Integrand size = 16, antiderivative size = 94 \[ \int \frac {x^3 (A+B x)}{(a+b x)^3} \, dx=\frac {(A b-3 a B) x}{b^4}+\frac {B x^2}{2 b^3}+\frac {a^3 (A b-a B)}{2 b^5 (a+b x)^2}-\frac {a^2 (3 A b-4 a B)}{b^5 (a+b x)}-\frac {3 a (A b-2 a B) \log (a+b x)}{b^5} \] Output:
(A*b-3*B*a)*x/b^4+1/2*B*x^2/b^3+1/2*a^3*(A*b-B*a)/b^5/(b*x+a)^2-a^2*(3*A*b -4*B*a)/b^5/(b*x+a)-3*a*(A*b-2*B*a)*ln(b*x+a)/b^5
Time = 0.04 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.91 \[ \int \frac {x^3 (A+B x)}{(a+b x)^3} \, dx=\frac {2 b (A b-3 a B) x+b^2 B x^2+\frac {a^3 (A b-a B)}{(a+b x)^2}+\frac {2 a^2 (-3 A b+4 a B)}{a+b x}+6 a (-A b+2 a B) \log (a+b x)}{2 b^5} \] Input:
Integrate[(x^3*(A + B*x))/(a + b*x)^3,x]
Output:
(2*b*(A*b - 3*a*B)*x + b^2*B*x^2 + (a^3*(A*b - a*B))/(a + b*x)^2 + (2*a^2* (-3*A*b + 4*a*B))/(a + b*x) + 6*a*(-(A*b) + 2*a*B)*Log[a + b*x])/(2*b^5)
Time = 0.25 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {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.
\(\displaystyle \int \frac {x^3 (A+B x)}{(a+b x)^3} \, dx\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \int \left (\frac {a^3 (a B-A b)}{b^4 (a+b x)^3}-\frac {a^2 (4 a B-3 A b)}{b^4 (a+b x)^2}+\frac {3 a (2 a B-A b)}{b^4 (a+b x)}+\frac {A b-3 a B}{b^4}+\frac {B x}{b^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {a^3 (A b-a B)}{2 b^5 (a+b x)^2}-\frac {a^2 (3 A b-4 a B)}{b^5 (a+b x)}-\frac {3 a (A b-2 a B) \log (a+b x)}{b^5}+\frac {x (A b-3 a B)}{b^4}+\frac {B x^2}{2 b^3}\) |
Input:
Int[(x^3*(A + B*x))/(a + b*x)^3,x]
Output:
((A*b - 3*a*B)*x)/b^4 + (B*x^2)/(2*b^3) + (a^3*(A*b - a*B))/(2*b^5*(a + b* x)^2) - (a^2*(3*A*b - 4*a*B))/(b^5*(a + b*x)) - (3*a*(A*b - 2*a*B)*Log[a + b*x])/b^5
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]))))
Time = 0.14 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.96
method | result | size |
default | \(\frac {\frac {1}{2} b B \,x^{2}+A b x -3 B a x}{b^{4}}+\frac {a^{3} \left (A b -B a \right )}{2 b^{5} \left (b x +a \right )^{2}}-\frac {a^{2} \left (3 A b -4 B a \right )}{b^{5} \left (b x +a \right )}-\frac {3 a \left (A b -2 B a \right ) \ln \left (b x +a \right )}{b^{5}}\) | \(90\) |
norman | \(\frac {\frac {\left (A b -2 B a \right ) x^{3}}{b^{2}}+\frac {B \,x^{4}}{2 b}-\frac {a^{2} \left (9 a b A -18 a^{2} B \right )}{2 b^{5}}-\frac {2 a \left (3 a b A -6 a^{2} B \right ) x}{b^{4}}}{\left (b x +a \right )^{2}}-\frac {3 a \left (A b -2 B a \right ) \ln \left (b x +a \right )}{b^{5}}\) | \(94\) |
risch | \(\frac {B \,x^{2}}{2 b^{3}}+\frac {A x}{b^{3}}-\frac {3 B a x}{b^{4}}+\frac {\left (-3 a^{2} b A +4 a^{3} B \right ) x -\frac {a^{3} \left (5 A b -7 B a \right )}{2 b}}{b^{4} \left (b x +a \right )^{2}}-\frac {3 a \ln \left (b x +a \right ) A}{b^{4}}+\frac {6 a^{2} \ln \left (b x +a \right ) B}{b^{5}}\) | \(98\) |
parallelrisch | \(-\frac {-B \,x^{4} b^{4}+6 A \ln \left (b x +a \right ) x^{2} a \,b^{3}-2 A \,x^{3} b^{4}-12 B \ln \left (b x +a \right ) x^{2} a^{2} b^{2}+4 B \,x^{3} a \,b^{3}+12 A \ln \left (b x +a \right ) x \,a^{2} b^{2}-24 B \ln \left (b x +a \right ) x \,a^{3} b +6 A \ln \left (b x +a \right ) a^{3} b +12 A x \,a^{2} b^{2}-12 B \ln \left (b x +a \right ) a^{4}-24 B x \,a^{3} b +9 A \,a^{3} b -18 B \,a^{4}}{2 b^{5} \left (b x +a \right )^{2}}\) | \(162\) |
Input:
int(x^3*(B*x+A)/(b*x+a)^3,x,method=_RETURNVERBOSE)
Output:
1/b^4*(1/2*b*B*x^2+A*b*x-3*B*a*x)+1/2*a^3*(A*b-B*a)/b^5/(b*x+a)^2-a^2*(3*A *b-4*B*a)/b^5/(b*x+a)-3*a*(A*b-2*B*a)*ln(b*x+a)/b^5
Time = 0.08 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.82 \[ \int \frac {x^3 (A+B x)}{(a+b x)^3} \, dx=\frac {B b^{4} x^{4} + 7 \, B a^{4} - 5 \, A a^{3} b - 2 \, {\left (2 \, B a b^{3} - A b^{4}\right )} x^{3} - {\left (11 \, B a^{2} b^{2} - 4 \, A a b^{3}\right )} x^{2} + 2 \, {\left (B a^{3} b - 2 \, A a^{2} b^{2}\right )} x + 6 \, {\left (2 \, B a^{4} - A a^{3} b + {\left (2 \, B a^{2} b^{2} - A a b^{3}\right )} x^{2} + 2 \, {\left (2 \, B a^{3} b - A a^{2} b^{2}\right )} x\right )} \log \left (b x + a\right )}{2 \, {\left (b^{7} x^{2} + 2 \, a b^{6} x + a^{2} b^{5}\right )}} \] Input:
integrate(x^3*(B*x+A)/(b*x+a)^3,x, algorithm="fricas")
Output:
1/2*(B*b^4*x^4 + 7*B*a^4 - 5*A*a^3*b - 2*(2*B*a*b^3 - A*b^4)*x^3 - (11*B*a ^2*b^2 - 4*A*a*b^3)*x^2 + 2*(B*a^3*b - 2*A*a^2*b^2)*x + 6*(2*B*a^4 - A*a^3 *b + (2*B*a^2*b^2 - A*a*b^3)*x^2 + 2*(2*B*a^3*b - A*a^2*b^2)*x)*log(b*x + a))/(b^7*x^2 + 2*a*b^6*x + a^2*b^5)
Time = 0.50 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.14 \[ \int \frac {x^3 (A+B x)}{(a+b x)^3} \, dx=\frac {B x^{2}}{2 b^{3}} + \frac {3 a \left (- A b + 2 B a\right ) \log {\left (a + b x \right )}}{b^{5}} + x \left (\frac {A}{b^{3}} - \frac {3 B a}{b^{4}}\right ) + \frac {- 5 A a^{3} b + 7 B a^{4} + x \left (- 6 A a^{2} b^{2} + 8 B a^{3} b\right )}{2 a^{2} b^{5} + 4 a b^{6} x + 2 b^{7} x^{2}} \] Input:
integrate(x**3*(B*x+A)/(b*x+a)**3,x)
Output:
B*x**2/(2*b**3) + 3*a*(-A*b + 2*B*a)*log(a + b*x)/b**5 + x*(A/b**3 - 3*B*a /b**4) + (-5*A*a**3*b + 7*B*a**4 + x*(-6*A*a**2*b**2 + 8*B*a**3*b))/(2*a** 2*b**5 + 4*a*b**6*x + 2*b**7*x**2)
Time = 0.03 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.15 \[ \int \frac {x^3 (A+B x)}{(a+b x)^3} \, dx=\frac {7 \, B a^{4} - 5 \, A a^{3} b + 2 \, {\left (4 \, B a^{3} b - 3 \, A a^{2} b^{2}\right )} x}{2 \, {\left (b^{7} x^{2} + 2 \, a b^{6} x + a^{2} b^{5}\right )}} + \frac {B b x^{2} - 2 \, {\left (3 \, B a - A b\right )} x}{2 \, b^{4}} + \frac {3 \, {\left (2 \, B a^{2} - A a b\right )} \log \left (b x + a\right )}{b^{5}} \] Input:
integrate(x^3*(B*x+A)/(b*x+a)^3,x, algorithm="maxima")
Output:
1/2*(7*B*a^4 - 5*A*a^3*b + 2*(4*B*a^3*b - 3*A*a^2*b^2)*x)/(b^7*x^2 + 2*a*b ^6*x + a^2*b^5) + 1/2*(B*b*x^2 - 2*(3*B*a - A*b)*x)/b^4 + 3*(2*B*a^2 - A*a *b)*log(b*x + a)/b^5
Time = 0.13 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.06 \[ \int \frac {x^3 (A+B x)}{(a+b x)^3} \, dx=\frac {3 \, {\left (2 \, B a^{2} - A a b\right )} \log \left ({\left | b x + a \right |}\right )}{b^{5}} + \frac {B b^{3} x^{2} - 6 \, B a b^{2} x + 2 \, A b^{3} x}{2 \, b^{6}} + \frac {7 \, B a^{4} - 5 \, A a^{3} b + 2 \, {\left (4 \, B a^{3} b - 3 \, A a^{2} b^{2}\right )} x}{2 \, {\left (b x + a\right )}^{2} b^{5}} \] Input:
integrate(x^3*(B*x+A)/(b*x+a)^3,x, algorithm="giac")
Output:
3*(2*B*a^2 - A*a*b)*log(abs(b*x + a))/b^5 + 1/2*(B*b^3*x^2 - 6*B*a*b^2*x + 2*A*b^3*x)/b^6 + 1/2*(7*B*a^4 - 5*A*a^3*b + 2*(4*B*a^3*b - 3*A*a^2*b^2)*x )/((b*x + a)^2*b^5)
Time = 0.05 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.15 \[ \int \frac {x^3 (A+B x)}{(a+b x)^3} \, dx=\frac {x\,\left (4\,B\,a^3-3\,A\,a^2\,b\right )+\frac {7\,B\,a^4-5\,A\,a^3\,b}{2\,b}}{a^2\,b^4+2\,a\,b^5\,x+b^6\,x^2}+x\,\left (\frac {A}{b^3}-\frac {3\,B\,a}{b^4}\right )+\frac {B\,x^2}{2\,b^3}+\frac {\ln \left (a+b\,x\right )\,\left (6\,B\,a^2-3\,A\,a\,b\right )}{b^5} \] Input:
int((x^3*(A + B*x))/(a + b*x)^3,x)
Output:
(x*(4*B*a^3 - 3*A*a^2*b) + (7*B*a^4 - 5*A*a^3*b)/(2*b))/(a^2*b^4 + b^6*x^2 + 2*a*b^5*x) + x*(A/b^3 - (3*B*a)/b^4) + (B*x^2)/(2*b^3) + (log(a + b*x)* (6*B*a^2 - 3*A*a*b))/b^5
Time = 0.16 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.64 \[ \int \frac {x^3 (A+B x)}{(a+b x)^3} \, dx=\frac {6 \,\mathrm {log}\left (b x +a \right ) a^{3}+6 \,\mathrm {log}\left (b x +a \right ) a^{2} b x -6 a^{2} b x -3 a \,b^{2} x^{2}+b^{3} x^{3}}{2 b^{4} \left (b x +a \right )} \] Input:
int(x^3*(B*x+A)/(b*x+a)^3,x)
Output:
(6*log(a + b*x)*a**3 + 6*log(a + b*x)*a**2*b*x - 6*a**2*b*x - 3*a*b**2*x** 2 + b**3*x**3)/(2*b**4*(a + b*x))