Integrand size = 27, antiderivative size = 102 \[ \int \frac {A+B x}{x \left (a^2+2 a b x+b^2 x^2\right )^3} \, dx=\frac {A b-a B}{5 a b (a+b x)^5}+\frac {A}{4 a^2 (a+b x)^4}+\frac {A}{3 a^3 (a+b x)^3}+\frac {A}{2 a^4 (a+b x)^2}+\frac {A}{a^5 (a+b x)}+\frac {A \log (x)}{a^6}-\frac {A \log (a+b x)}{a^6} \] Output:
1/5*(A*b-B*a)/a/b/(b*x+a)^5+1/4*A/a^2/(b*x+a)^4+1/3*A/a^3/(b*x+a)^3+1/2*A/ a^4/(b*x+a)^2+A/a^5/(b*x+a)+A*ln(x)/a^6-A*ln(b*x+a)/a^6
Time = 0.07 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.87 \[ \int \frac {A+B x}{x \left (a^2+2 a b x+b^2 x^2\right )^3} \, dx=\frac {\frac {a \left (137 a^4 A b-12 a^5 B+385 a^3 A b^2 x+470 a^2 A b^3 x^2+270 a A b^4 x^3+60 A b^5 x^4\right )}{b (a+b x)^5}+60 A \log (x)-60 A \log (a+b x)}{60 a^6} \] Input:
Integrate[(A + B*x)/(x*(a^2 + 2*a*b*x + b^2*x^2)^3),x]
Output:
((a*(137*a^4*A*b - 12*a^5*B + 385*a^3*A*b^2*x + 470*a^2*A*b^3*x^2 + 270*a* A*b^4*x^3 + 60*A*b^5*x^4))/(b*(a + b*x)^5) + 60*A*Log[x] - 60*A*Log[a + b* x])/(60*a^6)
Time = 0.50 (sec) , antiderivative size = 102, 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.
\(\displaystyle \int \frac {A+B x}{x \left (a^2+2 a b x+b^2 x^2\right )^3} \, dx\) |
\(\Big \downarrow \) 1184 |
\(\displaystyle b^6 \int \frac {A+B x}{b^6 x (a+b x)^6}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \frac {A+B x}{x (a+b x)^6}dx\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \int \left (-\frac {A b}{a^6 (a+b x)}+\frac {A}{a^6 x}-\frac {A b}{a^5 (a+b x)^2}-\frac {A b}{a^4 (a+b x)^3}-\frac {A b}{a^3 (a+b x)^4}-\frac {A b}{a^2 (a+b x)^5}+\frac {a B-A b}{a (a+b x)^6}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {A \log (a+b x)}{a^6}+\frac {A \log (x)}{a^6}+\frac {A}{a^5 (a+b x)}+\frac {A}{2 a^4 (a+b x)^2}+\frac {A}{3 a^3 (a+b x)^3}+\frac {A}{4 a^2 (a+b x)^4}+\frac {A b-a B}{5 a b (a+b x)^5}\) |
Input:
Int[(A + B*x)/(x*(a^2 + 2*a*b*x + b^2*x^2)^3),x]
Output:
(A*b - a*B)/(5*a*b*(a + b*x)^5) + A/(4*a^2*(a + b*x)^4) + A/(3*a^3*(a + b* x)^3) + A/(2*a^4*(a + b*x)^2) + A/(a^5*(a + b*x)) + (A*Log[x])/a^6 - (A*Lo g[a + b*x])/a^6
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 = 1.05 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.93
method | result | size |
default | \(-\frac {-A b +B a}{5 a b \left (b x +a \right )^{5}}-\frac {A \ln \left (b x +a \right )}{a^{6}}+\frac {A}{a^{5} \left (b x +a \right )}+\frac {A}{2 a^{4} \left (b x +a \right )^{2}}+\frac {A}{3 a^{3} \left (b x +a \right )^{3}}+\frac {A}{4 a^{2} \left (b x +a \right )^{4}}+\frac {A \ln \left (x \right )}{a^{6}}\) | \(95\) |
risch | \(\frac {\frac {b^{4} A \,x^{4}}{a^{5}}+\frac {9 b^{3} A \,x^{3}}{2 a^{4}}+\frac {47 b^{2} A \,x^{2}}{6 a^{3}}+\frac {77 x A b}{12 a^{2}}+\frac {137 A b -12 B a}{60 a b}}{\left (b x +a \right ) \left (b^{2} x^{2}+2 a b x +a^{2}\right )^{2}}-\frac {A \ln \left (b x +a \right )}{a^{6}}+\frac {A \ln \left (-x \right )}{a^{6}}\) | \(110\) |
norman | \(\frac {-\frac {\left (5 A b -B a \right ) x}{a^{2}}-\frac {b \left (15 A b -2 B a \right ) x^{2}}{a^{3}}-\frac {b^{2} \left (55 A b -6 B a \right ) x^{3}}{3 a^{4}}-\frac {b^{3} \left (125 A b -12 B a \right ) x^{4}}{12 a^{5}}-\frac {b^{4} \left (137 A b -12 B a \right ) x^{5}}{60 a^{6}}}{\left (b x +a \right )^{5}}+\frac {A \ln \left (x \right )}{a^{6}}-\frac {A \ln \left (b x +a \right )}{a^{6}}\) | \(123\) |
parallelrisch | \(\frac {-625 A a \,b^{4} x^{4}+120 B \,a^{4} b \,x^{2}-300 A \,a^{4} b x -300 A \ln \left (b x +a \right ) x \,a^{4} b +300 A \ln \left (x \right ) x \,a^{4} b +12 B a \,b^{4} x^{5}+300 A \ln \left (x \right ) x^{4} a \,b^{4}-300 A \ln \left (b x +a \right ) x^{4} a \,b^{4}+600 A \ln \left (x \right ) x^{3} a^{2} b^{3}-600 A \ln \left (b x +a \right ) x^{3} a^{2} b^{3}+600 A \ln \left (x \right ) x^{2} a^{3} b^{2}-600 A \ln \left (b x +a \right ) x^{2} a^{3} b^{2}+60 A \ln \left (x \right ) x^{5} b^{5}-60 A \ln \left (b x +a \right ) x^{5} b^{5}-137 A \,b^{5} x^{5}+60 B \,a^{2} b^{3} x^{4}-1100 A \,a^{2} b^{3} x^{3}+120 B \,a^{3} b^{2} x^{3}-900 A \,a^{3} b^{2} x^{2}+60 A \ln \left (x \right ) a^{5}-60 A \ln \left (b x +a \right ) a^{5}+60 B \,a^{5} x}{60 a^{6} \left (b^{2} x^{2}+2 a b x +a^{2}\right )^{2} \left (b x +a \right )}\) | \(296\) |
Input:
int((B*x+A)/x/(b^2*x^2+2*a*b*x+a^2)^3,x,method=_RETURNVERBOSE)
Output:
-1/5*(-A*b+B*a)/a/b/(b*x+a)^5-A*ln(b*x+a)/a^6+A/a^5/(b*x+a)+1/2*A/a^4/(b*x +a)^2+1/3*A/a^3/(b*x+a)^3+1/4*A/a^2/(b*x+a)^4+A*ln(x)/a^6
Leaf count of result is larger than twice the leaf count of optimal. 250 vs. \(2 (94) = 188\).
Time = 0.08 (sec) , antiderivative size = 250, normalized size of antiderivative = 2.45 \[ \int \frac {A+B x}{x \left (a^2+2 a b x+b^2 x^2\right )^3} \, dx=\frac {60 \, A a b^{5} x^{4} + 270 \, A a^{2} b^{4} x^{3} + 470 \, A a^{3} b^{3} x^{2} + 385 \, A a^{4} b^{2} x - 12 \, B a^{6} + 137 \, A a^{5} b - 60 \, {\left (A b^{6} x^{5} + 5 \, A a b^{5} x^{4} + 10 \, A a^{2} b^{4} x^{3} + 10 \, A a^{3} b^{3} x^{2} + 5 \, A a^{4} b^{2} x + A a^{5} b\right )} \log \left (b x + a\right ) + 60 \, {\left (A b^{6} x^{5} + 5 \, A a b^{5} x^{4} + 10 \, A a^{2} b^{4} x^{3} + 10 \, A a^{3} b^{3} x^{2} + 5 \, A a^{4} b^{2} x + A a^{5} b\right )} \log \left (x\right )}{60 \, {\left (a^{6} b^{6} x^{5} + 5 \, a^{7} b^{5} x^{4} + 10 \, a^{8} b^{4} x^{3} + 10 \, a^{9} b^{3} x^{2} + 5 \, a^{10} b^{2} x + a^{11} b\right )}} \] Input:
integrate((B*x+A)/x/(b^2*x^2+2*a*b*x+a^2)^3,x, algorithm="fricas")
Output:
1/60*(60*A*a*b^5*x^4 + 270*A*a^2*b^4*x^3 + 470*A*a^3*b^3*x^2 + 385*A*a^4*b ^2*x - 12*B*a^6 + 137*A*a^5*b - 60*(A*b^6*x^5 + 5*A*a*b^5*x^4 + 10*A*a^2*b ^4*x^3 + 10*A*a^3*b^3*x^2 + 5*A*a^4*b^2*x + A*a^5*b)*log(b*x + a) + 60*(A* b^6*x^5 + 5*A*a*b^5*x^4 + 10*A*a^2*b^4*x^3 + 10*A*a^3*b^3*x^2 + 5*A*a^4*b^ 2*x + A*a^5*b)*log(x))/(a^6*b^6*x^5 + 5*a^7*b^5*x^4 + 10*a^8*b^4*x^3 + 10* a^9*b^3*x^2 + 5*a^10*b^2*x + a^11*b)
Time = 0.41 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.38 \[ \int \frac {A+B x}{x \left (a^2+2 a b x+b^2 x^2\right )^3} \, dx=\frac {A \left (\log {\left (x \right )} - \log {\left (\frac {a}{b} + x \right )}\right )}{a^{6}} + \frac {137 A a^{4} b + 385 A a^{3} b^{2} x + 470 A a^{2} b^{3} x^{2} + 270 A a b^{4} x^{3} + 60 A b^{5} x^{4} - 12 B a^{5}}{60 a^{10} b + 300 a^{9} b^{2} x + 600 a^{8} b^{3} x^{2} + 600 a^{7} b^{4} x^{3} + 300 a^{6} b^{5} x^{4} + 60 a^{5} b^{6} x^{5}} \] Input:
integrate((B*x+A)/x/(b**2*x**2+2*a*b*x+a**2)**3,x)
Output:
A*(log(x) - log(a/b + x))/a**6 + (137*A*a**4*b + 385*A*a**3*b**2*x + 470*A *a**2*b**3*x**2 + 270*A*a*b**4*x**3 + 60*A*b**5*x**4 - 12*B*a**5)/(60*a**1 0*b + 300*a**9*b**2*x + 600*a**8*b**3*x**2 + 600*a**7*b**4*x**3 + 300*a**6 *b**5*x**4 + 60*a**5*b**6*x**5)
Time = 0.04 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.34 \[ \int \frac {A+B x}{x \left (a^2+2 a b x+b^2 x^2\right )^3} \, dx=\frac {60 \, A b^{5} x^{4} + 270 \, A a b^{4} x^{3} + 470 \, A a^{2} b^{3} x^{2} + 385 \, A a^{3} b^{2} x - 12 \, B a^{5} + 137 \, A a^{4} b}{60 \, {\left (a^{5} b^{6} x^{5} + 5 \, a^{6} b^{5} x^{4} + 10 \, a^{7} b^{4} x^{3} + 10 \, a^{8} b^{3} x^{2} + 5 \, a^{9} b^{2} x + a^{10} b\right )}} - \frac {A \log \left (b x + a\right )}{a^{6}} + \frac {A \log \left (x\right )}{a^{6}} \] Input:
integrate((B*x+A)/x/(b^2*x^2+2*a*b*x+a^2)^3,x, algorithm="maxima")
Output:
1/60*(60*A*b^5*x^4 + 270*A*a*b^4*x^3 + 470*A*a^2*b^3*x^2 + 385*A*a^3*b^2*x - 12*B*a^5 + 137*A*a^4*b)/(a^5*b^6*x^5 + 5*a^6*b^5*x^4 + 10*a^7*b^4*x^3 + 10*a^8*b^3*x^2 + 5*a^9*b^2*x + a^10*b) - A*log(b*x + a)/a^6 + A*log(x)/a^ 6
Time = 0.18 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.93 \[ \int \frac {A+B x}{x \left (a^2+2 a b x+b^2 x^2\right )^3} \, dx=-\frac {A \log \left ({\left | b x + a \right |}\right )}{a^{6}} + \frac {A \log \left ({\left | x \right |}\right )}{a^{6}} + \frac {60 \, A a b^{5} x^{4} + 270 \, A a^{2} b^{4} x^{3} + 470 \, A a^{3} b^{3} x^{2} + 385 \, A a^{4} b^{2} x - 12 \, B a^{6} + 137 \, A a^{5} b}{60 \, {\left (b x + a\right )}^{5} a^{6} b} \] Input:
integrate((B*x+A)/x/(b^2*x^2+2*a*b*x+a^2)^3,x, algorithm="giac")
Output:
-A*log(abs(b*x + a))/a^6 + A*log(abs(x))/a^6 + 1/60*(60*A*a*b^5*x^4 + 270* A*a^2*b^4*x^3 + 470*A*a^3*b^3*x^2 + 385*A*a^4*b^2*x - 12*B*a^6 + 137*A*a^5 *b)/((b*x + a)^5*a^6*b)
Time = 0.10 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.27 \[ \int \frac {A+B x}{x \left (a^2+2 a b x+b^2 x^2\right )^3} \, dx=\frac {\frac {137\,A\,b-12\,B\,a}{60\,a\,b}+\frac {77\,A\,b\,x}{12\,a^2}+\frac {47\,A\,b^2\,x^2}{6\,a^3}+\frac {9\,A\,b^3\,x^3}{2\,a^4}+\frac {A\,b^4\,x^4}{a^5}}{a^5+5\,a^4\,b\,x+10\,a^3\,b^2\,x^2+10\,a^2\,b^3\,x^3+5\,a\,b^4\,x^4+b^5\,x^5}-\frac {2\,A\,\mathrm {atanh}\left (\frac {2\,b\,x}{a}+1\right )}{a^6} \] Input:
int((A + B*x)/(x*(a^2 + b^2*x^2 + 2*a*b*x)^3),x)
Output:
((137*A*b - 12*B*a)/(60*a*b) + (77*A*b*x)/(12*a^2) + (47*A*b^2*x^2)/(6*a^3 ) + (9*A*b^3*x^3)/(2*a^4) + (A*b^4*x^4)/a^5)/(a^5 + b^5*x^5 + 5*a*b^4*x^4 + 10*a^3*b^2*x^2 + 10*a^2*b^3*x^3 + 5*a^4*b*x) - (2*A*atanh((2*b*x)/a + 1) )/a^6
Time = 0.29 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.93 \[ \int \frac {A+B x}{x \left (a^2+2 a b x+b^2 x^2\right )^3} \, dx=\frac {-12 \,\mathrm {log}\left (b x +a \right ) a^{4}-48 \,\mathrm {log}\left (b x +a \right ) a^{3} b x -72 \,\mathrm {log}\left (b x +a \right ) a^{2} b^{2} x^{2}-48 \,\mathrm {log}\left (b x +a \right ) a \,b^{3} x^{3}-12 \,\mathrm {log}\left (b x +a \right ) b^{4} x^{4}+12 \,\mathrm {log}\left (x \right ) a^{4}+48 \,\mathrm {log}\left (x \right ) a^{3} b x +72 \,\mathrm {log}\left (x \right ) a^{2} b^{2} x^{2}+48 \,\mathrm {log}\left (x \right ) a \,b^{3} x^{3}+12 \,\mathrm {log}\left (x \right ) b^{4} x^{4}+22 a^{4}+40 a^{3} b x +24 a^{2} b^{2} x^{2}-3 b^{4} x^{4}}{12 a^{5} \left (b^{4} x^{4}+4 a \,b^{3} x^{3}+6 a^{2} b^{2} x^{2}+4 a^{3} b x +a^{4}\right )} \] Input:
int((B*x+A)/x/(b^2*x^2+2*a*b*x+a^2)^3,x)
Output:
( - 12*log(a + b*x)*a**4 - 48*log(a + b*x)*a**3*b*x - 72*log(a + b*x)*a**2 *b**2*x**2 - 48*log(a + b*x)*a*b**3*x**3 - 12*log(a + b*x)*b**4*x**4 + 12* log(x)*a**4 + 48*log(x)*a**3*b*x + 72*log(x)*a**2*b**2*x**2 + 48*log(x)*a* b**3*x**3 + 12*log(x)*b**4*x**4 + 22*a**4 + 40*a**3*b*x + 24*a**2*b**2*x** 2 - 3*b**4*x**4)/(12*a**5*(a**4 + 4*a**3*b*x + 6*a**2*b**2*x**2 + 4*a*b**3 *x**3 + b**4*x**4))