Integrand size = 20, antiderivative size = 103 \[ \int \frac {(A+B x) (d+e x)^2}{(a+b x)^3} \, dx=\frac {B e^2 x}{b^3}-\frac {(A b-a B) (b d-a e)^2}{2 b^4 (a+b x)^2}-\frac {(b d-a e) (b B d+2 A b e-3 a B e)}{b^4 (a+b x)}+\frac {e (2 b B d+A b e-3 a B e) \log (a+b x)}{b^4} \] Output:
B*e^2*x/b^3-1/2*(A*b-B*a)*(-a*e+b*d)^2/b^4/(b*x+a)^2-(-a*e+b*d)*(2*A*b*e-3 *B*a*e+B*b*d)/b^4/(b*x+a)+e*(A*b*e-3*B*a*e+2*B*b*d)*ln(b*x+a)/b^4
Time = 0.04 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.36 \[ \int \frac {(A+B x) (d+e x)^2}{(a+b x)^3} \, dx=\frac {-A b (b d-a e) (3 a e+b (d+4 e x))+B \left (-5 a^3 e^2+2 a^2 b e (3 d-2 e x)+2 b^3 x \left (-d^2+e^2 x^2\right )+a b^2 \left (-d^2+8 d e x+4 e^2 x^2\right )\right )+2 e (2 b B d+A b e-3 a B e) (a+b x)^2 \log (a+b x)}{2 b^4 (a+b x)^2} \] Input:
Integrate[((A + B*x)*(d + e*x)^2)/(a + b*x)^3,x]
Output:
(-(A*b*(b*d - a*e)*(3*a*e + b*(d + 4*e*x))) + B*(-5*a^3*e^2 + 2*a^2*b*e*(3 *d - 2*e*x) + 2*b^3*x*(-d^2 + e^2*x^2) + a*b^2*(-d^2 + 8*d*e*x + 4*e^2*x^2 )) + 2*e*(2*b*B*d + A*b*e - 3*a*B*e)*(a + b*x)^2*Log[a + b*x])/(2*b^4*(a + b*x)^2)
Time = 0.27 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, 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.
\(\displaystyle \int \frac {(A+B x) (d+e x)^2}{(a+b x)^3} \, dx\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \int \left (\frac {e (-3 a B e+A b e+2 b B d)}{b^3 (a+b x)}+\frac {(b d-a e) (-3 a B e+2 A b e+b B d)}{b^3 (a+b x)^2}+\frac {(A b-a B) (b d-a e)^2}{b^3 (a+b x)^3}+\frac {B e^2}{b^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {(b d-a e) (-3 a B e+2 A b e+b B d)}{b^4 (a+b x)}-\frac {(A b-a B) (b d-a e)^2}{2 b^4 (a+b x)^2}+\frac {e \log (a+b x) (-3 a B e+A b e+2 b B d)}{b^4}+\frac {B e^2 x}{b^3}\) |
Input:
Int[((A + B*x)*(d + e*x)^2)/(a + b*x)^3,x]
Output:
(B*e^2*x)/b^3 - ((A*b - a*B)*(b*d - a*e)^2)/(2*b^4*(a + b*x)^2) - ((b*d - a*e)*(b*B*d + 2*A*b*e - 3*a*B*e))/(b^4*(a + b*x)) + (e*(2*b*B*d + A*b*e - 3*a*B*e)*Log[a + b*x])/b^4
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 = 0.21 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.52
method | result | size |
default | \(\frac {B \,e^{2} x}{b^{3}}-\frac {A \,a^{2} b \,e^{2}-2 A a \,b^{2} d e +A \,b^{3} d^{2}-B \,a^{3} e^{2}+2 B \,a^{2} b d e -B a \,b^{2} d^{2}}{2 b^{4} \left (b x +a \right )^{2}}-\frac {-2 A a b \,e^{2}+2 A \,b^{2} d e +3 B \,a^{2} e^{2}-4 B a b d e +b^{2} B \,d^{2}}{b^{4} \left (b x +a \right )}+\frac {e \left (A b e -3 B a e +2 B b d \right ) \ln \left (b x +a \right )}{b^{4}}\) | \(157\) |
norman | \(\frac {\frac {\left (2 A a b \,e^{2}-2 A \,b^{2} d e -6 B \,a^{2} e^{2}+4 B a b d e -b^{2} B \,d^{2}\right ) x}{b^{3}}+\frac {B \,e^{2} x^{3}}{b}+\frac {3 A \,a^{2} b \,e^{2}-2 A a \,b^{2} d e -A \,b^{3} d^{2}-9 B \,a^{3} e^{2}+6 B \,a^{2} b d e -B a \,b^{2} d^{2}}{2 b^{4}}}{\left (b x +a \right )^{2}}+\frac {e \left (A b e -3 B a e +2 B b d \right ) \ln \left (b x +a \right )}{b^{4}}\) | \(157\) |
risch | \(\frac {B \,e^{2} x}{b^{3}}+\frac {\left (2 A a b \,e^{2}-2 A \,b^{2} d e -3 B \,a^{2} e^{2}+4 B a b d e -b^{2} B \,d^{2}\right ) x +\frac {3 A \,a^{2} b \,e^{2}-2 A a \,b^{2} d e -A \,b^{3} d^{2}-5 B \,a^{3} e^{2}+6 B \,a^{2} b d e -B a \,b^{2} d^{2}}{2 b}}{b^{3} \left (b x +a \right )^{2}}+\frac {e^{2} \ln \left (b x +a \right ) A}{b^{3}}-\frac {3 e^{2} \ln \left (b x +a \right ) B a}{b^{4}}+\frac {2 e \ln \left (b x +a \right ) B d}{b^{3}}\) | \(173\) |
parallelrisch | \(\frac {2 A \ln \left (b x +a \right ) x^{2} b^{3} e^{2}-6 B \ln \left (b x +a \right ) x^{2} a \,b^{2} e^{2}+4 B \ln \left (b x +a \right ) x^{2} b^{3} d e +2 B \,e^{2} x^{3} b^{3}+4 A \ln \left (b x +a \right ) x a \,b^{2} e^{2}-12 B \ln \left (b x +a \right ) x \,a^{2} b \,e^{2}+8 B \ln \left (b x +a \right ) x a \,b^{2} d e +2 A \ln \left (b x +a \right ) a^{2} b \,e^{2}+4 A x a \,b^{2} e^{2}-4 A \,b^{3} d e x -6 B \ln \left (b x +a \right ) a^{3} e^{2}+4 B \ln \left (b x +a \right ) a^{2} b d e -12 B \,a^{2} b \,e^{2} x +8 B x a \,b^{2} d e -2 B \,b^{3} d^{2} x +3 A \,a^{2} b \,e^{2}-2 A a \,b^{2} d e -A \,b^{3} d^{2}-9 B \,a^{3} e^{2}+6 B \,a^{2} b d e -B a \,b^{2} d^{2}}{2 b^{4} \left (b x +a \right )^{2}}\) | \(283\) |
Input:
int((B*x+A)*(e*x+d)^2/(b*x+a)^3,x,method=_RETURNVERBOSE)
Output:
B*e^2*x/b^3-1/2*(A*a^2*b*e^2-2*A*a*b^2*d*e+A*b^3*d^2-B*a^3*e^2+2*B*a^2*b*d *e-B*a*b^2*d^2)/b^4/(b*x+a)^2-(-2*A*a*b*e^2+2*A*b^2*d*e+3*B*a^2*e^2-4*B*a* b*d*e+B*b^2*d^2)/b^4/(b*x+a)+e*(A*b*e-3*B*a*e+2*B*b*d)*ln(b*x+a)/b^4
Leaf count of result is larger than twice the leaf count of optimal. 258 vs. \(2 (101) = 202\).
Time = 0.09 (sec) , antiderivative size = 258, normalized size of antiderivative = 2.50 \[ \int \frac {(A+B x) (d+e x)^2}{(a+b x)^3} \, dx=\frac {2 \, B b^{3} e^{2} x^{3} + 4 \, B a b^{2} e^{2} x^{2} - {\left (B a b^{2} + A b^{3}\right )} d^{2} + 2 \, {\left (3 \, B a^{2} b - A a b^{2}\right )} d e - {\left (5 \, B a^{3} - 3 \, A a^{2} b\right )} e^{2} - 2 \, {\left (B b^{3} d^{2} - 2 \, {\left (2 \, B a b^{2} - A b^{3}\right )} d e + 2 \, {\left (B a^{2} b - A a b^{2}\right )} e^{2}\right )} x + 2 \, {\left (2 \, B a^{2} b d e - {\left (3 \, B a^{3} - A a^{2} b\right )} e^{2} + {\left (2 \, B b^{3} d e - {\left (3 \, B a b^{2} - A b^{3}\right )} e^{2}\right )} x^{2} + 2 \, {\left (2 \, B a b^{2} d e - {\left (3 \, B a^{2} b - A a b^{2}\right )} e^{2}\right )} x\right )} \log \left (b x + a\right )}{2 \, {\left (b^{6} x^{2} + 2 \, a b^{5} x + a^{2} b^{4}\right )}} \] Input:
integrate((B*x+A)*(e*x+d)^2/(b*x+a)^3,x, algorithm="fricas")
Output:
1/2*(2*B*b^3*e^2*x^3 + 4*B*a*b^2*e^2*x^2 - (B*a*b^2 + A*b^3)*d^2 + 2*(3*B* a^2*b - A*a*b^2)*d*e - (5*B*a^3 - 3*A*a^2*b)*e^2 - 2*(B*b^3*d^2 - 2*(2*B*a *b^2 - A*b^3)*d*e + 2*(B*a^2*b - A*a*b^2)*e^2)*x + 2*(2*B*a^2*b*d*e - (3*B *a^3 - A*a^2*b)*e^2 + (2*B*b^3*d*e - (3*B*a*b^2 - A*b^3)*e^2)*x^2 + 2*(2*B *a*b^2*d*e - (3*B*a^2*b - A*a*b^2)*e^2)*x)*log(b*x + a))/(b^6*x^2 + 2*a*b^ 5*x + a^2*b^4)
Time = 1.15 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.82 \[ \int \frac {(A+B x) (d+e x)^2}{(a+b x)^3} \, dx=\frac {B e^{2} x}{b^{3}} + \frac {3 A a^{2} b e^{2} - 2 A a b^{2} d e - A b^{3} d^{2} - 5 B a^{3} e^{2} + 6 B a^{2} b d e - B a b^{2} d^{2} + x \left (4 A a b^{2} e^{2} - 4 A b^{3} d e - 6 B a^{2} b e^{2} + 8 B a b^{2} d e - 2 B b^{3} d^{2}\right )}{2 a^{2} b^{4} + 4 a b^{5} x + 2 b^{6} x^{2}} - \frac {e \left (- A b e + 3 B a e - 2 B b d\right ) \log {\left (a + b x \right )}}{b^{4}} \] Input:
integrate((B*x+A)*(e*x+d)**2/(b*x+a)**3,x)
Output:
B*e**2*x/b**3 + (3*A*a**2*b*e**2 - 2*A*a*b**2*d*e - A*b**3*d**2 - 5*B*a**3 *e**2 + 6*B*a**2*b*d*e - B*a*b**2*d**2 + x*(4*A*a*b**2*e**2 - 4*A*b**3*d*e - 6*B*a**2*b*e**2 + 8*B*a*b**2*d*e - 2*B*b**3*d**2))/(2*a**2*b**4 + 4*a*b **5*x + 2*b**6*x**2) - e*(-A*b*e + 3*B*a*e - 2*B*b*d)*log(a + b*x)/b**4
Time = 0.05 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.65 \[ \int \frac {(A+B x) (d+e x)^2}{(a+b x)^3} \, dx=\frac {B e^{2} x}{b^{3}} - \frac {{\left (B a b^{2} + A b^{3}\right )} d^{2} - 2 \, {\left (3 \, B a^{2} b - A a b^{2}\right )} d e + {\left (5 \, B a^{3} - 3 \, A a^{2} b\right )} e^{2} + 2 \, {\left (B b^{3} d^{2} - 2 \, {\left (2 \, B a b^{2} - A b^{3}\right )} d e + {\left (3 \, B a^{2} b - 2 \, A a b^{2}\right )} e^{2}\right )} x}{2 \, {\left (b^{6} x^{2} + 2 \, a b^{5} x + a^{2} b^{4}\right )}} + \frac {{\left (2 \, B b d e - {\left (3 \, B a - A b\right )} e^{2}\right )} \log \left (b x + a\right )}{b^{4}} \] Input:
integrate((B*x+A)*(e*x+d)^2/(b*x+a)^3,x, algorithm="maxima")
Output:
B*e^2*x/b^3 - 1/2*((B*a*b^2 + A*b^3)*d^2 - 2*(3*B*a^2*b - A*a*b^2)*d*e + ( 5*B*a^3 - 3*A*a^2*b)*e^2 + 2*(B*b^3*d^2 - 2*(2*B*a*b^2 - A*b^3)*d*e + (3*B *a^2*b - 2*A*a*b^2)*e^2)*x)/(b^6*x^2 + 2*a*b^5*x + a^2*b^4) + (2*B*b*d*e - (3*B*a - A*b)*e^2)*log(b*x + a)/b^4
Time = 0.12 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.52 \[ \int \frac {(A+B x) (d+e x)^2}{(a+b x)^3} \, dx=\frac {B e^{2} x}{b^{3}} + \frac {{\left (2 \, B b d e - 3 \, B a e^{2} + A b e^{2}\right )} \log \left ({\left | b x + a \right |}\right )}{b^{4}} - \frac {B a b^{2} d^{2} + A b^{3} d^{2} - 6 \, B a^{2} b d e + 2 \, A a b^{2} d e + 5 \, B a^{3} e^{2} - 3 \, A a^{2} b e^{2} + 2 \, {\left (B b^{3} d^{2} - 4 \, B a b^{2} d e + 2 \, A b^{3} d e + 3 \, B a^{2} b e^{2} - 2 \, A a b^{2} e^{2}\right )} x}{2 \, {\left (b x + a\right )}^{2} b^{4}} \] Input:
integrate((B*x+A)*(e*x+d)^2/(b*x+a)^3,x, algorithm="giac")
Output:
B*e^2*x/b^3 + (2*B*b*d*e - 3*B*a*e^2 + A*b*e^2)*log(abs(b*x + a))/b^4 - 1/ 2*(B*a*b^2*d^2 + A*b^3*d^2 - 6*B*a^2*b*d*e + 2*A*a*b^2*d*e + 5*B*a^3*e^2 - 3*A*a^2*b*e^2 + 2*(B*b^3*d^2 - 4*B*a*b^2*d*e + 2*A*b^3*d*e + 3*B*a^2*b*e^ 2 - 2*A*a*b^2*e^2)*x)/((b*x + a)^2*b^4)
Time = 0.10 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.65 \[ \int \frac {(A+B x) (d+e x)^2}{(a+b x)^3} \, dx=\frac {\ln \left (a+b\,x\right )\,\left (A\,b\,e^2-3\,B\,a\,e^2+2\,B\,b\,d\,e\right )}{b^4}-\frac {x\,\left (3\,B\,a^2\,e^2-4\,B\,a\,b\,d\,e-2\,A\,a\,b\,e^2+B\,b^2\,d^2+2\,A\,b^2\,d\,e\right )+\frac {5\,B\,a^3\,e^2-6\,B\,a^2\,b\,d\,e-3\,A\,a^2\,b\,e^2+B\,a\,b^2\,d^2+2\,A\,a\,b^2\,d\,e+A\,b^3\,d^2}{2\,b}}{a^2\,b^3+2\,a\,b^4\,x+b^5\,x^2}+\frac {B\,e^2\,x}{b^3} \] Input:
int(((A + B*x)*(d + e*x)^2)/(a + b*x)^3,x)
Output:
(log(a + b*x)*(A*b*e^2 - 3*B*a*e^2 + 2*B*b*d*e))/b^4 - (x*(3*B*a^2*e^2 + B *b^2*d^2 - 2*A*a*b*e^2 + 2*A*b^2*d*e - 4*B*a*b*d*e) + (A*b^3*d^2 + 5*B*a^3 *e^2 - 3*A*a^2*b*e^2 + B*a*b^2*d^2 + 2*A*a*b^2*d*e - 6*B*a^2*b*d*e)/(2*b)) /(a^2*b^3 + b^5*x^2 + 2*a*b^4*x) + (B*e^2*x)/b^3
Time = 0.17 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.09 \[ \int \frac {(A+B x) (d+e x)^2}{(a+b x)^3} \, dx=\frac {-2 \,\mathrm {log}\left (b x +a \right ) a^{3} e^{2}+2 \,\mathrm {log}\left (b x +a \right ) a^{2} b d e -2 \,\mathrm {log}\left (b x +a \right ) a^{2} b \,e^{2} x +2 \,\mathrm {log}\left (b x +a \right ) a \,b^{2} d e x +2 a^{2} b \,e^{2} x -2 a \,b^{2} d e x +a \,b^{2} e^{2} x^{2}+b^{3} d^{2} x}{a \,b^{3} \left (b x +a \right )} \] Input:
int((B*x+A)*(e*x+d)^2/(b*x+a)^3,x)
Output:
( - 2*log(a + b*x)*a**3*e**2 + 2*log(a + b*x)*a**2*b*d*e - 2*log(a + b*x)* a**2*b*e**2*x + 2*log(a + b*x)*a*b**2*d*e*x + 2*a**2*b*e**2*x - 2*a*b**2*d *e*x + a*b**2*e**2*x**2 + b**3*d**2*x)/(a*b**3*(a + b*x))