Integrand size = 27, antiderivative size = 86 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2}{x^2} \, dx=-\frac {a^4 A}{x}+2 a^2 b (3 A b+2 a B) x+a b^2 (2 A b+3 a B) x^2+\frac {1}{3} b^3 (A b+4 a B) x^3+\frac {1}{4} b^4 B x^4+a^3 (4 A b+a B) \log (x) \] Output:
-a^4*A/x+2*a^2*b*(3*A*b+2*B*a)*x+a*b^2*(2*A*b+3*B*a)*x^2+1/3*b^3*(A*b+4*B* a)*x^3+1/4*b^4*B*x^4+a^3*(4*A*b+B*a)*ln(x)
Time = 0.04 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.99 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2}{x^2} \, dx=-\frac {a^4 A}{x}+4 a^3 b B x+3 a^2 b^2 x (2 A+B x)+\frac {2}{3} a b^3 x^2 (3 A+2 B x)+\frac {1}{12} b^4 x^3 (4 A+3 B x)+a^3 (4 A b+a B) \log (x) \] Input:
Integrate[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^2)/x^2,x]
Output:
-((a^4*A)/x) + 4*a^3*b*B*x + 3*a^2*b^2*x*(2*A + B*x) + (2*a*b^3*x^2*(3*A + 2*B*x))/3 + (b^4*x^3*(4*A + 3*B*x))/12 + a^3*(4*A*b + a*B)*Log[x]
Time = 0.42 (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.
\(\displaystyle \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^2 (A+B x)}{x^2} \, dx\) |
\(\Big \downarrow \) 1184 |
\(\displaystyle \frac {\int \frac {b^4 (a+b x)^4 (A+B x)}{x^2}dx}{b^4}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \frac {(a+b x)^4 (A+B x)}{x^2}dx\) |
\(\Big \downarrow \) 85 |
\(\displaystyle \int \left (\frac {a^4 A}{x^2}+\frac {a^3 (a B+4 A b)}{x}+2 a^2 b (2 a B+3 A b)+b^3 x^2 (4 a B+A b)+2 a b^2 x (3 a B+2 A b)+b^4 B x^3\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {a^4 A}{x}+a^3 \log (x) (a B+4 A b)+2 a^2 b x (2 a B+3 A b)+\frac {1}{3} b^3 x^3 (4 a B+A b)+a b^2 x^2 (3 a B+2 A b)+\frac {1}{4} b^4 B x^4\) |
Input:
Int[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^2)/x^2,x]
Output:
-((a^4*A)/x) + 2*a^2*b*(3*A*b + 2*a*B)*x + a*b^2*(2*A*b + 3*a*B)*x^2 + (b^ 3*(A*b + 4*a*B)*x^3)/3 + (b^4*B*x^4)/4 + a^3*(4*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 = 0.93 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.08
method | result | size |
default | \(\frac {b^{4} B \,x^{4}}{4}+\frac {A \,b^{4} x^{3}}{3}+\frac {4 B a \,b^{3} x^{3}}{3}+2 A a \,b^{3} x^{2}+3 B \,a^{2} b^{2} x^{2}+6 A \,a^{2} b^{2} x +4 B \,a^{3} b x +a^{3} \left (4 A b +B a \right ) \ln \left (x \right )-\frac {a^{4} A}{x}\) | \(93\) |
risch | \(\frac {b^{4} B \,x^{4}}{4}+\frac {A \,b^{4} x^{3}}{3}+\frac {4 B a \,b^{3} x^{3}}{3}+2 A a \,b^{3} x^{2}+3 B \,a^{2} b^{2} x^{2}+6 A \,a^{2} b^{2} x +4 B \,a^{3} b x -\frac {a^{4} A}{x}+4 A \ln \left (x \right ) a^{3} b +B \ln \left (x \right ) a^{4}\) | \(95\) |
norman | \(\frac {\left (\frac {1}{3} A \,b^{4}+\frac {4}{3} B a \,b^{3}\right ) x^{4}+\left (2 A a \,b^{3}+3 B \,a^{2} b^{2}\right ) x^{3}+\left (6 a^{2} A \,b^{2}+4 B \,a^{3} b \right ) x^{2}-a^{4} A +\frac {b^{4} B \,x^{5}}{4}}{x}+\left (4 A \,a^{3} b +a^{4} B \right ) \ln \left (x \right )\) | \(98\) |
parallelrisch | \(\frac {3 b^{4} B \,x^{5}+4 A \,b^{4} x^{4}+16 B a \,b^{3} x^{4}+24 A a \,b^{3} x^{3}+36 B \,a^{2} b^{2} x^{3}+48 A \ln \left (x \right ) x \,a^{3} b +72 A \,a^{2} b^{2} x^{2}+12 B \ln \left (x \right ) x \,a^{4}+48 B \,a^{3} b \,x^{2}-12 a^{4} A}{12 x}\) | \(104\) |
Input:
int((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^2/x^2,x,method=_RETURNVERBOSE)
Output:
1/4*b^4*B*x^4+1/3*A*b^4*x^3+4/3*B*a*b^3*x^3+2*A*a*b^3*x^2+3*B*a^2*b^2*x^2+ 6*A*a^2*b^2*x+4*B*a^3*b*x+a^3*(4*A*b+B*a)*ln(x)-a^4*A/x
Time = 0.06 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.17 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2}{x^2} \, dx=\frac {3 \, B b^{4} x^{5} - 12 \, A a^{4} + 4 \, {\left (4 \, B a b^{3} + A b^{4}\right )} x^{4} + 12 \, {\left (3 \, B a^{2} b^{2} + 2 \, A a b^{3}\right )} x^{3} + 24 \, {\left (2 \, B a^{3} b + 3 \, A a^{2} b^{2}\right )} x^{2} + 12 \, {\left (B a^{4} + 4 \, A a^{3} b\right )} x \log \left (x\right )}{12 \, x} \] Input:
integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^2/x^2,x, algorithm="fricas")
Output:
1/12*(3*B*b^4*x^5 - 12*A*a^4 + 4*(4*B*a*b^3 + A*b^4)*x^4 + 12*(3*B*a^2*b^2 + 2*A*a*b^3)*x^3 + 24*(2*B*a^3*b + 3*A*a^2*b^2)*x^2 + 12*(B*a^4 + 4*A*a^3 *b)*x*log(x))/x
Time = 0.12 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.09 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2}{x^2} \, dx=- \frac {A a^{4}}{x} + \frac {B b^{4} x^{4}}{4} + a^{3} \cdot \left (4 A b + B a\right ) \log {\left (x \right )} + x^{3} \left (\frac {A b^{4}}{3} + \frac {4 B a b^{3}}{3}\right ) + x^{2} \cdot \left (2 A a b^{3} + 3 B a^{2} b^{2}\right ) + x \left (6 A a^{2} b^{2} + 4 B a^{3} b\right ) \] Input:
integrate((B*x+A)*(b**2*x**2+2*a*b*x+a**2)**2/x**2,x)
Output:
-A*a**4/x + B*b**4*x**4/4 + a**3*(4*A*b + B*a)*log(x) + x**3*(A*b**4/3 + 4 *B*a*b**3/3) + x**2*(2*A*a*b**3 + 3*B*a**2*b**2) + x*(6*A*a**2*b**2 + 4*B* a**3*b)
Time = 0.03 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.09 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2}{x^2} \, dx=\frac {1}{4} \, B b^{4} x^{4} - \frac {A a^{4}}{x} + \frac {1}{3} \, {\left (4 \, B a b^{3} + A b^{4}\right )} x^{3} + {\left (3 \, B a^{2} b^{2} + 2 \, A a b^{3}\right )} x^{2} + 2 \, {\left (2 \, B a^{3} b + 3 \, A a^{2} b^{2}\right )} x + {\left (B a^{4} + 4 \, A a^{3} b\right )} \log \left (x\right ) \] Input:
integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^2/x^2,x, algorithm="maxima")
Output:
1/4*B*b^4*x^4 - A*a^4/x + 1/3*(4*B*a*b^3 + A*b^4)*x^3 + (3*B*a^2*b^2 + 2*A *a*b^3)*x^2 + 2*(2*B*a^3*b + 3*A*a^2*b^2)*x + (B*a^4 + 4*A*a^3*b)*log(x)
Time = 0.19 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.10 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2}{x^2} \, dx=\frac {1}{4} \, B b^{4} x^{4} + \frac {4}{3} \, B a b^{3} x^{3} + \frac {1}{3} \, A b^{4} x^{3} + 3 \, B a^{2} b^{2} x^{2} + 2 \, A a b^{3} x^{2} + 4 \, B a^{3} b x + 6 \, A a^{2} b^{2} x - \frac {A a^{4}}{x} + {\left (B a^{4} + 4 \, A a^{3} b\right )} \log \left ({\left | x \right |}\right ) \] Input:
integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^2/x^2,x, algorithm="giac")
Output:
1/4*B*b^4*x^4 + 4/3*B*a*b^3*x^3 + 1/3*A*b^4*x^3 + 3*B*a^2*b^2*x^2 + 2*A*a* b^3*x^2 + 4*B*a^3*b*x + 6*A*a^2*b^2*x - A*a^4/x + (B*a^4 + 4*A*a^3*b)*log( abs(x))
Time = 0.04 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.00 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2}{x^2} \, dx=x^3\,\left (\frac {A\,b^4}{3}+\frac {4\,B\,a\,b^3}{3}\right )+\ln \left (x\right )\,\left (B\,a^4+4\,A\,b\,a^3\right )-\frac {A\,a^4}{x}+\frac {B\,b^4\,x^4}{4}+2\,a^2\,b\,x\,\left (3\,A\,b+2\,B\,a\right )+a\,b^2\,x^2\,\left (2\,A\,b+3\,B\,a\right ) \] Input:
int(((A + B*x)*(a^2 + b^2*x^2 + 2*a*b*x)^2)/x^2,x)
Output:
x^3*((A*b^4)/3 + (4*B*a*b^3)/3) + log(x)*(B*a^4 + 4*A*a^3*b) - (A*a^4)/x + (B*b^4*x^4)/4 + 2*a^2*b*x*(3*A*b + 2*B*a) + a*b^2*x^2*(2*A*b + 3*B*a)
Time = 0.29 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.69 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2}{x^2} \, dx=\frac {60 \,\mathrm {log}\left (x \right ) a^{4} b x -12 a^{5}+120 a^{3} b^{2} x^{2}+60 a^{2} b^{3} x^{3}+20 a \,b^{4} x^{4}+3 b^{5} x^{5}}{12 x} \] Input:
int((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^2/x^2,x)
Output:
(60*log(x)*a**4*b*x - 12*a**5 + 120*a**3*b**2*x**2 + 60*a**2*b**3*x**3 + 2 0*a*b**4*x**4 + 3*b**5*x**5)/(12*x)