Integrand size = 27, antiderivative size = 133 \[ \int \frac {A+B x}{x^5 \left (a^2+2 a b x+b^2 x^2\right )} \, dx=-\frac {A}{4 a^2 x^4}+\frac {2 A b-a B}{3 a^3 x^3}-\frac {b (3 A b-2 a B)}{2 a^4 x^2}+\frac {b^2 (4 A b-3 a B)}{a^5 x}+\frac {b^3 (A b-a B)}{a^5 (a+b x)}+\frac {b^3 (5 A b-4 a B) \log (x)}{a^6}-\frac {b^3 (5 A b-4 a B) \log (a+b x)}{a^6} \] Output:
-1/4*A/a^2/x^4+1/3*(2*A*b-B*a)/a^3/x^3-1/2*b*(3*A*b-2*B*a)/a^4/x^2+b^2*(4* A*b-3*B*a)/a^5/x+b^3*(A*b-B*a)/a^5/(b*x+a)+b^3*(5*A*b-4*B*a)*ln(x)/a^6-b^3 *(5*A*b-4*B*a)*ln(b*x+a)/a^6
Time = 0.10 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.97 \[ \int \frac {A+B x}{x^5 \left (a^2+2 a b x+b^2 x^2\right )} \, dx=\frac {\frac {a \left (60 A b^4 x^4+6 a b^3 x^3 (5 A-8 B x)-a^4 (3 A+4 B x)+a^3 b x (5 A+8 B x)-2 a^2 b^2 x^2 (5 A+12 B x)\right )}{x^4 (a+b x)}+12 b^3 (5 A b-4 a B) \log (x)+12 b^3 (-5 A b+4 a B) \log (a+b x)}{12 a^6} \] Input:
Integrate[(A + B*x)/(x^5*(a^2 + 2*a*b*x + b^2*x^2)),x]
Output:
((a*(60*A*b^4*x^4 + 6*a*b^3*x^3*(5*A - 8*B*x) - a^4*(3*A + 4*B*x) + a^3*b* x*(5*A + 8*B*x) - 2*a^2*b^2*x^2*(5*A + 12*B*x)))/(x^4*(a + b*x)) + 12*b^3* (5*A*b - 4*a*B)*Log[x] + 12*b^3*(-5*A*b + 4*a*B)*Log[a + b*x])/(12*a^6)
Time = 0.57 (sec) , antiderivative size = 133, 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.
\(\displaystyle \int \frac {A+B x}{x^5 \left (a^2+2 a b x+b^2 x^2\right )} \, dx\) |
\(\Big \downarrow \) 1184 |
\(\displaystyle b^2 \int \frac {A+B x}{b^2 x^5 (a+b x)^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \frac {A+B x}{x^5 (a+b x)^2}dx\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \int \left (\frac {b^4 (4 a B-5 A b)}{a^6 (a+b x)}-\frac {b^3 (4 a B-5 A b)}{a^6 x}+\frac {b^4 (a B-A b)}{a^5 (a+b x)^2}+\frac {b^2 (3 a B-4 A b)}{a^5 x^2}-\frac {b (2 a B-3 A b)}{a^4 x^3}+\frac {a B-2 A b}{a^3 x^4}+\frac {A}{a^2 x^5}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {b^3 \log (x) (5 A b-4 a B)}{a^6}-\frac {b^3 (5 A b-4 a B) \log (a+b x)}{a^6}+\frac {b^3 (A b-a B)}{a^5 (a+b x)}+\frac {b^2 (4 A b-3 a B)}{a^5 x}-\frac {b (3 A b-2 a B)}{2 a^4 x^2}+\frac {2 A b-a B}{3 a^3 x^3}-\frac {A}{4 a^2 x^4}\) |
Input:
Int[(A + B*x)/(x^5*(a^2 + 2*a*b*x + b^2*x^2)),x]
Output:
-1/4*A/(a^2*x^4) + (2*A*b - a*B)/(3*a^3*x^3) - (b*(3*A*b - 2*a*B))/(2*a^4* x^2) + (b^2*(4*A*b - 3*a*B))/(a^5*x) + (b^3*(A*b - a*B))/(a^5*(a + b*x)) + (b^3*(5*A*b - 4*a*B)*Log[x])/a^6 - (b^3*(5*A*b - 4*a*B)*Log[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 = 1.07 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.95
method | result | size |
default | \(-\frac {b^{3} \left (5 A b -4 B a \right ) \ln \left (b x +a \right )}{a^{6}}+\frac {b^{3} \left (A b -B a \right )}{a^{5} \left (b x +a \right )}-\frac {A}{4 a^{2} x^{4}}-\frac {-2 A b +B a}{3 x^{3} a^{3}}-\frac {b \left (3 A b -2 B a \right )}{2 a^{4} x^{2}}+\frac {b^{3} \left (5 A b -4 B a \right ) \ln \left (x \right )}{a^{6}}+\frac {b^{2} \left (4 A b -3 B a \right )}{a^{5} x}\) | \(127\) |
norman | \(\frac {\frac {\left (5 A \,b^{5}-4 B a \,b^{4}\right ) x^{4}}{a^{5} b}-\frac {A}{4 a}+\frac {\left (5 A b -4 B a \right ) x}{12 a^{2}}-\frac {b \left (5 A b -4 B a \right ) x^{2}}{6 a^{3}}+\frac {b^{2} \left (5 A b -4 B a \right ) x^{3}}{2 a^{4}}}{x^{4} \left (b x +a \right )}+\frac {b^{3} \left (5 A b -4 B a \right ) \ln \left (x \right )}{a^{6}}-\frac {b^{3} \left (5 A b -4 B a \right ) \ln \left (b x +a \right )}{a^{6}}\) | \(138\) |
risch | \(\frac {\frac {b^{3} \left (5 A b -4 B a \right ) x^{4}}{a^{5}}+\frac {b^{2} \left (5 A b -4 B a \right ) x^{3}}{2 a^{4}}-\frac {b \left (5 A b -4 B a \right ) x^{2}}{6 a^{3}}+\frac {\left (5 A b -4 B a \right ) x}{12 a^{2}}-\frac {A}{4 a}}{x^{4} \left (b x +a \right )}-\frac {5 b^{4} \ln \left (b x +a \right ) A}{a^{6}}+\frac {4 b^{3} \ln \left (b x +a \right ) B}{a^{5}}+\frac {5 b^{4} \ln \left (-x \right ) A}{a^{6}}-\frac {4 b^{3} \ln \left (-x \right ) B}{a^{5}}\) | \(148\) |
parallelrisch | \(\frac {60 A \ln \left (x \right ) x^{5} b^{6}-60 A \ln \left (b x +a \right ) x^{5} b^{6}-48 B \ln \left (x \right ) x^{5} a \,b^{5}+48 B \ln \left (b x +a \right ) x^{5} a \,b^{5}+60 A \ln \left (x \right ) x^{4} a \,b^{5}-60 A \ln \left (b x +a \right ) x^{4} a \,b^{5}-48 B \ln \left (x \right ) x^{4} a^{2} b^{4}+48 B \ln \left (b x +a \right ) x^{4} a^{2} b^{4}+60 A a \,b^{5} x^{4}-48 B \,a^{2} b^{4} x^{4}+30 A \,a^{2} b^{4} x^{3}-24 B \,a^{3} b^{3} x^{3}-10 A \,a^{3} b^{3} x^{2}+8 B \,a^{4} b^{2} x^{2}+5 A \,a^{4} b^{2} x -4 B \,a^{5} b x -3 A \,a^{5} b}{12 a^{6} b \left (b x +a \right ) x^{4}}\) | \(229\) |
Input:
int((B*x+A)/x^5/(b^2*x^2+2*a*b*x+a^2),x,method=_RETURNVERBOSE)
Output:
-b^3*(5*A*b-4*B*a)*ln(b*x+a)/a^6+b^3*(A*b-B*a)/a^5/(b*x+a)-1/4*A/a^2/x^4-1 /3*(-2*A*b+B*a)/x^3/a^3-1/2*b*(3*A*b-2*B*a)/a^4/x^2+b^3*(5*A*b-4*B*a)*ln(x )/a^6+b^2*(4*A*b-3*B*a)/a^5/x
Time = 0.07 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.53 \[ \int \frac {A+B x}{x^5 \left (a^2+2 a b x+b^2 x^2\right )} \, dx=-\frac {3 \, A a^{5} + 12 \, {\left (4 \, B a^{2} b^{3} - 5 \, A a b^{4}\right )} x^{4} + 6 \, {\left (4 \, B a^{3} b^{2} - 5 \, A a^{2} b^{3}\right )} x^{3} - 2 \, {\left (4 \, B a^{4} b - 5 \, A a^{3} b^{2}\right )} x^{2} + {\left (4 \, B a^{5} - 5 \, A a^{4} b\right )} x - 12 \, {\left ({\left (4 \, B a b^{4} - 5 \, A b^{5}\right )} x^{5} + {\left (4 \, B a^{2} b^{3} - 5 \, A a b^{4}\right )} x^{4}\right )} \log \left (b x + a\right ) + 12 \, {\left ({\left (4 \, B a b^{4} - 5 \, A b^{5}\right )} x^{5} + {\left (4 \, B a^{2} b^{3} - 5 \, A a b^{4}\right )} x^{4}\right )} \log \left (x\right )}{12 \, {\left (a^{6} b x^{5} + a^{7} x^{4}\right )}} \] Input:
integrate((B*x+A)/x^5/(b^2*x^2+2*a*b*x+a^2),x, algorithm="fricas")
Output:
-1/12*(3*A*a^5 + 12*(4*B*a^2*b^3 - 5*A*a*b^4)*x^4 + 6*(4*B*a^3*b^2 - 5*A*a ^2*b^3)*x^3 - 2*(4*B*a^4*b - 5*A*a^3*b^2)*x^2 + (4*B*a^5 - 5*A*a^4*b)*x - 12*((4*B*a*b^4 - 5*A*b^5)*x^5 + (4*B*a^2*b^3 - 5*A*a*b^4)*x^4)*log(b*x + a ) + 12*((4*B*a*b^4 - 5*A*b^5)*x^5 + (4*B*a^2*b^3 - 5*A*a*b^4)*x^4)*log(x)) /(a^6*b*x^5 + a^7*x^4)
Time = 0.37 (sec) , antiderivative size = 243, normalized size of antiderivative = 1.83 \[ \int \frac {A+B x}{x^5 \left (a^2+2 a b x+b^2 x^2\right )} \, dx=\frac {- 3 A a^{4} + x^{4} \cdot \left (60 A b^{4} - 48 B a b^{3}\right ) + x^{3} \cdot \left (30 A a b^{3} - 24 B a^{2} b^{2}\right ) + x^{2} \left (- 10 A a^{2} b^{2} + 8 B a^{3} b\right ) + x \left (5 A a^{3} b - 4 B a^{4}\right )}{12 a^{6} x^{4} + 12 a^{5} b x^{5}} - \frac {b^{3} \left (- 5 A b + 4 B a\right ) \log {\left (x + \frac {- 5 A a b^{4} + 4 B a^{2} b^{3} - a b^{3} \left (- 5 A b + 4 B a\right )}{- 10 A b^{5} + 8 B a b^{4}} \right )}}{a^{6}} + \frac {b^{3} \left (- 5 A b + 4 B a\right ) \log {\left (x + \frac {- 5 A a b^{4} + 4 B a^{2} b^{3} + a b^{3} \left (- 5 A b + 4 B a\right )}{- 10 A b^{5} + 8 B a b^{4}} \right )}}{a^{6}} \] Input:
integrate((B*x+A)/x**5/(b**2*x**2+2*a*b*x+a**2),x)
Output:
(-3*A*a**4 + x**4*(60*A*b**4 - 48*B*a*b**3) + x**3*(30*A*a*b**3 - 24*B*a** 2*b**2) + x**2*(-10*A*a**2*b**2 + 8*B*a**3*b) + x*(5*A*a**3*b - 4*B*a**4)) /(12*a**6*x**4 + 12*a**5*b*x**5) - b**3*(-5*A*b + 4*B*a)*log(x + (-5*A*a*b **4 + 4*B*a**2*b**3 - a*b**3*(-5*A*b + 4*B*a))/(-10*A*b**5 + 8*B*a*b**4))/ a**6 + b**3*(-5*A*b + 4*B*a)*log(x + (-5*A*a*b**4 + 4*B*a**2*b**3 + a*b**3 *(-5*A*b + 4*B*a))/(-10*A*b**5 + 8*B*a*b**4))/a**6
Time = 0.03 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.14 \[ \int \frac {A+B x}{x^5 \left (a^2+2 a b x+b^2 x^2\right )} \, dx=-\frac {3 \, A a^{4} + 12 \, {\left (4 \, B a b^{3} - 5 \, A b^{4}\right )} x^{4} + 6 \, {\left (4 \, B a^{2} b^{2} - 5 \, A a b^{3}\right )} x^{3} - 2 \, {\left (4 \, B a^{3} b - 5 \, A a^{2} b^{2}\right )} x^{2} + {\left (4 \, B a^{4} - 5 \, A a^{3} b\right )} x}{12 \, {\left (a^{5} b x^{5} + a^{6} x^{4}\right )}} + \frac {{\left (4 \, B a b^{3} - 5 \, A b^{4}\right )} \log \left (b x + a\right )}{a^{6}} - \frac {{\left (4 \, B a b^{3} - 5 \, A b^{4}\right )} \log \left (x\right )}{a^{6}} \] Input:
integrate((B*x+A)/x^5/(b^2*x^2+2*a*b*x+a^2),x, algorithm="maxima")
Output:
-1/12*(3*A*a^4 + 12*(4*B*a*b^3 - 5*A*b^4)*x^4 + 6*(4*B*a^2*b^2 - 5*A*a*b^3 )*x^3 - 2*(4*B*a^3*b - 5*A*a^2*b^2)*x^2 + (4*B*a^4 - 5*A*a^3*b)*x)/(a^5*b* x^5 + a^6*x^4) + (4*B*a*b^3 - 5*A*b^4)*log(b*x + a)/a^6 - (4*B*a*b^3 - 5*A *b^4)*log(x)/a^6
Time = 0.20 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.18 \[ \int \frac {A+B x}{x^5 \left (a^2+2 a b x+b^2 x^2\right )} \, dx=-\frac {{\left (4 \, B a b^{3} - 5 \, A b^{4}\right )} \log \left ({\left | x \right |}\right )}{a^{6}} + \frac {{\left (4 \, B a b^{4} - 5 \, A b^{5}\right )} \log \left ({\left | b x + a \right |}\right )}{a^{6} b} - \frac {3 \, A a^{5} + 12 \, {\left (4 \, B a^{2} b^{3} - 5 \, A a b^{4}\right )} x^{4} + 6 \, {\left (4 \, B a^{3} b^{2} - 5 \, A a^{2} b^{3}\right )} x^{3} - 2 \, {\left (4 \, B a^{4} b - 5 \, A a^{3} b^{2}\right )} x^{2} + {\left (4 \, B a^{5} - 5 \, A a^{4} b\right )} x}{12 \, {\left (b x + a\right )} a^{6} x^{4}} \] Input:
integrate((B*x+A)/x^5/(b^2*x^2+2*a*b*x+a^2),x, algorithm="giac")
Output:
-(4*B*a*b^3 - 5*A*b^4)*log(abs(x))/a^6 + (4*B*a*b^4 - 5*A*b^5)*log(abs(b*x + a))/(a^6*b) - 1/12*(3*A*a^5 + 12*(4*B*a^2*b^3 - 5*A*a*b^4)*x^4 + 6*(4*B *a^3*b^2 - 5*A*a^2*b^3)*x^3 - 2*(4*B*a^4*b - 5*A*a^3*b^2)*x^2 + (4*B*a^5 - 5*A*a^4*b)*x)/((b*x + a)*a^6*x^4)
Time = 10.97 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.13 \[ \int \frac {A+B x}{x^5 \left (a^2+2 a b x+b^2 x^2\right )} \, dx=\frac {\frac {x\,\left (5\,A\,b-4\,B\,a\right )}{12\,a^2}-\frac {A}{4\,a}+\frac {b^2\,x^3\,\left (5\,A\,b-4\,B\,a\right )}{2\,a^4}+\frac {b^3\,x^4\,\left (5\,A\,b-4\,B\,a\right )}{a^5}-\frac {b\,x^2\,\left (5\,A\,b-4\,B\,a\right )}{6\,a^3}}{b\,x^5+a\,x^4}-\frac {2\,b^3\,\mathrm {atanh}\left (\frac {b^3\,\left (5\,A\,b-4\,B\,a\right )\,\left (a+2\,b\,x\right )}{a\,\left (5\,A\,b^4-4\,B\,a\,b^3\right )}\right )\,\left (5\,A\,b-4\,B\,a\right )}{a^6} \] Input:
int((A + B*x)/(x^5*(a^2 + b^2*x^2 + 2*a*b*x)),x)
Output:
((x*(5*A*b - 4*B*a))/(12*a^2) - A/(4*a) + (b^2*x^3*(5*A*b - 4*B*a))/(2*a^4 ) + (b^3*x^4*(5*A*b - 4*B*a))/a^5 - (b*x^2*(5*A*b - 4*B*a))/(6*a^3))/(a*x^ 4 + b*x^5) - (2*b^3*atanh((b^3*(5*A*b - 4*B*a)*(a + 2*b*x))/(a*(5*A*b^4 - 4*B*a*b^3)))*(5*A*b - 4*B*a))/a^6
Time = 0.27 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.49 \[ \int \frac {A+B x}{x^5 \left (a^2+2 a b x+b^2 x^2\right )} \, dx=\frac {-12 \,\mathrm {log}\left (b x +a \right ) b^{4} x^{4}+12 \,\mathrm {log}\left (x \right ) b^{4} x^{4}-3 a^{4}+4 a^{3} b x -6 a^{2} b^{2} x^{2}+12 a \,b^{3} x^{3}}{12 a^{5} x^{4}} \] Input:
int((B*x+A)/x^5/(b^2*x^2+2*a*b*x+a^2),x)
Output:
( - 12*log(a + b*x)*b**4*x**4 + 12*log(x)*b**4*x**4 - 3*a**4 + 4*a**3*b*x - 6*a**2*b**2*x**2 + 12*a*b**3*x**3)/(12*a**5*x**4)