Integrand size = 27, antiderivative size = 71 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2}{x^6} \, dx=-\frac {a^4 B}{4 x^4}-\frac {4 a^3 b B}{3 x^3}-\frac {3 a^2 b^2 B}{x^2}-\frac {4 a b^3 B}{x}-\frac {A (a+b x)^5}{5 a x^5}+b^4 B \log (x) \] Output:
-1/4*a^4*B/x^4-4/3*a^3*b*B/x^3-3*a^2*b^2*B/x^2-4*a*b^3*B/x-1/5*A*(b*x+a)^5 /a/x^5+b^4*B*ln(x)
Time = 0.05 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.23 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2}{x^6} \, dx=-\frac {60 A b^4 x^4+120 a b^3 x^3 (A+2 B x)+60 a^2 b^2 x^2 (2 A+3 B x)+20 a^3 b x (3 A+4 B x)+3 a^4 (4 A+5 B x)}{60 x^5}+b^4 B \log (x) \] Input:
Integrate[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^2)/x^6,x]
Output:
-1/60*(60*A*b^4*x^4 + 120*a*b^3*x^3*(A + 2*B*x) + 60*a^2*b^2*x^2*(2*A + 3* B*x) + 20*a^3*b*x*(3*A + 4*B*x) + 3*a^4*(4*A + 5*B*x))/x^5 + b^4*B*Log[x]
Time = 0.36 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.97, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {1184, 27, 87, 49, 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.
\(\displaystyle \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^2 (A+B x)}{x^6} \, dx\) |
\(\Big \downarrow \) 1184 |
\(\displaystyle \frac {\int \frac {b^4 (a+b x)^4 (A+B x)}{x^6}dx}{b^4}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \frac {(a+b x)^4 (A+B x)}{x^6}dx\) |
\(\Big \downarrow \) 87 |
\(\displaystyle B \int \frac {(a+b x)^4}{x^5}dx-\frac {A (a+b x)^5}{5 a x^5}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle B \int \left (\frac {a^4}{x^5}+\frac {4 b a^3}{x^4}+\frac {6 b^2 a^2}{x^3}+\frac {4 b^3 a}{x^2}+\frac {b^4}{x}\right )dx-\frac {A (a+b x)^5}{5 a x^5}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle B \left (-\frac {a^4}{4 x^4}-\frac {4 a^3 b}{3 x^3}-\frac {3 a^2 b^2}{x^2}-\frac {4 a b^3}{x}+b^4 \log (x)\right )-\frac {A (a+b x)^5}{5 a x^5}\) |
Input:
Int[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^2)/x^6,x]
Output:
-1/5*(A*(a + b*x)^5)/(a*x^5) + B*(-1/4*a^4/x^4 - (4*a^3*b)/(3*x^3) - (3*a^ 2*b^2)/x^2 - (4*a*b^3)/x + b^4*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[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 0] && IGtQ[m + n + 2, 0]
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)) Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege rQ[p] || !(IntegerQ[n] || !(EqQ[e, 0] || !(EqQ[c, 0] || LtQ[p, n]))))
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 = 0.92 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.21
method | result | size |
default | \(-\frac {a^{4} A}{5 x^{5}}-\frac {2 a^{2} b \left (3 A b +2 B a \right )}{3 x^{3}}-\frac {a \,b^{2} \left (2 A b +3 B a \right )}{x^{2}}-\frac {a^{3} \left (4 A b +B a \right )}{4 x^{4}}+b^{4} B \ln \left (x \right )-\frac {b^{3} \left (A b +4 B a \right )}{x}\) | \(86\) |
norman | \(\frac {\left (-A \,a^{3} b -\frac {1}{4} a^{4} B \right ) x +\left (-2 a^{2} A \,b^{2}-\frac {4}{3} B \,a^{3} b \right ) x^{2}+\left (-A \,b^{4}-4 B a \,b^{3}\right ) x^{4}+\left (-2 A a \,b^{3}-3 B \,a^{2} b^{2}\right ) x^{3}-\frac {a^{4} A}{5}}{x^{5}}+b^{4} B \ln \left (x \right )\) | \(96\) |
risch | \(\frac {\left (-A \,a^{3} b -\frac {1}{4} a^{4} B \right ) x +\left (-2 a^{2} A \,b^{2}-\frac {4}{3} B \,a^{3} b \right ) x^{2}+\left (-A \,b^{4}-4 B a \,b^{3}\right ) x^{4}+\left (-2 A a \,b^{3}-3 B \,a^{2} b^{2}\right ) x^{3}-\frac {a^{4} A}{5}}{x^{5}}+b^{4} B \ln \left (x \right )\) | \(96\) |
parallelrisch | \(-\frac {-60 B \,b^{4} \ln \left (x \right ) x^{5}+60 A \,b^{4} x^{4}+240 B a \,b^{3} x^{4}+120 A a \,b^{3} x^{3}+180 B \,a^{2} b^{2} x^{3}+120 A \,a^{2} b^{2} x^{2}+80 B \,a^{3} b \,x^{2}+60 A \,a^{3} b x +15 a^{4} B x +12 a^{4} A}{60 x^{5}}\) | \(102\) |
Input:
int((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^2/x^6,x,method=_RETURNVERBOSE)
Output:
-1/5*a^4*A/x^5-2/3*a^2*b*(3*A*b+2*B*a)/x^3-a*b^2*(2*A*b+3*B*a)/x^2-1/4*a^3 *(4*A*b+B*a)/x^4+b^4*B*ln(x)-b^3*(A*b+4*B*a)/x
Time = 0.07 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.42 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2}{x^6} \, dx=\frac {60 \, B b^{4} x^{5} \log \left (x\right ) - 12 \, A a^{4} - 60 \, {\left (4 \, B a b^{3} + A b^{4}\right )} x^{4} - 60 \, {\left (3 \, B a^{2} b^{2} + 2 \, A a b^{3}\right )} x^{3} - 40 \, {\left (2 \, B a^{3} b + 3 \, A a^{2} b^{2}\right )} x^{2} - 15 \, {\left (B a^{4} + 4 \, A a^{3} b\right )} x}{60 \, x^{5}} \] Input:
integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^2/x^6,x, algorithm="fricas")
Output:
1/60*(60*B*b^4*x^5*log(x) - 12*A*a^4 - 60*(4*B*a*b^3 + A*b^4)*x^4 - 60*(3* B*a^2*b^2 + 2*A*a*b^3)*x^3 - 40*(2*B*a^3*b + 3*A*a^2*b^2)*x^2 - 15*(B*a^4 + 4*A*a^3*b)*x)/x^5
Time = 0.97 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.48 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2}{x^6} \, dx=B b^{4} \log {\left (x \right )} + \frac {- 12 A a^{4} + x^{4} \left (- 60 A b^{4} - 240 B a b^{3}\right ) + x^{3} \left (- 120 A a b^{3} - 180 B a^{2} b^{2}\right ) + x^{2} \left (- 120 A a^{2} b^{2} - 80 B a^{3} b\right ) + x \left (- 60 A a^{3} b - 15 B a^{4}\right )}{60 x^{5}} \] Input:
integrate((B*x+A)*(b**2*x**2+2*a*b*x+a**2)**2/x**6,x)
Output:
B*b**4*log(x) + (-12*A*a**4 + x**4*(-60*A*b**4 - 240*B*a*b**3) + x**3*(-12 0*A*a*b**3 - 180*B*a**2*b**2) + x**2*(-120*A*a**2*b**2 - 80*B*a**3*b) + x* (-60*A*a**3*b - 15*B*a**4))/(60*x**5)
Time = 0.04 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.38 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2}{x^6} \, dx=B b^{4} \log \left (x\right ) - \frac {12 \, A a^{4} + 60 \, {\left (4 \, B a b^{3} + A b^{4}\right )} x^{4} + 60 \, {\left (3 \, B a^{2} b^{2} + 2 \, A a b^{3}\right )} x^{3} + 40 \, {\left (2 \, B a^{3} b + 3 \, A a^{2} b^{2}\right )} x^{2} + 15 \, {\left (B a^{4} + 4 \, A a^{3} b\right )} x}{60 \, x^{5}} \] Input:
integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^2/x^6,x, algorithm="maxima")
Output:
B*b^4*log(x) - 1/60*(12*A*a^4 + 60*(4*B*a*b^3 + A*b^4)*x^4 + 60*(3*B*a^2*b ^2 + 2*A*a*b^3)*x^3 + 40*(2*B*a^3*b + 3*A*a^2*b^2)*x^2 + 15*(B*a^4 + 4*A*a ^3*b)*x)/x^5
Time = 0.23 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.39 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2}{x^6} \, dx=B b^{4} \log \left ({\left | x \right |}\right ) - \frac {12 \, A a^{4} + 60 \, {\left (4 \, B a b^{3} + A b^{4}\right )} x^{4} + 60 \, {\left (3 \, B a^{2} b^{2} + 2 \, A a b^{3}\right )} x^{3} + 40 \, {\left (2 \, B a^{3} b + 3 \, A a^{2} b^{2}\right )} x^{2} + 15 \, {\left (B a^{4} + 4 \, A a^{3} b\right )} x}{60 \, x^{5}} \] Input:
integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^2/x^6,x, algorithm="giac")
Output:
B*b^4*log(abs(x)) - 1/60*(12*A*a^4 + 60*(4*B*a*b^3 + A*b^4)*x^4 + 60*(3*B* a^2*b^2 + 2*A*a*b^3)*x^3 + 40*(2*B*a^3*b + 3*A*a^2*b^2)*x^2 + 15*(B*a^4 + 4*A*a^3*b)*x)/x^5
Time = 10.61 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.32 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2}{x^6} \, dx=B\,b^4\,\ln \left (x\right )-\frac {x\,\left (\frac {B\,a^4}{4}+A\,b\,a^3\right )+\frac {A\,a^4}{5}+x^3\,\left (3\,B\,a^2\,b^2+2\,A\,a\,b^3\right )+x^2\,\left (\frac {4\,B\,a^3\,b}{3}+2\,A\,a^2\,b^2\right )+x^4\,\left (A\,b^4+4\,B\,a\,b^3\right )}{x^5} \] Input:
int(((A + B*x)*(a^2 + b^2*x^2 + 2*a*b*x)^2)/x^6,x)
Output:
B*b^4*log(x) - (x*((B*a^4)/4 + A*a^3*b) + (A*a^4)/5 + x^3*(3*B*a^2*b^2 + 2 *A*a*b^3) + x^2*(2*A*a^2*b^2 + (4*B*a^3*b)/3) + x^4*(A*b^4 + 4*B*a*b^3))/x ^5
Time = 0.28 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.83 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2}{x^6} \, dx=\frac {60 \,\mathrm {log}\left (x \right ) b^{5} x^{5}-12 a^{5}-75 a^{4} b x -200 a^{3} b^{2} x^{2}-300 a^{2} b^{3} x^{3}-300 a \,b^{4} x^{4}}{60 x^{5}} \] Input:
int((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^2/x^6,x)
Output:
(60*log(x)*b**5*x**5 - 12*a**5 - 75*a**4*b*x - 200*a**3*b**2*x**2 - 300*a* *2*b**3*x**3 - 300*a*b**4*x**4)/(60*x**5)