Integrand size = 27, antiderivative size = 90 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2}{x^3} \, dx=-\frac {a^4 A}{2 x^2}-\frac {a^3 (4 A b+a B)}{x}+2 a b^2 (2 A b+3 a B) x+\frac {1}{2} b^3 (A b+4 a B) x^2+\frac {1}{3} b^4 B x^3+2 a^2 b (3 A b+2 a B) \log (x) \] Output:
-1/2*a^4*A/x^2-a^3*(4*A*b+B*a)/x+2*a*b^2*(2*A*b+3*B*a)*x+1/2*b^3*(A*b+4*B* a)*x^2+1/3*b^4*B*x^3+2*a^2*b*(3*A*b+2*B*a)*ln(x)
Time = 0.04 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.96 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2}{x^3} \, dx=-\frac {4 a^3 A b}{x}+6 a^2 b^2 B x+2 a b^3 x (2 A+B x)-\frac {a^4 (A+2 B x)}{2 x^2}+\frac {1}{6} b^4 x^2 (3 A+2 B x)+2 a^2 b (3 A b+2 a B) \log (x) \] Input:
Integrate[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^2)/x^3,x]
Output:
(-4*a^3*A*b)/x + 6*a^2*b^2*B*x + 2*a*b^3*x*(2*A + B*x) - (a^4*(A + 2*B*x)) /(2*x^2) + (b^4*x^2*(3*A + 2*B*x))/6 + 2*a^2*b*(3*A*b + 2*a*B)*Log[x]
Time = 0.41 (sec) , antiderivative size = 90, 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.
\(\displaystyle \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^2 (A+B x)}{x^3} \, dx\) |
\(\Big \downarrow \) 1184 |
\(\displaystyle \frac {\int \frac {b^4 (a+b x)^4 (A+B x)}{x^3}dx}{b^4}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \frac {(a+b x)^4 (A+B x)}{x^3}dx\) |
\(\Big \downarrow \) 85 |
\(\displaystyle \int \left (\frac {a^4 A}{x^3}+\frac {a^3 (a B+4 A b)}{x^2}+\frac {2 a^2 b (2 a B+3 A b)}{x}+b^3 x (4 a B+A b)+2 a b^2 (3 a B+2 A b)+b^4 B x^2\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {a^4 A}{2 x^2}-\frac {a^3 (a B+4 A b)}{x}+2 a^2 b \log (x) (2 a B+3 A b)+\frac {1}{2} b^3 x^2 (4 a B+A b)+2 a b^2 x (3 a B+2 A b)+\frac {1}{3} b^4 B x^3\) |
Input:
Int[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^2)/x^3,x]
Output:
-1/2*(a^4*A)/x^2 - (a^3*(4*A*b + a*B))/x + 2*a*b^2*(2*A*b + 3*a*B)*x + (b^ 3*(A*b + 4*a*B)*x^2)/2 + (b^4*B*x^3)/3 + 2*a^2*b*(3*A*b + 2*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 = 0.84 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.00
method | result | size |
default | \(\frac {b^{4} B \,x^{3}}{3}+\frac {A \,b^{4} x^{2}}{2}+2 B a \,b^{3} x^{2}+4 a A \,b^{3} x +6 B \,a^{2} b^{2} x -\frac {a^{4} A}{2 x^{2}}+2 a^{2} b \left (3 A b +2 B a \right ) \ln \left (x \right )-\frac {a^{3} \left (4 A b +B a \right )}{x}\) | \(90\) |
risch | \(\frac {b^{4} B \,x^{3}}{3}+\frac {A \,b^{4} x^{2}}{2}+2 B a \,b^{3} x^{2}+4 a A \,b^{3} x +6 B \,a^{2} b^{2} x +\frac {\left (-4 A \,a^{3} b -a^{4} B \right ) x -\frac {a^{4} A}{2}}{x^{2}}+6 A \ln \left (x \right ) a^{2} b^{2}+4 B \ln \left (x \right ) a^{3} b\) | \(95\) |
norman | \(\frac {\left (\frac {1}{2} A \,b^{4}+2 B a \,b^{3}\right ) x^{4}+\left (4 A a \,b^{3}+6 B \,a^{2} b^{2}\right ) x^{3}+\left (-4 A \,a^{3} b -a^{4} B \right ) x -\frac {a^{4} A}{2}+\frac {b^{4} B \,x^{5}}{3}}{x^{2}}+\left (6 a^{2} A \,b^{2}+4 B \,a^{3} b \right ) \ln \left (x \right )\) | \(97\) |
parallelrisch | \(\frac {2 b^{4} B \,x^{5}+3 A \,b^{4} x^{4}+12 B a \,b^{3} x^{4}+36 A \ln \left (x \right ) x^{2} a^{2} b^{2}+24 A a \,b^{3} x^{3}+24 B \ln \left (x \right ) x^{2} a^{3} b +36 B \,a^{2} b^{2} x^{3}-24 A \,a^{3} b x -6 a^{4} B x -3 a^{4} A}{6 x^{2}}\) | \(104\) |
Input:
int((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^2/x^3,x,method=_RETURNVERBOSE)
Output:
1/3*b^4*B*x^3+1/2*A*b^4*x^2+2*B*a*b^3*x^2+4*a*A*b^3*x+6*B*a^2*b^2*x-1/2*a^ 4*A/x^2+2*a^2*b*(3*A*b+2*B*a)*ln(x)-a^3*(4*A*b+B*a)/x
Time = 0.06 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.12 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2}{x^3} \, dx=\frac {2 \, B b^{4} x^{5} - 3 \, A a^{4} + 3 \, {\left (4 \, B a b^{3} + A b^{4}\right )} x^{4} + 12 \, {\left (3 \, B a^{2} b^{2} + 2 \, A a b^{3}\right )} x^{3} + 12 \, {\left (2 \, B a^{3} b + 3 \, A a^{2} b^{2}\right )} x^{2} \log \left (x\right ) - 6 \, {\left (B a^{4} + 4 \, A a^{3} b\right )} x}{6 \, x^{2}} \] Input:
integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^2/x^3,x, algorithm="fricas")
Output:
1/6*(2*B*b^4*x^5 - 3*A*a^4 + 3*(4*B*a*b^3 + A*b^4)*x^4 + 12*(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*log(x) - 6*(B*a^4 + 4*A *a^3*b)*x)/x^2
Time = 0.21 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.08 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2}{x^3} \, dx=\frac {B b^{4} x^{3}}{3} + 2 a^{2} b \left (3 A b + 2 B a\right ) \log {\left (x \right )} + x^{2} \left (\frac {A b^{4}}{2} + 2 B a b^{3}\right ) + x \left (4 A a b^{3} + 6 B a^{2} b^{2}\right ) + \frac {- A a^{4} + x \left (- 8 A a^{3} b - 2 B a^{4}\right )}{2 x^{2}} \] Input:
integrate((B*x+A)*(b**2*x**2+2*a*b*x+a**2)**2/x**3,x)
Output:
B*b**4*x**3/3 + 2*a**2*b*(3*A*b + 2*B*a)*log(x) + x**2*(A*b**4/2 + 2*B*a*b **3) + x*(4*A*a*b**3 + 6*B*a**2*b**2) + (-A*a**4 + x*(-8*A*a**3*b - 2*B*a* *4))/(2*x**2)
Time = 0.03 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.07 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2}{x^3} \, dx=\frac {1}{3} \, B b^{4} x^{3} + \frac {1}{2} \, {\left (4 \, B a b^{3} + A b^{4}\right )} x^{2} + 2 \, {\left (3 \, B a^{2} b^{2} + 2 \, A a b^{3}\right )} x + 2 \, {\left (2 \, B a^{3} b + 3 \, A a^{2} b^{2}\right )} \log \left (x\right ) - \frac {A a^{4} + 2 \, {\left (B a^{4} + 4 \, A a^{3} b\right )} x}{2 \, x^{2}} \] Input:
integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^2/x^3,x, algorithm="maxima")
Output:
1/3*B*b^4*x^3 + 1/2*(4*B*a*b^3 + A*b^4)*x^2 + 2*(3*B*a^2*b^2 + 2*A*a*b^3)* x + 2*(2*B*a^3*b + 3*A*a^2*b^2)*log(x) - 1/2*(A*a^4 + 2*(B*a^4 + 4*A*a^3*b )*x)/x^2
Time = 0.20 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.07 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2}{x^3} \, dx=\frac {1}{3} \, B b^{4} x^{3} + 2 \, B a b^{3} x^{2} + \frac {1}{2} \, A b^{4} x^{2} + 6 \, B a^{2} b^{2} x + 4 \, A a b^{3} x + 2 \, {\left (2 \, B a^{3} b + 3 \, A a^{2} b^{2}\right )} \log \left ({\left | x \right |}\right ) - \frac {A a^{4} + 2 \, {\left (B a^{4} + 4 \, A a^{3} b\right )} x}{2 \, x^{2}} \] Input:
integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^2/x^3,x, algorithm="giac")
Output:
1/3*B*b^4*x^3 + 2*B*a*b^3*x^2 + 1/2*A*b^4*x^2 + 6*B*a^2*b^2*x + 4*A*a*b^3* x + 2*(2*B*a^3*b + 3*A*a^2*b^2)*log(abs(x)) - 1/2*(A*a^4 + 2*(B*a^4 + 4*A* a^3*b)*x)/x^2
Time = 10.73 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.01 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2}{x^3} \, dx=\ln \left (x\right )\,\left (4\,B\,a^3\,b+6\,A\,a^2\,b^2\right )-\frac {x\,\left (B\,a^4+4\,A\,b\,a^3\right )+\frac {A\,a^4}{2}}{x^2}+x^2\,\left (\frac {A\,b^4}{2}+2\,B\,a\,b^3\right )+\frac {B\,b^4\,x^3}{3}+2\,a\,b^2\,x\,\left (2\,A\,b+3\,B\,a\right ) \] Input:
int(((A + B*x)*(a^2 + b^2*x^2 + 2*a*b*x)^2)/x^3,x)
Output:
log(x)*(6*A*a^2*b^2 + 4*B*a^3*b) - (x*(B*a^4 + 4*A*a^3*b) + (A*a^4)/2)/x^2 + x^2*((A*b^4)/2 + 2*B*a*b^3) + (B*b^4*x^3)/3 + 2*a*b^2*x*(2*A*b + 3*B*a)
Time = 0.30 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.66 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2}{x^3} \, dx=\frac {60 \,\mathrm {log}\left (x \right ) a^{3} b^{2} x^{2}-3 a^{5}-30 a^{4} b x +60 a^{2} b^{3} x^{3}+15 a \,b^{4} x^{4}+2 b^{5} x^{5}}{6 x^{2}} \] Input:
int((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^2/x^3,x)
Output:
(60*log(x)*a**3*b**2*x**2 - 3*a**5 - 30*a**4*b*x + 60*a**2*b**3*x**3 + 15* a*b**4*x**4 + 2*b**5*x**5)/(6*x**2)