Integrand size = 27, antiderivative size = 86 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2}{x^5} \, dx=-\frac {a^4 A}{4 x^4}-\frac {a^3 (4 A b+a B)}{3 x^3}-\frac {a^2 b (3 A b+2 a B)}{x^2}-\frac {2 a b^2 (2 A b+3 a B)}{x}+b^4 B x+b^3 (A b+4 a B) \log (x) \] Output:
-1/4*a^4*A/x^4-1/3*a^3*(4*A*b+B*a)/x^3-a^2*b*(3*A*b+2*B*a)/x^2-2*a*b^2*(2* A*b+3*B*a)/x+b^4*B*x+b^3*(A*b+4*B*a)*ln(x)
Time = 0.04 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.99 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2}{x^5} \, dx=-\frac {4 a A b^3}{x}+b^4 B x-\frac {3 a^2 b^2 (A+2 B x)}{x^2}-\frac {2 a^3 b (2 A+3 B x)}{3 x^3}-\frac {a^4 (3 A+4 B x)}{12 x^4}+b^3 (A b+4 a B) \log (x) \] Input:
Integrate[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^2)/x^5,x]
Output:
(-4*a*A*b^3)/x + b^4*B*x - (3*a^2*b^2*(A + 2*B*x))/x^2 - (2*a^3*b*(2*A + 3 *B*x))/(3*x^3) - (a^4*(3*A + 4*B*x))/(12*x^4) + b^3*(A*b + 4*a*B)*Log[x]
Time = 0.41 (sec) , antiderivative size = 86, 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.
\(\displaystyle \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^2 (A+B x)}{x^5} \, dx\) |
\(\Big \downarrow \) 1184 |
\(\displaystyle \frac {\int \frac {b^4 (a+b x)^4 (A+B x)}{x^5}dx}{b^4}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \frac {(a+b x)^4 (A+B x)}{x^5}dx\) |
\(\Big \downarrow \) 85 |
\(\displaystyle \int \left (\frac {a^4 A}{x^5}+\frac {a^3 (a B+4 A b)}{x^4}+\frac {2 a^2 b (2 a B+3 A b)}{x^3}+\frac {b^3 (4 a B+A b)}{x}+\frac {2 a b^2 (3 a B+2 A b)}{x^2}+b^4 B\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {a^4 A}{4 x^4}-\frac {a^3 (a B+4 A b)}{3 x^3}-\frac {a^2 b (2 a B+3 A b)}{x^2}+b^3 \log (x) (4 a B+A b)-\frac {2 a b^2 (3 a B+2 A b)}{x}+b^4 B x\) |
Input:
Int[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^2)/x^5,x]
Output:
-1/4*(a^4*A)/x^4 - (a^3*(4*A*b + a*B))/(3*x^3) - (a^2*b*(3*A*b + 2*a*B))/x ^2 - (2*a*b^2*(2*A*b + 3*a*B))/x + b^4*B*x + b^3*(A*b + 4*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 = 83, normalized size of antiderivative = 0.97
method | result | size |
default | \(-\frac {a^{4} A}{4 x^{4}}-\frac {a^{3} \left (4 A b +B a \right )}{3 x^{3}}-\frac {a^{2} b \left (3 A b +2 B a \right )}{x^{2}}-\frac {2 a \,b^{2} \left (2 A b +3 B a \right )}{x}+B \,b^{4} x +b^{3} \left (A b +4 B a \right ) \ln \left (x \right )\) | \(83\) |
risch | \(B \,b^{4} x +\frac {\left (-4 A a \,b^{3}-6 B \,a^{2} b^{2}\right ) x^{3}+\left (-3 a^{2} A \,b^{2}-2 B \,a^{3} b \right ) x^{2}+\left (-\frac {4}{3} A \,a^{3} b -\frac {1}{3} a^{4} B \right ) x -\frac {a^{4} A}{4}}{x^{4}}+A \ln \left (x \right ) b^{4}+4 B \ln \left (x \right ) a \,b^{3}\) | \(93\) |
norman | \(\frac {\left (-\frac {4}{3} A \,a^{3} b -\frac {1}{3} a^{4} B \right ) x +\left (-4 A a \,b^{3}-6 B \,a^{2} b^{2}\right ) x^{3}+\left (-3 a^{2} A \,b^{2}-2 B \,a^{3} b \right ) x^{2}+b^{4} B \,x^{5}-\frac {a^{4} A}{4}}{x^{4}}+\left (A \,b^{4}+4 B a \,b^{3}\right ) \ln \left (x \right )\) | \(95\) |
parallelrisch | \(\frac {12 A \ln \left (x \right ) x^{4} b^{4}+48 B \ln \left (x \right ) x^{4} a \,b^{3}+12 b^{4} B \,x^{5}-48 A a \,b^{3} x^{3}-72 B \,a^{2} b^{2} x^{3}-36 A \,a^{2} b^{2} x^{2}-24 B \,a^{3} b \,x^{2}-16 A \,a^{3} b x -4 a^{4} B x -3 a^{4} A}{12 x^{4}}\) | \(104\) |
Input:
int((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^2/x^5,x,method=_RETURNVERBOSE)
Output:
-1/4*a^4*A/x^4-1/3*a^3*(4*A*b+B*a)/x^3-a^2*b*(3*A*b+2*B*a)/x^2-2*a*b^2*(2* A*b+3*B*a)/x+B*b^4*x+b^3*(A*b+4*B*a)*ln(x)
Time = 0.07 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.17 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2}{x^5} \, dx=\frac {12 \, B b^{4} x^{5} + 12 \, {\left (4 \, B a b^{3} + A b^{4}\right )} x^{4} \log \left (x\right ) - 3 \, A a^{4} - 24 \, {\left (3 \, B a^{2} b^{2} + 2 \, A a b^{3}\right )} x^{3} - 12 \, {\left (2 \, B a^{3} b + 3 \, A a^{2} b^{2}\right )} x^{2} - 4 \, {\left (B a^{4} + 4 \, A a^{3} b\right )} x}{12 \, x^{4}} \] Input:
integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^2/x^5,x, algorithm="fricas")
Output:
1/12*(12*B*b^4*x^5 + 12*(4*B*a*b^3 + A*b^4)*x^4*log(x) - 3*A*a^4 - 24*(3*B *a^2*b^2 + 2*A*a*b^3)*x^3 - 12*(2*B*a^3*b + 3*A*a^2*b^2)*x^2 - 4*(B*a^4 + 4*A*a^3*b)*x)/x^4
Time = 0.65 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.15 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2}{x^5} \, dx=B b^{4} x + b^{3} \left (A b + 4 B a\right ) \log {\left (x \right )} + \frac {- 3 A a^{4} + x^{3} \left (- 48 A a b^{3} - 72 B a^{2} b^{2}\right ) + x^{2} \left (- 36 A a^{2} b^{2} - 24 B a^{3} b\right ) + x \left (- 16 A a^{3} b - 4 B a^{4}\right )}{12 x^{4}} \] Input:
integrate((B*x+A)*(b**2*x**2+2*a*b*x+a**2)**2/x**5,x)
Output:
B*b**4*x + b**3*(A*b + 4*B*a)*log(x) + (-3*A*a**4 + x**3*(-48*A*a*b**3 - 7 2*B*a**2*b**2) + x**2*(-36*A*a**2*b**2 - 24*B*a**3*b) + x*(-16*A*a**3*b - 4*B*a**4))/(12*x**4)
Time = 0.04 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.10 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2}{x^5} \, dx=B b^{4} x + {\left (4 \, B a b^{3} + A b^{4}\right )} \log \left (x\right ) - \frac {3 \, A a^{4} + 24 \, {\left (3 \, B a^{2} b^{2} + 2 \, A a b^{3}\right )} x^{3} + 12 \, {\left (2 \, B a^{3} b + 3 \, A a^{2} b^{2}\right )} x^{2} + 4 \, {\left (B a^{4} + 4 \, A a^{3} b\right )} x}{12 \, x^{4}} \] Input:
integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^2/x^5,x, algorithm="maxima")
Output:
B*b^4*x + (4*B*a*b^3 + A*b^4)*log(x) - 1/12*(3*A*a^4 + 24*(3*B*a^2*b^2 + 2 *A*a*b^3)*x^3 + 12*(2*B*a^3*b + 3*A*a^2*b^2)*x^2 + 4*(B*a^4 + 4*A*a^3*b)*x )/x^4
Time = 0.16 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.12 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2}{x^5} \, dx=B b^{4} x + {\left (4 \, B a b^{3} + A b^{4}\right )} \log \left ({\left | x \right |}\right ) - \frac {3 \, A a^{4} + 24 \, {\left (3 \, B a^{2} b^{2} + 2 \, A a b^{3}\right )} x^{3} + 12 \, {\left (2 \, B a^{3} b + 3 \, A a^{2} b^{2}\right )} x^{2} + 4 \, {\left (B a^{4} + 4 \, A a^{3} b\right )} x}{12 \, x^{4}} \] Input:
integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^2/x^5,x, algorithm="giac")
Output:
B*b^4*x + (4*B*a*b^3 + A*b^4)*log(abs(x)) - 1/12*(3*A*a^4 + 24*(3*B*a^2*b^ 2 + 2*A*a*b^3)*x^3 + 12*(2*B*a^3*b + 3*A*a^2*b^2)*x^2 + 4*(B*a^4 + 4*A*a^3 *b)*x)/x^4
Time = 0.08 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.08 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2}{x^5} \, dx=\ln \left (x\right )\,\left (A\,b^4+4\,B\,a\,b^3\right )-\frac {x\,\left (\frac {B\,a^4}{3}+\frac {4\,A\,b\,a^3}{3}\right )+\frac {A\,a^4}{4}+x^2\,\left (2\,B\,a^3\,b+3\,A\,a^2\,b^2\right )+x^3\,\left (6\,B\,a^2\,b^2+4\,A\,a\,b^3\right )}{x^4}+B\,b^4\,x \] Input:
int(((A + B*x)*(a^2 + b^2*x^2 + 2*a*b*x)^2)/x^5,x)
Output:
log(x)*(A*b^4 + 4*B*a*b^3) - (x*((B*a^4)/3 + (4*A*a^3*b)/3) + (A*a^4)/4 + x^2*(3*A*a^2*b^2 + 2*B*a^3*b) + x^3*(6*B*a^2*b^2 + 4*A*a*b^3))/x^4 + B*b^4 *x
Time = 0.28 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.69 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2}{x^5} \, dx=\frac {60 \,\mathrm {log}\left (x \right ) a \,b^{4} x^{4}-3 a^{5}-20 a^{4} b x -60 a^{3} b^{2} x^{2}-120 a^{2} b^{3} x^{3}+12 b^{5} x^{5}}{12 x^{4}} \] Input:
int((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^2/x^5,x)
Output:
(60*log(x)*a*b**4*x**4 - 3*a**5 - 20*a**4*b*x - 60*a**3*b**2*x**2 - 120*a* *2*b**3*x**3 + 12*b**5*x**5)/(12*x**4)