Integrand size = 27, antiderivative size = 72 \[ \int \frac {A+B x}{x \left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\frac {A b-a B}{3 a b (a+b x)^3}+\frac {A}{2 a^2 (a+b x)^2}+\frac {A}{a^3 (a+b x)}+\frac {A \log (x)}{a^4}-\frac {A \log (a+b x)}{a^4} \] Output:
1/3*(A*b-B*a)/a/b/(b*x+a)^3+1/2*A/a^2/(b*x+a)^2+A/a^3/(b*x+a)+A*ln(x)/a^4- A*ln(b*x+a)/a^4
Time = 0.05 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.90 \[ \int \frac {A+B x}{x \left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\frac {\frac {a \left (11 a^2 A b-2 a^3 B+15 a A b^2 x+6 A b^3 x^2\right )}{b (a+b x)^3}+6 A \log (x)-6 A \log (a+b x)}{6 a^4} \] Input:
Integrate[(A + B*x)/(x*(a^2 + 2*a*b*x + b^2*x^2)^2),x]
Output:
((a*(11*a^2*A*b - 2*a^3*B + 15*a*A*b^2*x + 6*A*b^3*x^2))/(b*(a + b*x)^3) + 6*A*Log[x] - 6*A*Log[a + b*x])/(6*a^4)
Time = 0.40 (sec) , antiderivative size = 72, 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.
\(\displaystyle \int \frac {A+B x}{x \left (a^2+2 a b x+b^2 x^2\right )^2} \, dx\) |
\(\Big \downarrow \) 1184 |
\(\displaystyle b^4 \int \frac {A+B x}{b^4 x (a+b x)^4}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \frac {A+B x}{x (a+b x)^4}dx\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \int \left (-\frac {A b}{a^4 (a+b x)}+\frac {A}{a^4 x}-\frac {A b}{a^3 (a+b x)^2}-\frac {A b}{a^2 (a+b x)^3}+\frac {a B-A b}{a (a+b x)^4}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {A \log (a+b x)}{a^4}+\frac {A \log (x)}{a^4}+\frac {A}{a^3 (a+b x)}+\frac {A}{2 a^2 (a+b x)^2}+\frac {A b-a B}{3 a b (a+b x)^3}\) |
Input:
Int[(A + B*x)/(x*(a^2 + 2*a*b*x + b^2*x^2)^2),x]
Output:
(A*b - a*B)/(3*a*b*(a + b*x)^3) + A/(2*a^2*(a + b*x)^2) + A/(a^3*(a + b*x) ) + (A*Log[x])/a^4 - (A*Log[a + b*x])/a^4
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 = 1.03 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.96
method | result | size |
default | \(-\frac {-A b +B a}{3 a b \left (b x +a \right )^{3}}-\frac {A \ln \left (b x +a \right )}{a^{4}}+\frac {A}{a^{3} \left (b x +a \right )}+\frac {A}{2 a^{2} \left (b x +a \right )^{2}}+\frac {A \ln \left (x \right )}{a^{4}}\) | \(69\) |
norman | \(\frac {-\frac {\left (3 A b -B a \right ) x}{a^{2}}-\frac {b \left (9 A b -2 B a \right ) x^{2}}{2 a^{3}}-\frac {b^{2} \left (11 A b -2 B a \right ) x^{3}}{6 a^{4}}}{\left (b x +a \right )^{3}}+\frac {A \ln \left (x \right )}{a^{4}}-\frac {A \ln \left (b x +a \right )}{a^{4}}\) | \(83\) |
risch | \(\frac {\frac {b^{2} A \,x^{2}}{a^{3}}+\frac {5 x A b}{2 a^{2}}+\frac {11 A b -2 B a}{6 a b}}{\left (b x +a \right ) \left (b^{2} x^{2}+2 a b x +a^{2}\right )}+\frac {A \ln \left (-x \right )}{a^{4}}-\frac {A \ln \left (b x +a \right )}{a^{4}}\) | \(86\) |
parallelrisch | \(\frac {6 A \ln \left (x \right ) x^{3} b^{3}-6 A \ln \left (b x +a \right ) x^{3} b^{3}+18 A \ln \left (x \right ) x^{2} a \,b^{2}-18 A \ln \left (b x +a \right ) x^{2} a \,b^{2}-11 A \,b^{3} x^{3}+2 B a \,b^{2} x^{3}+18 A \ln \left (x \right ) x \,a^{2} b -18 A \ln \left (b x +a \right ) x \,a^{2} b -27 A a \,b^{2} x^{2}+6 B \,a^{2} b \,x^{2}+6 a^{3} A \ln \left (x \right )-6 A \ln \left (b x +a \right ) a^{3}-18 A \,a^{2} b x +6 B \,a^{3} x}{6 a^{4} \left (b^{2} x^{2}+2 a b x +a^{2}\right ) \left (b x +a \right )}\) | \(184\) |
Input:
int((B*x+A)/x/(b^2*x^2+2*a*b*x+a^2)^2,x,method=_RETURNVERBOSE)
Output:
-1/3*(-A*b+B*a)/a/b/(b*x+a)^3-A*ln(b*x+a)/a^4+A/a^3/(b*x+a)+1/2*A/a^2/(b*x +a)^2+A*ln(x)/a^4
Leaf count of result is larger than twice the leaf count of optimal. 156 vs. \(2 (68) = 136\).
Time = 0.07 (sec) , antiderivative size = 156, normalized size of antiderivative = 2.17 \[ \int \frac {A+B x}{x \left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\frac {6 \, A a b^{3} x^{2} + 15 \, A a^{2} b^{2} x - 2 \, B a^{4} + 11 \, A a^{3} b - 6 \, {\left (A b^{4} x^{3} + 3 \, A a b^{3} x^{2} + 3 \, A a^{2} b^{2} x + A a^{3} b\right )} \log \left (b x + a\right ) + 6 \, {\left (A b^{4} x^{3} + 3 \, A a b^{3} x^{2} + 3 \, A a^{2} b^{2} x + A a^{3} b\right )} \log \left (x\right )}{6 \, {\left (a^{4} b^{4} x^{3} + 3 \, a^{5} b^{3} x^{2} + 3 \, a^{6} b^{2} x + a^{7} b\right )}} \] Input:
integrate((B*x+A)/x/(b^2*x^2+2*a*b*x+a^2)^2,x, algorithm="fricas")
Output:
1/6*(6*A*a*b^3*x^2 + 15*A*a^2*b^2*x - 2*B*a^4 + 11*A*a^3*b - 6*(A*b^4*x^3 + 3*A*a*b^3*x^2 + 3*A*a^2*b^2*x + A*a^3*b)*log(b*x + a) + 6*(A*b^4*x^3 + 3 *A*a*b^3*x^2 + 3*A*a^2*b^2*x + A*a^3*b)*log(x))/(a^4*b^4*x^3 + 3*a^5*b^3*x ^2 + 3*a^6*b^2*x + a^7*b)
Time = 0.28 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.25 \[ \int \frac {A+B x}{x \left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\frac {A \left (\log {\left (x \right )} - \log {\left (\frac {a}{b} + x \right )}\right )}{a^{4}} + \frac {11 A a^{2} b + 15 A a b^{2} x + 6 A b^{3} x^{2} - 2 B a^{3}}{6 a^{6} b + 18 a^{5} b^{2} x + 18 a^{4} b^{3} x^{2} + 6 a^{3} b^{4} x^{3}} \] Input:
integrate((B*x+A)/x/(b**2*x**2+2*a*b*x+a**2)**2,x)
Output:
A*(log(x) - log(a/b + x))/a**4 + (11*A*a**2*b + 15*A*a*b**2*x + 6*A*b**3*x **2 - 2*B*a**3)/(6*a**6*b + 18*a**5*b**2*x + 18*a**4*b**3*x**2 + 6*a**3*b* *4*x**3)
Time = 0.03 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.26 \[ \int \frac {A+B x}{x \left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\frac {6 \, A b^{3} x^{2} + 15 \, A a b^{2} x - 2 \, B a^{3} + 11 \, A a^{2} b}{6 \, {\left (a^{3} b^{4} x^{3} + 3 \, a^{4} b^{3} x^{2} + 3 \, a^{5} b^{2} x + a^{6} b\right )}} - \frac {A \log \left (b x + a\right )}{a^{4}} + \frac {A \log \left (x\right )}{a^{4}} \] Input:
integrate((B*x+A)/x/(b^2*x^2+2*a*b*x+a^2)^2,x, algorithm="maxima")
Output:
1/6*(6*A*b^3*x^2 + 15*A*a*b^2*x - 2*B*a^3 + 11*A*a^2*b)/(a^3*b^4*x^3 + 3*a ^4*b^3*x^2 + 3*a^5*b^2*x + a^6*b) - A*log(b*x + a)/a^4 + A*log(x)/a^4
Time = 0.23 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.99 \[ \int \frac {A+B x}{x \left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=-\frac {A \log \left ({\left | b x + a \right |}\right )}{a^{4}} + \frac {A \log \left ({\left | x \right |}\right )}{a^{4}} + \frac {6 \, A a b^{3} x^{2} + 15 \, A a^{2} b^{2} x - 2 \, B a^{4} + 11 \, A a^{3} b}{6 \, {\left (b x + a\right )}^{3} a^{4} b} \] Input:
integrate((B*x+A)/x/(b^2*x^2+2*a*b*x+a^2)^2,x, algorithm="giac")
Output:
-A*log(abs(b*x + a))/a^4 + A*log(abs(x))/a^4 + 1/6*(6*A*a*b^3*x^2 + 15*A*a ^2*b^2*x - 2*B*a^4 + 11*A*a^3*b)/((b*x + a)^3*a^4*b)
Time = 10.46 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.17 \[ \int \frac {A+B x}{x \left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\frac {\frac {11\,A\,b-2\,B\,a}{6\,a\,b}+\frac {5\,A\,b\,x}{2\,a^2}+\frac {A\,b^2\,x^2}{a^3}}{a^3+3\,a^2\,b\,x+3\,a\,b^2\,x^2+b^3\,x^3}-\frac {2\,A\,\mathrm {atanh}\left (\frac {2\,b\,x}{a}+1\right )}{a^4} \] Input:
int((A + B*x)/(x*(a^2 + b^2*x^2 + 2*a*b*x)^2),x)
Output:
((11*A*b - 2*B*a)/(6*a*b) + (5*A*b*x)/(2*a^2) + (A*b^2*x^2)/a^3)/(a^3 + b^ 3*x^3 + 3*a*b^2*x^2 + 3*a^2*b*x) - (2*A*atanh((2*b*x)/a + 1))/a^4
Time = 0.30 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.35 \[ \int \frac {A+B x}{x \left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\frac {-2 \,\mathrm {log}\left (b x +a \right ) a^{2}-4 \,\mathrm {log}\left (b x +a \right ) a b x -2 \,\mathrm {log}\left (b x +a \right ) b^{2} x^{2}+2 \,\mathrm {log}\left (x \right ) a^{2}+4 \,\mathrm {log}\left (x \right ) a b x +2 \,\mathrm {log}\left (x \right ) b^{2} x^{2}+2 a^{2}-b^{2} x^{2}}{2 a^{3} \left (b^{2} x^{2}+2 a b x +a^{2}\right )} \] Input:
int((B*x+A)/x/(b^2*x^2+2*a*b*x+a^2)^2,x)
Output:
( - 2*log(a + b*x)*a**2 - 4*log(a + b*x)*a*b*x - 2*log(a + b*x)*b**2*x**2 + 2*log(x)*a**2 + 4*log(x)*a*b*x + 2*log(x)*b**2*x**2 + 2*a**2 - b**2*x**2 )/(2*a**3*(a**2 + 2*a*b*x + b**2*x**2))