Integrand size = 27, antiderivative size = 113 \[ \int \frac {A+B x}{x^4 \left (a^2+2 a b x+b^2 x^2\right )} \, dx=-\frac {A}{3 a^2 x^3}+\frac {2 A b-a B}{2 a^3 x^2}-\frac {b (3 A b-2 a B)}{a^4 x}-\frac {b^2 (A b-a B)}{a^4 (a+b x)}-\frac {b^2 (4 A b-3 a B) \log (x)}{a^5}+\frac {b^2 (4 A b-3 a B) \log (a+b x)}{a^5} \] Output:
-1/3*A/a^2/x^3+1/2*(2*A*b-B*a)/a^3/x^2-b*(3*A*b-2*B*a)/a^4/x-b^2*(A*b-B*a) /a^4/(b*x+a)-b^2*(4*A*b-3*B*a)*ln(x)/a^5+b^2*(4*A*b-3*B*a)*ln(b*x+a)/a^5
Time = 0.10 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.94 \[ \int \frac {A+B x}{x^4 \left (a^2+2 a b x+b^2 x^2\right )} \, dx=\frac {-\frac {2 a^3 A}{x^3}-\frac {3 a^2 (-2 A b+a B)}{x^2}+\frac {6 a b (-3 A b+2 a B)}{x}+\frac {6 a b^2 (-A b+a B)}{a+b x}+6 b^2 (-4 A b+3 a B) \log (x)+6 b^2 (4 A b-3 a B) \log (a+b x)}{6 a^5} \] Input:
Integrate[(A + B*x)/(x^4*(a^2 + 2*a*b*x + b^2*x^2)),x]
Output:
((-2*a^3*A)/x^3 - (3*a^2*(-2*A*b + a*B))/x^2 + (6*a*b*(-3*A*b + 2*a*B))/x + (6*a*b^2*(-(A*b) + a*B))/(a + b*x) + 6*b^2*(-4*A*b + 3*a*B)*Log[x] + 6*b ^2*(4*A*b - 3*a*B)*Log[a + b*x])/(6*a^5)
Time = 0.52 (sec) , antiderivative size = 113, 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, 86, 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.
\(\displaystyle \int \frac {A+B x}{x^4 \left (a^2+2 a b x+b^2 x^2\right )} \, dx\) |
\(\Big \downarrow \) 1184 |
\(\displaystyle b^2 \int \frac {A+B x}{b^2 x^4 (a+b x)^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \frac {A+B x}{x^4 (a+b x)^2}dx\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \int \left (-\frac {b^3 (3 a B-4 A b)}{a^5 (a+b x)}+\frac {b^2 (3 a B-4 A b)}{a^5 x}-\frac {b^3 (a B-A b)}{a^4 (a+b x)^2}-\frac {b (2 a B-3 A b)}{a^4 x^2}+\frac {a B-2 A b}{a^3 x^3}+\frac {A}{a^2 x^4}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {b^2 \log (x) (4 A b-3 a B)}{a^5}+\frac {b^2 (4 A b-3 a B) \log (a+b x)}{a^5}-\frac {b^2 (A b-a B)}{a^4 (a+b x)}-\frac {b (3 A b-2 a B)}{a^4 x}+\frac {2 A b-a B}{2 a^3 x^2}-\frac {A}{3 a^2 x^3}\) |
Input:
Int[(A + B*x)/(x^4*(a^2 + 2*a*b*x + b^2*x^2)),x]
Output:
-1/3*A/(a^2*x^3) + (2*A*b - a*B)/(2*a^3*x^2) - (b*(3*A*b - 2*a*B))/(a^4*x) - (b^2*(A*b - a*B))/(a^4*(a + b*x)) - (b^2*(4*A*b - 3*a*B)*Log[x])/a^5 + (b^2*(4*A*b - 3*a*B)*Log[a + b*x])/a^5
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_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_ .), x_] :> Int[ExpandIntegrand[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && ((ILtQ[n, 0] && ILtQ[p, 0]) || EqQ[p, 1 ] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))
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 = 1.24 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.96
method | result | size |
default | \(\frac {b^{2} \left (4 A b -3 B a \right ) \ln \left (b x +a \right )}{a^{5}}-\frac {b^{2} \left (A b -B a \right )}{a^{4} \left (b x +a \right )}-\frac {A}{3 a^{2} x^{3}}-\frac {-2 A b +B a}{2 x^{2} a^{3}}-\frac {b \left (3 A b -2 B a \right )}{a^{4} x}-\frac {b^{2} \left (4 A b -3 B a \right ) \ln \left (x \right )}{a^{5}}\) | \(109\) |
norman | \(\frac {\frac {b \left (4 A \,b^{3}-3 B a \,b^{2}\right ) x^{4}}{a^{5}}-\frac {A}{3 a}+\frac {\left (4 A b -3 B a \right ) x}{6 a^{2}}-\frac {b \left (4 A b -3 B a \right ) x^{2}}{2 a^{3}}}{x^{3} \left (b x +a \right )}+\frac {b^{2} \left (4 A b -3 B a \right ) \ln \left (b x +a \right )}{a^{5}}-\frac {b^{2} \left (4 A b -3 B a \right ) \ln \left (x \right )}{a^{5}}\) | \(116\) |
risch | \(\frac {-\frac {b^{2} \left (4 A b -3 B a \right ) x^{3}}{a^{4}}-\frac {b \left (4 A b -3 B a \right ) x^{2}}{2 a^{3}}+\frac {\left (4 A b -3 B a \right ) x}{6 a^{2}}-\frac {A}{3 a}}{x^{3} \left (b x +a \right )}-\frac {4 b^{3} \ln \left (x \right ) A}{a^{5}}+\frac {3 b^{2} \ln \left (x \right ) B}{a^{4}}+\frac {4 b^{3} \ln \left (-b x -a \right ) A}{a^{5}}-\frac {3 b^{2} \ln \left (-b x -a \right ) B}{a^{4}}\) | \(131\) |
parallelrisch | \(-\frac {24 A \ln \left (x \right ) x^{4} b^{4}-24 A \ln \left (b x +a \right ) x^{4} b^{4}-18 B \ln \left (x \right ) x^{4} a \,b^{3}+18 B \ln \left (b x +a \right ) x^{4} a \,b^{3}+24 A \ln \left (x \right ) x^{3} a \,b^{3}-24 A \ln \left (b x +a \right ) x^{3} a \,b^{3}-24 A \,b^{4} x^{4}-18 B \ln \left (x \right ) x^{3} a^{2} b^{2}+18 B \ln \left (b x +a \right ) x^{3} a^{2} b^{2}+18 B a \,b^{3} x^{4}+12 A \,a^{2} b^{2} x^{2}-9 B \,a^{3} b \,x^{2}-4 A \,a^{3} b x +3 a^{4} B x +2 a^{4} A}{6 a^{5} \left (b x +a \right ) x^{3}}\) | \(193\) |
Input:
int((B*x+A)/x^4/(b^2*x^2+2*a*b*x+a^2),x,method=_RETURNVERBOSE)
Output:
b^2*(4*A*b-3*B*a)*ln(b*x+a)/a^5-b^2*(A*b-B*a)/a^4/(b*x+a)-1/3*A/a^2/x^3-1/ 2*(-2*A*b+B*a)/x^2/a^3-b*(3*A*b-2*B*a)/a^4/x-b^2*(4*A*b-3*B*a)*ln(x)/a^5
Time = 0.07 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.58 \[ \int \frac {A+B x}{x^4 \left (a^2+2 a b x+b^2 x^2\right )} \, dx=-\frac {2 \, A a^{4} - 6 \, {\left (3 \, B a^{2} b^{2} - 4 \, A a b^{3}\right )} x^{3} - 3 \, {\left (3 \, B a^{3} b - 4 \, A a^{2} b^{2}\right )} x^{2} + {\left (3 \, B a^{4} - 4 \, A a^{3} b\right )} x + 6 \, {\left ({\left (3 \, B a b^{3} - 4 \, A b^{4}\right )} x^{4} + {\left (3 \, B a^{2} b^{2} - 4 \, A a b^{3}\right )} x^{3}\right )} \log \left (b x + a\right ) - 6 \, {\left ({\left (3 \, B a b^{3} - 4 \, A b^{4}\right )} x^{4} + {\left (3 \, B a^{2} b^{2} - 4 \, A a b^{3}\right )} x^{3}\right )} \log \left (x\right )}{6 \, {\left (a^{5} b x^{4} + a^{6} x^{3}\right )}} \] Input:
integrate((B*x+A)/x^4/(b^2*x^2+2*a*b*x+a^2),x, algorithm="fricas")
Output:
-1/6*(2*A*a^4 - 6*(3*B*a^2*b^2 - 4*A*a*b^3)*x^3 - 3*(3*B*a^3*b - 4*A*a^2*b ^2)*x^2 + (3*B*a^4 - 4*A*a^3*b)*x + 6*((3*B*a*b^3 - 4*A*b^4)*x^4 + (3*B*a^ 2*b^2 - 4*A*a*b^3)*x^3)*log(b*x + a) - 6*((3*B*a*b^3 - 4*A*b^4)*x^4 + (3*B *a^2*b^2 - 4*A*a*b^3)*x^3)*log(x))/(a^5*b*x^4 + a^6*x^3)
Leaf count of result is larger than twice the leaf count of optimal. 219 vs. \(2 (104) = 208\).
Time = 0.38 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.94 \[ \int \frac {A+B x}{x^4 \left (a^2+2 a b x+b^2 x^2\right )} \, dx=\frac {- 2 A a^{3} + x^{3} \left (- 24 A b^{3} + 18 B a b^{2}\right ) + x^{2} \left (- 12 A a b^{2} + 9 B a^{2} b\right ) + x \left (4 A a^{2} b - 3 B a^{3}\right )}{6 a^{5} x^{3} + 6 a^{4} b x^{4}} + \frac {b^{2} \left (- 4 A b + 3 B a\right ) \log {\left (x + \frac {- 4 A a b^{3} + 3 B a^{2} b^{2} - a b^{2} \left (- 4 A b + 3 B a\right )}{- 8 A b^{4} + 6 B a b^{3}} \right )}}{a^{5}} - \frac {b^{2} \left (- 4 A b + 3 B a\right ) \log {\left (x + \frac {- 4 A a b^{3} + 3 B a^{2} b^{2} + a b^{2} \left (- 4 A b + 3 B a\right )}{- 8 A b^{4} + 6 B a b^{3}} \right )}}{a^{5}} \] Input:
integrate((B*x+A)/x**4/(b**2*x**2+2*a*b*x+a**2),x)
Output:
(-2*A*a**3 + x**3*(-24*A*b**3 + 18*B*a*b**2) + x**2*(-12*A*a*b**2 + 9*B*a* *2*b) + x*(4*A*a**2*b - 3*B*a**3))/(6*a**5*x**3 + 6*a**4*b*x**4) + b**2*(- 4*A*b + 3*B*a)*log(x + (-4*A*a*b**3 + 3*B*a**2*b**2 - a*b**2*(-4*A*b + 3*B *a))/(-8*A*b**4 + 6*B*a*b**3))/a**5 - b**2*(-4*A*b + 3*B*a)*log(x + (-4*A* a*b**3 + 3*B*a**2*b**2 + a*b**2*(-4*A*b + 3*B*a))/(-8*A*b**4 + 6*B*a*b**3) )/a**5
Time = 0.04 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.13 \[ \int \frac {A+B x}{x^4 \left (a^2+2 a b x+b^2 x^2\right )} \, dx=-\frac {2 \, A a^{3} - 6 \, {\left (3 \, B a b^{2} - 4 \, A b^{3}\right )} x^{3} - 3 \, {\left (3 \, B a^{2} b - 4 \, A a b^{2}\right )} x^{2} + {\left (3 \, B a^{3} - 4 \, A a^{2} b\right )} x}{6 \, {\left (a^{4} b x^{4} + a^{5} x^{3}\right )}} - \frac {{\left (3 \, B a b^{2} - 4 \, A b^{3}\right )} \log \left (b x + a\right )}{a^{5}} + \frac {{\left (3 \, B a b^{2} - 4 \, A b^{3}\right )} \log \left (x\right )}{a^{5}} \] Input:
integrate((B*x+A)/x^4/(b^2*x^2+2*a*b*x+a^2),x, algorithm="maxima")
Output:
-1/6*(2*A*a^3 - 6*(3*B*a*b^2 - 4*A*b^3)*x^3 - 3*(3*B*a^2*b - 4*A*a*b^2)*x^ 2 + (3*B*a^3 - 4*A*a^2*b)*x)/(a^4*b*x^4 + a^5*x^3) - (3*B*a*b^2 - 4*A*b^3) *log(b*x + a)/a^5 + (3*B*a*b^2 - 4*A*b^3)*log(x)/a^5
Time = 0.19 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.18 \[ \int \frac {A+B x}{x^4 \left (a^2+2 a b x+b^2 x^2\right )} \, dx=\frac {{\left (3 \, B a b^{2} - 4 \, A b^{3}\right )} \log \left ({\left | x \right |}\right )}{a^{5}} - \frac {{\left (3 \, B a b^{3} - 4 \, A b^{4}\right )} \log \left ({\left | b x + a \right |}\right )}{a^{5} b} - \frac {2 \, A a^{4} - 6 \, {\left (3 \, B a^{2} b^{2} - 4 \, A a b^{3}\right )} x^{3} - 3 \, {\left (3 \, B a^{3} b - 4 \, A a^{2} b^{2}\right )} x^{2} + {\left (3 \, B a^{4} - 4 \, A a^{3} b\right )} x}{6 \, {\left (b x + a\right )} a^{5} x^{3}} \] Input:
integrate((B*x+A)/x^4/(b^2*x^2+2*a*b*x+a^2),x, algorithm="giac")
Output:
(3*B*a*b^2 - 4*A*b^3)*log(abs(x))/a^5 - (3*B*a*b^3 - 4*A*b^4)*log(abs(b*x + a))/(a^5*b) - 1/6*(2*A*a^4 - 6*(3*B*a^2*b^2 - 4*A*a*b^3)*x^3 - 3*(3*B*a^ 3*b - 4*A*a^2*b^2)*x^2 + (3*B*a^4 - 4*A*a^3*b)*x)/((b*x + a)*a^5*x^3)
Time = 0.11 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.16 \[ \int \frac {A+B x}{x^4 \left (a^2+2 a b x+b^2 x^2\right )} \, dx=\frac {2\,b^2\,\mathrm {atanh}\left (\frac {b^2\,\left (4\,A\,b-3\,B\,a\right )\,\left (a+2\,b\,x\right )}{a\,\left (4\,A\,b^3-3\,B\,a\,b^2\right )}\right )\,\left (4\,A\,b-3\,B\,a\right )}{a^5}-\frac {\frac {A}{3\,a}-\frac {x\,\left (4\,A\,b-3\,B\,a\right )}{6\,a^2}+\frac {b^2\,x^3\,\left (4\,A\,b-3\,B\,a\right )}{a^4}+\frac {b\,x^2\,\left (4\,A\,b-3\,B\,a\right )}{2\,a^3}}{b\,x^4+a\,x^3} \] Input:
int((A + B*x)/(x^4*(a^2 + b^2*x^2 + 2*a*b*x)),x)
Output:
(2*b^2*atanh((b^2*(4*A*b - 3*B*a)*(a + 2*b*x))/(a*(4*A*b^3 - 3*B*a*b^2)))* (4*A*b - 3*B*a))/a^5 - (A/(3*a) - (x*(4*A*b - 3*B*a))/(6*a^2) + (b^2*x^3*( 4*A*b - 3*B*a))/a^4 + (b*x^2*(4*A*b - 3*B*a))/(2*a^3))/(a*x^3 + b*x^4)
Time = 0.27 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.48 \[ \int \frac {A+B x}{x^4 \left (a^2+2 a b x+b^2 x^2\right )} \, dx=\frac {6 \,\mathrm {log}\left (b x +a \right ) b^{3} x^{3}-6 \,\mathrm {log}\left (x \right ) b^{3} x^{3}-2 a^{3}+3 a^{2} b x -6 a \,b^{2} x^{2}}{6 a^{4} x^{3}} \] Input:
int((B*x+A)/x^4/(b^2*x^2+2*a*b*x+a^2),x)
Output:
(6*log(a + b*x)*b**3*x**3 - 6*log(x)*b**3*x**3 - 2*a**3 + 3*a**2*b*x - 6*a *b**2*x**2)/(6*a**4*x**3)