Integrand size = 27, antiderivative size = 89 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2}{x^4} \, dx=-\frac {a^4 A}{3 x^3}-\frac {a^3 (4 A b+a B)}{2 x^2}-\frac {2 a^2 b (3 A b+2 a B)}{x}+b^3 (A b+4 a B) x+\frac {1}{2} b^4 B x^2+2 a b^2 (2 A b+3 a B) \log (x) \] Output:
-1/3*a^4*A/x^3-1/2*a^3*(4*A*b+B*a)/x^2-2*a^2*b*(3*A*b+2*B*a)/x+b^3*(A*b+4* B*a)*x+1/2*b^4*B*x^2+2*a*b^2*(2*A*b+3*B*a)*ln(x)
Time = 0.04 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.97 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2}{x^4} \, dx=-\frac {6 a^2 A b^2}{x}+4 a b^3 B x+\frac {1}{2} b^4 x (2 A+B x)-\frac {2 a^3 b (A+2 B x)}{x^2}-\frac {a^4 (2 A+3 B x)}{6 x^3}+2 a b^2 (2 A b+3 a B) \log (x) \] Input:
Integrate[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^2)/x^4,x]
Output:
(-6*a^2*A*b^2)/x + 4*a*b^3*B*x + (b^4*x*(2*A + B*x))/2 - (2*a^3*b*(A + 2*B *x))/x^2 - (a^4*(2*A + 3*B*x))/(6*x^3) + 2*a*b^2*(2*A*b + 3*a*B)*Log[x]
Time = 0.44 (sec) , antiderivative size = 89, 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.
\(\displaystyle \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^2 (A+B x)}{x^4} \, dx\) |
\(\Big \downarrow \) 1184 |
\(\displaystyle \frac {\int \frac {b^4 (a+b x)^4 (A+B x)}{x^4}dx}{b^4}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \frac {(a+b x)^4 (A+B x)}{x^4}dx\) |
\(\Big \downarrow \) 85 |
\(\displaystyle \int \left (\frac {a^4 A}{x^4}+\frac {a^3 (a B+4 A b)}{x^3}+\frac {2 a^2 b (2 a B+3 A b)}{x^2}+b^3 (4 a B+A b)+\frac {2 a b^2 (3 a B+2 A b)}{x}+b^4 B x\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {a^4 A}{3 x^3}-\frac {a^3 (a B+4 A b)}{2 x^2}-\frac {2 a^2 b (2 a B+3 A b)}{x}+b^3 x (4 a B+A b)+2 a b^2 \log (x) (3 a B+2 A b)+\frac {1}{2} b^4 B x^2\) |
Input:
Int[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^2)/x^4,x]
Output:
-1/3*(a^4*A)/x^3 - (a^3*(4*A*b + a*B))/(2*x^2) - (2*a^2*b*(3*A*b + 2*a*B)) /x + b^3*(A*b + 4*a*B)*x + (b^4*B*x^2)/2 + 2*a*b^2*(2*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 = 0.91 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.96
method | result | size |
default | \(\frac {B \,b^{4} x^{2}}{2}+A \,b^{4} x +4 B a \,b^{3} x -\frac {a^{4} A}{3 x^{3}}-\frac {a^{3} \left (4 A b +B a \right )}{2 x^{2}}+2 a \,b^{2} \left (2 A b +3 B a \right ) \ln \left (x \right )-\frac {2 a^{2} b \left (3 A b +2 B a \right )}{x}\) | \(85\) |
risch | \(\frac {B \,b^{4} x^{2}}{2}+A \,b^{4} x +4 B a \,b^{3} x +\frac {\left (-6 a^{2} A \,b^{2}-4 B \,a^{3} b \right ) x^{2}+\left (-2 A \,a^{3} b -\frac {1}{2} a^{4} B \right ) x -\frac {a^{4} A}{3}}{x^{3}}+4 A \ln \left (x \right ) a \,b^{3}+6 B \ln \left (x \right ) a^{2} b^{2}\) | \(93\) |
norman | \(\frac {\left (-2 A \,a^{3} b -\frac {1}{2} a^{4} B \right ) x +\left (A \,b^{4}+4 B a \,b^{3}\right ) x^{4}+\left (-6 a^{2} A \,b^{2}-4 B \,a^{3} b \right ) x^{2}-\frac {a^{4} A}{3}+\frac {b^{4} B \,x^{5}}{2}}{x^{3}}+\left (4 A a \,b^{3}+6 B \,a^{2} b^{2}\right ) \ln \left (x \right )\) | \(96\) |
parallelrisch | \(\frac {3 b^{4} B \,x^{5}+24 A \ln \left (x \right ) x^{3} a \,b^{3}+6 A \,b^{4} x^{4}+36 B \ln \left (x \right ) x^{3} a^{2} b^{2}+24 B a \,b^{3} x^{4}-36 A \,a^{2} b^{2} x^{2}-24 B \,a^{3} b \,x^{2}-12 A \,a^{3} b x -3 a^{4} B x -2 a^{4} A}{6 x^{3}}\) | \(104\) |
Input:
int((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^2/x^4,x,method=_RETURNVERBOSE)
Output:
1/2*B*b^4*x^2+A*b^4*x+4*B*a*b^3*x-1/3*a^4*A/x^3-1/2*a^3*(4*A*b+B*a)/x^2+2* a*b^2*(2*A*b+3*B*a)*ln(x)-2*a^2*b*(3*A*b+2*B*a)/x
Time = 0.06 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.13 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2}{x^4} \, dx=\frac {3 \, B b^{4} x^{5} - 2 \, A a^{4} + 6 \, {\left (4 \, B a b^{3} + A b^{4}\right )} x^{4} + 12 \, {\left (3 \, B a^{2} b^{2} + 2 \, A a b^{3}\right )} x^{3} \log \left (x\right ) - 12 \, {\left (2 \, B a^{3} b + 3 \, A a^{2} b^{2}\right )} x^{2} - 3 \, {\left (B a^{4} + 4 \, A a^{3} b\right )} x}{6 \, x^{3}} \] Input:
integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^2/x^4,x, algorithm="fricas")
Output:
1/6*(3*B*b^4*x^5 - 2*A*a^4 + 6*(4*B*a*b^3 + A*b^4)*x^4 + 12*(3*B*a^2*b^2 + 2*A*a*b^3)*x^3*log(x) - 12*(2*B*a^3*b + 3*A*a^2*b^2)*x^2 - 3*(B*a^4 + 4*A *a^3*b)*x)/x^3
Time = 0.39 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.11 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2}{x^4} \, dx=\frac {B b^{4} x^{2}}{2} + 2 a b^{2} \cdot \left (2 A b + 3 B a\right ) \log {\left (x \right )} + x \left (A b^{4} + 4 B a b^{3}\right ) + \frac {- 2 A a^{4} + x^{2} \left (- 36 A a^{2} b^{2} - 24 B a^{3} b\right ) + x \left (- 12 A a^{3} b - 3 B a^{4}\right )}{6 x^{3}} \] Input:
integrate((B*x+A)*(b**2*x**2+2*a*b*x+a**2)**2/x**4,x)
Output:
B*b**4*x**2/2 + 2*a*b**2*(2*A*b + 3*B*a)*log(x) + x*(A*b**4 + 4*B*a*b**3) + (-2*A*a**4 + x**2*(-36*A*a**2*b**2 - 24*B*a**3*b) + x*(-12*A*a**3*b - 3* B*a**4))/(6*x**3)
Time = 0.03 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.08 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2}{x^4} \, dx=\frac {1}{2} \, B b^{4} x^{2} + {\left (4 \, B a b^{3} + A b^{4}\right )} x + 2 \, {\left (3 \, B a^{2} b^{2} + 2 \, A a b^{3}\right )} \log \left (x\right ) - \frac {2 \, A a^{4} + 12 \, {\left (2 \, B a^{3} b + 3 \, A a^{2} b^{2}\right )} x^{2} + 3 \, {\left (B a^{4} + 4 \, A a^{3} b\right )} x}{6 \, x^{3}} \] Input:
integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^2/x^4,x, algorithm="maxima")
Output:
1/2*B*b^4*x^2 + (4*B*a*b^3 + A*b^4)*x + 2*(3*B*a^2*b^2 + 2*A*a*b^3)*log(x) - 1/6*(2*A*a^4 + 12*(2*B*a^3*b + 3*A*a^2*b^2)*x^2 + 3*(B*a^4 + 4*A*a^3*b) *x)/x^3
Time = 0.21 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.08 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2}{x^4} \, dx=\frac {1}{2} \, B b^{4} x^{2} + 4 \, B a b^{3} x + A b^{4} x + 2 \, {\left (3 \, B a^{2} b^{2} + 2 \, A a b^{3}\right )} \log \left ({\left | x \right |}\right ) - \frac {2 \, A a^{4} + 12 \, {\left (2 \, B a^{3} b + 3 \, A a^{2} b^{2}\right )} x^{2} + 3 \, {\left (B a^{4} + 4 \, A a^{3} b\right )} x}{6 \, x^{3}} \] Input:
integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^2/x^4,x, algorithm="giac")
Output:
1/2*B*b^4*x^2 + 4*B*a*b^3*x + A*b^4*x + 2*(3*B*a^2*b^2 + 2*A*a*b^3)*log(ab s(x)) - 1/6*(2*A*a^4 + 12*(2*B*a^3*b + 3*A*a^2*b^2)*x^2 + 3*(B*a^4 + 4*A*a ^3*b)*x)/x^3
Time = 10.62 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.06 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2}{x^4} \, dx=x\,\left (A\,b^4+4\,B\,a\,b^3\right )+\ln \left (x\right )\,\left (6\,B\,a^2\,b^2+4\,A\,a\,b^3\right )-\frac {x\,\left (\frac {B\,a^4}{2}+2\,A\,b\,a^3\right )+\frac {A\,a^4}{3}+x^2\,\left (4\,B\,a^3\,b+6\,A\,a^2\,b^2\right )}{x^3}+\frac {B\,b^4\,x^2}{2} \] Input:
int(((A + B*x)*(a^2 + b^2*x^2 + 2*a*b*x)^2)/x^4,x)
Output:
x*(A*b^4 + 4*B*a*b^3) + log(x)*(6*B*a^2*b^2 + 4*A*a*b^3) - (x*((B*a^4)/2 + 2*A*a^3*b) + (A*a^4)/3 + x^2*(6*A*a^2*b^2 + 4*B*a^3*b))/x^3 + (B*b^4*x^2) /2
Time = 0.29 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.66 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2}{x^4} \, dx=\frac {60 \,\mathrm {log}\left (x \right ) a^{2} b^{3} x^{3}-2 a^{5}-15 a^{4} b x -60 a^{3} b^{2} x^{2}+30 a \,b^{4} x^{4}+3 b^{5} x^{5}}{6 x^{3}} \] Input:
int((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^2/x^4,x)
Output:
(60*log(x)*a**2*b**3*x**3 - 2*a**5 - 15*a**4*b*x - 60*a**3*b**2*x**2 + 30* a*b**4*x**4 + 3*b**5*x**5)/(6*x**3)