Integrand size = 26, antiderivative size = 103 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^2}{(d+e x)^3} \, dx=-\frac {b^3 (3 b d-4 a e) x}{e^4}+\frac {b^4 x^2}{2 e^3}-\frac {(b d-a e)^4}{2 e^5 (d+e x)^2}+\frac {4 b (b d-a e)^3}{e^5 (d+e x)}+\frac {6 b^2 (b d-a e)^2 \log (d+e x)}{e^5} \]
-b^3*(-4*a*e+3*b*d)*x/e^4+1/2*b^4*x^2/e^3-1/2*(-a*e+b*d)^4/e^5/(e*x+d)^2+4 *b*(-a*e+b*d)^3/e^5/(e*x+d)+6*b^2*(-a*e+b*d)^2*ln(e*x+d)/e^5
Time = 0.04 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.62 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^2}{(d+e x)^3} \, dx=\frac {-a^4 e^4-4 a^3 b e^3 (d+2 e x)+6 a^2 b^2 d e^2 (3 d+4 e x)+4 a b^3 e \left (-5 d^3-4 d^2 e x+4 d e^2 x^2+2 e^3 x^3\right )+b^4 \left (7 d^4+2 d^3 e x-11 d^2 e^2 x^2-4 d e^3 x^3+e^4 x^4\right )+12 b^2 (b d-a e)^2 (d+e x)^2 \log (d+e x)}{2 e^5 (d+e x)^2} \]
(-(a^4*e^4) - 4*a^3*b*e^3*(d + 2*e*x) + 6*a^2*b^2*d*e^2*(3*d + 4*e*x) + 4* a*b^3*e*(-5*d^3 - 4*d^2*e*x + 4*d*e^2*x^2 + 2*e^3*x^3) + b^4*(7*d^4 + 2*d^ 3*e*x - 11*d^2*e^2*x^2 - 4*d*e^3*x^3 + e^4*x^4) + 12*b^2*(b*d - a*e)^2*(d + e*x)^2*Log[d + e*x])/(2*e^5*(d + e*x)^2)
Time = 0.28 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {1098, 27, 49, 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.
\(\displaystyle \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^2}{(d+e x)^3} \, dx\) |
\(\Big \downarrow \) 1098 |
\(\displaystyle \frac {\int \frac {b^4 (a+b x)^4}{(d+e x)^3}dx}{b^4}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \frac {(a+b x)^4}{(d+e x)^3}dx\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \int \left (-\frac {b^3 (3 b d-4 a e)}{e^4}+\frac {6 b^2 (b d-a e)^2}{e^4 (d+e x)}-\frac {4 b (b d-a e)^3}{e^4 (d+e x)^2}+\frac {(a e-b d)^4}{e^4 (d+e x)^3}+\frac {b^4 x}{e^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {b^3 x (3 b d-4 a e)}{e^4}+\frac {6 b^2 (b d-a e)^2 \log (d+e x)}{e^5}+\frac {4 b (b d-a e)^3}{e^5 (d+e x)}-\frac {(b d-a e)^4}{2 e^5 (d+e x)^2}+\frac {b^4 x^2}{2 e^3}\) |
-((b^3*(3*b*d - 4*a*e)*x)/e^4) + (b^4*x^2)/(2*e^3) - (b*d - a*e)^4/(2*e^5* (d + e*x)^2) + (4*b*(b*d - a*e)^3)/(e^5*(d + e*x)) + (6*b^2*(b*d - a*e)^2* Log[d + e*x])/e^5
3.15.72.3.1 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_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 0] && IGtQ[m + n + 2, 0]
Int[((d_.) + (e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_ Symbol] :> Simp[1/c^p Int[(d + e*x)^m*(b/2 + c*x)^(2*p), x], x] /; FreeQ[ {a, b, c, d, e, m}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]
Time = 2.23 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.67
method | result | size |
default | \(\frac {b^{3} \left (\frac {1}{2} b e \,x^{2}+4 a e x -3 b d x \right )}{e^{4}}-\frac {4 b \left (a^{3} e^{3}-3 a^{2} b d \,e^{2}+3 a \,b^{2} d^{2} e -b^{3} d^{3}\right )}{e^{5} \left (e x +d \right )}-\frac {e^{4} a^{4}-4 b \,e^{3} d \,a^{3}+6 b^{2} e^{2} d^{2} a^{2}-4 a \,b^{3} d^{3} e +b^{4} d^{4}}{2 e^{5} \left (e x +d \right )^{2}}+\frac {6 b^{2} \left (a^{2} e^{2}-2 a b d e +b^{2} d^{2}\right ) \ln \left (e x +d \right )}{e^{5}}\) | \(172\) |
norman | \(\frac {-\frac {e^{4} a^{4}+4 b \,e^{3} d \,a^{3}-18 b^{2} e^{2} d^{2} a^{2}+36 a \,b^{3} d^{3} e -18 b^{4} d^{4}}{2 e^{5}}+\frac {b^{4} x^{4}}{2 e}-\frac {2 \left (2 a^{3} b \,e^{3}-6 a^{2} b^{2} d \,e^{2}+12 d^{2} e a \,b^{3}-6 b^{4} d^{3}\right ) x}{e^{4}}+\frac {2 b^{3} \left (2 a e -b d \right ) x^{3}}{e^{2}}}{\left (e x +d \right )^{2}}+\frac {6 b^{2} \left (a^{2} e^{2}-2 a b d e +b^{2} d^{2}\right ) \ln \left (e x +d \right )}{e^{5}}\) | \(178\) |
risch | \(\frac {b^{4} x^{2}}{2 e^{3}}+\frac {4 b^{3} a x}{e^{3}}-\frac {3 b^{4} d x}{e^{4}}+\frac {\left (-4 a^{3} b \,e^{3}+12 a^{2} b^{2} d \,e^{2}-12 d^{2} e a \,b^{3}+4 b^{4} d^{3}\right ) x -\frac {e^{4} a^{4}+4 b \,e^{3} d \,a^{3}-18 b^{2} e^{2} d^{2} a^{2}+20 a \,b^{3} d^{3} e -7 b^{4} d^{4}}{2 e}}{e^{4} \left (e x +d \right )^{2}}+\frac {6 b^{2} \ln \left (e x +d \right ) a^{2}}{e^{3}}-\frac {12 b^{3} \ln \left (e x +d \right ) a d}{e^{4}}+\frac {6 b^{4} \ln \left (e x +d \right ) d^{2}}{e^{5}}\) | \(192\) |
parallelrisch | \(\frac {b^{4} x^{4} e^{4}+12 \ln \left (e x +d \right ) x^{2} a^{2} b^{2} e^{4}-24 \ln \left (e x +d \right ) x^{2} a \,b^{3} d \,e^{3}+12 \ln \left (e x +d \right ) x^{2} b^{4} d^{2} e^{2}+8 x^{3} a \,b^{3} e^{4}-4 x^{3} b^{4} d \,e^{3}+24 \ln \left (e x +d \right ) x \,a^{2} b^{2} d \,e^{3}-48 \ln \left (e x +d \right ) x a \,b^{3} d^{2} e^{2}+24 \ln \left (e x +d \right ) x \,b^{4} d^{3} e +12 \ln \left (e x +d \right ) a^{2} b^{2} d^{2} e^{2}-24 \ln \left (e x +d \right ) a \,b^{3} d^{3} e +12 \ln \left (e x +d \right ) b^{4} d^{4}-8 x \,a^{3} b \,e^{4}+24 x \,a^{2} b^{2} d \,e^{3}-48 x a \,b^{3} d^{2} e^{2}+24 x \,b^{4} d^{3} e -e^{4} a^{4}-4 b \,e^{3} d \,a^{3}+18 b^{2} e^{2} d^{2} a^{2}-36 a \,b^{3} d^{3} e +18 b^{4} d^{4}}{2 e^{5} \left (e x +d \right )^{2}}\) | \(307\) |
b^3/e^4*(1/2*b*e*x^2+4*a*e*x-3*b*d*x)-4*b/e^5*(a^3*e^3-3*a^2*b*d*e^2+3*a*b ^2*d^2*e-b^3*d^3)/(e*x+d)-1/2*(a^4*e^4-4*a^3*b*d*e^3+6*a^2*b^2*d^2*e^2-4*a *b^3*d^3*e+b^4*d^4)/e^5/(e*x+d)^2+6*b^2/e^5*(a^2*e^2-2*a*b*d*e+b^2*d^2)*ln (e*x+d)
Leaf count of result is larger than twice the leaf count of optimal. 291 vs. \(2 (99) = 198\).
Time = 0.28 (sec) , antiderivative size = 291, normalized size of antiderivative = 2.83 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^2}{(d+e x)^3} \, dx=\frac {b^{4} e^{4} x^{4} + 7 \, b^{4} d^{4} - 20 \, a b^{3} d^{3} e + 18 \, a^{2} b^{2} d^{2} e^{2} - 4 \, a^{3} b d e^{3} - a^{4} e^{4} - 4 \, {\left (b^{4} d e^{3} - 2 \, a b^{3} e^{4}\right )} x^{3} - {\left (11 \, b^{4} d^{2} e^{2} - 16 \, a b^{3} d e^{3}\right )} x^{2} + 2 \, {\left (b^{4} d^{3} e - 8 \, a b^{3} d^{2} e^{2} + 12 \, a^{2} b^{2} d e^{3} - 4 \, a^{3} b e^{4}\right )} x + 12 \, {\left (b^{4} d^{4} - 2 \, a b^{3} d^{3} e + a^{2} b^{2} d^{2} e^{2} + {\left (b^{4} d^{2} e^{2} - 2 \, a b^{3} d e^{3} + a^{2} b^{2} e^{4}\right )} x^{2} + 2 \, {\left (b^{4} d^{3} e - 2 \, a b^{3} d^{2} e^{2} + a^{2} b^{2} d e^{3}\right )} x\right )} \log \left (e x + d\right )}{2 \, {\left (e^{7} x^{2} + 2 \, d e^{6} x + d^{2} e^{5}\right )}} \]
1/2*(b^4*e^4*x^4 + 7*b^4*d^4 - 20*a*b^3*d^3*e + 18*a^2*b^2*d^2*e^2 - 4*a^3 *b*d*e^3 - a^4*e^4 - 4*(b^4*d*e^3 - 2*a*b^3*e^4)*x^3 - (11*b^4*d^2*e^2 - 1 6*a*b^3*d*e^3)*x^2 + 2*(b^4*d^3*e - 8*a*b^3*d^2*e^2 + 12*a^2*b^2*d*e^3 - 4 *a^3*b*e^4)*x + 12*(b^4*d^4 - 2*a*b^3*d^3*e + a^2*b^2*d^2*e^2 + (b^4*d^2*e ^2 - 2*a*b^3*d*e^3 + a^2*b^2*e^4)*x^2 + 2*(b^4*d^3*e - 2*a*b^3*d^2*e^2 + a ^2*b^2*d*e^3)*x)*log(e*x + d))/(e^7*x^2 + 2*d*e^6*x + d^2*e^5)
Time = 0.67 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.80 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^2}{(d+e x)^3} \, dx=\frac {b^{4} x^{2}}{2 e^{3}} + \frac {6 b^{2} \left (a e - b d\right )^{2} \log {\left (d + e x \right )}}{e^{5}} + x \left (\frac {4 a b^{3}}{e^{3}} - \frac {3 b^{4} d}{e^{4}}\right ) + \frac {- a^{4} e^{4} - 4 a^{3} b d e^{3} + 18 a^{2} b^{2} d^{2} e^{2} - 20 a b^{3} d^{3} e + 7 b^{4} d^{4} + x \left (- 8 a^{3} b e^{4} + 24 a^{2} b^{2} d e^{3} - 24 a b^{3} d^{2} e^{2} + 8 b^{4} d^{3} e\right )}{2 d^{2} e^{5} + 4 d e^{6} x + 2 e^{7} x^{2}} \]
b**4*x**2/(2*e**3) + 6*b**2*(a*e - b*d)**2*log(d + e*x)/e**5 + x*(4*a*b**3 /e**3 - 3*b**4*d/e**4) + (-a**4*e**4 - 4*a**3*b*d*e**3 + 18*a**2*b**2*d**2 *e**2 - 20*a*b**3*d**3*e + 7*b**4*d**4 + x*(-8*a**3*b*e**4 + 24*a**2*b**2* d*e**3 - 24*a*b**3*d**2*e**2 + 8*b**4*d**3*e))/(2*d**2*e**5 + 4*d*e**6*x + 2*e**7*x**2)
Time = 0.20 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.85 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^2}{(d+e x)^3} \, dx=\frac {7 \, b^{4} d^{4} - 20 \, a b^{3} d^{3} e + 18 \, a^{2} b^{2} d^{2} e^{2} - 4 \, a^{3} b d e^{3} - a^{4} e^{4} + 8 \, {\left (b^{4} d^{3} e - 3 \, a b^{3} d^{2} e^{2} + 3 \, a^{2} b^{2} d e^{3} - a^{3} b e^{4}\right )} x}{2 \, {\left (e^{7} x^{2} + 2 \, d e^{6} x + d^{2} e^{5}\right )}} + \frac {b^{4} e x^{2} - 2 \, {\left (3 \, b^{4} d - 4 \, a b^{3} e\right )} x}{2 \, e^{4}} + \frac {6 \, {\left (b^{4} d^{2} - 2 \, a b^{3} d e + a^{2} b^{2} e^{2}\right )} \log \left (e x + d\right )}{e^{5}} \]
1/2*(7*b^4*d^4 - 20*a*b^3*d^3*e + 18*a^2*b^2*d^2*e^2 - 4*a^3*b*d*e^3 - a^4 *e^4 + 8*(b^4*d^3*e - 3*a*b^3*d^2*e^2 + 3*a^2*b^2*d*e^3 - a^3*b*e^4)*x)/(e ^7*x^2 + 2*d*e^6*x + d^2*e^5) + 1/2*(b^4*e*x^2 - 2*(3*b^4*d - 4*a*b^3*e)*x )/e^4 + 6*(b^4*d^2 - 2*a*b^3*d*e + a^2*b^2*e^2)*log(e*x + d)/e^5
Time = 0.26 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.78 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^2}{(d+e x)^3} \, dx=\frac {6 \, {\left (b^{4} d^{2} - 2 \, a b^{3} d e + a^{2} b^{2} e^{2}\right )} \log \left ({\left | e x + d \right |}\right )}{e^{5}} + \frac {b^{4} e^{3} x^{2} - 6 \, b^{4} d e^{2} x + 8 \, a b^{3} e^{3} x}{2 \, e^{6}} + \frac {7 \, b^{4} d^{4} - 20 \, a b^{3} d^{3} e + 18 \, a^{2} b^{2} d^{2} e^{2} - 4 \, a^{3} b d e^{3} - a^{4} e^{4} + 8 \, {\left (b^{4} d^{3} e - 3 \, a b^{3} d^{2} e^{2} + 3 \, a^{2} b^{2} d e^{3} - a^{3} b e^{4}\right )} x}{2 \, {\left (e x + d\right )}^{2} e^{5}} \]
6*(b^4*d^2 - 2*a*b^3*d*e + a^2*b^2*e^2)*log(abs(e*x + d))/e^5 + 1/2*(b^4*e ^3*x^2 - 6*b^4*d*e^2*x + 8*a*b^3*e^3*x)/e^6 + 1/2*(7*b^4*d^4 - 20*a*b^3*d^ 3*e + 18*a^2*b^2*d^2*e^2 - 4*a^3*b*d*e^3 - a^4*e^4 + 8*(b^4*d^3*e - 3*a*b^ 3*d^2*e^2 + 3*a^2*b^2*d*e^3 - a^3*b*e^4)*x)/((e*x + d)^2*e^5)
Time = 0.09 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.90 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^2}{(d+e x)^3} \, dx=x\,\left (\frac {4\,a\,b^3}{e^3}-\frac {3\,b^4\,d}{e^4}\right )-\frac {\frac {a^4\,e^4+4\,a^3\,b\,d\,e^3-18\,a^2\,b^2\,d^2\,e^2+20\,a\,b^3\,d^3\,e-7\,b^4\,d^4}{2\,e}-x\,\left (-4\,a^3\,b\,e^3+12\,a^2\,b^2\,d\,e^2-12\,a\,b^3\,d^2\,e+4\,b^4\,d^3\right )}{d^2\,e^4+2\,d\,e^5\,x+e^6\,x^2}+\frac {b^4\,x^2}{2\,e^3}+\frac {\ln \left (d+e\,x\right )\,\left (6\,a^2\,b^2\,e^2-12\,a\,b^3\,d\,e+6\,b^4\,d^2\right )}{e^5} \]
x*((4*a*b^3)/e^3 - (3*b^4*d)/e^4) - ((a^4*e^4 - 7*b^4*d^4 - 18*a^2*b^2*d^2 *e^2 + 20*a*b^3*d^3*e + 4*a^3*b*d*e^3)/(2*e) - x*(4*b^4*d^3 - 4*a^3*b*e^3 + 12*a^2*b^2*d*e^2 - 12*a*b^3*d^2*e))/(d^2*e^4 + e^6*x^2 + 2*d*e^5*x) + (b ^4*x^2)/(2*e^3) + (log(d + e*x)*(6*b^4*d^2 + 6*a^2*b^2*e^2 - 12*a*b^3*d*e) )/e^5