^4*(2*A*b + 5*a*B)*x^3 + (b^5*(A*b + 6*a*B)*x^4)/4 + (b^6*B*x^5)/5 + 3*a^4*b*(5*A*b + 2*a*B)*Log[x]
3.6.47.3.1 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 = 0.10 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.05
method | result | size |
default | \(\frac {b^{6} B \,x^{5}}{5}+\frac {A \,b^{6} x^{4}}{4}+\frac {3 B a \,b^{5} x^{4}}{2}+2 A a \,b^{5} x^{3}+5 B \,a^{2} b^{4} x^{3}+\frac {15 A \,a^{2} b^{4} x^{2}}{2}+10 B \,a^{3} b^{3} x^{2}+20 A \,a^{3} b^{3} x +15 B \,a^{4} b^{2} x +3 a^{4} b \left (5 A b +2 B a \right ) \ln \left (x \right )-\frac {a^{6} A}{2 x^{2}}-\frac {a^{5} \left (6 A b +B a \right )}{x}\) | \(138\) |
norman | \(\frac {\left (\frac {1}{4} A \,b^{6}+\frac {3}{2} B a \,b^{5}\right ) x^{6}+\left (\frac {15}{2} A \,b^{4} a^{2}+10 B \,a^{3} b^{3}\right ) x^{4}+\left (2 A a \,b^{5}+5 B \,b^{4} a^{2}\right ) x^{5}+\left (20 A \,a^{3} b^{3}+15 B \,a^{4} b^{2}\right ) x^{3}+\left (-6 A \,a^{5} b -B \,a^{6}\right ) x -\frac {A \,a^{6}}{2}+\frac {b^{6} B \,x^{7}}{5}}{x^{2}}+\left (15 A \,a^{4} b^{2}+6 B \,a^{5} b \right ) \ln \left (x \right )\) | \(143\) |
risch | \(\frac {b^{6} B \,x^{5}}{5}+\frac {A \,b^{6} x^{4}}{4}+\frac {3 B a \,b^{5} x^{4}}{2}+2 A a \,b^{5} x^{3}+5 B \,a^{2} b^{4} x^{3}+\frac {15 A \,a^{2} b^{4} x^{2}}{2}+10 B \,a^{3} b^{3} x^{2}+20 A \,a^{3} b^{3} x +15 B \,a^{4} b^{2} x +\frac {\left (-6 A \,a^{5} b -B \,a^{6}\right ) x -\frac {A \,a^{6}}{2}}{x^{2}}+15 A \ln \left (x \right ) a^{4} b^{2}+6 B \ln \left (x \right ) a^{5} b\) | \(143\) |
parallelrisch | \(\frac {4 b^{6} B \,x^{7}+5 A \,b^{6} x^{6}+30 x^{6} B a \,b^{5}+40 a A \,b^{5} x^{5}+100 x^{5} B \,b^{4} a^{2}+150 a^{2} A \,b^{4} x^{4}+200 x^{4} B \,a^{3} b^{3}+300 A \ln \left (x \right ) x^{2} a^{4} b^{2}+400 a^{3} A \,b^{3} x^{3}+120 B \ln \left (x \right ) x^{2} a^{5} b +300 x^{3} B \,a^{4} b^{2}-120 a^{5} A b x -20 x B \,a^{6}-10 A \,a^{6}}{20 x^{2}}\) | \(152\) |
1/5*b^6*B*x^5+1/4*A*b^6*x^4+3/2*B*a*b^5*x^4+2*A*a*b^5*x^3+5*B*a^2*b^4*x^3+ 15/2*A*a^2*b^4*x^2+10*B*a^3*b^3*x^2+20*A*a^3*b^3*x+15*B*a^4*b^2*x+3*a^4*b* (5*A*b+2*B*a)*ln(x)-1/2*a^6*A/x^2-a^5*(6*A*b+B*a)/x
Time = 0.41 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.14 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^3}{x^3} \, dx=\frac {4 \, B b^{6} x^{7} - 10 \, A a^{6} + 5 \, {\left (6 \, B a b^{5} + A b^{6}\right )} x^{6} + 20 \, {\left (5 \, B a^{2} b^{4} + 2 \, A a b^{5}\right )} x^{5} + 50 \, {\left (4 \, B a^{3} b^{3} + 3 \, A a^{2} b^{4}\right )} x^{4} + 100 \, {\left (3 \, B a^{4} b^{2} + 4 \, A a^{3} b^{3}\right )} x^{3} + 60 \, {\left (2 \, B a^{5} b + 5 \, A a^{4} b^{2}\right )} x^{2} \log \left (x\right ) - 20 \, {\left (B a^{6} + 6 \, A a^{5} b\right )} x}{20 \, x^{2}} \]
1/20*(4*B*b^6*x^7 - 10*A*a^6 + 5*(6*B*a*b^5 + A*b^6)*x^6 + 20*(5*B*a^2*b^4 + 2*A*a*b^5)*x^5 + 50*(4*B*a^3*b^3 + 3*A*a^2*b^4)*x^4 + 100*(3*B*a^4*b^2 + 4*A*a^3*b^3)*x^3 + 60*(2*B*a^5*b + 5*A*a^4*b^2)*x^2*log(x) - 20*(B*a^6 + 6*A*a^5*b)*x)/x^2
Time = 0.28 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.13 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^3}{x^3} \, dx=\frac {B b^{6} x^{5}}{5} + 3 a^{4} b \left (5 A b + 2 B a\right ) \log {\left (x \right )} + x^{4} \left (\frac {A b^{6}}{4} + \frac {3 B a b^{5}}{2}\right ) + x^{3} \cdot \left (2 A a b^{5} + 5 B a^{2} b^{4}\right ) + x^{2} \cdot \left (\frac {15 A a^{2} b^{4}}{2} + 10 B a^{3} b^{3}\right ) + x \left (20 A a^{3} b^{3} + 15 B a^{4} b^{2}\right ) + \frac {- A a^{6} + x \left (- 12 A a^{5} b - 2 B a^{6}\right )}{2 x^{2}} \]
B*b**6*x**5/5 + 3*a**4*b*(5*A*b + 2*B*a)*log(x) + x**4*(A*b**6/4 + 3*B*a*b **5/2) + x**3*(2*A*a*b**5 + 5*B*a**2*b**4) + x**2*(15*A*a**2*b**4/2 + 10*B *a**3*b**3) + x*(20*A*a**3*b**3 + 15*B*a**4*b**2) + (-A*a**6 + x*(-12*A*a* *5*b - 2*B*a**6))/(2*x**2)
Time = 0.20 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.09 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^3}{x^3} \, dx=\frac {1}{5} \, B b^{6} x^{5} + \frac {1}{4} \, {\left (6 \, B a b^{5} + A b^{6}\right )} x^{4} + {\left (5 \, B a^{2} b^{4} + 2 \, A a b^{5}\right )} x^{3} + \frac {5}{2} \, {\left (4 \, B a^{3} b^{3} + 3 \, A a^{2} b^{4}\right )} x^{2} + 5 \, {\left (3 \, B a^{4} b^{2} + 4 \, A a^{3} b^{3}\right )} x + 3 \, {\left (2 \, B a^{5} b + 5 \, A a^{4} b^{2}\right )} \log \left (x\right ) - \frac {A a^{6} + 2 \, {\left (B a^{6} + 6 \, A a^{5} b\right )} x}{2 \, x^{2}} \]
1/5*B*b^6*x^5 + 1/4*(6*B*a*b^5 + A*b^6)*x^4 + (5*B*a^2*b^4 + 2*A*a*b^5)*x^ 3 + 5/2*(4*B*a^3*b^3 + 3*A*a^2*b^4)*x^2 + 5*(3*B*a^4*b^2 + 4*A*a^3*b^3)*x + 3*(2*B*a^5*b + 5*A*a^4*b^2)*log(x) - 1/2*(A*a^6 + 2*(B*a^6 + 6*A*a^5*b)* x)/x^2
Time = 0.27 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.10 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^3}{x^3} \, dx=\frac {1}{5} \, B b^{6} x^{5} + \frac {3}{2} \, B a b^{5} x^{4} + \frac {1}{4} \, A b^{6} x^{4} + 5 \, B a^{2} b^{4} x^{3} + 2 \, A a b^{5} x^{3} + 10 \, B a^{3} b^{3} x^{2} + \frac {15}{2} \, A a^{2} b^{4} x^{2} + 15 \, B a^{4} b^{2} x + 20 \, A a^{3} b^{3} x + 3 \, {\left (2 \, B a^{5} b + 5 \, A a^{4} b^{2}\right )} \log \left ({\left | x \right |}\right ) - \frac {A a^{6} + 2 \, {\left (B a^{6} + 6 \, A a^{5} b\right )} x}{2 \, x^{2}} \]
1/5*B*b^6*x^5 + 3/2*B*a*b^5*x^4 + 1/4*A*b^6*x^4 + 5*B*a^2*b^4*x^3 + 2*A*a* b^5*x^3 + 10*B*a^3*b^3*x^2 + 15/2*A*a^2*b^4*x^2 + 15*B*a^4*b^2*x + 20*A*a^ 3*b^3*x + 3*(2*B*a^5*b + 5*A*a^4*b^2)*log(abs(x)) - 1/2*(A*a^6 + 2*(B*a^6 + 6*A*a^5*b)*x)/x^2
Time = 9.97 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.99 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^3}{x^3} \, dx=\ln \left (x\right )\,\left (6\,B\,a^5\,b+15\,A\,a^4\,b^2\right )-\frac {x\,\left (B\,a^6+6\,A\,b\,a^5\right )+\frac {A\,a^6}{2}}{x^2}+x^4\,\left (\frac {A\,b^6}{4}+\frac {3\,B\,a\,b^5}{2}\right )+\frac {B\,b^6\,x^5}{5}+\frac {5\,a^2\,b^3\,x^2\,\left (3\,A\,b+4\,B\,a\right )}{2}+5\,a^3\,b^2\,x\,\left (4\,A\,b+3\,B\,a\right )+a\,b^4\,x^3\,\left (2\,A\,b+5\,B\,a\right ) \]