Integrand size = 23, antiderivative size = 89 \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^2} \, dx=\frac {(b C-2 a D) x}{b^3}+\frac {D x^2}{2 b^2}-\frac {A b^3-a \left (b^2 B-a b C+a^2 D\right )}{b^4 (a+b x)}+\frac {\left (b^2 B-2 a b C+3 a^2 D\right ) \log (a+b x)}{b^4} \] Output:
(C*b-2*D*a)*x/b^3+1/2*D*x^2/b^2-(A*b^3-a*(B*b^2-C*a*b+D*a^2))/b^4/(b*x+a)+ (B*b^2-2*C*a*b+3*D*a^2)*ln(b*x+a)/b^4
Time = 0.03 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.99 \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^2} \, dx=\frac {(b C-2 a D) x}{b^3}+\frac {D x^2}{2 b^2}+\frac {-A b^3+a b^2 B-a^2 b C+a^3 D}{b^4 (a+b x)}+\frac {\left (b^2 B-2 a b C+3 a^2 D\right ) \log (a+b x)}{b^4} \] Input:
Integrate[(A + B*x + C*x^2 + D*x^3)/(a + b*x)^2,x]
Output:
((b*C - 2*a*D)*x)/b^3 + (D*x^2)/(2*b^2) + (-(A*b^3) + a*b^2*B - a^2*b*C + a^3*D)/(b^4*(a + b*x)) + ((b^2*B - 2*a*b*C + 3*a^2*D)*Log[a + b*x])/b^4
Time = 0.29 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {2389, 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.
\(\displaystyle \int \frac {A+B x+C x^2+D x^3}{(a+b x)^2} \, dx\) |
\(\Big \downarrow \) 2389 |
\(\displaystyle \int \left (\frac {A b^3-a \left (a^2 D-a b C+b^2 B\right )}{b^3 (a+b x)^2}+\frac {3 a^2 D-2 a b C+b^2 B}{b^3 (a+b x)}+\frac {b C-2 a D}{b^3}+\frac {D x}{b^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {A b^3-a \left (a^2 D-a b C+b^2 B\right )}{b^4 (a+b x)}+\frac {\log (a+b x) \left (3 a^2 D-2 a b C+b^2 B\right )}{b^4}+\frac {x (b C-2 a D)}{b^3}+\frac {D x^2}{2 b^2}\) |
Input:
Int[(A + B*x + C*x^2 + D*x^3)/(a + b*x)^2,x]
Output:
((b*C - 2*a*D)*x)/b^3 + (D*x^2)/(2*b^2) - (A*b^3 - a*(b^2*B - a*b*C + a^2* D))/(b^4*(a + b*x)) + ((b^2*B - 2*a*b*C + 3*a^2*D)*Log[a + b*x])/b^4
Int[(Pq_)*((a_) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[ExpandIntegrand [Pq*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, n}, x] && PolyQ[Pq, x] && (IGtQ[p , 0] || EqQ[n, 1])
Time = 0.30 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.98
method | result | size |
default | \(\frac {\frac {1}{2} D x^{2} b +C b x -2 D a x}{b^{3}}-\frac {b^{3} A -a \,b^{2} B +a^{2} b C -a^{3} D}{b^{4} \left (b x +a \right )}+\frac {\left (B \,b^{2}-2 C a b +3 D a^{2}\right ) \ln \left (b x +a \right )}{b^{4}}\) | \(87\) |
norman | \(\frac {-\frac {b^{3} A -a \,b^{2} B +2 a^{2} b C -3 a^{3} D}{b^{4}}+\frac {D x^{3}}{2 b}+\frac {\left (2 C b -3 D a \right ) x^{2}}{2 b^{2}}}{b x +a}+\frac {\left (B \,b^{2}-2 C a b +3 D a^{2}\right ) \ln \left (b x +a \right )}{b^{4}}\) | \(95\) |
parallelrisch | \(-\frac {-D x^{3} b^{3}-2 B \ln \left (b x +a \right ) x \,b^{3}+4 C \ln \left (b x +a \right ) x a \,b^{2}-2 C \,x^{2} b^{3}-6 D \ln \left (b x +a \right ) x \,a^{2} b +3 D x^{2} a \,b^{2}-2 B \ln \left (b x +a \right ) a \,b^{2}+4 C \ln \left (b x +a \right ) a^{2} b -6 D \ln \left (b x +a \right ) a^{3}+2 b^{3} A -2 a \,b^{2} B +4 a^{2} b C -6 a^{3} D}{2 b^{4} \left (b x +a \right )}\) | \(147\) |
Input:
int((D*x^3+C*x^2+B*x+A)/(b*x+a)^2,x,method=_RETURNVERBOSE)
Output:
1/b^3*(1/2*D*x^2*b+C*b*x-2*D*a*x)-(A*b^3-B*a*b^2+C*a^2*b-D*a^3)/b^4/(b*x+a )+(B*b^2-2*C*a*b+3*D*a^2)*ln(b*x+a)/b^4
Time = 0.07 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.54 \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^2} \, dx=\frac {D b^{3} x^{3} + 2 \, D a^{3} - 2 \, C a^{2} b + 2 \, B a b^{2} - 2 \, A b^{3} - {\left (3 \, D a b^{2} - 2 \, C b^{3}\right )} x^{2} - 2 \, {\left (2 \, D a^{2} b - C a b^{2}\right )} x + 2 \, {\left (3 \, D a^{3} - 2 \, C a^{2} b + B a b^{2} + {\left (3 \, D a^{2} b - 2 \, C a b^{2} + B b^{3}\right )} x\right )} \log \left (b x + a\right )}{2 \, {\left (b^{5} x + a b^{4}\right )}} \] Input:
integrate((D*x^3+C*x^2+B*x+A)/(b*x+a)^2,x, algorithm="fricas")
Output:
1/2*(D*b^3*x^3 + 2*D*a^3 - 2*C*a^2*b + 2*B*a*b^2 - 2*A*b^3 - (3*D*a*b^2 - 2*C*b^3)*x^2 - 2*(2*D*a^2*b - C*a*b^2)*x + 2*(3*D*a^3 - 2*C*a^2*b + B*a*b^ 2 + (3*D*a^2*b - 2*C*a*b^2 + B*b^3)*x)*log(b*x + a))/(b^5*x + a*b^4)
Time = 0.25 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.98 \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^2} \, dx=\frac {D x^{2}}{2 b^{2}} + x \left (\frac {C}{b^{2}} - \frac {2 D a}{b^{3}}\right ) + \frac {- A b^{3} + B a b^{2} - C a^{2} b + D a^{3}}{a b^{4} + b^{5} x} + \frac {\left (B b^{2} - 2 C a b + 3 D a^{2}\right ) \log {\left (a + b x \right )}}{b^{4}} \] Input:
integrate((D*x**3+C*x**2+B*x+A)/(b*x+a)**2,x)
Output:
D*x**2/(2*b**2) + x*(C/b**2 - 2*D*a/b**3) + (-A*b**3 + B*a*b**2 - C*a**2*b + D*a**3)/(a*b**4 + b**5*x) + (B*b**2 - 2*C*a*b + 3*D*a**2)*log(a + b*x)/ b**4
Time = 0.03 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.02 \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^2} \, dx=\frac {D a^{3} - C a^{2} b + B a b^{2} - A b^{3}}{b^{5} x + a b^{4}} + \frac {D b x^{2} - 2 \, {\left (2 \, D a - C b\right )} x}{2 \, b^{3}} + \frac {{\left (3 \, D a^{2} - 2 \, C a b + B b^{2}\right )} \log \left (b x + a\right )}{b^{4}} \] Input:
integrate((D*x^3+C*x^2+B*x+A)/(b*x+a)^2,x, algorithm="maxima")
Output:
(D*a^3 - C*a^2*b + B*a*b^2 - A*b^3)/(b^5*x + a*b^4) + 1/2*(D*b*x^2 - 2*(2* D*a - C*b)*x)/b^3 + (3*D*a^2 - 2*C*a*b + B*b^2)*log(b*x + a)/b^4
Time = 0.12 (sec) , antiderivative size = 178, normalized size of antiderivative = 2.00 \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^2} \, dx=-\frac {1}{2} \, D {\left (\frac {{\left (b x + a\right )}^{2} {\left (\frac {6 \, a}{b x + a} - 1\right )}}{b^{4}} + \frac {6 \, a^{2} \log \left (\frac {{\left | b x + a \right |}}{{\left (b x + a\right )}^{2} {\left | b \right |}}\right )}{b^{4}} - \frac {2 \, a^{3}}{{\left (b x + a\right )} b^{4}}\right )} + C {\left (\frac {2 \, a \log \left (\frac {{\left | b x + a \right |}}{{\left (b x + a\right )}^{2} {\left | b \right |}}\right )}{b^{3}} + \frac {b x + a}{b^{3}} - \frac {a^{2}}{{\left (b x + a\right )} b^{3}}\right )} - \frac {B {\left (\frac {\log \left (\frac {{\left | b x + a \right |}}{{\left (b x + a\right )}^{2} {\left | b \right |}}\right )}{b} - \frac {a}{{\left (b x + a\right )} b}\right )}}{b} - \frac {A}{{\left (b x + a\right )} b} \] Input:
integrate((D*x^3+C*x^2+B*x+A)/(b*x+a)^2,x, algorithm="giac")
Output:
-1/2*D*((b*x + a)^2*(6*a/(b*x + a) - 1)/b^4 + 6*a^2*log(abs(b*x + a)/((b*x + a)^2*abs(b)))/b^4 - 2*a^3/((b*x + a)*b^4)) + C*(2*a*log(abs(b*x + a)/(( b*x + a)^2*abs(b)))/b^3 + (b*x + a)/b^3 - a^2/((b*x + a)*b^3)) - B*(log(ab s(b*x + a)/((b*x + a)^2*abs(b)))/b - a/((b*x + a)*b))/b - A/((b*x + a)*b)
Timed out. \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^2} \, dx=\int \frac {A+B\,x+C\,x^2+x^3\,D}{{\left (a+b\,x\right )}^2} \,d x \] Input:
int((A + B*x + C*x^2 + x^3*D)/(a + b*x)^2,x)
Output:
int((A + B*x + C*x^2 + x^3*D)/(a + b*x)^2, x)
Time = 0.14 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.49 \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^2} \, dx=\frac {6 \,\mathrm {log}\left (b x +a \right ) a^{3} d -4 \,\mathrm {log}\left (b x +a \right ) a^{2} b c +6 \,\mathrm {log}\left (b x +a \right ) a^{2} b d x +2 \,\mathrm {log}\left (b x +a \right ) a \,b^{3}-4 \,\mathrm {log}\left (b x +a \right ) a \,b^{2} c x +2 \,\mathrm {log}\left (b x +a \right ) b^{4} x -6 a^{2} b d x +4 a \,b^{2} c x -3 a \,b^{2} d \,x^{2}+2 b^{3} c \,x^{2}+b^{3} d \,x^{3}}{2 b^{4} \left (b x +a \right )} \] Input:
int((D*x^3+C*x^2+B*x+A)/(b*x+a)^2,x)
Output:
(6*log(a + b*x)*a**3*d - 4*log(a + b*x)*a**2*b*c + 6*log(a + b*x)*a**2*b*d *x + 2*log(a + b*x)*a*b**3 - 4*log(a + b*x)*a*b**2*c*x + 2*log(a + b*x)*b* *4*x - 6*a**2*b*d*x + 4*a*b**2*c*x - 3*a*b**2*d*x**2 + 2*b**3*c*x**2 + b** 3*d*x**3)/(2*b**4*(a + b*x))