Integrand size = 27, antiderivative size = 134 \[ \int \frac {x^5 (A+B x)}{a^2+2 a b x+b^2 x^2} \, dx=-\frac {a^3 (4 A b-5 a B) x}{b^6}+\frac {a^2 (3 A b-4 a B) x^2}{2 b^5}-\frac {a (2 A b-3 a B) x^3}{3 b^4}+\frac {(A b-2 a B) x^4}{4 b^3}+\frac {B x^5}{5 b^2}+\frac {a^5 (A b-a B)}{b^7 (a+b x)}+\frac {a^4 (5 A b-6 a B) \log (a+b x)}{b^7} \] Output:
-a^3*(4*A*b-5*B*a)*x/b^6+1/2*a^2*(3*A*b-4*B*a)*x^2/b^5-1/3*a*(2*A*b-3*B*a) *x^3/b^4+1/4*(A*b-2*B*a)*x^4/b^3+1/5*B*x^5/b^2+a^5*(A*b-B*a)/b^7/(b*x+a)+a ^4*(5*A*b-6*B*a)*ln(b*x+a)/b^7
Time = 0.07 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.95 \[ \int \frac {x^5 (A+B x)}{a^2+2 a b x+b^2 x^2} \, dx=\frac {60 a^3 b (-4 A b+5 a B) x-30 a^2 b^2 (-3 A b+4 a B) x^2+20 a b^3 (-2 A b+3 a B) x^3+15 b^4 (A b-2 a B) x^4+12 b^5 B x^5+\frac {60 a^5 (A b-a B)}{a+b x}+60 a^4 (5 A b-6 a B) \log (a+b x)}{60 b^7} \] Input:
Integrate[(x^5*(A + B*x))/(a^2 + 2*a*b*x + b^2*x^2),x]
Output:
(60*a^3*b*(-4*A*b + 5*a*B)*x - 30*a^2*b^2*(-3*A*b + 4*a*B)*x^2 + 20*a*b^3* (-2*A*b + 3*a*B)*x^3 + 15*b^4*(A*b - 2*a*B)*x^4 + 12*b^5*B*x^5 + (60*a^5*( A*b - a*B))/(a + b*x) + 60*a^4*(5*A*b - 6*a*B)*Log[a + b*x])/(60*b^7)
Time = 0.62 (sec) , antiderivative size = 134, 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.
\(\displaystyle \int \frac {x^5 (A+B x)}{a^2+2 a b x+b^2 x^2} \, dx\) |
\(\Big \downarrow \) 1184 |
\(\displaystyle b^2 \int \frac {x^5 (A+B x)}{b^2 (a+b x)^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \frac {x^5 (A+B x)}{(a+b x)^2}dx\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \int \left (\frac {a^5 (a B-A b)}{b^6 (a+b x)^2}-\frac {a^4 (6 a B-5 A b)}{b^6 (a+b x)}+\frac {a^3 (5 a B-4 A b)}{b^6}-\frac {a^2 x (4 a B-3 A b)}{b^5}+\frac {a x^2 (3 a B-2 A b)}{b^4}+\frac {x^3 (A b-2 a B)}{b^3}+\frac {B x^4}{b^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {a^5 (A b-a B)}{b^7 (a+b x)}+\frac {a^4 (5 A b-6 a B) \log (a+b x)}{b^7}-\frac {a^3 x (4 A b-5 a B)}{b^6}+\frac {a^2 x^2 (3 A b-4 a B)}{2 b^5}-\frac {a x^3 (2 A b-3 a B)}{3 b^4}+\frac {x^4 (A b-2 a B)}{4 b^3}+\frac {B x^5}{5 b^2}\) |
Input:
Int[(x^5*(A + B*x))/(a^2 + 2*a*b*x + b^2*x^2),x]
Output:
-((a^3*(4*A*b - 5*a*B)*x)/b^6) + (a^2*(3*A*b - 4*a*B)*x^2)/(2*b^5) - (a*(2 *A*b - 3*a*B)*x^3)/(3*b^4) + ((A*b - 2*a*B)*x^4)/(4*b^3) + (B*x^5)/(5*b^2) + (a^5*(A*b - a*B))/(b^7*(a + b*x)) + (a^4*(5*A*b - 6*a*B)*Log[a + b*x])/ b^7
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 = 136, normalized size of antiderivative = 1.01
method | result | size |
norman | \(\frac {\frac {a \left (5 A \,a^{4} b -6 B \,a^{5}\right )}{b^{7}}+\frac {B \,x^{6}}{5 b}+\frac {\left (5 A b -6 B a \right ) x^{5}}{20 b^{2}}-\frac {a \left (5 A b -6 B a \right ) x^{4}}{12 b^{3}}+\frac {a^{2} \left (5 A b -6 B a \right ) x^{3}}{6 b^{4}}-\frac {a^{3} \left (5 A b -6 B a \right ) x^{2}}{2 b^{5}}}{b x +a}+\frac {a^{4} \left (5 A b -6 B a \right ) \ln \left (b x +a \right )}{b^{7}}\) | \(136\) |
default | \(-\frac {-\frac {1}{5} b^{4} B \,x^{5}-\frac {1}{4} A \,b^{4} x^{4}+\frac {1}{2} B a \,b^{3} x^{4}+\frac {2}{3} A a \,b^{3} x^{3}-B \,a^{2} b^{2} x^{3}-\frac {3}{2} A \,a^{2} b^{2} x^{2}+2 B \,a^{3} b \,x^{2}+4 A \,a^{3} b x -5 a^{4} B x}{b^{6}}+\frac {a^{5} \left (A b -B a \right )}{b^{7} \left (b x +a \right )}+\frac {a^{4} \left (5 A b -6 B a \right ) \ln \left (b x +a \right )}{b^{7}}\) | \(139\) |
risch | \(\frac {B \,x^{5}}{5 b^{2}}+\frac {A \,x^{4}}{4 b^{2}}-\frac {B a \,x^{4}}{2 b^{3}}-\frac {2 A a \,x^{3}}{3 b^{3}}+\frac {B \,a^{2} x^{3}}{b^{4}}+\frac {3 A \,a^{2} x^{2}}{2 b^{4}}-\frac {2 B \,a^{3} x^{2}}{b^{5}}-\frac {4 A \,a^{3} x}{b^{5}}+\frac {5 a^{4} B x}{b^{6}}+\frac {a^{5} A}{b^{6} \left (b x +a \right )}-\frac {a^{6} B}{b^{7} \left (b x +a \right )}+\frac {5 a^{4} \ln \left (b x +a \right ) A}{b^{6}}-\frac {6 a^{5} \ln \left (b x +a \right ) B}{b^{7}}\) | \(156\) |
parallelrisch | \(\frac {12 b^{6} B \,x^{6}+15 A \,b^{6} x^{5}-18 B a \,b^{5} x^{5}-25 A a \,b^{5} x^{4}+30 B \,a^{2} b^{4} x^{4}+50 A \,a^{2} b^{4} x^{3}-60 B \,a^{3} b^{3} x^{3}+300 A \ln \left (b x +a \right ) x \,a^{4} b^{2}-150 A \,a^{3} b^{3} x^{2}-360 B \ln \left (b x +a \right ) x \,a^{5} b +180 B \,a^{4} b^{2} x^{2}+300 A \ln \left (b x +a \right ) a^{5} b -360 B \ln \left (b x +a \right ) a^{6}+300 A \,a^{5} b -360 B \,a^{6}}{60 b^{7} \left (b x +a \right )}\) | \(180\) |
Input:
int(x^5*(B*x+A)/(b^2*x^2+2*a*b*x+a^2),x,method=_RETURNVERBOSE)
Output:
(a*(5*A*a^4*b-6*B*a^5)/b^7+1/5*B/b*x^6+1/20*(5*A*b-6*B*a)/b^2*x^5-1/12*a*( 5*A*b-6*B*a)/b^3*x^4+1/6*a^2*(5*A*b-6*B*a)/b^4*x^3-1/2*a^3*(5*A*b-6*B*a)/b ^5*x^2)/(b*x+a)+a^4*(5*A*b-6*B*a)*ln(b*x+a)/b^7
Time = 0.07 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.40 \[ \int \frac {x^5 (A+B x)}{a^2+2 a b x+b^2 x^2} \, dx=\frac {12 \, B b^{6} x^{6} - 60 \, B a^{6} + 60 \, A a^{5} b - 3 \, {\left (6 \, B a b^{5} - 5 \, A b^{6}\right )} x^{5} + 5 \, {\left (6 \, B a^{2} b^{4} - 5 \, A a b^{5}\right )} x^{4} - 10 \, {\left (6 \, B a^{3} b^{3} - 5 \, A a^{2} b^{4}\right )} x^{3} + 30 \, {\left (6 \, B a^{4} b^{2} - 5 \, A a^{3} b^{3}\right )} x^{2} + 60 \, {\left (5 \, B a^{5} b - 4 \, A a^{4} b^{2}\right )} x - 60 \, {\left (6 \, B a^{6} - 5 \, A a^{5} b + {\left (6 \, B a^{5} b - 5 \, A a^{4} b^{2}\right )} x\right )} \log \left (b x + a\right )}{60 \, {\left (b^{8} x + a b^{7}\right )}} \] Input:
integrate(x^5*(B*x+A)/(b^2*x^2+2*a*b*x+a^2),x, algorithm="fricas")
Output:
1/60*(12*B*b^6*x^6 - 60*B*a^6 + 60*A*a^5*b - 3*(6*B*a*b^5 - 5*A*b^6)*x^5 + 5*(6*B*a^2*b^4 - 5*A*a*b^5)*x^4 - 10*(6*B*a^3*b^3 - 5*A*a^2*b^4)*x^3 + 30 *(6*B*a^4*b^2 - 5*A*a^3*b^3)*x^2 + 60*(5*B*a^5*b - 4*A*a^4*b^2)*x - 60*(6* B*a^6 - 5*A*a^5*b + (6*B*a^5*b - 5*A*a^4*b^2)*x)*log(b*x + a))/(b^8*x + a* b^7)
Time = 0.29 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.07 \[ \int \frac {x^5 (A+B x)}{a^2+2 a b x+b^2 x^2} \, dx=\frac {B x^{5}}{5 b^{2}} - \frac {a^{4} \left (- 5 A b + 6 B a\right ) \log {\left (a + b x \right )}}{b^{7}} + x^{4} \left (\frac {A}{4 b^{2}} - \frac {B a}{2 b^{3}}\right ) + x^{3} \left (- \frac {2 A a}{3 b^{3}} + \frac {B a^{2}}{b^{4}}\right ) + x^{2} \cdot \left (\frac {3 A a^{2}}{2 b^{4}} - \frac {2 B a^{3}}{b^{5}}\right ) + x \left (- \frac {4 A a^{3}}{b^{5}} + \frac {5 B a^{4}}{b^{6}}\right ) + \frac {A a^{5} b - B a^{6}}{a b^{7} + b^{8} x} \] Input:
integrate(x**5*(B*x+A)/(b**2*x**2+2*a*b*x+a**2),x)
Output:
B*x**5/(5*b**2) - a**4*(-5*A*b + 6*B*a)*log(a + b*x)/b**7 + x**4*(A/(4*b** 2) - B*a/(2*b**3)) + x**3*(-2*A*a/(3*b**3) + B*a**2/b**4) + x**2*(3*A*a**2 /(2*b**4) - 2*B*a**3/b**5) + x*(-4*A*a**3/b**5 + 5*B*a**4/b**6) + (A*a**5* b - B*a**6)/(a*b**7 + b**8*x)
Time = 0.03 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.11 \[ \int \frac {x^5 (A+B x)}{a^2+2 a b x+b^2 x^2} \, dx=-\frac {B a^{6} - A a^{5} b}{b^{8} x + a b^{7}} + \frac {12 \, B b^{4} x^{5} - 15 \, {\left (2 \, B a b^{3} - A b^{4}\right )} x^{4} + 20 \, {\left (3 \, B a^{2} b^{2} - 2 \, A a b^{3}\right )} x^{3} - 30 \, {\left (4 \, B a^{3} b - 3 \, A a^{2} b^{2}\right )} x^{2} + 60 \, {\left (5 \, B a^{4} - 4 \, A a^{3} b\right )} x}{60 \, b^{6}} - \frac {{\left (6 \, B a^{5} - 5 \, A a^{4} b\right )} \log \left (b x + a\right )}{b^{7}} \] Input:
integrate(x^5*(B*x+A)/(b^2*x^2+2*a*b*x+a^2),x, algorithm="maxima")
Output:
-(B*a^6 - A*a^5*b)/(b^8*x + a*b^7) + 1/60*(12*B*b^4*x^5 - 15*(2*B*a*b^3 - A*b^4)*x^4 + 20*(3*B*a^2*b^2 - 2*A*a*b^3)*x^3 - 30*(4*B*a^3*b - 3*A*a^2*b^ 2)*x^2 + 60*(5*B*a^4 - 4*A*a^3*b)*x)/b^6 - (6*B*a^5 - 5*A*a^4*b)*log(b*x + a)/b^7
Time = 0.21 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.13 \[ \int \frac {x^5 (A+B x)}{a^2+2 a b x+b^2 x^2} \, dx=-\frac {{\left (6 \, B a^{5} - 5 \, A a^{4} b\right )} \log \left ({\left | b x + a \right |}\right )}{b^{7}} - \frac {B a^{6} - A a^{5} b}{{\left (b x + a\right )} b^{7}} + \frac {12 \, B b^{8} x^{5} - 30 \, B a b^{7} x^{4} + 15 \, A b^{8} x^{4} + 60 \, B a^{2} b^{6} x^{3} - 40 \, A a b^{7} x^{3} - 120 \, B a^{3} b^{5} x^{2} + 90 \, A a^{2} b^{6} x^{2} + 300 \, B a^{4} b^{4} x - 240 \, A a^{3} b^{5} x}{60 \, b^{10}} \] Input:
integrate(x^5*(B*x+A)/(b^2*x^2+2*a*b*x+a^2),x, algorithm="giac")
Output:
-(6*B*a^5 - 5*A*a^4*b)*log(abs(b*x + a))/b^7 - (B*a^6 - A*a^5*b)/((b*x + a )*b^7) + 1/60*(12*B*b^8*x^5 - 30*B*a*b^7*x^4 + 15*A*b^8*x^4 + 60*B*a^2*b^6 *x^3 - 40*A*a*b^7*x^3 - 120*B*a^3*b^5*x^2 + 90*A*a^2*b^6*x^2 + 300*B*a^4*b ^4*x - 240*A*a^3*b^5*x)/b^10
Time = 0.08 (sec) , antiderivative size = 279, normalized size of antiderivative = 2.08 \[ \int \frac {x^5 (A+B x)}{a^2+2 a b x+b^2 x^2} \, dx=x^2\,\left (\frac {a\,\left (\frac {2\,a\,\left (\frac {A}{b^2}-\frac {2\,B\,a}{b^3}\right )}{b}+\frac {B\,a^2}{b^4}\right )}{b}-\frac {a^2\,\left (\frac {A}{b^2}-\frac {2\,B\,a}{b^3}\right )}{2\,b^2}\right )+x^4\,\left (\frac {A}{4\,b^2}-\frac {B\,a}{2\,b^3}\right )-x^3\,\left (\frac {2\,a\,\left (\frac {A}{b^2}-\frac {2\,B\,a}{b^3}\right )}{3\,b}+\frac {B\,a^2}{3\,b^4}\right )-x\,\left (\frac {2\,a\,\left (\frac {2\,a\,\left (\frac {2\,a\,\left (\frac {A}{b^2}-\frac {2\,B\,a}{b^3}\right )}{b}+\frac {B\,a^2}{b^4}\right )}{b}-\frac {a^2\,\left (\frac {A}{b^2}-\frac {2\,B\,a}{b^3}\right )}{b^2}\right )}{b}-\frac {a^2\,\left (\frac {2\,a\,\left (\frac {A}{b^2}-\frac {2\,B\,a}{b^3}\right )}{b}+\frac {B\,a^2}{b^4}\right )}{b^2}\right )-\frac {\ln \left (a+b\,x\right )\,\left (6\,B\,a^5-5\,A\,a^4\,b\right )}{b^7}+\frac {B\,x^5}{5\,b^2}-\frac {B\,a^6-A\,a^5\,b}{b\,\left (x\,b^7+a\,b^6\right )} \] Input:
int((x^5*(A + B*x))/(a^2 + b^2*x^2 + 2*a*b*x),x)
Output:
x^2*((a*((2*a*(A/b^2 - (2*B*a)/b^3))/b + (B*a^2)/b^4))/b - (a^2*(A/b^2 - ( 2*B*a)/b^3))/(2*b^2)) + x^4*(A/(4*b^2) - (B*a)/(2*b^3)) - x^3*((2*a*(A/b^2 - (2*B*a)/b^3))/(3*b) + (B*a^2)/(3*b^4)) - x*((2*a*((2*a*((2*a*(A/b^2 - ( 2*B*a)/b^3))/b + (B*a^2)/b^4))/b - (a^2*(A/b^2 - (2*B*a)/b^3))/b^2))/b - ( a^2*((2*a*(A/b^2 - (2*B*a)/b^3))/b + (B*a^2)/b^4))/b^2) - (log(a + b*x)*(6 *B*a^5 - 5*A*a^4*b))/b^7 + (B*x^5)/(5*b^2) - (B*a^6 - A*a^5*b)/(b*(a*b^6 + b^7*x))
Time = 0.28 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.47 \[ \int \frac {x^5 (A+B x)}{a^2+2 a b x+b^2 x^2} \, dx=\frac {-60 \,\mathrm {log}\left (b x +a \right ) a^{5}+60 a^{4} b x -30 a^{3} b^{2} x^{2}+20 a^{2} b^{3} x^{3}-15 a \,b^{4} x^{4}+12 b^{5} x^{5}}{60 b^{6}} \] Input:
int(x^5*(B*x+A)/(b^2*x^2+2*a*b*x+a^2),x)
Output:
( - 60*log(a + b*x)*a**5 + 60*a**4*b*x - 30*a**3*b**2*x**2 + 20*a**2*b**3* x**3 - 15*a*b**4*x**4 + 12*b**5*x**5)/(60*b**6)