Integrand size = 16, antiderivative size = 90 \[ \int \frac {x^3 (A+B x)}{(a+b x)^2} \, dx=-\frac {a (2 A b-3 a B) x}{b^4}+\frac {(A b-2 a B) x^2}{2 b^3}+\frac {B x^3}{3 b^2}+\frac {a^3 (A b-a B)}{b^5 (a+b x)}+\frac {a^2 (3 A b-4 a B) \log (a+b x)}{b^5} \] Output:
-a*(2*A*b-3*B*a)*x/b^4+1/2*(A*b-2*B*a)*x^2/b^3+1/3*B*x^3/b^2+a^3*(A*b-B*a) /b^5/(b*x+a)+a^2*(3*A*b-4*B*a)*ln(b*x+a)/b^5
Time = 0.04 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.97 \[ \int \frac {x^3 (A+B x)}{(a+b x)^2} \, dx=\frac {6 a b (-2 A b+3 a B) x+3 b^2 (A b-2 a B) x^2+2 b^3 B x^3+\frac {6 a^3 (A b-a B)}{a+b x}+6 a^2 (3 A b-4 a B) \log (a+b x)}{6 b^5} \] Input:
Integrate[(x^3*(A + B*x))/(a + b*x)^2,x]
Output:
(6*a*b*(-2*A*b + 3*a*B)*x + 3*b^2*(A*b - 2*a*B)*x^2 + 2*b^3*B*x^3 + (6*a^3 *(A*b - a*B))/(a + b*x) + 6*a^2*(3*A*b - 4*a*B)*Log[a + b*x])/(6*b^5)
Time = 0.24 (sec) , antiderivative size = 90, 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.
\(\displaystyle \int \frac {x^3 (A+B x)}{(a+b x)^2} \, dx\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \int \left (\frac {a^3 (a B-A b)}{b^4 (a+b x)^2}-\frac {a^2 (4 a B-3 A b)}{b^4 (a+b x)}+\frac {a (3 a B-2 A b)}{b^4}+\frac {x (A b-2 a B)}{b^3}+\frac {B x^2}{b^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {a^3 (A b-a B)}{b^5 (a+b x)}+\frac {a^2 (3 A b-4 a B) \log (a+b x)}{b^5}-\frac {a x (2 A b-3 a B)}{b^4}+\frac {x^2 (A b-2 a B)}{2 b^3}+\frac {B x^3}{3 b^2}\) |
Input:
Int[(x^3*(A + B*x))/(a + b*x)^2,x]
Output:
-((a*(2*A*b - 3*a*B)*x)/b^4) + ((A*b - 2*a*B)*x^2)/(2*b^3) + (B*x^3)/(3*b^ 2) + (a^3*(A*b - a*B))/(b^5*(a + b*x)) + (a^2*(3*A*b - 4*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 = 0.13 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.00
method | result | size |
default | \(-\frac {-\frac {1}{3} B \,b^{2} x^{3}-\frac {1}{2} A \,b^{2} x^{2}+B a b \,x^{2}+2 a A b x -3 B \,a^{2} x}{b^{4}}+\frac {a^{3} \left (A b -B a \right )}{b^{5} \left (b x +a \right )}+\frac {a^{2} \left (3 A b -4 B a \right ) \ln \left (b x +a \right )}{b^{5}}\) | \(90\) |
norman | \(\frac {\frac {a \left (3 a^{2} b A -4 a^{3} B \right )}{b^{5}}+\frac {B \,x^{4}}{3 b}+\frac {\left (3 A b -4 B a \right ) x^{3}}{6 b^{2}}-\frac {a \left (3 A b -4 B a \right ) x^{2}}{2 b^{3}}}{b x +a}+\frac {a^{2} \left (3 A b -4 B a \right ) \ln \left (b x +a \right )}{b^{5}}\) | \(96\) |
risch | \(\frac {B \,x^{3}}{3 b^{2}}+\frac {A \,x^{2}}{2 b^{2}}-\frac {B a \,x^{2}}{b^{3}}-\frac {2 a A x}{b^{3}}+\frac {3 B \,a^{2} x}{b^{4}}+\frac {a^{3} A}{b^{4} \left (b x +a \right )}-\frac {a^{4} B}{b^{5} \left (b x +a \right )}+\frac {3 a^{2} \ln \left (b x +a \right ) A}{b^{4}}-\frac {4 a^{3} \ln \left (b x +a \right ) B}{b^{5}}\) | \(109\) |
parallelrisch | \(\frac {2 B \,x^{4} b^{4}+3 A \,x^{3} b^{4}-4 B \,x^{3} a \,b^{3}+18 A \ln \left (b x +a \right ) x \,a^{2} b^{2}-9 A \,x^{2} a \,b^{3}-24 B \ln \left (b x +a \right ) x \,a^{3} b +12 B \,x^{2} a^{2} b^{2}+18 A \ln \left (b x +a \right ) a^{3} b -24 B \ln \left (b x +a \right ) a^{4}+18 A \,a^{3} b -24 B \,a^{4}}{6 b^{5} \left (b x +a \right )}\) | \(132\) |
Input:
int(x^3*(B*x+A)/(b*x+a)^2,x,method=_RETURNVERBOSE)
Output:
-1/b^4*(-1/3*B*b^2*x^3-1/2*A*b^2*x^2+B*a*b*x^2+2*a*A*b*x-3*B*a^2*x)+a^3*(A *b-B*a)/b^5/(b*x+a)+a^2*(3*A*b-4*B*a)*ln(b*x+a)/b^5
Time = 0.07 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.56 \[ \int \frac {x^3 (A+B x)}{(a+b x)^2} \, dx=\frac {2 \, B b^{4} x^{4} - 6 \, B a^{4} + 6 \, A a^{3} b - {\left (4 \, B a b^{3} - 3 \, A b^{4}\right )} x^{3} + 3 \, {\left (4 \, B a^{2} b^{2} - 3 \, A a b^{3}\right )} x^{2} + 6 \, {\left (3 \, B a^{3} b - 2 \, A a^{2} b^{2}\right )} x - 6 \, {\left (4 \, B a^{4} - 3 \, A a^{3} b + {\left (4 \, B a^{3} b - 3 \, A a^{2} b^{2}\right )} x\right )} \log \left (b x + a\right )}{6 \, {\left (b^{6} x + a b^{5}\right )}} \] Input:
integrate(x^3*(B*x+A)/(b*x+a)^2,x, algorithm="fricas")
Output:
1/6*(2*B*b^4*x^4 - 6*B*a^4 + 6*A*a^3*b - (4*B*a*b^3 - 3*A*b^4)*x^3 + 3*(4* B*a^2*b^2 - 3*A*a*b^3)*x^2 + 6*(3*B*a^3*b - 2*A*a^2*b^2)*x - 6*(4*B*a^4 - 3*A*a^3*b + (4*B*a^3*b - 3*A*a^2*b^2)*x)*log(b*x + a))/(b^6*x + a*b^5)
Time = 0.22 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.02 \[ \int \frac {x^3 (A+B x)}{(a+b x)^2} \, dx=\frac {B x^{3}}{3 b^{2}} - \frac {a^{2} \left (- 3 A b + 4 B a\right ) \log {\left (a + b x \right )}}{b^{5}} + x^{2} \left (\frac {A}{2 b^{2}} - \frac {B a}{b^{3}}\right ) + x \left (- \frac {2 A a}{b^{3}} + \frac {3 B a^{2}}{b^{4}}\right ) + \frac {A a^{3} b - B a^{4}}{a b^{5} + b^{6} x} \] Input:
integrate(x**3*(B*x+A)/(b*x+a)**2,x)
Output:
B*x**3/(3*b**2) - a**2*(-3*A*b + 4*B*a)*log(a + b*x)/b**5 + x**2*(A/(2*b** 2) - B*a/b**3) + x*(-2*A*a/b**3 + 3*B*a**2/b**4) + (A*a**3*b - B*a**4)/(a* b**5 + b**6*x)
Time = 0.03 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.12 \[ \int \frac {x^3 (A+B x)}{(a+b x)^2} \, dx=-\frac {B a^{4} - A a^{3} b}{b^{6} x + a b^{5}} + \frac {2 \, B b^{2} x^{3} - 3 \, {\left (2 \, B a b - A b^{2}\right )} x^{2} + 6 \, {\left (3 \, B a^{2} - 2 \, A a b\right )} x}{6 \, b^{4}} - \frac {{\left (4 \, B a^{3} - 3 \, A a^{2} b\right )} \log \left (b x + a\right )}{b^{5}} \] Input:
integrate(x^3*(B*x+A)/(b*x+a)^2,x, algorithm="maxima")
Output:
-(B*a^4 - A*a^3*b)/(b^6*x + a*b^5) + 1/6*(2*B*b^2*x^3 - 3*(2*B*a*b - A*b^2 )*x^2 + 6*(3*B*a^2 - 2*A*a*b)*x)/b^4 - (4*B*a^3 - 3*A*a^2*b)*log(b*x + a)/ b^5
Time = 0.11 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.60 \[ \int \frac {x^3 (A+B x)}{(a+b x)^2} \, dx=\frac {{\left (b x + a\right )}^{3} {\left (2 \, B - \frac {3 \, {\left (4 \, B a b - A b^{2}\right )}}{{\left (b x + a\right )} b} + \frac {18 \, {\left (2 \, B a^{2} b^{2} - A a b^{3}\right )}}{{\left (b x + a\right )}^{2} b^{2}}\right )}}{6 \, b^{5}} + \frac {{\left (4 \, B a^{3} - 3 \, A a^{2} b\right )} \log \left (\frac {{\left | b x + a \right |}}{{\left (b x + a\right )}^{2} {\left | b \right |}}\right )}{b^{5}} - \frac {\frac {B a^{4} b^{3}}{b x + a} - \frac {A a^{3} b^{4}}{b x + a}}{b^{8}} \] Input:
integrate(x^3*(B*x+A)/(b*x+a)^2,x, algorithm="giac")
Output:
1/6*(b*x + a)^3*(2*B - 3*(4*B*a*b - A*b^2)/((b*x + a)*b) + 18*(2*B*a^2*b^2 - A*a*b^3)/((b*x + a)^2*b^2))/b^5 + (4*B*a^3 - 3*A*a^2*b)*log(abs(b*x + a )/((b*x + a)^2*abs(b)))/b^5 - (B*a^4*b^3/(b*x + a) - A*a^3*b^4/(b*x + a))/ b^8
Time = 0.04 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.28 \[ \int \frac {x^3 (A+B x)}{(a+b x)^2} \, dx=x^2\,\left (\frac {A}{2\,b^2}-\frac {B\,a}{b^3}\right )-x\,\left (\frac {2\,a\,\left (\frac {A}{b^2}-\frac {2\,B\,a}{b^3}\right )}{b}+\frac {B\,a^2}{b^4}\right )-\frac {\ln \left (a+b\,x\right )\,\left (4\,B\,a^3-3\,A\,a^2\,b\right )}{b^5}+\frac {B\,x^3}{3\,b^2}-\frac {B\,a^4-A\,a^3\,b}{b\,\left (x\,b^5+a\,b^4\right )} \] Input:
int((x^3*(A + B*x))/(a + b*x)^2,x)
Output:
x^2*(A/(2*b^2) - (B*a)/b^3) - x*((2*a*(A/b^2 - (2*B*a)/b^3))/b + (B*a^2)/b ^4) - (log(a + b*x)*(4*B*a^3 - 3*A*a^2*b))/b^5 + (B*x^3)/(3*b^2) - (B*a^4 - A*a^3*b)/(b*(a*b^4 + b^5*x))
Time = 0.16 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.46 \[ \int \frac {x^3 (A+B x)}{(a+b x)^2} \, dx=\frac {-6 \,\mathrm {log}\left (b x +a \right ) a^{3}+6 a^{2} b x -3 a \,b^{2} x^{2}+2 b^{3} x^{3}}{6 b^{4}} \] Input:
int(x^3*(B*x+A)/(b*x+a)^2,x)
Output:
( - 6*log(a + b*x)*a**3 + 6*a**2*b*x - 3*a*b**2*x**2 + 2*b**3*x**3)/(6*b** 4)