Integrand size = 27, antiderivative size = 134 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^3}{x^4} \, dx=-\frac {a^6 A}{3 x^3}-\frac {a^5 (6 A b+a B)}{2 x^2}-\frac {3 a^4 b (5 A b+2 a B)}{x}+5 a^2 b^3 (3 A b+4 a B) x+\frac {3}{2} a b^4 (2 A b+5 a B) x^2+\frac {1}{3} b^5 (A b+6 a B) x^3+\frac {1}{4} b^6 B x^4+5 a^3 b^2 (4 A b+3 a B) \log (x) \] Output:
-1/3*a^6*A/x^3-1/2*a^5*(6*A*b+B*a)/x^2-3*a^4*b*(5*A*b+2*B*a)/x+5*a^2*b^3*( 3*A*b+4*B*a)*x+3/2*a*b^4*(2*A*b+5*B*a)*x^2+1/3*b^5*(A*b+6*B*a)*x^3+1/4*b^6 *B*x^4+5*a^3*b^2*(4*A*b+3*B*a)*ln(x)
Time = 0.23 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.95 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^3}{x^4} \, dx=-\frac {15 a^4 A b^2}{x}+20 a^3 b^3 B x+\frac {15}{2} a^2 b^4 x (2 A+B x)-\frac {3 a^5 b (A+2 B x)}{x^2}+a b^5 x^2 (3 A+2 B x)-\frac {a^6 (2 A+3 B x)}{6 x^3}+\frac {1}{12} b^6 x^3 (4 A+3 B x)+5 a^3 b^2 (4 A b+3 a B) \log (x) \] Input:
Integrate[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^3)/x^4,x]
Output:
(-15*a^4*A*b^2)/x + 20*a^3*b^3*B*x + (15*a^2*b^4*x*(2*A + B*x))/2 - (3*a^5 *b*(A + 2*B*x))/x^2 + a*b^5*x^2*(3*A + 2*B*x) - (a^6*(2*A + 3*B*x))/(6*x^3 ) + (b^6*x^3*(4*A + 3*B*x))/12 + 5*a^3*b^2*(4*A*b + 3*a*B)*Log[x]
Time = 0.51 (sec) , antiderivative size = 134, 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.
\(\displaystyle \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^3 (A+B x)}{x^4} \, dx\) |
\(\Big \downarrow \) 1184 |
\(\displaystyle \frac {\int \frac {b^6 (a+b x)^6 (A+B x)}{x^4}dx}{b^6}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \frac {(a+b x)^6 (A+B x)}{x^4}dx\) |
\(\Big \downarrow \) 85 |
\(\displaystyle \int \left (\frac {a^6 A}{x^4}+\frac {a^5 (a B+6 A b)}{x^3}+\frac {3 a^4 b (2 a B+5 A b)}{x^2}+\frac {5 a^3 b^2 (3 a B+4 A b)}{x}+5 a^2 b^3 (4 a B+3 A b)+b^5 x^2 (6 a B+A b)+3 a b^4 x (5 a B+2 A b)+b^6 B x^3\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {a^6 A}{3 x^3}-\frac {a^5 (a B+6 A b)}{2 x^2}-\frac {3 a^4 b (2 a B+5 A b)}{x}+5 a^3 b^2 \log (x) (3 a B+4 A b)+5 a^2 b^3 x (4 a B+3 A b)+\frac {1}{3} b^5 x^3 (6 a B+A b)+\frac {3}{2} a b^4 x^2 (5 a B+2 A b)+\frac {1}{4} b^6 B x^4\) |
Input:
Int[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^3)/x^4,x]
Output:
-1/3*(a^6*A)/x^3 - (a^5*(6*A*b + a*B))/(2*x^2) - (3*a^4*b*(5*A*b + 2*a*B)) /x + 5*a^2*b^3*(3*A*b + 4*a*B)*x + (3*a*b^4*(2*A*b + 5*a*B)*x^2)/2 + (b^5* (A*b + 6*a*B)*x^3)/3 + (b^6*B*x^4)/4 + 5*a^3*b^2*(4*A*b + 3*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 = 0.87 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.00
method | result | size |
default | \(\frac {b^{6} B \,x^{4}}{4}+\frac {A \,b^{6} x^{3}}{3}+2 B a \,b^{5} x^{3}+3 A a \,b^{5} x^{2}+\frac {15 B \,a^{2} b^{4} x^{2}}{2}+15 A \,a^{2} b^{4} x +20 B \,a^{3} b^{3} x -\frac {a^{6} A}{3 x^{3}}-\frac {a^{5} \left (6 A b +B a \right )}{2 x^{2}}+5 a^{3} b^{2} \left (4 A b +3 B a \right ) \ln \left (x \right )-\frac {3 a^{4} b \left (5 A b +2 B a \right )}{x}\) | \(134\) |
risch | \(\frac {b^{6} B \,x^{4}}{4}+\frac {A \,b^{6} x^{3}}{3}+2 B a \,b^{5} x^{3}+3 A a \,b^{5} x^{2}+\frac {15 B \,a^{2} b^{4} x^{2}}{2}+15 A \,a^{2} b^{4} x +20 B \,a^{3} b^{3} x +\frac {\left (-15 A \,a^{4} b^{2}-6 B \,a^{5} b \right ) x^{2}+\left (-3 A \,a^{5} b -\frac {1}{2} B \,a^{6}\right ) x -\frac {A \,a^{6}}{3}}{x^{3}}+20 A \ln \left (x \right ) a^{3} b^{3}+15 B \ln \left (x \right ) a^{4} b^{2}\) | \(142\) |
norman | \(\frac {\left (\frac {1}{3} A \,b^{6}+2 B a \,b^{5}\right ) x^{6}+\left (3 A a \,b^{5}+\frac {15}{2} B \,a^{2} b^{4}\right ) x^{5}+\left (-3 A \,a^{5} b -\frac {1}{2} B \,a^{6}\right ) x +\left (15 A \,a^{2} b^{4}+20 B \,a^{3} b^{3}\right ) x^{4}+\left (-15 A \,a^{4} b^{2}-6 B \,a^{5} b \right ) x^{2}-\frac {A \,a^{6}}{3}+\frac {b^{6} B \,x^{7}}{4}}{x^{3}}+\left (20 A \,a^{3} b^{3}+15 B \,a^{4} b^{2}\right ) \ln \left (x \right )\) | \(143\) |
parallelrisch | \(\frac {3 b^{6} B \,x^{7}+4 A \,b^{6} x^{6}+24 B a \,b^{5} x^{6}+36 A a \,b^{5} x^{5}+90 B \,a^{2} b^{4} x^{5}+240 A \ln \left (x \right ) x^{3} a^{3} b^{3}+180 A \,a^{2} b^{4} x^{4}+180 B \ln \left (x \right ) x^{3} a^{4} b^{2}+240 B \,a^{3} b^{3} x^{4}-180 A \,a^{4} b^{2} x^{2}-72 B \,a^{5} b \,x^{2}-36 A \,a^{5} b x -6 B \,a^{6} x -4 A \,a^{6}}{12 x^{3}}\) | \(152\) |
Input:
int((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^3/x^4,x,method=_RETURNVERBOSE)
Output:
1/4*b^6*B*x^4+1/3*A*b^6*x^3+2*B*a*b^5*x^3+3*A*a*b^5*x^2+15/2*B*a^2*b^4*x^2 +15*A*a^2*b^4*x+20*B*a^3*b^3*x-1/3*a^6*A/x^3-1/2*a^5*(6*A*b+B*a)/x^2+5*a^3 *b^2*(4*A*b+3*B*a)*ln(x)-3*a^4*b*(5*A*b+2*B*a)/x
Time = 0.07 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.11 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^3}{x^4} \, dx=\frac {3 \, B b^{6} x^{7} - 4 \, A a^{6} + 4 \, {\left (6 \, B a b^{5} + A b^{6}\right )} x^{6} + 18 \, {\left (5 \, B a^{2} b^{4} + 2 \, A a b^{5}\right )} x^{5} + 60 \, {\left (4 \, B a^{3} b^{3} + 3 \, A a^{2} b^{4}\right )} x^{4} + 60 \, {\left (3 \, B a^{4} b^{2} + 4 \, A a^{3} b^{3}\right )} x^{3} \log \left (x\right ) - 36 \, {\left (2 \, B a^{5} b + 5 \, A a^{4} b^{2}\right )} x^{2} - 6 \, {\left (B a^{6} + 6 \, A a^{5} b\right )} x}{12 \, x^{3}} \] Input:
integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^3/x^4,x, algorithm="fricas")
Output:
1/12*(3*B*b^6*x^7 - 4*A*a^6 + 4*(6*B*a*b^5 + A*b^6)*x^6 + 18*(5*B*a^2*b^4 + 2*A*a*b^5)*x^5 + 60*(4*B*a^3*b^3 + 3*A*a^2*b^4)*x^4 + 60*(3*B*a^4*b^2 + 4*A*a^3*b^3)*x^3*log(x) - 36*(2*B*a^5*b + 5*A*a^4*b^2)*x^2 - 6*(B*a^6 + 6* A*a^5*b)*x)/x^3
Time = 0.43 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.12 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^3}{x^4} \, dx=\frac {B b^{6} x^{4}}{4} + 5 a^{3} b^{2} \cdot \left (4 A b + 3 B a\right ) \log {\left (x \right )} + x^{3} \left (\frac {A b^{6}}{3} + 2 B a b^{5}\right ) + x^{2} \cdot \left (3 A a b^{5} + \frac {15 B a^{2} b^{4}}{2}\right ) + x \left (15 A a^{2} b^{4} + 20 B a^{3} b^{3}\right ) + \frac {- 2 A a^{6} + x^{2} \left (- 90 A a^{4} b^{2} - 36 B a^{5} b\right ) + x \left (- 18 A a^{5} b - 3 B a^{6}\right )}{6 x^{3}} \] Input:
integrate((B*x+A)*(b**2*x**2+2*a*b*x+a**2)**3/x**4,x)
Output:
B*b**6*x**4/4 + 5*a**3*b**2*(4*A*b + 3*B*a)*log(x) + x**3*(A*b**6/3 + 2*B* a*b**5) + x**2*(3*A*a*b**5 + 15*B*a**2*b**4/2) + x*(15*A*a**2*b**4 + 20*B* a**3*b**3) + (-2*A*a**6 + x**2*(-90*A*a**4*b**2 - 36*B*a**5*b) + x*(-18*A* a**5*b - 3*B*a**6))/(6*x**3)
Time = 0.03 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.08 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^3}{x^4} \, dx=\frac {1}{4} \, B b^{6} x^{4} + \frac {1}{3} \, {\left (6 \, B a b^{5} + A b^{6}\right )} x^{3} + \frac {3}{2} \, {\left (5 \, B a^{2} b^{4} + 2 \, A a b^{5}\right )} x^{2} + 5 \, {\left (4 \, B a^{3} b^{3} + 3 \, A a^{2} b^{4}\right )} x + 5 \, {\left (3 \, B a^{4} b^{2} + 4 \, A a^{3} b^{3}\right )} \log \left (x\right ) - \frac {2 \, A a^{6} + 18 \, {\left (2 \, B a^{5} b + 5 \, A a^{4} b^{2}\right )} x^{2} + 3 \, {\left (B a^{6} + 6 \, A a^{5} b\right )} x}{6 \, x^{3}} \] Input:
integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^3/x^4,x, algorithm="maxima")
Output:
1/4*B*b^6*x^4 + 1/3*(6*B*a*b^5 + A*b^6)*x^3 + 3/2*(5*B*a^2*b^4 + 2*A*a*b^5 )*x^2 + 5*(4*B*a^3*b^3 + 3*A*a^2*b^4)*x + 5*(3*B*a^4*b^2 + 4*A*a^3*b^3)*lo g(x) - 1/6*(2*A*a^6 + 18*(2*B*a^5*b + 5*A*a^4*b^2)*x^2 + 3*(B*a^6 + 6*A*a^ 5*b)*x)/x^3
Time = 0.19 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.08 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^3}{x^4} \, dx=\frac {1}{4} \, B b^{6} x^{4} + 2 \, B a b^{5} x^{3} + \frac {1}{3} \, A b^{6} x^{3} + \frac {15}{2} \, B a^{2} b^{4} x^{2} + 3 \, A a b^{5} x^{2} + 20 \, B a^{3} b^{3} x + 15 \, A a^{2} b^{4} x + 5 \, {\left (3 \, B a^{4} b^{2} + 4 \, A a^{3} b^{3}\right )} \log \left ({\left | x \right |}\right ) - \frac {2 \, A a^{6} + 18 \, {\left (2 \, B a^{5} b + 5 \, A a^{4} b^{2}\right )} x^{2} + 3 \, {\left (B a^{6} + 6 \, A a^{5} b\right )} x}{6 \, x^{3}} \] Input:
integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^3/x^4,x, algorithm="giac")
Output:
1/4*B*b^6*x^4 + 2*B*a*b^5*x^3 + 1/3*A*b^6*x^3 + 15/2*B*a^2*b^4*x^2 + 3*A*a *b^5*x^2 + 20*B*a^3*b^3*x + 15*A*a^2*b^4*x + 5*(3*B*a^4*b^2 + 4*A*a^3*b^3) *log(abs(x)) - 1/6*(2*A*a^6 + 18*(2*B*a^5*b + 5*A*a^4*b^2)*x^2 + 3*(B*a^6 + 6*A*a^5*b)*x)/x^3
Time = 0.06 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.01 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^3}{x^4} \, dx=x^3\,\left (\frac {A\,b^6}{3}+2\,B\,a\,b^5\right )-\frac {x\,\left (\frac {B\,a^6}{2}+3\,A\,b\,a^5\right )+\frac {A\,a^6}{3}+x^2\,\left (6\,B\,a^5\,b+15\,A\,a^4\,b^2\right )}{x^3}+\ln \left (x\right )\,\left (15\,B\,a^4\,b^2+20\,A\,a^3\,b^3\right )+\frac {B\,b^6\,x^4}{4}+5\,a^2\,b^3\,x\,\left (3\,A\,b+4\,B\,a\right )+\frac {3\,a\,b^4\,x^2\,\left (2\,A\,b+5\,B\,a\right )}{2} \] Input:
int(((A + B*x)*(a^2 + b^2*x^2 + 2*a*b*x)^3)/x^4,x)
Output:
x^3*((A*b^6)/3 + 2*B*a*b^5) - (x*((B*a^6)/2 + 3*A*a^5*b) + (A*a^6)/3 + x^2 *(15*A*a^4*b^2 + 6*B*a^5*b))/x^3 + log(x)*(20*A*a^3*b^3 + 15*B*a^4*b^2) + (B*b^6*x^4)/4 + 5*a^2*b^3*x*(3*A*b + 4*B*a) + (3*a*b^4*x^2*(2*A*b + 5*B*a) )/2
Time = 0.26 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.60 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^3}{x^4} \, dx=\frac {420 \,\mathrm {log}\left (x \right ) a^{4} b^{3} x^{3}-4 a^{7}-42 a^{6} b x -252 a^{5} b^{2} x^{2}+420 a^{3} b^{4} x^{4}+126 a^{2} b^{5} x^{5}+28 a \,b^{6} x^{6}+3 b^{7} x^{7}}{12 x^{3}} \] Input:
int((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^3/x^4,x)
Output:
(420*log(x)*a**4*b**3*x**3 - 4*a**7 - 42*a**6*b*x - 252*a**5*b**2*x**2 + 4 20*a**3*b**4*x**4 + 126*a**2*b**5*x**5 + 28*a*b**6*x**6 + 3*b**7*x**7)/(12 *x**3)