Integrand size = 23, antiderivative size = 119 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )}{(d+e x)^3} \, dx=\frac {B c x}{e^3}+\frac {(B d-A e) \left (c d^2-b d e+a e^2\right )}{2 e^4 (d+e x)^2}-\frac {3 B c d^2-B e (2 b d-a e)-A e (2 c d-b e)}{e^4 (d+e x)}-\frac {(3 B c d-b B e-A c e) \log (d+e x)}{e^4} \] Output:
B*c*x/e^3+1/2*(-A*e+B*d)*(a*e^2-b*d*e+c*d^2)/e^4/(e*x+d)^2-(3*B*c*d^2-B*e* (-a*e+2*b*d)-A*e*(-b*e+2*c*d))/e^4/(e*x+d)-(-A*c*e-B*b*e+3*B*c*d)*ln(e*x+d )/e^4
Time = 0.07 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.13 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )}{(d+e x)^3} \, dx=\frac {B c x}{e^3}+\frac {B c d^3-b B d^2 e-A c d^2 e+A b d e^2+a B d e^2-a A e^3}{2 e^4 (d+e x)^2}+\frac {-3 B c d^2+2 b B d e+2 A c d e-A b e^2-a B e^2}{e^4 (d+e x)}+\frac {(-3 B c d+b B e+A c e) \log (d+e x)}{e^4} \] Input:
Integrate[((A + B*x)*(a + b*x + c*x^2))/(d + e*x)^3,x]
Output:
(B*c*x)/e^3 + (B*c*d^3 - b*B*d^2*e - A*c*d^2*e + A*b*d*e^2 + a*B*d*e^2 - a *A*e^3)/(2*e^4*(d + e*x)^2) + (-3*B*c*d^2 + 2*b*B*d*e + 2*A*c*d*e - A*b*e^ 2 - a*B*e^2)/(e^4*(d + e*x)) + ((-3*B*c*d + b*B*e + A*c*e)*Log[d + e*x])/e ^4
Time = 0.51 (sec) , antiderivative size = 119, 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 = {1195, 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.
\(\displaystyle \int \frac {(A+B x) \left (a+b x+c x^2\right )}{(d+e x)^3} \, dx\) |
\(\Big \downarrow \) 1195 |
\(\displaystyle \int \left (\frac {-B e (2 b d-a e)-A e (2 c d-b e)+3 B c d^2}{e^3 (d+e x)^2}+\frac {(A e-B d) \left (a e^2-b d e+c d^2\right )}{e^3 (d+e x)^3}+\frac {A c e+b B e-3 B c d}{e^3 (d+e x)}+\frac {B c}{e^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {-B e (2 b d-a e)-A e (2 c d-b e)+3 B c d^2}{e^4 (d+e x)}+\frac {(B d-A e) \left (a e^2-b d e+c d^2\right )}{2 e^4 (d+e x)^2}-\frac {\log (d+e x) (-A c e-b B e+3 B c d)}{e^4}+\frac {B c x}{e^3}\) |
Input:
Int[((A + B*x)*(a + b*x + c*x^2))/(d + e*x)^3,x]
Output:
(B*c*x)/e^3 + ((B*d - A*e)*(c*d^2 - b*d*e + a*e^2))/(2*e^4*(d + e*x)^2) - (3*B*c*d^2 - B*e*(2*b*d - a*e) - A*e*(2*c*d - b*e))/(e^4*(d + e*x)) - ((3* B*c*d - b*B*e - A*c*e)*Log[d + e*x])/e^4
Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_.) + (b_.)*(x _) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n}, x ] && IGtQ[p, 0]
Time = 0.85 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.10
method | result | size |
norman | \(\frac {\frac {B c \,x^{3}}{e}-\frac {A a \,e^{3}+A b d \,e^{2}-3 A c \,d^{2} e +B a d \,e^{2}-3 B b \,d^{2} e +9 B c \,d^{3}}{2 e^{4}}-\frac {\left (A b \,e^{2}-2 A c d e +B a \,e^{2}-2 B b d e +6 B c \,d^{2}\right ) x}{e^{3}}}{\left (e x +d \right )^{2}}+\frac {\left (A c e +B b e -3 B c d \right ) \ln \left (e x +d \right )}{e^{4}}\) | \(131\) |
default | \(\frac {B c x}{e^{3}}+\frac {\left (A c e +B b e -3 B c d \right ) \ln \left (e x +d \right )}{e^{4}}-\frac {A b \,e^{2}-2 A c d e +B a \,e^{2}-2 B b d e +3 B c \,d^{2}}{e^{4} \left (e x +d \right )}-\frac {A a \,e^{3}-A b d \,e^{2}+A c \,d^{2} e -B a d \,e^{2}+B b \,d^{2} e -B c \,d^{3}}{2 e^{4} \left (e x +d \right )^{2}}\) | \(133\) |
risch | \(\frac {B c x}{e^{3}}+\frac {\left (-A b \,e^{2}+2 A c d e -B a \,e^{2}+2 B b d e -3 B c \,d^{2}\right ) x -\frac {A a \,e^{3}+A b d \,e^{2}-3 A c \,d^{2} e +B a d \,e^{2}-3 B b \,d^{2} e +5 B c \,d^{3}}{2 e}}{e^{3} \left (e x +d \right )^{2}}+\frac {\ln \left (e x +d \right ) A c}{e^{3}}+\frac {\ln \left (e x +d \right ) B b}{e^{3}}-\frac {3 \ln \left (e x +d \right ) B c d}{e^{4}}\) | \(144\) |
parallelrisch | \(\frac {2 A \ln \left (e x +d \right ) x^{2} c \,e^{3}+2 B \ln \left (e x +d \right ) x^{2} b \,e^{3}-6 B \ln \left (e x +d \right ) x^{2} c d \,e^{2}+2 B c \,x^{3} e^{3}+4 A \ln \left (e x +d \right ) x c d \,e^{2}+4 B \ln \left (e x +d \right ) x b d \,e^{2}-12 B \ln \left (e x +d \right ) x c \,d^{2} e +2 A \ln \left (e x +d \right ) c \,d^{2} e -2 A x b \,e^{3}+4 A x c d \,e^{2}+2 B \ln \left (e x +d \right ) b \,d^{2} e -6 B \ln \left (e x +d \right ) c \,d^{3}-2 B x a \,e^{3}+4 B x b d \,e^{2}-12 B x c \,d^{2} e -A a \,e^{3}-A b d \,e^{2}+3 A c \,d^{2} e -B a d \,e^{2}+3 B b \,d^{2} e -9 B c \,d^{3}}{2 e^{4} \left (e x +d \right )^{2}}\) | \(248\) |
Input:
int((B*x+A)*(c*x^2+b*x+a)/(e*x+d)^3,x,method=_RETURNVERBOSE)
Output:
(B*c*x^3/e-1/2*(A*a*e^3+A*b*d*e^2-3*A*c*d^2*e+B*a*d*e^2-3*B*b*d^2*e+9*B*c* d^3)/e^4-(A*b*e^2-2*A*c*d*e+B*a*e^2-2*B*b*d*e+6*B*c*d^2)/e^3*x)/(e*x+d)^2+ 1/e^4*(A*c*e+B*b*e-3*B*c*d)*ln(e*x+d)
Time = 0.08 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.71 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )}{(d+e x)^3} \, dx=\frac {2 \, B c e^{3} x^{3} + 4 \, B c d e^{2} x^{2} - 5 \, B c d^{3} - A a e^{3} + 3 \, {\left (B b + A c\right )} d^{2} e - {\left (B a + A b\right )} d e^{2} - 2 \, {\left (2 \, B c d^{2} e - 2 \, {\left (B b + A c\right )} d e^{2} + {\left (B a + A b\right )} e^{3}\right )} x - 2 \, {\left (3 \, B c d^{3} - {\left (B b + A c\right )} d^{2} e + {\left (3 \, B c d e^{2} - {\left (B b + A c\right )} e^{3}\right )} x^{2} + 2 \, {\left (3 \, B c d^{2} e - {\left (B b + A c\right )} d e^{2}\right )} x\right )} \log \left (e x + d\right )}{2 \, {\left (e^{6} x^{2} + 2 \, d e^{5} x + d^{2} e^{4}\right )}} \] Input:
integrate((B*x+A)*(c*x^2+b*x+a)/(e*x+d)^3,x, algorithm="fricas")
Output:
1/2*(2*B*c*e^3*x^3 + 4*B*c*d*e^2*x^2 - 5*B*c*d^3 - A*a*e^3 + 3*(B*b + A*c) *d^2*e - (B*a + A*b)*d*e^2 - 2*(2*B*c*d^2*e - 2*(B*b + A*c)*d*e^2 + (B*a + A*b)*e^3)*x - 2*(3*B*c*d^3 - (B*b + A*c)*d^2*e + (3*B*c*d*e^2 - (B*b + A* c)*e^3)*x^2 + 2*(3*B*c*d^2*e - (B*b + A*c)*d*e^2)*x)*log(e*x + d))/(e^6*x^ 2 + 2*d*e^5*x + d^2*e^4)
Time = 1.85 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.36 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )}{(d+e x)^3} \, dx=\frac {B c x}{e^{3}} + \frac {- A a e^{3} - A b d e^{2} + 3 A c d^{2} e - B a d e^{2} + 3 B b d^{2} e - 5 B c d^{3} + x \left (- 2 A b e^{3} + 4 A c d e^{2} - 2 B a e^{3} + 4 B b d e^{2} - 6 B c d^{2} e\right )}{2 d^{2} e^{4} + 4 d e^{5} x + 2 e^{6} x^{2}} + \frac {\left (A c e + B b e - 3 B c d\right ) \log {\left (d + e x \right )}}{e^{4}} \] Input:
integrate((B*x+A)*(c*x**2+b*x+a)/(e*x+d)**3,x)
Output:
B*c*x/e**3 + (-A*a*e**3 - A*b*d*e**2 + 3*A*c*d**2*e - B*a*d*e**2 + 3*B*b*d **2*e - 5*B*c*d**3 + x*(-2*A*b*e**3 + 4*A*c*d*e**2 - 2*B*a*e**3 + 4*B*b*d* e**2 - 6*B*c*d**2*e))/(2*d**2*e**4 + 4*d*e**5*x + 2*e**6*x**2) + (A*c*e + B*b*e - 3*B*c*d)*log(d + e*x)/e**4
Time = 0.04 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.14 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )}{(d+e x)^3} \, dx=-\frac {5 \, B c d^{3} + A a e^{3} - 3 \, {\left (B b + A c\right )} d^{2} e + {\left (B a + A b\right )} d e^{2} + 2 \, {\left (3 \, B c d^{2} e - 2 \, {\left (B b + A c\right )} d e^{2} + {\left (B a + A b\right )} e^{3}\right )} x}{2 \, {\left (e^{6} x^{2} + 2 \, d e^{5} x + d^{2} e^{4}\right )}} + \frac {B c x}{e^{3}} - \frac {{\left (3 \, B c d - {\left (B b + A c\right )} e\right )} \log \left (e x + d\right )}{e^{4}} \] Input:
integrate((B*x+A)*(c*x^2+b*x+a)/(e*x+d)^3,x, algorithm="maxima")
Output:
-1/2*(5*B*c*d^3 + A*a*e^3 - 3*(B*b + A*c)*d^2*e + (B*a + A*b)*d*e^2 + 2*(3 *B*c*d^2*e - 2*(B*b + A*c)*d*e^2 + (B*a + A*b)*e^3)*x)/(e^6*x^2 + 2*d*e^5* x + d^2*e^4) + B*c*x/e^3 - (3*B*c*d - (B*b + A*c)*e)*log(e*x + d)/e^4
Time = 0.15 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.11 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )}{(d+e x)^3} \, dx=\frac {B c x}{e^{3}} - \frac {{\left (3 \, B c d - B b e - A c e\right )} \log \left ({\left | e x + d \right |}\right )}{e^{4}} - \frac {5 \, B c d^{3} - 3 \, B b d^{2} e - 3 \, A c d^{2} e + B a d e^{2} + A b d e^{2} + A a e^{3} + 2 \, {\left (3 \, B c d^{2} e - 2 \, B b d e^{2} - 2 \, A c d e^{2} + B a e^{3} + A b e^{3}\right )} x}{2 \, {\left (e x + d\right )}^{2} e^{4}} \] Input:
integrate((B*x+A)*(c*x^2+b*x+a)/(e*x+d)^3,x, algorithm="giac")
Output:
B*c*x/e^3 - (3*B*c*d - B*b*e - A*c*e)*log(abs(e*x + d))/e^4 - 1/2*(5*B*c*d ^3 - 3*B*b*d^2*e - 3*A*c*d^2*e + B*a*d*e^2 + A*b*d*e^2 + A*a*e^3 + 2*(3*B* c*d^2*e - 2*B*b*d*e^2 - 2*A*c*d*e^2 + B*a*e^3 + A*b*e^3)*x)/((e*x + d)^2*e ^4)
Time = 12.29 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.19 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )}{(d+e x)^3} \, dx=\frac {\ln \left (d+e\,x\right )\,\left (A\,c\,e+B\,b\,e-3\,B\,c\,d\right )}{e^4}-\frac {x\,\left (A\,b\,e^2+B\,a\,e^2+3\,B\,c\,d^2-2\,A\,c\,d\,e-2\,B\,b\,d\,e\right )+\frac {A\,a\,e^3+5\,B\,c\,d^3+A\,b\,d\,e^2+B\,a\,d\,e^2-3\,A\,c\,d^2\,e-3\,B\,b\,d^2\,e}{2\,e}}{d^2\,e^3+2\,d\,e^4\,x+e^5\,x^2}+\frac {B\,c\,x}{e^3} \] Input:
int(((A + B*x)*(a + b*x + c*x^2))/(d + e*x)^3,x)
Output:
(log(d + e*x)*(A*c*e + B*b*e - 3*B*c*d))/e^4 - (x*(A*b*e^2 + B*a*e^2 + 3*B *c*d^2 - 2*A*c*d*e - 2*B*b*d*e) + (A*a*e^3 + 5*B*c*d^3 + A*b*d*e^2 + B*a*d *e^2 - 3*A*c*d^2*e - 3*B*b*d^2*e)/(2*e))/(d^2*e^3 + e^5*x^2 + 2*d*e^4*x) + (B*c*x)/e^3
Time = 0.21 (sec) , antiderivative size = 261, normalized size of antiderivative = 2.19 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )}{(d+e x)^3} \, dx=\frac {2 \,\mathrm {log}\left (e x +d \right ) a c \,d^{3} e +4 \,\mathrm {log}\left (e x +d \right ) a c \,d^{2} e^{2} x +2 \,\mathrm {log}\left (e x +d \right ) a c d \,e^{3} x^{2}+2 \,\mathrm {log}\left (e x +d \right ) b^{2} d^{3} e +4 \,\mathrm {log}\left (e x +d \right ) b^{2} d^{2} e^{2} x +2 \,\mathrm {log}\left (e x +d \right ) b^{2} d \,e^{3} x^{2}-6 \,\mathrm {log}\left (e x +d \right ) b c \,d^{4}-12 \,\mathrm {log}\left (e x +d \right ) b c \,d^{3} e x -6 \,\mathrm {log}\left (e x +d \right ) b c \,d^{2} e^{2} x^{2}-a^{2} d \,e^{3}+2 a b \,e^{4} x^{2}+a c \,d^{3} e -2 a c d \,e^{3} x^{2}+b^{2} d^{3} e -2 b^{2} d \,e^{3} x^{2}-3 b c \,d^{4}+6 b c \,d^{2} e^{2} x^{2}+2 b c d \,e^{3} x^{3}}{2 d \,e^{4} \left (e^{2} x^{2}+2 d e x +d^{2}\right )} \] Input:
int((B*x+A)*(c*x^2+b*x+a)/(e*x+d)^3,x)
Output:
(2*log(d + e*x)*a*c*d**3*e + 4*log(d + e*x)*a*c*d**2*e**2*x + 2*log(d + e* x)*a*c*d*e**3*x**2 + 2*log(d + e*x)*b**2*d**3*e + 4*log(d + e*x)*b**2*d**2 *e**2*x + 2*log(d + e*x)*b**2*d*e**3*x**2 - 6*log(d + e*x)*b*c*d**4 - 12*l og(d + e*x)*b*c*d**3*e*x - 6*log(d + e*x)*b*c*d**2*e**2*x**2 - a**2*d*e**3 + 2*a*b*e**4*x**2 + a*c*d**3*e - 2*a*c*d*e**3*x**2 + b**2*d**3*e - 2*b**2 *d*e**3*x**2 - 3*b*c*d**4 + 6*b*c*d**2*e**2*x**2 + 2*b*c*d*e**3*x**3)/(2*d *e**4*(d**2 + 2*d*e*x + e**2*x**2))