Integrand size = 27, antiderivative size = 133 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^3}{x^2} \, dx=-\frac {a^6 A}{x}+3 a^4 b (5 A b+2 a B) x+\frac {5}{2} a^3 b^2 (4 A b+3 a B) x^2+\frac {5}{3} a^2 b^3 (3 A b+4 a B) x^3+\frac {3}{4} a b^4 (2 A b+5 a B) x^4+\frac {1}{5} b^5 (A b+6 a B) x^5+\frac {1}{6} b^6 B x^6+a^5 (6 A b+a B) \log (x) \] Output:
-a^6*A/x+3*a^4*b*(5*A*b+2*B*a)*x+5/2*a^3*b^2*(4*A*b+3*B*a)*x^2+5/3*a^2*b^3 *(3*A*b+4*B*a)*x^3+3/4*a*b^4*(2*A*b+5*B*a)*x^4+1/5*b^5*(A*b+6*B*a)*x^5+1/6 *b^6*B*x^6+a^5*(6*A*b+B*a)*ln(x)
Time = 0.07 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.97 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^3}{x^2} \, dx=-\frac {a^6 A}{x}+6 a^5 b B x+\frac {15}{2} a^4 b^2 x (2 A+B x)+\frac {10}{3} a^3 b^3 x^2 (3 A+2 B x)+\frac {5}{4} a^2 b^4 x^3 (4 A+3 B x)+\frac {3}{10} a b^5 x^4 (5 A+4 B x)+\frac {1}{30} b^6 x^5 (6 A+5 B x)+a^5 (6 A b+a B) \log (x) \] Input:
Integrate[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^3)/x^2,x]
Output:
-((a^6*A)/x) + 6*a^5*b*B*x + (15*a^4*b^2*x*(2*A + B*x))/2 + (10*a^3*b^3*x^ 2*(3*A + 2*B*x))/3 + (5*a^2*b^4*x^3*(4*A + 3*B*x))/4 + (3*a*b^5*x^4*(5*A + 4*B*x))/10 + (b^6*x^5*(6*A + 5*B*x))/30 + a^5*(6*A*b + a*B)*Log[x]
Time = 0.51 (sec) , antiderivative size = 133, 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.
\(\displaystyle \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^3 (A+B x)}{x^2} \, dx\) |
\(\Big \downarrow \) 1184 |
\(\displaystyle \frac {\int \frac {b^6 (a+b x)^6 (A+B x)}{x^2}dx}{b^6}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \frac {(a+b x)^6 (A+B x)}{x^2}dx\) |
\(\Big \downarrow \) 85 |
\(\displaystyle \int \left (\frac {a^6 A}{x^2}+\frac {a^5 (a B+6 A b)}{x}+3 a^4 b (2 a B+5 A b)+5 a^3 b^2 x (3 a B+4 A b)+5 a^2 b^3 x^2 (4 a B+3 A b)+b^5 x^4 (6 a B+A b)+3 a b^4 x^3 (5 a B+2 A b)+b^6 B x^5\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {a^6 A}{x}+a^5 \log (x) (a B+6 A b)+3 a^4 b x (2 a B+5 A b)+\frac {5}{2} a^3 b^2 x^2 (3 a B+4 A b)+\frac {5}{3} a^2 b^3 x^3 (4 a B+3 A b)+\frac {1}{5} b^5 x^5 (6 a B+A b)+\frac {3}{4} a b^4 x^4 (5 a B+2 A b)+\frac {1}{6} b^6 B x^6\) |
Input:
Int[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^3)/x^2,x]
Output:
-((a^6*A)/x) + 3*a^4*b*(5*A*b + 2*a*B)*x + (5*a^3*b^2*(4*A*b + 3*a*B)*x^2) /2 + (5*a^2*b^3*(3*A*b + 4*a*B)*x^3)/3 + (3*a*b^4*(2*A*b + 5*a*B)*x^4)/4 + (b^5*(A*b + 6*a*B)*x^5)/5 + (b^6*B*x^6)/6 + a^5*(6*A*b + a*B)*Log[x]
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 = 0.94 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.06
method | result | size |
default | \(\frac {b^{6} B \,x^{6}}{6}+\frac {A \,b^{6} x^{5}}{5}+\frac {6 B a \,b^{5} x^{5}}{5}+\frac {3 A a \,b^{5} x^{4}}{2}+\frac {15 B \,a^{2} b^{4} x^{4}}{4}+5 A \,a^{2} b^{4} x^{3}+\frac {20 B \,a^{3} b^{3} x^{3}}{3}+10 A \,a^{3} b^{3} x^{2}+\frac {15 B \,a^{4} b^{2} x^{2}}{2}+15 A \,a^{4} b^{2} x +6 B \,a^{5} b x +a^{5} \left (6 A b +B a \right ) \ln \left (x \right )-\frac {a^{6} A}{x}\) | \(141\) |
risch | \(\frac {b^{6} B \,x^{6}}{6}+\frac {A \,b^{6} x^{5}}{5}+\frac {6 B a \,b^{5} x^{5}}{5}+\frac {3 A a \,b^{5} x^{4}}{2}+\frac {15 B \,a^{2} b^{4} x^{4}}{4}+5 A \,a^{2} b^{4} x^{3}+\frac {20 B \,a^{3} b^{3} x^{3}}{3}+10 A \,a^{3} b^{3} x^{2}+\frac {15 B \,a^{4} b^{2} x^{2}}{2}+15 A \,a^{4} b^{2} x +6 B \,a^{5} b x -\frac {a^{6} A}{x}+6 A \ln \left (x \right ) a^{5} b +B \ln \left (x \right ) a^{6}\) | \(143\) |
norman | \(\frac {\left (\frac {1}{5} A \,b^{6}+\frac {6}{5} B a \,b^{5}\right ) x^{6}+\left (\frac {3}{2} A a \,b^{5}+\frac {15}{4} B \,a^{2} b^{4}\right ) x^{5}+\left (5 A \,a^{2} b^{4}+\frac {20}{3} B \,a^{3} b^{3}\right ) x^{4}+\left (10 A \,a^{3} b^{3}+\frac {15}{2} B \,a^{4} b^{2}\right ) x^{3}+\left (15 A \,a^{4} b^{2}+6 B \,a^{5} b \right ) x^{2}-A \,a^{6}+\frac {b^{6} B \,x^{7}}{6}}{x}+\left (6 A \,a^{5} b +B \,a^{6}\right ) \ln \left (x \right )\) | \(144\) |
parallelrisch | \(\frac {10 b^{6} B \,x^{7}+12 A \,b^{6} x^{6}+72 B a \,b^{5} x^{6}+90 A a \,b^{5} x^{5}+225 B \,a^{2} b^{4} x^{5}+300 A \,a^{2} b^{4} x^{4}+400 B \,a^{3} b^{3} x^{4}+600 A \,a^{3} b^{3} x^{3}+450 B \,a^{4} b^{2} x^{3}+360 A \ln \left (x \right ) x \,a^{5} b +900 A \,a^{4} b^{2} x^{2}+60 B \ln \left (x \right ) x \,a^{6}+360 B \,a^{5} b \,x^{2}-60 A \,a^{6}}{60 x}\) | \(152\) |
Input:
int((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^3/x^2,x,method=_RETURNVERBOSE)
Output:
1/6*b^6*B*x^6+1/5*A*b^6*x^5+6/5*B*a*b^5*x^5+3/2*A*a*b^5*x^4+15/4*B*a^2*b^4 *x^4+5*A*a^2*b^4*x^3+20/3*B*a^3*b^3*x^3+10*A*a^3*b^3*x^2+15/2*B*a^4*b^2*x^ 2+15*A*a^4*b^2*x+6*B*a^5*b*x+a^5*(6*A*b+B*a)*ln(x)-a^6*A/x
Time = 0.07 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.12 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^3}{x^2} \, dx=\frac {10 \, B b^{6} x^{7} - 60 \, A a^{6} + 12 \, {\left (6 \, B a b^{5} + A b^{6}\right )} x^{6} + 45 \, {\left (5 \, B a^{2} b^{4} + 2 \, A a b^{5}\right )} x^{5} + 100 \, {\left (4 \, B a^{3} b^{3} + 3 \, A a^{2} b^{4}\right )} x^{4} + 150 \, {\left (3 \, B a^{4} b^{2} + 4 \, A a^{3} b^{3}\right )} x^{3} + 180 \, {\left (2 \, B a^{5} b + 5 \, A a^{4} b^{2}\right )} x^{2} + 60 \, {\left (B a^{6} + 6 \, A a^{5} b\right )} x \log \left (x\right )}{60 \, x} \] Input:
integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^3/x^2,x, algorithm="fricas")
Output:
1/60*(10*B*b^6*x^7 - 60*A*a^6 + 12*(6*B*a*b^5 + A*b^6)*x^6 + 45*(5*B*a^2*b ^4 + 2*A*a*b^5)*x^5 + 100*(4*B*a^3*b^3 + 3*A*a^2*b^4)*x^4 + 150*(3*B*a^4*b ^2 + 4*A*a^3*b^3)*x^3 + 180*(2*B*a^5*b + 5*A*a^4*b^2)*x^2 + 60*(B*a^6 + 6* A*a^5*b)*x*log(x))/x
Time = 0.14 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.11 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^3}{x^2} \, dx=- \frac {A a^{6}}{x} + \frac {B b^{6} x^{6}}{6} + a^{5} \cdot \left (6 A b + B a\right ) \log {\left (x \right )} + x^{5} \left (\frac {A b^{6}}{5} + \frac {6 B a b^{5}}{5}\right ) + x^{4} \cdot \left (\frac {3 A a b^{5}}{2} + \frac {15 B a^{2} b^{4}}{4}\right ) + x^{3} \cdot \left (5 A a^{2} b^{4} + \frac {20 B a^{3} b^{3}}{3}\right ) + x^{2} \cdot \left (10 A a^{3} b^{3} + \frac {15 B a^{4} b^{2}}{2}\right ) + x \left (15 A a^{4} b^{2} + 6 B a^{5} b\right ) \] Input:
integrate((B*x+A)*(b**2*x**2+2*a*b*x+a**2)**3/x**2,x)
Output:
-A*a**6/x + B*b**6*x**6/6 + a**5*(6*A*b + B*a)*log(x) + x**5*(A*b**6/5 + 6 *B*a*b**5/5) + x**4*(3*A*a*b**5/2 + 15*B*a**2*b**4/4) + x**3*(5*A*a**2*b** 4 + 20*B*a**3*b**3/3) + x**2*(10*A*a**3*b**3 + 15*B*a**4*b**2/2) + x*(15*A *a**4*b**2 + 6*B*a**5*b)
Time = 0.03 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.08 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^3}{x^2} \, dx=\frac {1}{6} \, B b^{6} x^{6} - \frac {A a^{6}}{x} + \frac {1}{5} \, {\left (6 \, B a b^{5} + A b^{6}\right )} x^{5} + \frac {3}{4} \, {\left (5 \, B a^{2} b^{4} + 2 \, A a b^{5}\right )} x^{4} + \frac {5}{3} \, {\left (4 \, B a^{3} b^{3} + 3 \, A a^{2} b^{4}\right )} x^{3} + \frac {5}{2} \, {\left (3 \, B a^{4} b^{2} + 4 \, A a^{3} b^{3}\right )} x^{2} + 3 \, {\left (2 \, B a^{5} b + 5 \, A a^{4} b^{2}\right )} x + {\left (B a^{6} + 6 \, A a^{5} b\right )} \log \left (x\right ) \] Input:
integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^3/x^2,x, algorithm="maxima")
Output:
1/6*B*b^6*x^6 - A*a^6/x + 1/5*(6*B*a*b^5 + A*b^6)*x^5 + 3/4*(5*B*a^2*b^4 + 2*A*a*b^5)*x^4 + 5/3*(4*B*a^3*b^3 + 3*A*a^2*b^4)*x^3 + 5/2*(3*B*a^4*b^2 + 4*A*a^3*b^3)*x^2 + 3*(2*B*a^5*b + 5*A*a^4*b^2)*x + (B*a^6 + 6*A*a^5*b)*lo g(x)
Time = 0.20 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.08 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^3}{x^2} \, dx=\frac {1}{6} \, B b^{6} x^{6} + \frac {6}{5} \, B a b^{5} x^{5} + \frac {1}{5} \, A b^{6} x^{5} + \frac {15}{4} \, B a^{2} b^{4} x^{4} + \frac {3}{2} \, A a b^{5} x^{4} + \frac {20}{3} \, B a^{3} b^{3} x^{3} + 5 \, A a^{2} b^{4} x^{3} + \frac {15}{2} \, B a^{4} b^{2} x^{2} + 10 \, A a^{3} b^{3} x^{2} + 6 \, B a^{5} b x + 15 \, A a^{4} b^{2} x - \frac {A a^{6}}{x} + {\left (B a^{6} + 6 \, A a^{5} b\right )} \log \left ({\left | x \right |}\right ) \] Input:
integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^3/x^2,x, algorithm="giac")
Output:
1/6*B*b^6*x^6 + 6/5*B*a*b^5*x^5 + 1/5*A*b^6*x^5 + 15/4*B*a^2*b^4*x^4 + 3/2 *A*a*b^5*x^4 + 20/3*B*a^3*b^3*x^3 + 5*A*a^2*b^4*x^3 + 15/2*B*a^4*b^2*x^2 + 10*A*a^3*b^3*x^2 + 6*B*a^5*b*x + 15*A*a^4*b^2*x - A*a^6/x + (B*a^6 + 6*A* a^5*b)*log(abs(x))
Time = 10.51 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.95 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^3}{x^2} \, dx=x^5\,\left (\frac {A\,b^6}{5}+\frac {6\,B\,a\,b^5}{5}\right )+\ln \left (x\right )\,\left (B\,a^6+6\,A\,b\,a^5\right )-\frac {A\,a^6}{x}+\frac {B\,b^6\,x^6}{6}+\frac {5\,a^3\,b^2\,x^2\,\left (4\,A\,b+3\,B\,a\right )}{2}+\frac {5\,a^2\,b^3\,x^3\,\left (3\,A\,b+4\,B\,a\right )}{3}+3\,a^4\,b\,x\,\left (5\,A\,b+2\,B\,a\right )+\frac {3\,a\,b^4\,x^4\,\left (2\,A\,b+5\,B\,a\right )}{4} \] Input:
int(((A + B*x)*(a^2 + b^2*x^2 + 2*a*b*x)^3)/x^2,x)
Output:
x^5*((A*b^6)/5 + (6*B*a*b^5)/5) + log(x)*(B*a^6 + 6*A*a^5*b) - (A*a^6)/x + (B*b^6*x^6)/6 + (5*a^3*b^2*x^2*(4*A*b + 3*B*a))/2 + (5*a^2*b^3*x^3*(3*A*b + 4*B*a))/3 + 3*a^4*b*x*(5*A*b + 2*B*a) + (3*a*b^4*x^4*(2*A*b + 5*B*a))/4
Time = 0.26 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.61 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^3}{x^2} \, dx=\frac {420 \,\mathrm {log}\left (x \right ) a^{6} b x -60 a^{7}+1260 a^{5} b^{2} x^{2}+1050 a^{4} b^{3} x^{3}+700 a^{3} b^{4} x^{4}+315 a^{2} b^{5} x^{5}+84 a \,b^{6} x^{6}+10 b^{7} x^{7}}{60 x} \] Input:
int((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^3/x^2,x)
Output:
(420*log(x)*a**6*b*x - 60*a**7 + 1260*a**5*b**2*x**2 + 1050*a**4*b**3*x**3 + 700*a**3*b**4*x**4 + 315*a**2*b**5*x**5 + 84*a*b**6*x**6 + 10*b**7*x**7 )/(60*x)