Integrand size = 23, antiderivative size = 93 \[ \int \frac {A+B x+C x^2+D x^3}{(c+d x)^3} \, dx=\frac {D x}{d^3}-\frac {c^2 C d-B c d^2+A d^3-c^3 D}{2 d^4 (c+d x)^2}+\frac {2 c C d-B d^2-3 c^2 D}{d^4 (c+d x)}+\frac {(C d-3 c D) \log (c+d x)}{d^4} \] Output:
D*x/d^3-1/2*(A*d^3-B*c*d^2+C*c^2*d-D*c^3)/d^4/(d*x+c)^2+(-B*d^2+2*C*c*d-3* D*c^2)/d^4/(d*x+c)+(C*d-3*D*c)*ln(d*x+c)/d^4
Time = 0.02 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.00 \[ \int \frac {A+B x+C x^2+D x^3}{(c+d x)^3} \, dx=\frac {D x}{d^3}+\frac {-c^2 C d+B c d^2-A d^3+c^3 D}{2 d^4 (c+d x)^2}+\frac {2 c C d-B d^2-3 c^2 D}{d^4 (c+d x)}+\frac {(C d-3 c D) \log (c+d x)}{d^4} \] Input:
Integrate[(A + B*x + C*x^2 + D*x^3)/(c + d*x)^3,x]
Output:
(D*x)/d^3 + (-(c^2*C*d) + B*c*d^2 - A*d^3 + c^3*D)/(2*d^4*(c + d*x)^2) + ( 2*c*C*d - B*d^2 - 3*c^2*D)/(d^4*(c + d*x)) + ((C*d - 3*c*D)*Log[c + d*x])/ d^4
Time = 0.31 (sec) , antiderivative size = 93, 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.
\(\displaystyle \int \frac {A+B x+C x^2+D x^3}{(c+d x)^3} \, dx\) |
\(\Big \downarrow \) 2389 |
\(\displaystyle \int \left (\frac {A d^3-B c d^2+c^3 (-D)+c^2 C d}{d^3 (c+d x)^3}+\frac {B d^2+3 c^2 D-2 c C d}{d^3 (c+d x)^2}+\frac {C d-3 c D}{d^3 (c+d x)}+\frac {D}{d^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {A d^3-B c d^2+c^3 (-D)+c^2 C d}{2 d^4 (c+d x)^2}+\frac {-B d^2-3 c^2 D+2 c C d}{d^4 (c+d x)}+\frac {(C d-3 c D) \log (c+d x)}{d^4}+\frac {D x}{d^3}\) |
Input:
Int[(A + B*x + C*x^2 + D*x^3)/(c + d*x)^3,x]
Output:
(D*x)/d^3 - (c^2*C*d - B*c*d^2 + A*d^3 - c^3*D)/(2*d^4*(c + d*x)^2) + (2*c *C*d - B*d^2 - 3*c^2*D)/(d^4*(c + d*x)) + ((C*d - 3*c*D)*Log[c + d*x])/d^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 = 0.31 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.97
method | result | size |
norman | \(\frac {\frac {D x^{3}}{d}-\frac {A \,d^{3}+B c \,d^{2}-3 C \,c^{2} d +9 D c^{3}}{2 d^{4}}-\frac {\left (B \,d^{2}-2 C c d +6 D c^{2}\right ) x}{d^{3}}}{\left (x d +c \right )^{2}}+\frac {\left (C d -3 D c \right ) \ln \left (x d +c \right )}{d^{4}}\) | \(90\) |
default | \(\frac {D x}{d^{3}}-\frac {B \,d^{2}-2 C c d +3 D c^{2}}{d^{4} \left (x d +c \right )}+\frac {\left (C d -3 D c \right ) \ln \left (x d +c \right )}{d^{4}}-\frac {A \,d^{3}-B c \,d^{2}+C \,c^{2} d -D c^{3}}{2 d^{4} \left (x d +c \right )^{2}}\) | \(92\) |
parallelrisch | \(-\frac {-2 C \ln \left (x d +c \right ) x^{2} d^{3}+6 D \ln \left (x d +c \right ) x^{2} c \,d^{2}-2 D x^{3} d^{3}-4 C \ln \left (x d +c \right ) x c \,d^{2}+12 D \ln \left (x d +c \right ) x \,c^{2} d +2 B x \,d^{3}-2 C \ln \left (x d +c \right ) c^{2} d -4 C x c \,d^{2}+6 D \ln \left (x d +c \right ) c^{3}+12 D x \,c^{2} d +A \,d^{3}+B c \,d^{2}-3 C \,c^{2} d +9 D c^{3}}{2 d^{4} \left (x d +c \right )^{2}}\) | \(154\) |
Input:
int((D*x^3+C*x^2+B*x+A)/(d*x+c)^3,x,method=_RETURNVERBOSE)
Output:
(D*x^3/d-1/2*(A*d^3+B*c*d^2-3*C*c^2*d+9*D*c^3)/d^4-(B*d^2-2*C*c*d+6*D*c^2) /d^3*x)/(d*x+c)^2+(C*d-3*D*c)*ln(d*x+c)/d^4
Time = 0.07 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.65 \[ \int \frac {A+B x+C x^2+D x^3}{(c+d x)^3} \, dx=\frac {2 \, D d^{3} x^{3} + 4 \, D c d^{2} x^{2} - 5 \, D c^{3} + 3 \, C c^{2} d - B c d^{2} - A d^{3} - 2 \, {\left (2 \, D c^{2} d - 2 \, C c d^{2} + B d^{3}\right )} x - 2 \, {\left (3 \, D c^{3} - C c^{2} d + {\left (3 \, D c d^{2} - C d^{3}\right )} x^{2} + 2 \, {\left (3 \, D c^{2} d - C c d^{2}\right )} x\right )} \log \left (d x + c\right )}{2 \, {\left (d^{6} x^{2} + 2 \, c d^{5} x + c^{2} d^{4}\right )}} \] Input:
integrate((D*x^3+C*x^2+B*x+A)/(d*x+c)^3,x, algorithm="fricas")
Output:
1/2*(2*D*d^3*x^3 + 4*D*c*d^2*x^2 - 5*D*c^3 + 3*C*c^2*d - B*c*d^2 - A*d^3 - 2*(2*D*c^2*d - 2*C*c*d^2 + B*d^3)*x - 2*(3*D*c^3 - C*c^2*d + (3*D*c*d^2 - C*d^3)*x^2 + 2*(3*D*c^2*d - C*c*d^2)*x)*log(d*x + c))/(d^6*x^2 + 2*c*d^5* x + c^2*d^4)
Time = 0.79 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.10 \[ \int \frac {A+B x+C x^2+D x^3}{(c+d x)^3} \, dx=\frac {D x}{d^{3}} + \frac {- A d^{3} - B c d^{2} + 3 C c^{2} d - 5 D c^{3} + x \left (- 2 B d^{3} + 4 C c d^{2} - 6 D c^{2} d\right )}{2 c^{2} d^{4} + 4 c d^{5} x + 2 d^{6} x^{2}} - \frac {\left (- C d + 3 D c\right ) \log {\left (c + d x \right )}}{d^{4}} \] Input:
integrate((D*x**3+C*x**2+B*x+A)/(d*x+c)**3,x)
Output:
D*x/d**3 + (-A*d**3 - B*c*d**2 + 3*C*c**2*d - 5*D*c**3 + x*(-2*B*d**3 + 4* C*c*d**2 - 6*D*c**2*d))/(2*c**2*d**4 + 4*c*d**5*x + 2*d**6*x**2) - (-C*d + 3*D*c)*log(c + d*x)/d**4
Time = 0.03 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.09 \[ \int \frac {A+B x+C x^2+D x^3}{(c+d x)^3} \, dx=-\frac {5 \, D c^{3} - 3 \, C c^{2} d + B c d^{2} + A d^{3} + 2 \, {\left (3 \, D c^{2} d - 2 \, C c d^{2} + B d^{3}\right )} x}{2 \, {\left (d^{6} x^{2} + 2 \, c d^{5} x + c^{2} d^{4}\right )}} + \frac {D x}{d^{3}} - \frac {{\left (3 \, D c - C d\right )} \log \left (d x + c\right )}{d^{4}} \] Input:
integrate((D*x^3+C*x^2+B*x+A)/(d*x+c)^3,x, algorithm="maxima")
Output:
-1/2*(5*D*c^3 - 3*C*c^2*d + B*c*d^2 + A*d^3 + 2*(3*D*c^2*d - 2*C*c*d^2 + B *d^3)*x)/(d^6*x^2 + 2*c*d^5*x + c^2*d^4) + D*x/d^3 - (3*D*c - C*d)*log(d*x + c)/d^4
Time = 0.12 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.95 \[ \int \frac {A+B x+C x^2+D x^3}{(c+d x)^3} \, dx=\frac {D x}{d^{3}} - \frac {{\left (3 \, D c - C d\right )} \log \left ({\left | d x + c \right |}\right )}{d^{4}} - \frac {5 \, D c^{3} - 3 \, C c^{2} d + B c d^{2} + A d^{3} + 2 \, {\left (3 \, D c^{2} d - 2 \, C c d^{2} + B d^{3}\right )} x}{2 \, {\left (d x + c\right )}^{2} d^{4}} \] Input:
integrate((D*x^3+C*x^2+B*x+A)/(d*x+c)^3,x, algorithm="giac")
Output:
D*x/d^3 - (3*D*c - C*d)*log(abs(d*x + c))/d^4 - 1/2*(5*D*c^3 - 3*C*c^2*d + B*c*d^2 + A*d^3 + 2*(3*D*c^2*d - 2*C*c*d^2 + B*d^3)*x)/((d*x + c)^2*d^4)
Time = 3.28 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.62 \[ \int \frac {A+B x+C x^2+D x^3}{(c+d x)^3} \, dx=\frac {\frac {3\,C\,c^2}{2\,d^3}+\frac {2\,C\,c\,x}{d^2}}{c^2+2\,c\,d\,x+d^2\,x^2}-\frac {\frac {B\,c}{2\,d^2}+\frac {B\,x}{d}}{c^2+2\,c\,d\,x+d^2\,x^2}-\frac {A}{2\,d\,\left (c^2+2\,c\,d\,x+d^2\,x^2\right )}-\frac {D\,\left (3\,c\,\ln \left (c+d\,x\right )-d\,x+\frac {3\,c^2}{c+d\,x}-\frac {c^3}{2\,{\left (c+d\,x\right )}^2}\right )}{d^4}+\frac {C\,\ln \left (c+d\,x\right )}{d^3} \] Input:
int((A + B*x + C*x^2 + x^3*D)/(c + d*x)^3,x)
Output:
((3*C*c^2)/(2*d^3) + (2*C*c*x)/d^2)/(c^2 + d^2*x^2 + 2*c*d*x) - ((B*c)/(2* d^2) + (B*x)/d)/(c^2 + d^2*x^2 + 2*c*d*x) - A/(2*d*(c^2 + d^2*x^2 + 2*c*d* x)) - (D*(3*c*log(c + d*x) - d*x + (3*c^2)/(c + d*x) - c^3/(2*(c + d*x)^2) ))/d^4 + (C*log(c + d*x))/d^3
Time = 0.14 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.16 \[ \int \frac {A+B x+C x^2+D x^3}{(c+d x)^3} \, dx=\frac {-4 \,\mathrm {log}\left (d x +c \right ) c^{4}-8 \,\mathrm {log}\left (d x +c \right ) c^{3} d x -4 \,\mathrm {log}\left (d x +c \right ) c^{2} d^{2} x^{2}-a c \,d^{2}+b \,d^{3} x^{2}-2 c^{4}+4 c^{2} d^{2} x^{2}+2 c \,d^{3} x^{3}}{2 c \,d^{3} \left (d^{2} x^{2}+2 c d x +c^{2}\right )} \] Input:
int((D*x^3+C*x^2+B*x+A)/(d*x+c)^3,x)
Output:
( - 4*log(c + d*x)*c**4 - 8*log(c + d*x)*c**3*d*x - 4*log(c + d*x)*c**2*d* *2*x**2 - a*c*d**2 + b*d**3*x**2 - 2*c**4 + 4*c**2*d**2*x**2 + 2*c*d**3*x* *3)/(2*c*d**3*(c**2 + 2*c*d*x + d**2*x**2))