Integrand size = 29, antiderivative size = 106 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )}{(d+e x)^3} \, dx=\frac {b^2 B x}{e^3}+\frac {(b d-a e)^2 (B d-A e)}{2 e^4 (d+e x)^2}-\frac {(b d-a e) (3 b B d-2 A b e-a B e)}{e^4 (d+e x)}-\frac {b (3 b B d-A b e-2 a B e) \log (d+e x)}{e^4} \]
b^2*B*x/e^3+1/2*(-a*e+b*d)^2*(-A*e+B*d)/e^4/(e*x+d)^2-(-a*e+b*d)*(-2*A*b*e -B*a*e+3*B*b*d)/e^4/(e*x+d)-b*(-A*b*e-2*B*a*e+3*B*b*d)*ln(e*x+d)/e^4
Time = 0.05 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.35 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )}{(d+e x)^3} \, dx=-\frac {a^2 e^2 (A e+B (d+2 e x))+2 a b e (A e (d+2 e x)-B d (3 d+4 e x))-b^2 \left (A d e (3 d+4 e x)+B \left (-5 d^3-4 d^2 e x+4 d e^2 x^2+2 e^3 x^3\right )\right )+2 b (3 b B d-A b e-2 a B e) (d+e x)^2 \log (d+e x)}{2 e^4 (d+e x)^2} \]
-1/2*(a^2*e^2*(A*e + B*(d + 2*e*x)) + 2*a*b*e*(A*e*(d + 2*e*x) - B*d*(3*d + 4*e*x)) - b^2*(A*d*e*(3*d + 4*e*x) + B*(-5*d^3 - 4*d^2*e*x + 4*d*e^2*x^2 + 2*e^3*x^3)) + 2*b*(3*b*B*d - A*b*e - 2*a*B*e)*(d + e*x)^2*Log[d + e*x]) /(e^4*(d + e*x)^2)
Time = 0.31 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {1184, 27, 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.
\(\displaystyle \int \frac {\left (a^2+2 a b x+b^2 x^2\right ) (A+B x)}{(d+e x)^3} \, dx\) |
\(\Big \downarrow \) 1184 |
\(\displaystyle \frac {\int \frac {b^2 (a+b x)^2 (A+B x)}{(d+e x)^3}dx}{b^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \frac {(a+b x)^2 (A+B x)}{(d+e x)^3}dx\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \int \left (\frac {b (2 a B e+A b e-3 b B d)}{e^3 (d+e x)}+\frac {(a e-b d) (a B e+2 A b e-3 b B d)}{e^3 (d+e x)^2}+\frac {(a e-b d)^2 (A e-B d)}{e^3 (d+e x)^3}+\frac {b^2 B}{e^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {(b d-a e) (-a B e-2 A b e+3 b B d)}{e^4 (d+e x)}+\frac {(b d-a e)^2 (B d-A e)}{2 e^4 (d+e x)^2}-\frac {b \log (d+e x) (-2 a B e-A b e+3 b B d)}{e^4}+\frac {b^2 B x}{e^3}\) |
(b^2*B*x)/e^3 + ((b*d - a*e)^2*(B*d - A*e))/(2*e^4*(d + e*x)^2) - ((b*d - a*e)*(3*b*B*d - 2*A*b*e - a*B*e))/(e^4*(d + e*x)) - (b*(3*b*B*d - A*b*e - 2*a*B*e)*Log[d + e*x])/e^4
3.17.67.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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]))))
Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_) + (b_.)*(x_ ) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[1/c^p Int[(d + e*x)^m*(f + g*x )^n*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n}, x] && E qQ[b^2 - 4*a*c, 0] && IntegerQ[p]
Time = 0.21 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.46
method | result | size |
norman | \(\frac {\frac {B \,b^{2} x^{3}}{e}-\frac {A \,a^{2} e^{3}+2 A a b d \,e^{2}-3 A \,b^{2} d^{2} e +B \,a^{2} d \,e^{2}-6 B a b \,d^{2} e +9 B \,b^{2} d^{3}}{2 e^{4}}-\frac {\left (2 A a b \,e^{2}-2 A \,b^{2} d e +a^{2} B \,e^{2}-4 B a b d e +6 B \,b^{2} d^{2}\right ) x}{e^{3}}}{\left (e x +d \right )^{2}}+\frac {b \left (A b e +2 B a e -3 B b d \right ) \ln \left (e x +d \right )}{e^{4}}\) | \(155\) |
default | \(\frac {b^{2} B x}{e^{3}}-\frac {2 A a b \,e^{2}-2 A \,b^{2} d e +a^{2} B \,e^{2}-4 B a b d e +3 B \,b^{2} d^{2}}{e^{4} \left (e x +d \right )}+\frac {b \left (A b e +2 B a e -3 B b d \right ) \ln \left (e x +d \right )}{e^{4}}-\frac {A \,a^{2} e^{3}-2 A a b d \,e^{2}+A \,b^{2} d^{2} e -B \,a^{2} d \,e^{2}+2 B a b \,d^{2} e -B \,b^{2} d^{3}}{2 e^{4} \left (e x +d \right )^{2}}\) | \(157\) |
risch | \(\frac {b^{2} B x}{e^{3}}+\frac {\left (-2 A a b \,e^{2}+2 A \,b^{2} d e -a^{2} B \,e^{2}+4 B a b d e -3 B \,b^{2} d^{2}\right ) x -\frac {A \,a^{2} e^{3}+2 A a b d \,e^{2}-3 A \,b^{2} d^{2} e +B \,a^{2} d \,e^{2}-6 B a b \,d^{2} e +5 B \,b^{2} d^{3}}{2 e}}{e^{3} \left (e x +d \right )^{2}}+\frac {b^{2} \ln \left (e x +d \right ) A}{e^{3}}+\frac {2 b \ln \left (e x +d \right ) B a}{e^{3}}-\frac {3 b^{2} \ln \left (e x +d \right ) B d}{e^{4}}\) | \(171\) |
parallelrisch | \(\frac {2 A \ln \left (e x +d \right ) x^{2} b^{2} e^{3}+4 B \ln \left (e x +d \right ) x^{2} a b \,e^{3}-6 B \ln \left (e x +d \right ) x^{2} b^{2} d \,e^{2}+2 B \,b^{2} x^{3} e^{3}+4 A \ln \left (e x +d \right ) x \,b^{2} d \,e^{2}+8 B \ln \left (e x +d \right ) x a b d \,e^{2}-12 B \ln \left (e x +d \right ) x \,b^{2} d^{2} e +2 A \ln \left (e x +d \right ) b^{2} d^{2} e -4 A x a b \,e^{3}+4 A x \,b^{2} d \,e^{2}+4 B \ln \left (e x +d \right ) a b \,d^{2} e -6 B \ln \left (e x +d \right ) b^{2} d^{3}-2 B x \,a^{2} e^{3}+8 B x a b d \,e^{2}-12 B x \,b^{2} d^{2} e -A \,a^{2} e^{3}-2 A a b d \,e^{2}+3 A \,b^{2} d^{2} e -B \,a^{2} d \,e^{2}+6 B a b \,d^{2} e -9 B \,b^{2} d^{3}}{2 e^{4} \left (e x +d \right )^{2}}\) | \(283\) |
(B*b^2/e*x^3-1/2*(A*a^2*e^3+2*A*a*b*d*e^2-3*A*b^2*d^2*e+B*a^2*d*e^2-6*B*a* b*d^2*e+9*B*b^2*d^3)/e^4-(2*A*a*b*e^2-2*A*b^2*d*e+B*a^2*e^2-4*B*a*b*d*e+6* B*b^2*d^2)/e^3*x)/(e*x+d)^2+b/e^4*(A*b*e+2*B*a*e-3*B*b*d)*ln(e*x+d)
Leaf count of result is larger than twice the leaf count of optimal. 247 vs. \(2 (104) = 208\).
Time = 0.30 (sec) , antiderivative size = 247, normalized size of antiderivative = 2.33 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )}{(d+e x)^3} \, dx=\frac {2 \, B b^{2} e^{3} x^{3} + 4 \, B b^{2} d e^{2} x^{2} - 5 \, B b^{2} d^{3} - A a^{2} e^{3} + 3 \, {\left (2 \, B a b + A b^{2}\right )} d^{2} e - {\left (B a^{2} + 2 \, A a b\right )} d e^{2} - 2 \, {\left (2 \, B b^{2} d^{2} e - 2 \, {\left (2 \, B a b + A b^{2}\right )} d e^{2} + {\left (B a^{2} + 2 \, A a b\right )} e^{3}\right )} x - 2 \, {\left (3 \, B b^{2} d^{3} - {\left (2 \, B a b + A b^{2}\right )} d^{2} e + {\left (3 \, B b^{2} d e^{2} - {\left (2 \, B a b + A b^{2}\right )} e^{3}\right )} x^{2} + 2 \, {\left (3 \, B b^{2} d^{2} e - {\left (2 \, B a b + A b^{2}\right )} d e^{2}\right )} x\right )} \log \left (e x + d\right )}{2 \, {\left (e^{6} x^{2} + 2 \, d e^{5} x + d^{2} e^{4}\right )}} \]
1/2*(2*B*b^2*e^3*x^3 + 4*B*b^2*d*e^2*x^2 - 5*B*b^2*d^3 - A*a^2*e^3 + 3*(2* B*a*b + A*b^2)*d^2*e - (B*a^2 + 2*A*a*b)*d*e^2 - 2*(2*B*b^2*d^2*e - 2*(2*B *a*b + A*b^2)*d*e^2 + (B*a^2 + 2*A*a*b)*e^3)*x - 2*(3*B*b^2*d^3 - (2*B*a*b + A*b^2)*d^2*e + (3*B*b^2*d*e^2 - (2*B*a*b + A*b^2)*e^3)*x^2 + 2*(3*B*b^2 *d^2*e - (2*B*a*b + A*b^2)*d*e^2)*x)*log(e*x + d))/(e^6*x^2 + 2*d*e^5*x + d^2*e^4)
Time = 1.09 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.76 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )}{(d+e x)^3} \, dx=\frac {B b^{2} x}{e^{3}} + \frac {b \left (A b e + 2 B a e - 3 B b d\right ) \log {\left (d + e x \right )}}{e^{4}} + \frac {- A a^{2} e^{3} - 2 A a b d e^{2} + 3 A b^{2} d^{2} e - B a^{2} d e^{2} + 6 B a b d^{2} e - 5 B b^{2} d^{3} + x \left (- 4 A a b e^{3} + 4 A b^{2} d e^{2} - 2 B a^{2} e^{3} + 8 B a b d e^{2} - 6 B b^{2} d^{2} e\right )}{2 d^{2} e^{4} + 4 d e^{5} x + 2 e^{6} x^{2}} \]
B*b**2*x/e**3 + b*(A*b*e + 2*B*a*e - 3*B*b*d)*log(d + e*x)/e**4 + (-A*a**2 *e**3 - 2*A*a*b*d*e**2 + 3*A*b**2*d**2*e - B*a**2*d*e**2 + 6*B*a*b*d**2*e - 5*B*b**2*d**3 + x*(-4*A*a*b*e**3 + 4*A*b**2*d*e**2 - 2*B*a**2*e**3 + 8*B *a*b*d*e**2 - 6*B*b**2*d**2*e))/(2*d**2*e**4 + 4*d*e**5*x + 2*e**6*x**2)
Time = 0.23 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.57 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )}{(d+e x)^3} \, dx=\frac {B b^{2} x}{e^{3}} - \frac {5 \, B b^{2} d^{3} + A a^{2} e^{3} - 3 \, {\left (2 \, B a b + A b^{2}\right )} d^{2} e + {\left (B a^{2} + 2 \, A a b\right )} d e^{2} + 2 \, {\left (3 \, B b^{2} d^{2} e - 2 \, {\left (2 \, B a b + A b^{2}\right )} d e^{2} + {\left (B a^{2} + 2 \, A a b\right )} e^{3}\right )} x}{2 \, {\left (e^{6} x^{2} + 2 \, d e^{5} x + d^{2} e^{4}\right )}} - \frac {{\left (3 \, B b^{2} d - {\left (2 \, B a b + A b^{2}\right )} e\right )} \log \left (e x + d\right )}{e^{4}} \]
B*b^2*x/e^3 - 1/2*(5*B*b^2*d^3 + A*a^2*e^3 - 3*(2*B*a*b + A*b^2)*d^2*e + ( B*a^2 + 2*A*a*b)*d*e^2 + 2*(3*B*b^2*d^2*e - 2*(2*B*a*b + A*b^2)*d*e^2 + (B *a^2 + 2*A*a*b)*e^3)*x)/(e^6*x^2 + 2*d*e^5*x + d^2*e^4) - (3*B*b^2*d - (2* B*a*b + A*b^2)*e)*log(e*x + d)/e^4
Time = 0.28 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.50 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )}{(d+e x)^3} \, dx=\frac {B b^{2} x}{e^{3}} - \frac {{\left (3 \, B b^{2} d - 2 \, B a b e - A b^{2} e\right )} \log \left ({\left | e x + d \right |}\right )}{e^{4}} - \frac {5 \, B b^{2} d^{3} - 6 \, B a b d^{2} e - 3 \, A b^{2} d^{2} e + B a^{2} d e^{2} + 2 \, A a b d e^{2} + A a^{2} e^{3} + 2 \, {\left (3 \, B b^{2} d^{2} e - 4 \, B a b d e^{2} - 2 \, A b^{2} d e^{2} + B a^{2} e^{3} + 2 \, A a b e^{3}\right )} x}{2 \, {\left (e x + d\right )}^{2} e^{4}} \]
B*b^2*x/e^3 - (3*B*b^2*d - 2*B*a*b*e - A*b^2*e)*log(abs(e*x + d))/e^4 - 1/ 2*(5*B*b^2*d^3 - 6*B*a*b*d^2*e - 3*A*b^2*d^2*e + B*a^2*d*e^2 + 2*A*a*b*d*e ^2 + A*a^2*e^3 + 2*(3*B*b^2*d^2*e - 4*B*a*b*d*e^2 - 2*A*b^2*d*e^2 + B*a^2* e^3 + 2*A*a*b*e^3)*x)/((e*x + d)^2*e^4)
Time = 0.14 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.60 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )}{(d+e x)^3} \, dx=\frac {\ln \left (d+e\,x\right )\,\left (A\,b^2\,e-3\,B\,b^2\,d+2\,B\,a\,b\,e\right )}{e^4}-\frac {x\,\left (B\,a^2\,e^2-4\,B\,a\,b\,d\,e+2\,A\,a\,b\,e^2+3\,B\,b^2\,d^2-2\,A\,b^2\,d\,e\right )+\frac {B\,a^2\,d\,e^2+A\,a^2\,e^3-6\,B\,a\,b\,d^2\,e+2\,A\,a\,b\,d\,e^2+5\,B\,b^2\,d^3-3\,A\,b^2\,d^2\,e}{2\,e}}{d^2\,e^3+2\,d\,e^4\,x+e^5\,x^2}+\frac {B\,b^2\,x}{e^3} \]