Integrand size = 27, antiderivative size = 121 \[ \int \frac {x^4 (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\frac {(A b-4 a B) x}{b^5}+\frac {B x^2}{2 b^4}-\frac {a^4 (A b-a B)}{3 b^6 (a+b x)^3}+\frac {a^3 (4 A b-5 a B)}{2 b^6 (a+b x)^2}-\frac {2 a^2 (3 A b-5 a B)}{b^6 (a+b x)}-\frac {2 a (2 A b-5 a B) \log (a+b x)}{b^6} \] Output:
(A*b-4*B*a)*x/b^5+1/2*B*x^2/b^4-1/3*a^4*(A*b-B*a)/b^6/(b*x+a)^3+1/2*a^3*(4 *A*b-5*B*a)/b^6/(b*x+a)^2-2*a^2*(3*A*b-5*B*a)/b^6/(b*x+a)-2*a*(2*A*b-5*B*a )*ln(b*x+a)/b^6
Time = 0.05 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.07 \[ \int \frac {x^4 (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\frac {(A b-4 a B) x}{b^5}+\frac {B x^2}{2 b^4}+\frac {-a^4 A b+a^5 B}{3 b^6 (a+b x)^3}+\frac {4 a^3 A b-5 a^4 B}{2 b^6 (a+b x)^2}+\frac {2 \left (-3 a^2 A b+5 a^3 B\right )}{b^6 (a+b x)}+\frac {2 \left (-2 a A b+5 a^2 B\right ) \log (a+b x)}{b^6} \] Input:
Integrate[(x^4*(A + B*x))/(a^2 + 2*a*b*x + b^2*x^2)^2,x]
Output:
((A*b - 4*a*B)*x)/b^5 + (B*x^2)/(2*b^4) + (-(a^4*A*b) + a^5*B)/(3*b^6*(a + b*x)^3) + (4*a^3*A*b - 5*a^4*B)/(2*b^6*(a + b*x)^2) + (2*(-3*a^2*A*b + 5* a^3*B))/(b^6*(a + b*x)) + (2*(-2*a*A*b + 5*a^2*B)*Log[a + b*x])/b^6
Time = 0.56 (sec) , antiderivative size = 121, 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.
\(\displaystyle \int \frac {x^4 (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx\) |
\(\Big \downarrow \) 1184 |
\(\displaystyle b^4 \int \frac {x^4 (A+B x)}{b^4 (a+b x)^4}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \frac {x^4 (A+B x)}{(a+b x)^4}dx\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \int \left (-\frac {a^4 (a B-A b)}{b^5 (a+b x)^4}+\frac {a^3 (5 a B-4 A b)}{b^5 (a+b x)^3}-\frac {2 a^2 (5 a B-3 A b)}{b^5 (a+b x)^2}+\frac {2 a (5 a B-2 A b)}{b^5 (a+b x)}+\frac {A b-4 a B}{b^5}+\frac {B x}{b^4}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {a^4 (A b-a B)}{3 b^6 (a+b x)^3}+\frac {a^3 (4 A b-5 a B)}{2 b^6 (a+b x)^2}-\frac {2 a^2 (3 A b-5 a B)}{b^6 (a+b x)}-\frac {2 a (2 A b-5 a B) \log (a+b x)}{b^6}+\frac {x (A b-4 a B)}{b^5}+\frac {B x^2}{2 b^4}\) |
Input:
Int[(x^4*(A + B*x))/(a^2 + 2*a*b*x + b^2*x^2)^2,x]
Output:
((A*b - 4*a*B)*x)/b^5 + (B*x^2)/(2*b^4) - (a^4*(A*b - a*B))/(3*b^6*(a + b* x)^3) + (a^3*(4*A*b - 5*a*B))/(2*b^6*(a + b*x)^2) - (2*a^2*(3*A*b - 5*a*B) )/(b^6*(a + b*x)) - (2*a*(2*A*b - 5*a*B)*Log[a + b*x])/b^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 = 1.10 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.95
method | result | size |
default | \(\frac {\frac {1}{2} B b \,x^{2}+A b x -4 B a x}{b^{5}}-\frac {a^{4} \left (A b -B a \right )}{3 b^{6} \left (b x +a \right )^{3}}-\frac {2 a^{2} \left (3 A b -5 B a \right )}{b^{6} \left (b x +a \right )}-\frac {2 a \left (2 A b -5 B a \right ) \ln \left (b x +a \right )}{b^{6}}+\frac {a^{3} \left (4 A b -5 B a \right )}{2 b^{6} \left (b x +a \right )^{2}}\) | \(115\) |
norman | \(\frac {\frac {B \,x^{5}}{2 b}-\frac {a^{3} \left (22 a b A -55 a^{2} B \right )}{3 b^{6}}+\frac {\left (2 A b -5 B a \right ) x^{4}}{2 b^{2}}-\frac {3 a \left (4 a b A -10 a^{2} B \right ) x^{2}}{b^{4}}-\frac {3 a^{2} \left (6 a b A -15 a^{2} B \right ) x}{b^{5}}}{\left (b x +a \right )^{3}}-\frac {2 a \left (2 A b -5 B a \right ) \ln \left (b x +a \right )}{b^{6}}\) | \(120\) |
risch | \(\frac {B \,x^{2}}{2 b^{4}}+\frac {A x}{b^{4}}-\frac {4 B a x}{b^{5}}+\frac {\left (-6 a^{2} A \,b^{2}+10 B \,a^{3} b \right ) x^{2}-\frac {5 a^{3} \left (4 A b -7 B a \right ) x}{2}-\frac {a^{4} \left (26 A b -47 B a \right )}{6 b}}{b^{5} \left (b x +a \right ) \left (b^{2} x^{2}+2 a b x +a^{2}\right )}-\frac {4 a \ln \left (b x +a \right ) A}{b^{5}}+\frac {10 a^{2} \ln \left (b x +a \right ) B}{b^{6}}\) | \(136\) |
parallelrisch | \(-\frac {-3 B \,x^{5} b^{5}+24 A \ln \left (b x +a \right ) x^{3} a \,b^{4}-6 A \,b^{5} x^{4}-60 B \ln \left (b x +a \right ) x^{3} a^{2} b^{3}+15 B a \,b^{4} x^{4}+72 A \ln \left (b x +a \right ) x^{2} a^{2} b^{3}-180 B \ln \left (b x +a \right ) x^{2} a^{3} b^{2}+72 A \ln \left (b x +a \right ) x \,a^{3} b^{2}+72 A \,a^{2} b^{3} x^{2}-180 B \ln \left (b x +a \right ) x \,a^{4} b -180 B \,a^{3} b^{2} x^{2}+24 A \ln \left (b x +a \right ) a^{4} b +108 A \,a^{3} b^{2} x -60 B \ln \left (b x +a \right ) a^{5}-270 B \,a^{4} b x +44 A \,a^{4} b -110 B \,a^{5}}{6 b^{6} \left (b^{2} x^{2}+2 a b x +a^{2}\right ) \left (b x +a \right )}\) | \(240\) |
Input:
int(x^4*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^2,x,method=_RETURNVERBOSE)
Output:
1/b^5*(1/2*B*b*x^2+A*b*x-4*B*a*x)-1/3*a^4*(A*b-B*a)/b^6/(b*x+a)^3-2*a^2*(3 *A*b-5*B*a)/b^6/(b*x+a)-2*a*(2*A*b-5*B*a)*ln(b*x+a)/b^6+1/2*a^3*(4*A*b-5*B *a)/b^6/(b*x+a)^2
Time = 0.08 (sec) , antiderivative size = 231, normalized size of antiderivative = 1.91 \[ \int \frac {x^4 (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\frac {3 \, B b^{5} x^{5} + 47 \, B a^{5} - 26 \, A a^{4} b - 3 \, {\left (5 \, B a b^{4} - 2 \, A b^{5}\right )} x^{4} - 9 \, {\left (7 \, B a^{2} b^{3} - 2 \, A a b^{4}\right )} x^{3} - 9 \, {\left (B a^{3} b^{2} + 2 \, A a^{2} b^{3}\right )} x^{2} + 27 \, {\left (3 \, B a^{4} b - 2 \, A a^{3} b^{2}\right )} x + 12 \, {\left (5 \, B a^{5} - 2 \, A a^{4} b + {\left (5 \, B a^{2} b^{3} - 2 \, A a b^{4}\right )} x^{3} + 3 \, {\left (5 \, B a^{3} b^{2} - 2 \, A a^{2} b^{3}\right )} x^{2} + 3 \, {\left (5 \, B a^{4} b - 2 \, A a^{3} b^{2}\right )} x\right )} \log \left (b x + a\right )}{6 \, {\left (b^{9} x^{3} + 3 \, a b^{8} x^{2} + 3 \, a^{2} b^{7} x + a^{3} b^{6}\right )}} \] Input:
integrate(x^4*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^2,x, algorithm="fricas")
Output:
1/6*(3*B*b^5*x^5 + 47*B*a^5 - 26*A*a^4*b - 3*(5*B*a*b^4 - 2*A*b^5)*x^4 - 9 *(7*B*a^2*b^3 - 2*A*a*b^4)*x^3 - 9*(B*a^3*b^2 + 2*A*a^2*b^3)*x^2 + 27*(3*B *a^4*b - 2*A*a^3*b^2)*x + 12*(5*B*a^5 - 2*A*a^4*b + (5*B*a^2*b^3 - 2*A*a*b ^4)*x^3 + 3*(5*B*a^3*b^2 - 2*A*a^2*b^3)*x^2 + 3*(5*B*a^4*b - 2*A*a^3*b^2)* x)*log(b*x + a))/(b^9*x^3 + 3*a*b^8*x^2 + 3*a^2*b^7*x + a^3*b^6)
Time = 0.73 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.19 \[ \int \frac {x^4 (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\frac {B x^{2}}{2 b^{4}} + \frac {2 a \left (- 2 A b + 5 B a\right ) \log {\left (a + b x \right )}}{b^{6}} + x \left (\frac {A}{b^{4}} - \frac {4 B a}{b^{5}}\right ) + \frac {- 26 A a^{4} b + 47 B a^{5} + x^{2} \left (- 36 A a^{2} b^{3} + 60 B a^{3} b^{2}\right ) + x \left (- 60 A a^{3} b^{2} + 105 B a^{4} b\right )}{6 a^{3} b^{6} + 18 a^{2} b^{7} x + 18 a b^{8} x^{2} + 6 b^{9} x^{3}} \] Input:
integrate(x**4*(B*x+A)/(b**2*x**2+2*a*b*x+a**2)**2,x)
Output:
B*x**2/(2*b**4) + 2*a*(-2*A*b + 5*B*a)*log(a + b*x)/b**6 + x*(A/b**4 - 4*B *a/b**5) + (-26*A*a**4*b + 47*B*a**5 + x**2*(-36*A*a**2*b**3 + 60*B*a**3*b **2) + x*(-60*A*a**3*b**2 + 105*B*a**4*b))/(6*a**3*b**6 + 18*a**2*b**7*x + 18*a*b**8*x**2 + 6*b**9*x**3)
Time = 0.04 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.18 \[ \int \frac {x^4 (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\frac {47 \, B a^{5} - 26 \, A a^{4} b + 12 \, {\left (5 \, B a^{3} b^{2} - 3 \, A a^{2} b^{3}\right )} x^{2} + 15 \, {\left (7 \, B a^{4} b - 4 \, A a^{3} b^{2}\right )} x}{6 \, {\left (b^{9} x^{3} + 3 \, a b^{8} x^{2} + 3 \, a^{2} b^{7} x + a^{3} b^{6}\right )}} + \frac {B b x^{2} - 2 \, {\left (4 \, B a - A b\right )} x}{2 \, b^{5}} + \frac {2 \, {\left (5 \, B a^{2} - 2 \, A a b\right )} \log \left (b x + a\right )}{b^{6}} \] Input:
integrate(x^4*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^2,x, algorithm="maxima")
Output:
1/6*(47*B*a^5 - 26*A*a^4*b + 12*(5*B*a^3*b^2 - 3*A*a^2*b^3)*x^2 + 15*(7*B* a^4*b - 4*A*a^3*b^2)*x)/(b^9*x^3 + 3*a*b^8*x^2 + 3*a^2*b^7*x + a^3*b^6) + 1/2*(B*b*x^2 - 2*(4*B*a - A*b)*x)/b^5 + 2*(5*B*a^2 - 2*A*a*b)*log(b*x + a) /b^6
Time = 0.19 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.02 \[ \int \frac {x^4 (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\frac {2 \, {\left (5 \, B a^{2} - 2 \, A a b\right )} \log \left ({\left | b x + a \right |}\right )}{b^{6}} + \frac {B b^{4} x^{2} - 8 \, B a b^{3} x + 2 \, A b^{4} x}{2 \, b^{8}} + \frac {47 \, B a^{5} - 26 \, A a^{4} b + 12 \, {\left (5 \, B a^{3} b^{2} - 3 \, A a^{2} b^{3}\right )} x^{2} + 15 \, {\left (7 \, B a^{4} b - 4 \, A a^{3} b^{2}\right )} x}{6 \, {\left (b x + a\right )}^{3} b^{6}} \] Input:
integrate(x^4*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^2,x, algorithm="giac")
Output:
2*(5*B*a^2 - 2*A*a*b)*log(abs(b*x + a))/b^6 + 1/2*(B*b^4*x^2 - 8*B*a*b^3*x + 2*A*b^4*x)/b^8 + 1/6*(47*B*a^5 - 26*A*a^4*b + 12*(5*B*a^3*b^2 - 3*A*a^2 *b^3)*x^2 + 15*(7*B*a^4*b - 4*A*a^3*b^2)*x)/((b*x + a)^3*b^6)
Time = 0.09 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.17 \[ \int \frac {x^4 (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=x\,\left (\frac {A}{b^4}-\frac {4\,B\,a}{b^5}\right )+\frac {x\,\left (\frac {35\,B\,a^4}{2}-10\,A\,a^3\,b\right )-x^2\,\left (6\,A\,a^2\,b^2-10\,B\,a^3\,b\right )+\frac {47\,B\,a^5-26\,A\,a^4\,b}{6\,b}}{a^3\,b^5+3\,a^2\,b^6\,x+3\,a\,b^7\,x^2+b^8\,x^3}+\frac {B\,x^2}{2\,b^4}+\frac {\ln \left (a+b\,x\right )\,\left (10\,B\,a^2-4\,A\,a\,b\right )}{b^6} \] Input:
int((x^4*(A + B*x))/(a^2 + b^2*x^2 + 2*a*b*x)^2,x)
Output:
x*(A/b^4 - (4*B*a)/b^5) + (x*((35*B*a^4)/2 - 10*A*a^3*b) - x^2*(6*A*a^2*b^ 2 - 10*B*a^3*b) + (47*B*a^5 - 26*A*a^4*b)/(6*b))/(a^3*b^5 + b^8*x^3 + 3*a^ 2*b^6*x + 3*a*b^7*x^2) + (B*x^2)/(2*b^4) + (log(a + b*x)*(10*B*a^2 - 4*A*a *b))/b^6
Time = 0.28 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.80 \[ \int \frac {x^4 (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\frac {12 \,\mathrm {log}\left (b x +a \right ) a^{4}+24 \,\mathrm {log}\left (b x +a \right ) a^{3} b x +12 \,\mathrm {log}\left (b x +a \right ) a^{2} b^{2} x^{2}+6 a^{4}-12 a^{2} b^{2} x^{2}-4 a \,b^{3} x^{3}+b^{4} x^{4}}{2 b^{5} \left (b^{2} x^{2}+2 a b x +a^{2}\right )} \] Input:
int(x^4*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^2,x)
Output:
(12*log(a + b*x)*a**4 + 24*log(a + b*x)*a**3*b*x + 12*log(a + b*x)*a**2*b* *2*x**2 + 6*a**4 - 12*a**2*b**2*x**2 - 4*a*b**3*x**3 + b**4*x**4)/(2*b**5* (a**2 + 2*a*b*x + b**2*x**2))